Laws of Motion, Friction & Dynamics of Circular Motion

In each of the five cases the net force acting on the body is zero.

(a) The raindrop is falling with a constant speed along a constant direction, so its acceleration is zero. Hence, by Newton's second law, the net force F (= ma) on the drop is zero. Its weight is balanced by the viscous drag of air and the buoyant force.

(b) The cork is floating on water, so its weight is balanced by the upward buoyant force due to water. Hence, the net force on the cork is zero.

(c) The kite is maintained at rest in the sky. Hence, as required by Newton's first (or second) law, the net force on it is zero.

(d) The car is moving on the rough road with a constant velocity, that is, with zero acceleration. Hence, the net force F (= ma) on it is zero. The force exerted by the accelerator is balanced by the force of friction between the tyres and the rough road.

(e) No field — gravitational, electric or magnetic — is acting on the electron, and it is far away from all material objects. Hence, the net force on it is zero.

Given,

• Mass of the pebble, m = 0.05 kg
• Acceleration due to gravity, g = 9.8 m s^-2

The magnitude and the direction of the net force in each case has to be found.

Air resistance is ignored, so the only force acting on the pebble at every point of its motion is the gravity force. It always acts vertically downwards, towards the centre of the earth, and its magnitude is

$$\begin{aligned}\text F = \text{mg} = 0.05 \times 9.8 \\[1em] = 0.49\ \text N\end{aligned}$$

(a) During the upward motion, the net force is 0.49 N, directed vertically downwards.

(b) During the downward motion, the net force is again 0.49 N, directed vertically downwards.

(c) At the highest point the velocity of the pebble is momentarily zero, but its acceleration is not zero. The net force is still 0.49 N, directed vertically downwards.

No, the answers do not change if the pebble is thrown at an angle of 45° with the horizontal. Ignoring air resistance, gravity is the only force acting on it, and it is 0.49 N vertically downwards at every point of the trajectory. At the highest point of this trajectory only the vertical component of velocity becomes zero; the horizontal component remains constant.

Given,

• Mass of the stone, m = 0.1 kg
• Acceleration due to gravity, g = 9.8 m s^-2

Air resistance is neglected, so once the stone is dropped the only force acting on it is the gravity force,

$$\begin{aligned}\text F = \text{mg} = 0.1 \times 9.8 \\[1em] = 0.98\ \text N\end{aligned}$$

(a) By Newton's second law, the net force on the stone is F = ma, where a = g = 9.8 m s^-2. Hence, the net force is 0.98 N, directed vertically downwards.

(b) The motion of the train with a constant velocity exerts no force on the stone. Hence, the net force is the same as in case (a), that is, 0.98 N, directed vertically downwards.

(c) While the stone is being carried by the accelerating train, an additional force

$$\begin{aligned}\text F' = \text{ma} = 0.1 \times 1 \\[1em] = 0.1\ \text N\end{aligned}$$

acts on it along the direction of motion of the train. But as soon as the stone is dropped, the force F' no longer acts on it. Hence, the net force on the stone is again 0.98 N, directed vertically downwards.

(d) The stone lying on the floor is at rest relative to the train, so the gravity force mg on it is balanced by the normal reaction of the floor. The stone is, however, accelerated along with the train. Hence, the net force acting on it is

$$\begin{aligned}\text F = \text{ma} = 0.1 \times 1 \\[1em] = 0.1\ \text N\end{aligned}$$

directed horizontally, along the direction of motion of the train. This force is provided by the force of friction between the stone and the floor of the train.

(i) T

Reason — On a smooth horizontal table the particle describes a horizontal circle, so the gravity force mg on the particle and the normal reaction of the table balance each other and have no component towards the centre. The net force on the particle directed towards the centre is therefore the centripetal force required to keep it in circular motion, and this force is provided entirely by the tension in the string (there is neither friction nor any other horizontal force). Hence,

$\text{Net force} = \text{Tension in the string} = \text T$

Given,

• Retarding force, F = −50 N
• Mass of the body, m = 20 kg
• Initial speed, u = 15 m s^-1
• Final speed, v = 0 (the body stops)

The time t taken by the body to stop has to be calculated.

By Newton's second law of motion, the acceleration produced in the body is

$\text a = \frac{\text F}{\text m}$

Substituting the values,

$$\begin{aligned}\text a = \frac{-50}{20} \\[1em] = -2.5\ \text{m s}^{-2}\end{aligned}$$

The negative sign shows that it is a retardation. Now, from the first equation of motion,

$\text v = \text u + \text{at}$

$$\begin{aligned}\text t = \frac{\text v - \text u}{\text a} = \frac{0 - 15}{-2.5} \\[1em] = \frac{-15}{-2.5} \\[1em] = 6\ \text s\end{aligned}$$

Hence, the body takes 6 s to stop.

Given,

• Mass of the body, m = 3.0 kg
• Initial speed, u = 2.0 m s^-1
• Final speed, v = 3.5 m s^-1
• Time interval, t = 25 s

The magnitude and the direction of the force have to be calculated.

From the first equation of motion, the acceleration produced in the body is

$\text a = \frac{\text v - \text u}{\text t}$

Substituting the values,

$$\begin{aligned}\text a = \frac{3.5 - 2.0}{25} = \frac{1.5}{25} \\[1em] = 0.06\ \text{m s}^{-2}\end{aligned}$$

By Newton's second law of motion, the force acting upon the body is

$$\begin{aligned}\text F = \text{ma} = 3.0 \times 0.06 \\[1em] = 0.18\ \text N\end{aligned}$$

Since the force increases the speed of the body without changing the direction of motion, it acts along the direction of motion.

Hence, the force is 0.18 N, acting along the direction of motion of the body.

Given,

• Mass of the body, m = 5 kg
• Two mutually perpendicular forces, F₁ = 8 N and F₂ = 6 N

The magnitude and the direction of the acceleration have to be calculated.

The two forces act at right angles to each other, so by the parallelogram law of vector addition the magnitude of their resultant is

$\text F = \sqrt{\text F_1^2 + \text F_2^2 + 2\text F_1\text F_2 \cos 90^\circ}$

Substituting the values,

$$\begin{aligned}\text F = \sqrt{(8)^2 + (6)^2 + 0} \\[1em] = \sqrt{64 + 36} = \sqrt{100} \\[1em] = 10\ \text N\end{aligned}$$

By Newton's second law of motion, the magnitude of the acceleration produced is

$$\begin{aligned}\text a = \frac{\text F}{\text m} = \frac{10}{5} \\[1em] = 2\ \text{m s}^{-2}\end{aligned}$$

If the resultant force makes an angle θ with the 8 N force, then

$$\begin{aligned}\cos \theta = \frac{\text {OA}}{\text {OC}}= \frac{\text F_1}{\text F} = \frac{8}{10} \\[1em] = 0.8\end{aligned}$$

$\theta = \cos^{-1}(0.8) = 37^\circ$

Hence, the acceleration produced is 2 m s^-2, directed at nearly 37° with the direction of the 8 N force.

Given,

• Initial speed, u = 36 km/h = $36 \times \frac{5}{18}$ = 10 m s^-1
• Final speed, v = 0
• Time taken to stop, t = 4 s
• Mass of the vehicle = 400 kg and mass of the driver = 65 kg

The average retarding force on the vehicle has to be calculated.

The total mass which is retarded is

$$\begin{aligned}\text m = 400 + 65 \\[1em] = 465\ \text{kg}\end{aligned}$$

From the first equation of motion, v = u + at, the retardation produced is

$$\begin{aligned}\text a = \frac{\text v - \text u}{\text t} = \frac{0 - 10}{4} \\[1em] = -2.5\ \text{m s}^{-2}\end{aligned}$$

By Newton's second law of motion, the retarding force applied on the three-wheeler is

$$\begin{aligned}\text F = \text{ma} = 465 \times (-2.5) \\[1em] = -1162.5\ \text N\end{aligned}$$

The negative sign shows that the force is opposite to the direction of motion.

Hence, the average retarding force on the vehicle is 1162.5 N, acting opposite to the direction of motion.

Given,

• Lift-off mass of the rocket, m = 20,000 kg
• Initial upward acceleration, a = 5.0 m s^-2
• Acceleration due to gravity, g = 9.8 m s^-2

The initial thrust (force) of the blast has to be calculated.

The rocket moves up against gravity. Hence, the blast has to produce an acceleration which is the sum of the acceleration due to gravity and the observed upward acceleration,

$$\begin{aligned}\text{Total acceleration} = \text g + \text a = 9.8 + 5.0 \\[1em] = 14.8\ \text{m s}^{-2}\end{aligned}$$

By Newton's second law of motion, the initial thrust of the blast is

$\text F = \text m(\text g + \text a)$

Substituting the values,

$$\begin{aligned}\text F = 20000 \times 14.8 \\[1em] = 296000\ \text N \\[1em] = 2.96 \times 10^5\ \text N\end{aligned}$$

Hence, the initial thrust of the blast is 2.96 × 10⁵ N, directed vertically upwards.

Given,

• Mass of the body, m = 0.40 kg
• Initial velocity, u = +10 m s^-1 (towards the north, taken as the positive direction)
• Force, F = −8.0 N (towards the south)
• The force acts from t = 0 to t = 30 s, and x = 0 at t = 0

The position of the body at t = −5 s, 25 s and 100 s has to be predicted.

By Newton's second law of motion, the acceleration produced during 0 < t < 30 s is

$$\begin{aligned}\text a = \frac{\text F}{\text m} = \frac{-8.0}{0.40} \\[1em] = -20\ \text{m s}^{-2}\end{aligned}$$

At t = −5 s : The force has not yet been applied, so the body moves uniformly with the speed of 10 m s^-1. Going backwards from x = 0 at t = 0,

$$\begin{aligned}\text x = \text{ut} = 10 \times (-5) \\[1em] = -50\ \text m\end{aligned}$$

At t = 25 s : This instant lies within the interval during which the force acts, so the motion is uniformly accelerated. Using $\text x = \text{ut} + \frac{1}{2}\text{at}^2$,

$$\begin{aligned}\text x = 10 \times 25 + \frac{1}{2} \times (-20) \times (25)^2 \\[1em] = 250 - 10 \times 625 \\[1em] = 250 - 6250 \\[1em] = -6000\ \text m\end{aligned}$$

At t = 100 s : The motion has to be considered in two parts.

From t = 0 to t = 30 s (accelerated motion) :

$$\begin{aligned}\text x_1 = 10 \times 30 + \frac{1}{2} \times (-20) \times (30)^2 \\[1em] = 300 - 10 \times 900 \\[1em] = 300 - 9000 \\[1em] = -8700\ \text m\end{aligned}$$

The velocity at t = 30 s is

$$\begin{aligned}\text v = \text u + \text{at} = 10 + (-20) \times 30 \\[1em] = 10 - 600 \\[1em] = -590\ \text{m s}^{-1}\end{aligned}$$

From t = 30 s to t = 100 s (uniform motion, since the force no longer acts) : here Δt = 70 s, so

$$\begin{aligned}\text x_2 = \text v \times \Delta \text t = (-590) \times 70 \\[1em] = -41300\ \text m\end{aligned}$$

Therefore, the total displacement from t = 0 to t = 100 s is

$$\begin{aligned}\text x = \text x_1 + \text x_2 = -8700 - 41300 \\[1em] = -50000\ \text m = -50\ \text{km}\end{aligned}$$

Hence, the position of the body is 50 m to the south of the origin at t = −5 s, 6 km to the south at t = 25 s and 50 km to the south at t = 100 s.

Given,

• Acceleration of the truck, a = 2.0 m s^-2, starting from rest (u = 0)
• The stone is dropped at t = 10 s from a height of 6.0 m
• Acceleration due to gravity, g = 9.8 m s^-2

The velocity and the acceleration of the stone at t = 11 s have to be calculated.

Horizontal motion : At the instant of dropping, the stone shares the horizontal velocity of the truck. From v = u + at,

$$\begin{aligned}\text v_x = 0 + 2.0 \times 10 \\[1em] = 20\ \text{m s}^{-1}\end{aligned}$$

After the stone is dropped no horizontal force acts on it (air resistance is neglected), so this horizontal velocity remains unchanged. Hence, at t = 11 s,

$\text v_x = 20\ \text{m s}^{-1}$

Vertical motion : At the instant of dropping the vertical velocity of the stone is zero, and it falls freely for a time

$\text t' = 11 - 10 = 1\ \text s$

Using v = u + gt',

$$\begin{aligned}\text v_y = 0 + 9.8 \times 1 \\[1em] = 9.8\ \text{m s}^{-1}\ (\text{vertically downwards})\end{aligned}$$

(a) The two components are mutually perpendicular, so the magnitude of the resultant velocity is

$$\begin{aligned}\text v = \sqrt{\text v_x^2 + \text v_y^2} = \sqrt{(20)^2 + (9.8)^2} \\[1em] = \sqrt{400 + 96.04} = \sqrt{496.04} \\[1em] = 22.3\ \text{m s}^{-1}\end{aligned}$$

If θ is the angle made by this velocity with the horizontal, then

$$\begin{aligned}\tan \theta = \frac{\text v_y}{\text v_x} = \frac{9.8}{20} \\[1em] = 0.49\end{aligned}$$

$\theta = \tan^{-1}(0.49) = 26.1^\circ$

(b) After the stone is dropped, the only force acting on it is the gravity force. Hence, its acceleration is 9.8 m s^-2, directed vertically downwards.

Hence, the velocity of the stone at t = 11 s is 22.3 m s^-1 at 26.1° below the horizontal, and its acceleration is 9.8 m s^-2 vertically downwards.

Given,

• Mass of the bob, m = 0.1 kg
• Length of the string, l = 2 m
• Speed of the bob at the mean position, v = 1 m s^-1

The trajectory of the bob after the string is cut has to be described in each case.

As soon as the string is cut, the tension vanishes and the only force acting on the bob is the gravity force, acting vertically downwards. The subsequent path therefore depends entirely on the velocity the bob possesses at that instant.

(a) At an extreme position : The bob is momentarily at rest at an extreme position, so its velocity is zero. Under the gravity force alone, it therefore falls vertically downwards along a straight line.

(b) At the mean position : At the mean position the velocity of the bob is 1 m s^-1, directed horizontally along the tangent to the arc. The bob thus has a horizontal velocity and a vertical acceleration g, which is exactly the condition of a horizontal projection. Hence, it describes a parabolic path and finally strikes the floor.

Given,

• Mass of the man, m = 70 kg
• Acceleration due to gravity, g = 9.8 m s^-2

The reading of the weighing machine, which is the normal reaction R exerted by the machine on the man, has to be calculated in each case.

(a) The lift is moving upwards with a uniform speed, so its acceleration is zero and the man is in equilibrium. Hence,

$$\begin{aligned}\text R = \text{mg} = 70 \times 9.8 \\[1em] = 686\ \text N = 70\ \text{kg-wt}\end{aligned}$$

(b) The lift is moving downwards with an acceleration a = 5 m s^-2. The net downward force on the man is mg − R, so

$$\begin{aligned}\text R = \text m(\text g - \text a) = 70 \times (9.8 - 5.0) \\[1em] = 70 \times 4.8 \\[1em] = 336\ \text N = 34.3\ \text{kg-wt}\end{aligned}$$

(c) The lift is moving upwards with an acceleration a = 5.0 m s^-2. The net upward force on the man is R − mg, so

$$\begin{aligned}\text R = \text m(\text g + \text a) = 70 \times (9.8 + 5.0) \\[1em] = 70 \times 14.8 \\[1em] = 1036\ \text N = 105.7\ \text{kg-wt}\end{aligned}$$

(d) If the lift mechanism fails, the lift and the man both fall freely with the acceleration due to gravity, that is, a = g. Hence,

$$\begin{aligned}\text R = \text m(\text g - \text g) \\[1em] = 0\end{aligned}$$

Hence, the readings are 686 N, 336 N, 1036 N and zero respectively. In the last case the man is in the state of weightlessness.

Given,

• Mass of the particle, m = 4 kg
• The position-time graph is as shown

(a) For t < 0 : The graph lies along the time axis, showing that the particle remains at rest at x = 0. Since there is no acceleration, by Newton's second law the force on the particle is zero.

For 0 < t < 4 s : The graph is a straight line of constant slope, so the particle moves with a constant velocity

$$\begin{aligned}\text v = \frac{\Delta \text x}{\Delta \text t} = \frac{3}{4} \\[1em] = 0.75\ \text{m s}^{-1}\end{aligned}$$

As the velocity is constant, the acceleration is zero and hence the force on the particle is zero.

For t > 4 s : The graph is again parallel to the time axis, showing that the position of the particle does not change with time, that is, the particle is at rest. Hence, the force on the particle is zero.

(b) Impulse is equal to the change in momentum produced.

Impulse at t = 0 : Just before t = 0 the particle is at rest (u = 0) and just after t = 0 it has a constant velocity v = 0.75 m s^-1. Therefore,

$\text{Impulse} = \text{mv} - \text{mu} = \text m(\text v - \text u)$

$$\begin{aligned}= 4 \times (0.75 - 0) \\[1em] = 3\ \text{kg m s}^{-1}\end{aligned}$$

Impulse at t = 4 s : Just before t = 4 s the particle moves with u = 0.75 m s^-1 and just after t = 4 s it is at rest (v = 0). Therefore,

$$\begin{aligned}\text{Impulse} = \text m(\text v - \text u) = 4 \times (0 - 0.75) \\[1em] = -3\ \text{kg m s}^{-1}\end{aligned}$$

Hence, the force on the particle is zero in all the three intervals, and the impulses imparted at t = 0 and at t = 4 s are 3 kg m s^-1 and −3 kg m s^-1 respectively.

Given,

• Mass of body A, m₁ = 10 kg
• Mass of body B, m₂ = 20 kg
• Applied horizontal force, F = 600 N
• The horizontal surface is smooth

The tension in the string in each case has to be calculated.

Since the two bodies are tied by a light string, they move together with a common acceleration a, and the tension is the same throughout the string.

(i) When F is applied to A : The net force on A is F − T and the net force on B is T. By Newton's second law,

$\text F - \text T = \text m_1\text a \quad \text{...............(1)}$

$\text T = \text m_2\text a \quad \text{...............(2)}$

Adding the two equations,

$\text F = (\text m_1 + \text m_2)\text a \quad \Rightarrow \quad \text a = \frac{\text F}{\text m_1 + \text m_2}$

Substituting this in equation (2),

$$\begin{aligned}\text T = \frac{\text m_2\text F}{\text m_1 + \text m_2} = \frac{20 \times 600}{10 + 20} \\[1em] = \frac{12000}{30} \\[1em] = 400\ \text N\end{aligned}$$

(ii) When F is applied to B : Now the net force on B is F − T' and the net force on A is T'. Proceeding in the same way,

$$\begin{aligned}\text T' = \frac{\text m_1\text F}{\text m_1 + \text m_2} = \frac{10 \times 600}{10 + 20} \\[1em] = \frac{6000}{30} \\[1em] = 200\ \text N\end{aligned}$$

Hence, the tension in the string is 400 N when the force is applied to A and 200 N when it is applied to B.

Given,

• Masses, m₁ = 12 kg and m₂ = 8 kg
• Acceleration due to gravity, g = 9.8 m s^-2

The acceleration of the masses and the tension in the string have to be calculated.

Since the string is light and inextensible, the tension T is the same throughout and the two masses have accelerations of the same magnitude a. As m₁ is heavier, it moves downwards and m₂ moves upwards.

For m₁, the net downward force is m₁g − T. By Newton's second law,

$\text m_1\text g - \text T = \text m_1\text a \quad \text{...............(1)}$

For m₂, the net upward force is T − m₂g. Again by Newton's second law,

$\text T - \text m_2\text g = \text m_2\text a \quad \text{...............(2)}$

Adding equations (1) and (2),

$$\begin{aligned}(\text m_1 - \text m_2)\text g = (\text m_1 + \text m_2)\text a \\[1em] \text a = \frac{(\text m_1 - \text m_2)\text g}{\text m_1 + \text m_2}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text a = \frac{(12 - 8) \times 9.8}{12 + 8} = \frac{4 \times 9.8}{20} \\[1em] = \frac{39.2}{20} \\[1em] = 1.96\ \text{m s}^{-2}\end{aligned}$$

Substituting this value of a in equation (1),

$$\begin{aligned}\text T = \text m_1(\text g - \text a) = 12 \times (9.8 - 1.96) \\[1em] = 12 \times 7.84 \\[1em] = 94.08\ \text N\end{aligned}$$

Hence, the acceleration of the masses is 1.96 m s^-2 and the tension in the string is nearly 94 N.

Let m₁ and m₂ be the masses of the two products of disintegration and $\vec{\text v}_1$ and $\vec{\text v}_2$ their respective velocities.

Before disintegration the nucleus is at rest in the laboratory frame of reference, so the total linear momentum of the system is zero,

$\vec{\text p}_{\text{initial}} = 0$

After disintegration, the total linear momentum of the system is

$\vec{\text p}_{\text{final}} = \text m_1\vec{\text v}_1 + \text m_2\vec{\text v}_2$

The disintegration is caused by internal forces only, so no external force acts on the system. Hence, by the principle of conservation of linear momentum,

$\text m_1\vec{\text v}_1 + \text m_2\vec{\text v}_2 = 0$

$\vec{\text v}_1 = -\frac{\text m_2}{\text m_1}\vec{\text v}_2$

The masses m₁ and m₂ are positive quantities, so the negative sign shows that $\vec{\text v}_1$ and $\vec{\text v}_2$ are oppositely directed.

Hence, the two products of disintegration must move in opposite directions.

Given,

• Mass of each ball, m = 0.05 kg
• Speed of each ball before collision = 6 m s^-1
• Speed of each ball after collision = 6 m s^-1, the direction being reversed

The impulse imparted to each ball due to the other has to be calculated.

Let the two balls be A and B, and let the direction of motion of A before the collision be taken as positive.

For ball A,

$$\begin{aligned}\text{Initial momentum} = \text{mu} = 0.05 \times 6 \\[1em] = 0.3\ \text{kg m s}^{-1}\end{aligned}$$

$$\begin{aligned}\text{Final momentum} = \text{mv} = 0.05 \times (-6) \\[1em] = -0.3\ \text{kg m s}^{-1}\end{aligned}$$

By the impulse-momentum theorem, the impulse imparted to ball A by ball B is equal to the change in its momentum,

$$\begin{aligned}\text J = \text{mv} - \text{mu} = -0.3 - 0.3 \\[1em] = -0.6\ \text{kg m s}^{-1}\end{aligned}$$

By Newton's third law of motion, an equal and opposite impulse of +0.6 kg m s^-1 is imparted to ball B by ball A.

Hence, the impulse imparted to each ball is 0.6 kg m s^-1 in magnitude, the two impulses being oppositely directed.

Given,

• Mass of the shell, m₁ = 0.020 kg
• Mass of the gun, m₂ = 100 kg
• Muzzle speed of the shell, v₁ = 80 m s^-1

The recoil speed of the gun has to be calculated.

Initially both the gun and the shell are at rest, so the initial linear momentum of the system is zero. No external force acts on the system, so by the principle of conservation of linear momentum,

$$\begin{aligned}\text m_1\text v_1 + \text m_2\text v_2 = 0 \\[1em] \text v_2 = -\frac{\text m_1\text v_1}{\text m_2}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text v_2 = -\frac{0.020 \times 80}{100} = -\frac{1.6}{100} \\[1em] = -0.016\ \text{m s}^{-1}\end{aligned}$$

The negative sign shows that the gun moves in a direction opposite to that of the shell.

Hence, the recoil speed of the gun is 0.016 m s^-1.

Given,

• Mass of the ball, m = 0.15 kg
• Speed of the ball, u = 54 km/h = $54 \times \frac{5}{18}$ = 15 m s^-1, unchanged by the deflection
• Angle of deflection = 45°

The impulse imparted to the ball has to be calculated.

Suppose the ball moving along AB with a speed u strikes the bat at B and is deflected along BC with the same speed. The angle by which the ball is deflected is ∠ABC = 2θ = 45°, so θ = 22.5°.

Resolving the initial and the final velocities into components parallel and normal to the bat :

The components along the bat are each u sin θ and they are in the same direction. Hence, the change in momentum along the bat is

$\text{mu}\sin \theta - \text{mu}\sin \theta = 0$

The components normal to the bat are just reversed. Therefore,

$\text{Momentum before impact} = \text{mu}\cos \theta$

$\text{Momentum after impact} = -\text{mu}\cos \theta$

By the impulse-momentum theorem, the impulse imparted to the ball by the bat is the change in this normal momentum,

$$\begin{aligned}\text J = \text{mu}\cos \theta - (-\text{mu}\cos \theta) \\[1em] = 2\text{mu}\cos \theta\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text J = 2 \times 0.15 \times 15 \times \cos 22.5^\circ \\[1em] = 4.5 \times 0.924 \\[1em] = 4.16\ \text{kg m s}^{-1}\end{aligned}$$

Hence, the impulse imparted to the ball is nearly 4.2 kg m s^-1, directed normal to the bat.

Given,

• Mass of the stone, m = 0.25 kg
• Radius of the circle, r = 1.5 m
• Number of revolutions, n = 40 in t = 1 min = 60 s
• Maximum tension the string can withstand, T_max = 200 N

The tension in the string and the maximum permissible speed have to be calculated.

The stone is whirled in a horizontal plane, so gravity has no effect on the motion. The necessary centripetal force is obtained wholly from the tension in the string,

$\text T = \text{mr}\omega^2$

where ω is the angular speed of the stone. Now,

$$\begin{aligned}\omega = \frac{2\pi \text n}{\text t} = \frac{2 \times 3.14 \times 40}{60} \\[1em] = \frac{251.2}{60} \\[1em] = 4.19\ \text{rad s}^{-1}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text T = 0.25 \times 1.5 \times (4.19)^2 \\[1em] = 0.375 \times 17.56 \\[1em] = 6.6\ \text N\end{aligned}$$

Let v_max be the maximum speed at which the tension just becomes 200 N. Then,

$$\begin{aligned}\frac{\text{mv}_{max}^2}{\text r} = \text T_{max} \\[1em] \text v_{max} = \sqrt{\frac{\text T_{max}\ \text r}{\text m}}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text v_{max} = \sqrt{\frac{200 \times 1.5}{0.25}} = \sqrt{\frac{300}{0.25}} \\[1em] = \sqrt{1200} \\[1em] = 34.6\ \text{m s}^{-1}\end{aligned}$$

Hence, the tension in the string is 6.6 N and the maximum permissible speed is nearly 34.6 m s^-1.

(b) the stone flies off tangentially from the instant the string breaks.

Reason — While the stone is being whirled round, the tension in the string supplies the centripetal force which continuously changes the direction of the velocity of the stone. At every instant the velocity of the stone is directed along the tangent to the circle at that point.

The moment the string breaks, the tension and hence the centripetal force vanishes instantly, so no net force acts on the stone towards the centre. By Newton's first law of motion the stone must then continue to move along the direction of its instantaneous velocity. Hence, it flies off along the tangent to the circular path at the point where the string breaks.

Alternative (a) is wrong because a radially outward motion would require a radially outward force, which does not exist, and alternative (c) is wrong because the direction of the instantaneous velocity is tangential whatever the speed of the stone may be.

(a) A horse cannot pull a cart and run in empty space. In order to pull the cart, the horse presses the ground obliquely backwards with its feet. By Newton's third law of motion the ground exerts an equal and opposite reaction on the horse, and it is the horizontal component of this reaction that pushes the horse-cart system forward. In empty space there is no ground, so no such frictional reaction is available and no external force acts on the horse-cart system. Hence, the system cannot move.

(b) Passengers are thrown forward when a speeding bus stops suddenly. When the bus stops suddenly, the lower part of the body of a passenger, which is in contact with the bus, comes to rest along with it. The upper part of the body, however, continues to remain in motion due to the inertia of motion. Hence, the passenger is thrown forward from the seat.

(c) It is easier to pull a lawn mower than to push it. When the mower is pulled by a force F applied at an angle θ to the vertical, the vertical component F cos θ acts opposite to the weight, so the normal reaction becomes R₁ = mg − F cos θ. When it is pushed, this component acts along the weight, so the normal reaction becomes R₂ = mg + F cos θ. Since the force of friction is directly proportional to the normal reaction, the friction in pushing is greater than that in pulling. Hence, it is easier to pull the mower than to push it.

(d) A cricketer moves his hands backward while holding a catch. By moving his hands backward the cricketer increases the time in which the momentum of the ball is reduced to zero. From Newton's second law the force is equal to the rate of change of momentum,

$\text F = \frac{\Delta \text p}{\Delta \text t}$

The change in momentum Δp is fixed, so increasing Δt reduces the force exerted on his hands. Hence, his hands are not injured.

Given,

• Mass of block B, m = 0.5 kg
• Mass of block A, M = 10 kg
• Length of block A, l = 50 cm = 0.5 m
• Applied force, F = 50 N
• All the surfaces are frictionless

The time elapsed before block B falls down has to be calculated.

Since all the surfaces are frictionless, no horizontal force is transmitted from A to B. By Newton's second law, the acceleration of block A is

$$\begin{aligned}\text a_A = \frac{\text F}{\text M} = \frac{50}{10} \\[1em] = 5\ \text{m s}^{-2}\end{aligned}$$

For block B there is no horizontal force at all, so

$\text a_B = 0$

Therefore, the acceleration of A relative to B is

$$\begin{aligned}\text a_{AB} = \text a_A - \text a_B = 5 - 0 \\[1em] = 5\ \text{m s}^{-2}\end{aligned}$$

Block B falls off when A has slipped through its whole length relative to B, that is, when the relative displacement is l = 0.5 m. Both start from rest, so using $\text s = \text{ut} + \frac{1}{2}\text{at}^2$ with u = 0,

$$\begin{aligned}\text l = \frac{1}{2}\text a_{AB}\text t^2 \\[1em] \text t = \sqrt{\frac{2\text l}{\text a_{AB}}}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text t = \sqrt{\frac{2 \times 0.5}{5}} = \sqrt{0.2} \\[1em] = 0.45\ \text s\end{aligned}$$

Hence, block B falls down after 0.45 s.

Given,

• Masses, m₁ = 5 kg and m₂ = 5 kg
• The platform, the pulleys and the strings are smooth and light
• Acceleration due to gravity, g = 10 m s^-2

The string carrying the tension T passes over the fixed pulley P₁ and then round the movable pulley P₂, so if a is the acceleration of the block m₁, the pulley P₂ and hence the block m₂ move with an acceleration $\frac{\text a}{2}$. Let T' be the tension in the string joining m₂ to the pulley P₂.

The platform is smooth, so the vertical forces on m₂ have no role to play. For the block m₂, by Newton's second law,

$\text T' = \text m_2\left(\frac{\text a}{2}\right) \quad \text{...............(1)}$

The pulley P₂ is light, so the net force on it must be zero,

$2\text T - \text T' = 0 \quad \Rightarrow \quad \text T' = 2\text T \quad \text{...............(2)}$

For the hanging block m₁,

$\text m_1\text g - \text T = \text m_1\text a \quad \text{...............(3)}$

(a) Multiplying equation (3) by 2,

$2\text m_1\text g - 2\text T = 2\text m_1\text a \quad \text{...............(4)}$

Adding equations (1), (2) and (4),

$2\text m_1\text g = \text m_2\left(\frac{\text a}{2}\right) + 2\text m_1\text a$

Here m₁ = m₂ = m = 5 kg, so

$$\begin{aligned}2\text{mg} = \frac{\text{ma}}{2} + 2\text{ma} = \frac{5\text{ma}}{2} \\[1em] \text a = \frac{4\text g}{5}\end{aligned}$$

Substituting g = 10 m s^-2,

$$\begin{aligned}\text a = \frac{4 \times 10}{5} \\[1em] = 8\ \text{m s}^{-2}\end{aligned}$$

Hence, the block m₁ moves with an acceleration of 8 m s^-2 and the block m₂ with an acceleration of 4 m s^-2.

(b) Substituting this value of a in equation (1),

$$\begin{aligned}\text T' = 5 \times \frac{8}{2} \\[1em] = 20\ \text N\end{aligned}$$

Hence, the tension in the string connecting m₂ to the pulley P₂ is 20 N.

(c) From equation (2), the tension in the string passing over P₁ is

$$\begin{aligned}\text T = \frac{\text T'}{2} = \frac{20}{2} \\[1em] = 10\ \text N\end{aligned}$$

The two segments of this string at the pulley P₁ are mutually perpendicular, one horizontal and the other vertical, and each pulls the pulley with a force T. Hence, the resultant force exerted on the axle of the pulley by the clamp is

$\text R = \sqrt{\text T^2 + \text T^2} = \text T\sqrt{2}$

Substituting the value of T,

$$\begin{aligned}\text R = 10 \times 1.414 \\[1em] = 14.14\ \text N\end{aligned}$$

Hence, the force exerted by the clamp on the pulley P₁ is 14.14 N.

Given,

• Masses, m₁ = 1 kg, m₂ = 2 kg and m₃ = 4 kg
• The strings are light and inextensible, and the pulleys are light and frictionless
• Acceleration due to gravity, g = 10 m s^-2

The block m₁ hangs from one side of the fixed pulley P₁, and the movable pulley P₂ hangs from the other side. The blocks m₂ and m₃ hang from the two sides of P₂.

Let a₁ be the absolute acceleration of m₁ (upwards), so that the pulley P₂ descends with the same acceleration a₁. Let a₂ be the acceleration of m₂ and m₃ relative to P₂. Then m₂ has an absolute upward acceleration (a₂ − a₁) and m₃ has an absolute downward acceleration (a₂ + a₁).

Let T₁ be the tension in the string over P₁ and T₂ that in the string over P₂. Applying Newton's second law,

$\text T_1 - \text m_1\text g = \text m_1\text a_1 \quad \text{...............(1)}$

$\text T_2 - \text m_2\text g = \text m_2(\text a_2 - \text a_1) \quad \text{...............(2)}$

$\text m_3\text g - \text T_2 = \text m_3(\text a_2 + \text a_1) \quad \text{...............(3)}$

Since the pulley P₂ is weightless,

$\text T_1 - 2\text T_2 = 0 \quad \Rightarrow \quad \text T_1 = 2\text T_2 \quad \text{...............(4)}$

Adding equations (2) and (3),

$\text m_3\text g - \text m_2\text g = \text m_2\text a_2 - \text m_2\text a_1 + \text m_3\text a_2 + \text m_3\text a_1$

Substituting m₂ = 2 kg and m₃ = 4 kg,

$$\begin{aligned}4\text g - 2\text g = 2\text a_2 - 2\text a_1 + 4\text a_2 + 4\text a_1 \\[1em] 2\text g = 6\text a_2 + 2\text a_1 \\[1em] \text g = 3\text a_2 + \text a_1 \quad \text{...............(5)}\end{aligned}$$

Substituting T₁ from equation (4) in equation (1),

$2\text T_2 - \text m_1\text g = \text m_1\text a_1 \quad \text{...............(6)}$

Adding equation (6) and equation (3) multiplied by 2,

$2\text m_3\text g - \text m_1\text g = 2\text m_3\text a_2 + 2\text m_3\text a_1 + \text m_1\text a_1$

Substituting m₁ = 1 kg and m₃ = 4 kg,

$$\begin{aligned}8\text g - \text g = 8\text a_2 + 8\text a_1 + \text a_1 \\[1em] 7\text g = 8\text a_2 + 9\text a_1 \quad \text{...............(7)}\end{aligned}$$

From equation (5), a₁ = g − 3a₂. Substituting this in equation (7),

$$\begin{aligned}7\text g = 8\text a_2 + 9(\text g - 3\text a_2) \\[1em] 7\text g = 8\text a_2 + 9\text g - 27\text a_2 \\[1em] 19\text a_2 = 2\text g \quad \Rightarrow \quad \text a_2 = \frac{2\text g}{19}\end{aligned}$$

Substituting g = 10 m s^-2,

$$\begin{aligned}\text a_2 = \frac{2 \times 10}{19} = \frac{20}{19} \\[1em] = 1.05\ \text{m s}^{-2}\end{aligned}$$

From equation (5),

$$\begin{aligned}\text a_1 = \text g - 3\text a_2 = 10 - 3 \times 1.05 \\[1em] = 10 - 3.16 \\[1em] = 6.84\ \text{m s}^{-2}\end{aligned}$$

Therefore, the absolute accelerations of the blocks are

$\text{For } \text m_1 : \text a_1 = 6.84\ \text{m s}^{-2}\ (\text{upwards})$

$\text{For } \text m_2 : \text a_2 - \text a_1 = 1.05 - 6.84 = -5.79\ \text{m s}^{-2},\ \text{that is, } 5.79\ \text{m s}^{-2}\ (\text{downwards})$

$\text{For } \text m_3 : \text a_2 + \text a_1 = 1.05 + 6.84 = 7.89\ \text{m s}^{-2}\ (\text{downwards})$

From equation (1),

$$\begin{aligned}\text T_1 = \text m_1(\text g + \text a_1) = 1 \times (10 + 6.84) \\[1em] = 16.84\ \text N\end{aligned}$$

From equation (4),

$$\begin{aligned}\text T_2 = \frac{\text T_1}{2} = \frac{16.84}{2} \\[1em] = 8.42\ \text N\end{aligned}$$

Hence, the tensions in the two strings are 16.84 N and 8.42 N respectively.

Note: The question states m₃ = 4 kg, but the solution printed in the textbook substitutes m₃ = 3 kg and therefore obtains a₂ = 0.69 m s^-2, a₁ = 6.6 m s^-2, T₁ = 16.6 N and T₂ = 8.3 N. The above solution used the value of m₃ given in the question.

Given,

• Mass of the person, m₁ = 50 kg
• Mass of the box, m₂ = 25 kg
• The string is massless and inextensible, and the pulley is light and frictionless
• The person and the box remain stationary

The force applied by the person on the rope has to be calculated.

Let T be the tension in the rope and R the normal reaction between the floor of the box and the person. By Newton's third law of motion, the person presses the floor with a force R downwards and the floor pushes the person up with the same force R.

Since the person and the box are both stationary, the net force on each is zero.

For the box, the upward forces are the tension T and the reaction R exerted by the person on the floor acts downwards. Hence,

$\text T - \text R = \text m_2\text g \quad \text{...............(1)}$

For the person, the upward forces are the tension T (he pulls the rope downwards, so the rope pulls him upwards) and the normal reaction R of the floor. Hence,

$\text T + \text R = \text m_1\text g \quad \text{...............(2)}$

Adding equations (1) and (2),

$$\begin{aligned}2\text T = (\text m_1 + \text m_2)\text g \\[1em] \text T = \frac{(\text m_1 + \text m_2)\text g}{2}\end{aligned}$$

Substituting the values with g = 9.8 m s^-2,

$$\begin{aligned}\text T = \frac{(50 + 25) \times 9.8}{2} = \frac{75 \times 9.8}{2} \\[1em] = \frac{735}{2} \\[1em] = 367.5\ \text N\end{aligned}$$

By Newton's third law of motion, the force applied by the person on the rope downwards is equal in magnitude to the tension in the rope.

Hence, the person applies a force of 367.5 N on the rope in the downward direction.

Given,

• Mass of block A, m₁ = 100 kg
• Mass of block B, m₂ = 150 kg
• Coefficient of friction between A and B, μ₁ = 0.25
• Coefficient of friction between B and the floor, μ₂ = 0.2
• Acceleration due to gravity, g = 10 m s^-2

The horizontal force F needed to move block B has to be calculated.

Free body diagram of block A : The forces are the weight m₁g downwards, the normal reaction N₁ of B upwards, the tension T of the string and the frictional force f₁ exerted by B.

Free body diagram of block B : The forces are the weight m₂g downwards, the normal reaction N₁ of A pressing down on it, the normal reaction N₂ of the floor upwards, the applied force F, and the frictional forces f₁ (due to A) and f₂ (due to the floor).

For the vertical equilibrium of block A, there is no vertical motion, so

$\text N_1 = \text m_1\text g \quad \text{...............(1)}$

For the vertical equilibrium of block B,

$$\begin{aligned}\text N_2 - \text N_1 - \text m_2\text g = 0 \\[1em] \text N_2 = \text N_1 + \text m_2\text g = (\text m_1 + \text m_2)\text g \quad \text{...............(2)}\end{aligned}$$

Block B is just made to move, so the frictional forces have their limiting values. For the horizontal equilibrium of block B,

$\text F = \text f_1 + \text f_2 = \mu_1\text N_1 + \mu_2\text N_2$

$\text F = \mu_1\text m_1\text g + \mu_2(\text m_1 + \text m_2)\text g$

Substituting the values,

$$\begin{aligned}\text F = 0.25 \times 100 \times 10 + 0.2 \times (100 + 150) \times 10 \\[1em] = 250 + 0.2 \times 2500 \\[1em] = 250 + 500 \\[1em] = 750\ \text N\end{aligned}$$

Hence, a horizontal force of 750 N is needed to move the block B.

Let the ladder AB of weight W and length l rest with its lower end A on the ground and its upper end B against the wall, making an angle θ with the horizontal. Since the ladder is uniform, its weight W acts vertically downwards at its mid-point, where its centre of gravity is situated.

Let N₁ and N₂ be the normal contact forces at A and B, and f₁ and f₂ the frictional forces at A and B respectively. At the minimum angle of inclination the ladder is just about to slip, so the frictional forces have their limiting values,

$\text f_1 = \mu_1\text N_1 \quad \text{and} \quad \text f_2 = \mu_2\text N_2$

Vertical equilibrium : The upward forces are N₁ and f₂, and the downward force is W,

$$\begin{aligned}\text N_1 + \text f_2 = \text W \\[1em] \text N_1 + \mu_2\text N_2 = \text W \quad \text{...............(1)}\end{aligned}$$

Horizontal equilibrium : The wall pushes the ladder with N₂ and the friction at the ground opposes it,

$\text N_2 = \text f_1 = \mu_1\text N_1 \quad \text{...............(2)}$

Substituting N₂ from equation (2) in equation (1),

$$\begin{aligned}\text N_1 + \mu_2\mu_1\text N_1 = \text W \\[1em] \text N_1(1 + \mu_1\mu_2) = \text W \\[1em] \text N_1 = \frac{\text W}{1 + \mu_1\mu_2} \quad \text{...............(3)}\end{aligned}$$

From equation (2),

$\text N_2 = \frac{\mu_1\text W}{1 + \mu_1\mu_2} \quad \text{...............(4)}$

Rotational equilibrium : Taking moments of all the forces about the lower end A. The forces N₁ and f₁ act at A itself, so their moments about A are zero. The weight W tends to rotate the ladder clockwise, while N₂ and f₂ tend to rotate it anticlockwise. The perpendicular distances are $\frac{ l}{2}\cos \theta$ for W, $l$ sin θ for N₂ and l cos θ for f₂. Hence, by the principle of moments,

$-\text W \times \frac{ l}{2}\cos \theta + \text N_2 \times l\sin \theta + \mu_2\text N_2 \times \text l\cos \theta = 0$

Substituting N₂ from equation (4) and dividing throughout by $l$,

$-\frac{\text W}{2}\cos \theta + \frac{\mu_1\text W}{1 + \mu_1\mu_2}\sin \theta + \frac{\mu_1\mu_2\text W}{1 + \mu_1\mu_2}\cos \theta = 0$

Dividing throughout by W and rearranging,

$\frac{1}{2}\cos \theta - \frac{\mu_1\mu_2}{1 + \mu_1\mu_2}\cos \theta = \frac{\mu_1}{1 + \mu_1\mu_2}\sin \theta$

$\frac{\sin \theta}{\cos \theta} = \frac{1 + \mu_1\mu_2}{2\mu_1} - \frac{\mu_1\mu_2}{\mu_1}$

$\tan \theta = \frac{1 + \mu_1\mu_2 - 2\mu_1\mu_2}{2\mu_1} = \frac{1 - \mu_1\mu_2}{2\mu_1}$

Therefore,

$\theta = \tan^{-1}\left(\frac{1 - \mu_1\mu_2}{2\mu_1}\right)$

Hence proved.

Given,

• Length of the beam AB, l = 5 m
• Weight of the beam, W₁ = 20 kg, acting at its mid-point G (2.5 m from A)
• W acts at 1 m from A
• Each cable can hold a maximum tension of 50 kg
• The cable AC makes 37° and the cable BD makes 53° with the beam, BE being along the beam

Which cable breaks first, and the value of W at which it breaks, has to be found.

The components T₁ cos 37° of T₁, T₂ cos 53° of T₂ and the tension T₃ along BE are all parallel to the length of the beam, so they do not test the strength of the cables AC and BD. Only the vertical components T₁ sin 37° and T₂ sin 53° enter the moment equations.

Taking moments of the forces about A : The weights W and W₁ rotate the beam clockwise while T₂ sin 53° rotates it anticlockwise. Hence,

$-\text W \times 1 - 20 \times 2.5 + \text T_2\sin 53^\circ \times 5 = 0$

$\text T_2 = \frac{\text W + 50}{5\sin 53^\circ} \quad \text{...............(1)}$

Taking moments of the forces about B :

$-\text T_1\sin 37^\circ \times 5 + \text W \times 4 + 20 \times 2.5 = 0$

$\text T_1 = \frac{4\text W + 50}{5\sin 37^\circ} \quad \text{...............(2)}$

Comparing equations (1) and (2), the numerator of T₁ increases four times as fast with W as that of T₂, and sin 37° is smaller than sin 53°. Hence, as W is slowly increased, T₁ increases much more rapidly than T₂, and so the cable AC breaks first.

Putting T₁ = 50 kg in equation (2) and taking sin 37° = 0.6018,

$$\begin{aligned}50 = \frac{4\text W + 50}{5 \times 0.6018} \\[1em] 4\text W + 50 = 50 \times 5 \times 0.6018 \\[1em] 4\text W + 50 = 150.45 \\[1em] \text W = \frac{100.45}{4} \\[1em] = 25\ \text{kg-wt}\end{aligned}$$

Hence, the cable AC will break first, and it will break when W becomes nearly 25 kg-wt.

Given,

• Speed of the water, v = 15 m s^-1
• Area of cross-section of the tube, A = 10^-2 m²
• Density of water, ρ = 1.0 × 10³ kg m^-3
• The water does not rebound from the wall

The force exerted on the wall has to be calculated.

The volume of water striking the wall per second is Av, so the mass of water striking the wall per second is

$\text m = \text{Av}\rho$

Substituting the values,

$$\begin{aligned}\text m = 10^{-2} \times 15 \times 1.0 \times 10^3 \\[1em] = 150\ \text{kg s}^{-1}\end{aligned}$$

The momentum carried to the wall per second is

$$\begin{aligned}\text{mv} = 150 \times 15 \\[1em] = 2250\ \text{kg m s}^{-2}\end{aligned}$$

Since the water does not rebound from the wall, the momentum of the water after the impact is zero. Hence, the rate of change of momentum is 2250 kg m s^-2. By Newton's second law of motion, the force is equal to the rate of change of momentum,

$\text F = 2250\ \text N$

Hence, the force exerted on the wall by the impact of water is 2250 N.

Given,

• Ten one-rupee coins are placed one above the other
• Mass of each coin = m kg

The magnitude and the direction of the force in each case has to be found.

(a) Above the 7th coin (counted from the bottom) there are three coins, namely the 8th, the 9th and the 10th. The force on the 7th coin due to all the coins on its top is the sum of their weights,

$\text F = 3\text{mg}\ \text N$

This force acts vertically downwards.

(b) The 8th coin already supports the weights of the two coins above it (the 9th and the 10th) and has its own weight as well. Hence, the force exerted on the 7th coin by the 8th coin is the sum of these three weights,

$\text F = (2\text m + \text m)\text g = 3\text{mg}\ \text N$

This force also acts vertically downwards.

(c) The 6th coin experiences a downward force due to the weights of the four coins above it, that is, the 7th, 8th, 9th and 10th coins,

$\text F = 4\text{mg}\ \text N$

By Newton's third law of motion, the reaction of the 6th coin on the 7th coin is

$\text R = -\text F = -4\text{mg}\ \text N$

The negative sign shows that this reaction is directed vertically upwards.

Given,

• Mass of the block, m = 25 kg
• Mass of the man, M = 50 kg
• Acceleration due to gravity, g = 9.8 m s^-2
• The floor yields to a normal force of 700 N

The action on the floor by the man in the two cases has to be calculated.

The force required to lift the block is its weight,

$$\begin{aligned}\text F = \text{mg} = 25 \times 9.8 \\[1em] = 245\ \text N\end{aligned}$$

and the weight of the man himself is

$$\begin{aligned}\text W = \text{Mg} = 50 \times 9.8 \\[1em] = 490\ \text N\end{aligned}$$

Case (a) : Here the man pulls the rope upwards, so he exerts the force in a direction opposite to his weight. He has to overcome his own weight and also exert the force to raise the block. Hence, the action on the floor is

$$\begin{aligned}\text W + \text F = 490 + 245 \\[1em] = 735\ \text N\end{aligned}$$

Case (b) : Here the man pulls the rope in the direction of his weight, that is, his weight helps him in exerting the force. Hence, the action on the floor is

$$\begin{aligned}\text W - \text F = 490 - 245 \\[1em] = 245\ \text N\end{aligned}$$

Since the floor yields to a normal force of 700 N, the action of 735 N in case (a) would make the floor yield, whereas the action of 245 N in case (b) would not.

Hence, the man should adopt mode (b) to lift the block without yielding of the floor.

Given,

• Mass of the monkey, m = 40 kg
• Maximum tension the rope can withstand = 600 N
• Acceleration due to gravity, g = 10 m s^-2
• The mass of the rope is ignored

The case in which the rope breaks has to be found.

Let T be the tension in the rope, acting upwards, and mg the gravity force on the monkey, acting downwards.

(a) Climbing up with an acceleration of 6 m s^-2 : The net force on the monkey must be upwards,

$\text T - \text{mg} = \text{ma}_1 \quad \Rightarrow \quad \text T = \text m(\text g + \text a_1)$

$$\begin{aligned}\text T = 40 \times (10 + 6) \\[1em] = 640\ \text N\end{aligned}$$

(b) Climbing down with an acceleration of 4 m s^-2 : The net force on the monkey must be downwards,

$\text{mg} - \text T = \text{ma}_2 \quad \Rightarrow \quad \text T = \text m(\text g - \text a_2)$

$$\begin{aligned}\text T = 40 \times (10 - 4) \\[1em] = 240\ \text N\end{aligned}$$

(c) Climbing up with a uniform speed of 5 m s^-1 : The acceleration is zero, so

$$\begin{aligned}\text T = \text{mg} = 40 \times 10 \\[1em] = 400\ \text N\end{aligned}$$

(d) Falling down nearly freely under gravity : Here a = g, so

$$\begin{aligned}\text T = \text m(\text g - \text g) \\[1em] = 0\end{aligned}$$

The rope breaks only when the tension exceeds 600 N, and this happens only in case (a), where the tension is 640 N.

Hence, the rope will break in case (a), that is, when the monkey climbs up with an acceleration of 6 m s^-2.

Given,

• Mass of the man, m = 65 kg
• Acceleration of the belt, a = 1.0 m s^-2
• Coefficient of static friction, μ_s = 0.2
• Acceleration due to gravity, g = 9.8 m s^-2

(a) The man is stationary relative to the belt, so his acceleration is the same as that of the belt. The gravity force mg on him is balanced by the equal and opposite normal reaction R of the belt. Hence, the net force on the man is only due to his acceleration,

$$\begin{aligned}\text F = \text{ma} = 65 \times 1.0 \\[1em] = 65\ \text N\end{aligned}$$

Hence, the net force on the man is 65 N, directed along the direction of motion of the belt. This force is provided by the force of friction between his shoes and the belt.

(b) The limiting frictional force between the man's shoes and the belt is

$\text f_s = \mu_s\text R = \mu_s\text{mg}$

$$\begin{aligned}\text f_s = 0.2 \times 65 \times 9.8 \\[1em] = 127.4\ \text N\end{aligned}$$

The man can continue to remain stationary relative to the belt only so long as the force needed to accelerate him does not exceed this limiting friction. At the maximum acceleration a_max,

$$\begin{aligned}\text{ma}_{max} = \text f_s \\[1em] \text a_{max} = \frac{\text f_s}{\text m} = \frac{127.4}{65} \\[1em] = 1.96\ \text{m s}^{-2}\end{aligned}$$

Hence, the man can remain stationary relative to the belt up to an acceleration of 1.96 m s^-2.

Given,

• Mass of body A, m₁ = 5 kg and mass of body B, m₂ = 10 kg
• Coefficient of friction between the bodies and the table, μ_s = 0.15
• Applied force, F = 200 N, horizontally at A
• Acceleration due to gravity, g = 9.8 m s^-2

(a) When the force F is applied at A towards the right, the limiting frictional force on the two bodies together, directed towards the left, is

$$\begin{aligned}\text f_s =\text f_1+ \text f_2= \mu_s(\text m_1 + \text m_2)\text g = 0.15 \times (5 + 10) \times 9.8 \\[1em] = 0.15 \times 147 \\[1em] = 22\ \text N\end{aligned}$$

Hence, the net force exerted on the partition is

$$\begin{aligned}\text F' = \text F - \text f_s = 200 - 22 \\[1em] = 178\ \text N,\ \text{towards the right}\end{aligned}$$

By Newton's third law of motion, the reaction of the partition is 178 N, directed towards the left.

(b) The force of limiting friction on body A alone is

$$\begin{aligned}(\text f_1) = \mu_s\text m_1\text g = 0.15 \times 5 \times 9.8 \\[1em] = 7.35\ \text N\end{aligned}$$

Hence, the net force exerted by body A on body B is

$$\begin{aligned}\text F'' = \text F - (\text f_1) = 200 - 7.35 \\[1em] = 192.65\ \text N,\ \text{towards the right}\end{aligned}$$

By Newton's third law of motion, the reaction of body B on body A is 192.65 N, directed towards the left.

If the partition is removed : The two bodies move together towards the right under the net force F' = 178 N, with an acceleration

$$\begin{aligned}\text a = \frac{\text F'}{\text m_1 + \text m_2} = \frac{178}{15} \\[1em] = 11.9\ \text{m s}^{-2}\end{aligned}$$

The force producing this motion in A alone is

$$\begin{aligned}\text m_1\text a = 5 \times 11.9 \\[1em] = 59.5\ \text N\end{aligned}$$

Hence, the net force now exerted by A on B is

$$\begin{aligned}\text F'' - \text m_1\text a = 192.65 - 59.5 \\[1em] = 133.15\ \text N,\ \text{towards the right}\end{aligned}$$

and the reaction of B on A is 133.15 N towards the left.

Hence, the answer to part (b) does change when the bodies are in motion.

Given,

• Mass of the block, m = 15 kg
• Coefficient of friction between the block and the trolley, μ_s = 0.18
• Acceleration of the trolley, a = 0.5 m s^-2 for 20 s, and uniform velocity thereafter
• Acceleration due to gravity, g = 9.8 m s^-2

The force required to give the block the same acceleration as the trolley is

$$\begin{aligned}\text F = \text{ma} = 15 \times 0.5 \\[1em] = 7.5\ \text N\end{aligned}$$

The weight mg of the block is balanced by the normal reaction R of the trolley. The limiting frictional force between the block and the trolley is

$$\begin{aligned}\text f_s = \mu_s\text{mg} = 0.18 \times 15 \times 9.8 \\[1em] = 26.5\ \text N\end{aligned}$$

(a) As viewed by a stationary observer on the ground : As the trolley accelerates forward it exerts a reactionary force F' (= F = 7.5 N) on the block in the backward direction. As the block tends to move under this force, the force of friction comes into play opposite to F'. Since the limiting friction (26.5 N) is much larger than 7.5 N, the frictional force, being self-adjusting, simply adjusts itself to 7.5 N in the direction of motion of the trolley, and the block does not slide. Hence, during the accelerated motion of the trolley the block moves along with the trolley with the same acceleration of 0.5 m s^-2, appearing to be at rest relative to the trolley.

When the trolley later moves with a uniform velocity, no force F is needed, and the block shows no tendency to move relative to the trolley. Hence, the force of friction does not come into play at all, and the block moves along with the trolley with the same uniform velocity.

(b) As viewed by an observer moving with the trolley : During the first 20 s this observer has an accelerated motion, so his is a non-inertial frame in which the law of inertia is not valid. To him the block appears stationary, and he explains this by invoking a pseudo force of 7.5 N acting backwards on the block, which is balanced by the frictional force of 7.5 N acting forwards. Once the trolley moves with a uniform velocity, the observer becomes an inertial observer and the block still appears to him to be at rest, with no force acting on it.

Given,

• Mass of the box, m = 40 kg
• Distance of the box from the open end, s = 5 m
• Coefficient of friction, μ_s = 0.15
• Acceleration of the truck, a = 2 m s^-2, starting from rest
• Acceleration due to gravity, g = 9.8 m s^-2

The distance from the starting point at which the box falls off the truck has to be calculated.

The force on the box due to the accelerated motion of the truck is

$$\begin{aligned}\text F = \text{ma} = 40 \times 2 \\[1em] = 80\ \text N\end{aligned}$$

As the truck accelerates in the forward direction, a reactionary force F' (= F = 80 N) acts on the box in the backward direction. As the box tends to move under this force, the force of friction comes into play in the forward direction. The limiting frictional force between the box and the surface below it is

$$\begin{aligned}\text f_s = \mu_s\text{mg} = 0.15 \times 40 \times 9.8 \\[1em] = 58.8\ \text N\end{aligned}$$

Since F' is greater than f_s, the box slides backwards. The net force acting on the box in the backward direction is

$$\begin{aligned}\text F' - \text f_s = 80 - 58.8 \\[1em] = 21.2\ \text N\end{aligned}$$

The acceleration produced in the box relative to the truck, in the backward direction, is

$$\begin{aligned}\text a' = \frac{21.2}{40} \\[1em] = 0.53\ \text{m s}^{-2}\end{aligned}$$

If t is the time in which the box slides through 5 m relative to the truck and falls off, then using $\text s = \text{ut} + \frac{1}{2}\text a'\text t^2$ with u = 0,

$$\begin{aligned}5 = \frac{1}{2} \times 0.53 \times \text t^2 \\[1em] \text t^2 = \frac{10}{0.53} = 18.87 \\[1em] \text t = 4.34\ \text s\end{aligned}$$

The distance covered by the truck from the starting point in this time is

$$\begin{aligned}\text s = \text{ut} + \frac{1}{2}\text{at}^2 = 0 + \frac{1}{2} \times 2 \times (4.34)^2 \\[1em] = 18.8\ \text m\end{aligned}$$

Hence, the box falls off the truck at a distance of 18.8 m from the starting point.

Given,

• Mass of the smaller block, m = 2 kg
• Mass of the triangular block, M' = 5 kg
• Angle of the triangular block, θ = 30°
• All the surfaces are frictionless, and the strings and the pulleys are light

The mass M of the hanging block has to be calculated.

Let the hanging block M descend with an acceleration a. Since the string is light and inextensible and passes over a light pulley, the triangular block together with the smaller block placed on it moves horizontally with the same acceleration a, and the smaller block moves along with the triangular block without slipping.

Condition that the smaller block does not slip : In the frame of the accelerating triangular block, a pseudo force ma acts on the smaller block in the backward direction. Resolving the pseudo force and the gravity force along the inclined surface, the smaller block does not slip when these components balance,

$\text{ma}\cos \theta = \text{mg}\sin \theta$

Dividing throughout by m cos θ,

$\text a = \text g\tan \theta \quad \text{...............(1)}$

Substituting θ = 30°,

$\text a = \text g\tan 30^\circ = \frac{\text g}{\sqrt{3}}$

For the hanging block M : Let T be the tension in the string. The net downward force on M is Mg − T, so by Newton's second law,

$\text{Mg} - \text T = \text{Ma} \quad \text{...............(2)}$

For the triangular block together with the smaller block : This combination, of total mass (M' + m), is pulled horizontally by the tension T alone, the floor being frictionless. Hence,

$\text T = (\text M' + \text m)\text a \quad \text{...............(3)}$

Substituting the values,

$\text T = (5 + 2)\text a = 7\text a$

Adding equations (2) and (3),

$\text{Mg} = \text{Ma} + 7\text a = (\text M + 7)\text a$

$\text a = \frac{\text{Mg}}{\text M + 7} \quad \text{...............(4)}$

Comparing equations (1) and (4),

$\frac{\text{Mg}}{\text M + 7} = \text g\tan 30^\circ = \frac{\text g}{\sqrt{3}}$

Cancelling g and cross-multiplying,

$$\begin{aligned}\sqrt{3}\text M = \text M + 7 \\[1em] \text M(\sqrt{3} - 1) = 7 \\[1em] \text M = \frac{7}{\sqrt{3} - 1}\end{aligned}$$

Rationalising the denominator,

$$\begin{aligned}\text M = \frac{7}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{7(\sqrt{3} + 1)}{3 - 1} \\[1em] = \frac{7(\sqrt{3} + 1)}{2}\end{aligned}$$

Substituting $\sqrt{3} = 1.732$,

$$\begin{aligned}\text M = \frac{7 \times 2.732}{2} = \frac{19.124}{2} \\[1em] = 9.56\ \text{kg}\end{aligned}$$

Hence, the mass of the hanging block must be $\frac{7(\sqrt{3} + 1)}{2}$ kg, that is, nearly 9.56 kg.

Given,

• Weight of the sphere = W
• Radius of the sphere, r = 3 m
• Length of the string, BO = 5 m
• The wall is frictionless

The tension in the string and the normal reaction of the wall have to be found in terms of W.

The forces acting on the sphere are its weight W, acting vertically downwards at the centre O, the tension T along the string, and the normal reaction N of the wall, acting horizontally. The wall is frictionless, so N has no vertical component.

The sphere is uniform and rests in equilibrium under these three forces, so their lines of action must be concurrent. Hence, the line of the string passes through the centre O of the sphere.

Let A be the point at which the sphere touches the wall and B the point on the wall at which the string is attached. Since the radius drawn to the point of contact is perpendicular to the wall, AO is horizontal and

$\text{AO} = \text r = 3\ \text m$

while AB lies along the wall and is therefore vertical. Thus, the triangle ABO is right-angled at A, and

$$\begin{aligned}\text{AB} = \sqrt{(\text{BO})^2 - (\text{AO})^2} = \sqrt{(5)^2 - (3)^2} \\[1em] = \sqrt{25 - 9} = \sqrt{16} \\[1em] = 4\ \text m\end{aligned}$$

Let θ be the angle which the string makes with the horizontal at O. From the triangle ABO,

$\sin \theta = \frac{\text{AB}}{\text{BO}} = \frac{4}{5} \quad \text{and} \quad \cos \theta = \frac{\text{AO}}{\text{BO}} = \frac{3}{5}$

Resolving the tension into its vertical and horizontal components, for the equilibrium of the sphere,

$\text T\sin \theta = \text W \quad \text{...............(1)}$

$\text T\cos \theta = \text N \quad \text{...............(2)}$

From equation (1),

$$\begin{aligned}\text T = \frac{\text W}{\sin \theta} = \frac{\text W}{4/5} \\[1em] = \frac{5\text W}{4} = 1.25\ \text W\end{aligned}$$

Substituting this value in equation (2),

$$\begin{aligned}\text N = \text T\cos \theta = \frac{5\text W}{4} \times \frac{3}{5} \\[1em] = \frac{3\text W}{4} = 0.75\ \text W\end{aligned}$$

Hence, the tension in the string is $\frac{5\text W}{4}$ and the normal reaction of the wall is $\frac{3\text W}{4}$.

Given,

• Mass of the block, m = 15 kg, so its weight W = 15 kg-wt
• Length of the chord A = 4 m and length of the chord B = 3 m
• The angle between the chords A and B at their meeting point is 90°

The tensions in the chords A and B have to be calculated.

The chord C carries the whole weight of the block, so the tension in it is 15 kg-wt, acting vertically downwards at the point where the three chords meet.

Let P and Q be the points at which the chords A and B are fixed to the horizontal ceiling, and let C be the point at which the three chords meet. Then

$\text{PC} = 4\ \text m, \quad \text{QC} = 3\ \text m \quad \text{and} \quad \angle \text{PCQ} = 90^\circ$

Since the triangle PQC is right-angled at C,

$$\begin{aligned}\text{PQ} = \sqrt{(\text{PC})^2 + (\text{QC})^2} = \sqrt{(4)^2 + (3)^2} \\[1em] = \sqrt{16 + 9} = \sqrt{25} \\[1em] = 5\ \text m\end{aligned}$$

PQ lies along the ceiling and is therefore horizontal. Let θ = ∠QPC be the angle which the chord A makes with the horizontal, and α = ∠PQC the angle which the chord B makes with the horizontal. From the right-angled triangle PQC,

$\sin \theta = \frac{\text{QC}}{\text{PQ}} = \frac{3}{5} \quad \text{and} \quad \cos \theta = \frac{\text{PC}}{\text{PQ}} = \frac{4}{5}$

$\sin \alpha = \frac{\text{PC}}{\text{PQ}} = \frac{4}{5} \quad \text{and} \quad \cos \alpha = \frac{\text{QC}}{\text{PQ}} = \frac{3}{5}$

The point C is in equilibrium under the three tensions. Resolving the tensions T_A and T_B into horizontal and vertical components,

Horizontal components :

$\text T_A\cos \theta = \text T_B\cos \alpha \quad \text{...............(1)}$

Vertical components :

$\text T_A\sin \theta + \text T_B\sin \alpha = \text W \quad \text{...............(2)}$

Substituting the values in equation (1),

$$\begin{aligned}\text T_A \times \frac{4}{5} = \text T_B \times \frac{3}{5} \\[1em] 4\text T_A = 3\text T_B \quad \Rightarrow \quad \text T_B = \frac{4}{3}\text T_A\end{aligned}$$

Substituting this in equation (2),

$$\begin{aligned}\text T_A \times \frac{3}{5} + \frac{4}{3}\text T_A \times \frac{4}{5} = 15 \\[1em] \frac{3\text T_A}{5} + \frac{16\text T_A}{15} = 15 \\[1em] \frac{9\text T_A + 16\text T_A}{15} = 15 \\[1em] \frac{25\text T_A}{15} = 15 \\[1em] \text T_A = \frac{15 \times 15}{25} \\[1em] = 9\ \text{kg-wt}\end{aligned}$$

Therefore,

$$\begin{aligned}\text T_B = \frac{4}{3} \times 9 \\[1em] = 12\ \text{kg-wt}\end{aligned}$$

Hence, the tension in the chord A is 9 kg-wt and that in the chord B is 12 kg-wt.

The result may be checked by noting that the chords A and B are at right angles to each other, so the resultant of the two tensions is

$$\begin{aligned}\sqrt{(\text T_A)^2 + (\text T_B)^2} = \sqrt{(9)^2 + (12)^2} = \sqrt{81 + 144} \\[1em] = \sqrt{225} = 15\ \text{kg-wt}\end{aligned}$$

which is exactly the weight of the block, as required for equilibrium.

8N

Reason — For the block P to be at rest, the net force acting on it must be zero. Taking the forces along the vertical line and giving the upward forces the positive sign,

$$\begin{aligned}10 + 15 - 2 - \text F - 15 = 0 \\[1em] 25 - 17 - \text F = 0 \\[1em] \text F = 8\ \text N\end{aligned}$$

90°

Reason — The magnitude of the linear momentum is

$$\begin{aligned}\text p = \sqrt{\text p_x^2 + \text p_y^2} = \sqrt{(2\cos \text t)^2 + (2\sin \text t)^2} \\[1em] = 2\sqrt{\cos^2 \text t + \sin^2 \text t} \\[1em] = 2\end{aligned}$$

which is a constant. Differentiating the components with respect to time, the components of the force are

$\text F_x = \frac{\text{dp}_x}{\text{dt}} = -2\sin \text t \quad \text{and} \quad \text F_y = \frac{\text{dp}_y}{\text{dt}} = 2\cos \text t$

Taking the scalar product,

$$\begin{aligned}\vec{\text F} \cdot \vec{\text p} = (-2\sin \text t)(2\cos \text t) + (2\cos \text t)(2\sin \text t) \\[1em] = 0\end{aligned}$$

Since the scalar product is zero and neither vector is a null vector, cos θ = 0 and hence θ = 90°. This is expected, because the magnitude of the momentum is constant and only its direction changes.

34.6 kg

Reason — By Newton's second law of motion, F = ma, so

$$\begin{aligned}\text m = \frac{\text F}{\text a} = \frac{90}{2.6} \\[1em] = 34.6\ \text{kg}\end{aligned}$$

15 km

Reason — For an aircraft executing a horizontal loop with its wings banked at an angle θ,

$\tan \theta = \frac{\text v^2}{\text{rg}} \quad \Rightarrow \quad \text r = \frac{\text v^2}{\text g\tan \theta}$

Here v = 720 km h^-1 = $720 \times \frac{5}{18}$ = 200 m s^-1,

tan 15° = 0.27 and

g = 10 m s^-2.

Substituting the values,

$$\begin{aligned}\text r = \frac{200 \times 200}{10 \times 0.27} = \frac{40000}{2.7} \\[1em] = 14814\ \text m \\[1em] = 14.8\ \text{km} \approx 15\ \text{km}\end{aligned}$$

2 ms^-2

Reason — The centripetal acceleration is

$$\begin{aligned}\text a_c = \frac{\text v^2}{\text r} = \frac{(10)^2}{50} = \frac{100}{50} \\[1em] = 2\ \text{m s}^{-2}\end{aligned}$$

2.92 ms^-2

Reason — The cyclist completes one revolution in T = 20 s, so the linear speed is

$$\begin{aligned}\text v = \frac{2\pi \text r}{\text T} = \frac{2 \times 3.14 \times 30}{20} \\[1em] = \frac{188.4}{20} \\[1em] = 9.42\ \text{m s}^{-1}\end{aligned}$$

Therefore, the centripetal acceleration is

$$\begin{aligned}\text a_c = \frac{\text v^2}{\text r} = \frac{(9.42)^2}{30} = \frac{88.74}{30} \\[1em] = 2.96\ \text{m s}^{-2}\end{aligned}$$

which is closest to the value given in this option.

Note: The exact value is 2.96 m s^-2, while the option printed in the textbook reads 2.92 m s^-2. The nearest available alternative has been chosen.

18750 N

Reason — The centripetal force acting on the car is

$$\begin{aligned}\text F_c = \frac{\text{mv}^2}{\text r} = \frac{1500 \times (25)^2}{50} \\[1em] = \frac{1500 \times 625}{50} \\[1em] = 18750\ \text N\end{aligned}$$

5600 ms^-1

Reason — For a satellite in a circular orbit, the gravitational force provides the necessary centripetal force. Hence, if g' is the acceleration due to gravity at that place and R the radius of the orbit,

$\frac{\text{mv}^2}{\text R} = \text{mg}' \quad \Rightarrow \quad \text v = \sqrt{\text g'\text R}$

Substituting g' = 4.9 m s^-2 and R = 6.4 × 10⁶ m,

$$\begin{aligned}\text v = \sqrt{4.9 \times 6.4 \times 10^6} = \sqrt{31.36 \times 10^6} \\[1em] = 5.6 \times 10^3 \\[1em] = 5600\ \text{m s}^{-1}\end{aligned}$$

inertial force due to change in velocity

Reason — When the train comes to a stop, the lower part of the body of the passenger, being in contact with the floor, comes to rest along with the train. The upper part of his body, however, continues to remain in motion due to the inertia of motion, and so he falls forward. In the accelerated (non-inertial) frame of the train this effect is described by an inertial (pseudo) force arising from the change in velocity of the train.

linear momentum

Reason — Consider two bodies which exert forces on each other. By Newton's third law of motion the force exerted by the first on the second is equal and opposite to the force exerted by the second on the first,

$\vec{\text F}_{12} = -\vec{\text F}_{21}$

By Newton's second law in its alternative form, each force equals the rate of change of momentum of the body on which it acts,

$\frac{\text{d}\vec{\text p}_1}{\text{dt}} = -\frac{\text{d}\vec{\text p}_2}{\text{dt}} \quad \Rightarrow \quad \frac{\text d}{\text{dt}}(\vec{\text p}_1 + \vec{\text p}_2) = 0$

Hence, the total linear momentum of the system remains constant, which is the law of conservation of linear momentum.

1 : 4

Reason — By Newton's second law of motion, the initial acceleration is

$\text a = \frac{\text F}{\text m}$

When the force is doubled and the mass is reduced to half,

$\text a' = \frac{2\text F}{\text m/2} = \frac{4\text F}{\text m} = 4\text a$

Therefore,

$\frac{\text a}{\text a'} = \frac{\text a}{4\text a} = \frac{1}{4}$

impulsive forces

Reason — A large force which acts on a body for a very short duration of time is called an impulsive force. Examples are the force exerted by a bat on a ball, or the force between two colliding billiard balls. Such a force cannot be measured easily at every instant, so its effect is measured by the impulse, which is equal to the change in momentum produced.

contact force

Reason — Friction comes into play only when the surfaces of two bodies are in actual contact with each other, and it acts parallel to the surfaces in contact. Hence, it is a contact force. On the modern theory it arises from the strong adhesive forces between the molecules of the two surfaces at the points of actual contact.

Frictional force

Reason — Friction is a non-conservative force. The work done against friction is not recoverable, since it is dissipated as heat and noise, and the work done depends on the path followed. The gravitational, electrostatic and nuclear forces are conservative, because the work done by them in taking a body from one point to another does not depend on the path followed.

The contact area between the surfaces

Reason — By the third law of limiting friction, the limiting frictional force is independent of the apparent area of contact between the two surfaces. It depends only on the normal reaction and on the nature of the materials in contact. The modern theory explains this by noting that with an increase in the contact area the adhesive force also increases in the same ratio, so that the adhesive pressure responsible for friction remains constant. The other three factors listed do affect the frictional force.

greater than kinetic friction

Reason — The limiting friction, which is the maximum value of static friction, is greater than the kinetic friction between the same pair of surfaces, so that μ_s > μ_k. This is why a larger force is required to set a body in motion than to keep it moving once the motion has started.

centripetal acceleration

Reason — In uniform circular motion the magnitude of the velocity remains constant, but its direction changes continuously. A change in the direction of velocity is itself a change in velocity, so the body is accelerated. This acceleration is directed radially inwards, towards the centre of the circular path, and is called centripetal acceleration. Its magnitude is $\frac{\text v^2}{\text r}$.

towards the centre

Reason — The centripetal force is the force which is required to keep a body moving in a circular path, and it is always directed along the radius towards the centre of the circular path. It is therefore perpendicular to the instantaneous velocity of the body, and so it changes only the direction of the velocity and not its magnitude.

halved

Reason — The centripetal force required is

$\text F_c = \frac{\text{mv}^2}{\text r}$

For a constant velocity, $\text F_c \propto \frac{1}{\text r}$. Hence, if the radius is doubled, the centripetal force required becomes half of its initial value.

ML T^-2

Reason — In Newton's third law of motion, action and reaction are the two forces which two bodies exert on each other. Hence, action is a force, and its dimensions are the dimensions of force,

$$\begin{aligned}[\text{Force}] = [\text{mass}] \times [\text{acceleration}] = \text M \times \text{LT}^{-2} \\[1em] = \text{MLT}^{-2}\end{aligned}$$

proportional to t

Reason — By Newton's second law of motion in its alternative form, the net force is the rate of change of momentum,

$\text F = \frac{\text{dp}}{\text{dt}}$

Differentiating p = a + bt² with respect to time, and remembering that a and b are constants,

$$\begin{aligned}\text F = \frac{\text d}{\text{dt}}(\text a + \text{bt}^2) \\[1em] = \frac{\text{da}}{\text{dt}} + \text b\frac{\text d}{\text{dt}}(\text t^2) \\[1em] = 0 + \text b(2\text t) \\[1em] = 2\text{bt}\end{aligned}$$

Since 2b is a constant, F ∝ t.

$\frac{40}{3}$ N

Reason — The two blocks are in contact on a smooth platform, so they move together with a common acceleration. By Newton's second law,

$$\begin{aligned}\text a = \frac{\text F}{\text m_A + \text m_B} = \frac{20}{5 + 10} = \frac{20}{15} \\[1em] = \frac{4}{3}\ \text{m s}^{-2}\end{aligned}$$

The insect is trapped between the two blocks, so the force it experiences is the contact force N between them. Considering the 10 kg block, this contact force alone accelerates it,

$$\begin{aligned}\text N = \text m_B\text a = 10 \times \frac{4}{3} \\[1em] = \frac{40}{3}\ \text N\end{aligned}$$

law of inertia

Reason — Newton's first law states that a body continues in its state of rest or of uniform motion in a straight line unless it is compelled by an external force to change that state. This is precisely the property of inertia, by virtue of which a body is unable to change its state of rest or of uniform motion by itself. Hence, the first law is also called the law of inertia.

$-(3\hat{i} + 4\hat{k})$

Reason — For a body to be in equilibrium under the action of three concurrent forces, their vector sum must be zero,

$\vec{\text A} + \vec{\text B} + \vec{\text C} = 0 \quad \Rightarrow \quad \vec{\text C} = -(\vec{\text A} + \vec{\text B})$

Adding the corresponding components,

$$\begin{aligned}\vec{\text A} + \vec{\text B} = (1 + 2)\hat{i} + (1 - 1)\hat{j} + (1 + 3)\hat{k} \\[1em] = 3\hat{i} + 0\hat{j} + 4\hat{k}\end{aligned}$$

Therefore,

$\vec{\text C} = -(3\hat{i} + 4\hat{k})$

120°

Reason — By the parallelogram law of vector addition, the magnitude of the resultant of two forces A and B inclined at an angle θ is

$\text R = \sqrt{\text A^2 + \text B^2 + 2\text{AB}\cos \theta}$

Here A = B = R = 2 N. Squaring both sides and substituting,

$$\begin{aligned}(2)^2 = (2)^2 + (2)^2 + 2 \times 2 \times 2 \times \cos \theta \\[1em] 4 = 8 + 8\cos \theta \\[1em] 8\cos \theta = -4 \quad \Rightarrow \quad \cos \theta = -\frac{1}{2}\end{aligned}$$

Therefore, θ = 120°.

$\left(\frac{m}{M+m}\right)v$

Reason — The bullet gets embedded in the block, so the two move together after the impact. No external force acts on the system, so by the principle of conservation of linear momentum,

$\text{mv} + \text M \times 0 = (\text M + \text m)\text v'$

$\text v' = \left(\frac{\text m}{\text M + \text m}\right)\text v$

impulse

Reason — The area under a force-time graph is the product of force and the time interval for which it acts. By definition, this product is the impulse of the force,

$\text J = \text F \times \Delta \text t$

For a variable force, the impulse is obtained by integration, $\text J = \displaystyle\int \text F\text{dt}$, which again is the area enclosed by the force-time graph.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When the bus suddenly starts moving forward, the lower part of the passenger's body, being in contact with the bus, moves along with it, while the upper part tends to remain at rest. Hence, he feels a backward push and leans backwards.

Reason (R) is also correct: By Newton's first law of motion, a body continues in its state of rest unless compelled by an external force to change that state. It is this inertia of rest of the passenger's body that resists the change in motion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: This is a statement of Newton's first law of motion, which says that a body continues in its state of rest or of uniform motion in a straight line unless an external force compels it to change that state.

Reason (R) is also correct: By Newton's second law of motion, F = ma, so for a given mass the acceleration produced is directly proportional to the net force acting on the body.

However, the Reason is a statement of the second law, while the Assertion is a statement of the first law. It states the relation between force and acceleration but does not by itself explain why a body continues to move with a uniform velocity.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: This is a statement of Newton's first law of motion. If the net external force on a body is zero, its acceleration is zero, so it remains at rest or continues to move with a uniform velocity in a straight line.

Reason (R) is also correct: This is a statement of Newton's third law of motion, according to which to every action there is an equal and opposite reaction.

However, the third law deals with the pair of forces which two bodies exert on each other, whereas the Assertion concerns the motion of a single body under zero net force. Hence, the Reason does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For the same pair of surfaces, the coefficient of static friction is greater than the coefficient of kinetic friction, that is, μ_s > μ_k. The limiting friction, which is the maximum value of the static friction, is therefore greater than the kinetic friction.

Reason (R) is also correct: Since a larger force is needed to overcome the limiting friction and start the motion than to overcome the kinetic friction and maintain it, it is easier to keep an object moving than to start it moving from rest.

This everyday observation is exactly the consequence of μ_s being greater than μ_k, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: By the third law of limiting friction, the frictional force is independent of the apparent area of contact between the two surfaces. It depends only on the normal reaction and on the nature of the materials in contact.

Reason (R) is correct: The limiting frictional force is directly proportional to the normal reaction, f_s = μ_sR, which is the first law of limiting friction.

Therefore, assertion is false but reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: An object moving in a circular path with a constant speed does have an acceleration. Although the magnitude of its velocity is constant, the direction of the velocity changes continuously, and a change in direction is itself a change in velocity.

Reason (R) is correct: This changing direction of velocity gives rise to the centripetal acceleration, of magnitude $\frac{\text v^2}{\text r}$, which is directed radially inwards towards the centre of the circle.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A car moving in a circular path at a constant speed is accelerated, because its velocity, being a vector, changes continuously in direction.

Reason (R) is also correct: In circular motion the velocity is always along the tangent to the path, so its direction changes from point to point. This continuous change of direction is precisely what produces the centripetal acceleration.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: At the lowest point of the swing the tension in the string has to balance the weight of the bob and, in addition, to provide the centripetal force required for the circular motion, so that $\text T = \text{mg} + \frac{\text{mv}^2}{\text r}$. Since the speed is also greatest there, the tension is maximum at the lowest point.

Reason (R) is incorrect: At the lowest point, the gravitational force acts vertically downwards. However, the centripetal force (the net force required to keep the bob in a circular path) is directed towards the center of the circle, which is vertically upwards. Therefore, they act in opposite directions, not the same direction.

Therefore, assertion is true but reason is false

Note: The answer key is incorrect. The assertion is true, but the reason is false because gravity acts away from the centre at the lowest point. Hence, option 3 is correct.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a car suddenly stops, the lower part of the body of a passenger comes to rest along with the car, while the upper part continues to move forward. Hence, the passengers lurch forward.

Reason (R) is also correct: By virtue of the inertia of motion, the body of a passenger resists any change in its state of motion, and this is exactly why the passengers are thrown forward.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A satellite orbiting the earth is in a continuous state of free fall towards the earth. It does not strike the earth only because its tangential velocity carries it forward at just the rate at which it falls, so that it keeps moving in its orbit.

Reason (R) is also correct: The gravitational force exerted by the earth on the satellite provides exactly the centripetal force needed for the circular motion. Since gravity is the only force acting on it, the satellite is in free fall.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: An object moving in a circular path with a constant speed has a centripetal force acting on it, directed towards the centre of the circle. Hence, the net force on it is not zero.

Reason (R) is correct: The object in circular motion is accelerating towards the centre of the circle, the centripetal acceleration being $\frac{\text v^2}{\text r}$. By Newton's second law, a non-zero acceleration means a non-zero net force.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: On a banked road the outer edge is raised above the inner edge. If the road is banked at the correct angle for a given speed, the vehicle can negotiate the curve even if the road is perfectly smooth, that is, without relying on friction at all. The condition is $\text v = \sqrt{\text{rg}\tan \theta}$.

Reason (R) is also correct: When the road is banked, the normal reaction N is no longer vertical. Its horizontal component N sin θ is directed towards the centre of the curve and provides the necessary centripetal force, while the vertical component N cos θ balances the weight.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The frictional force on a body always acts in a direction opposite to the direction in which the body moves, or tends to move, relative to the other surface.

Reason (R) is also correct: Friction is a resistive force which, by its very nature, opposes the relative motion between two surfaces in contact. This is exactly why it acts opposite to the direction of motion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: By the third law of kinetic friction, the kinetic frictional force is relatively independent of the relative speed of the surfaces, at least within a certain range of speeds. For most practical purposes the kinetic friction is taken to be constant regardless of speed.

Reason (R) is correct: The frictional force is directly proportional to the normal reaction between the two surfaces, f_k = μ_kR.

Therefore, assertion is false but reason is true.

Note: The answer key printed in the textbook marks this question as "assertion is true but reason is false". The textbook's own Hints, however, state that kinetic friction is independent of velocity and that the frictional force is proportional to the normal force, which supports the choice made above.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In circular motion the velocity of the object is always along the tangent to the path, while the centripetal force is always directed along the radius towards the centre. The radius is perpendicular to the tangent, so the centripetal force is always perpendicular to the direction of motion.

Reason (R) is also correct: Because the force is perpendicular to the velocity, it has no component along the direction of motion and therefore cannot change the speed. It only changes the direction of the velocity, which is exactly why it must remain perpendicular to the motion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: An astronaut in a space station orbiting the earth does feel weightless, because both the astronaut and the station are in free fall, accelerating towards the earth at the same rate. Hence, there is no normal reaction between the astronaut and the floor of the station.

Reason (R) is false: Centrifugal force is not a real force at all; it is a pseudo force which appears only in a rotating (non-inertial) frame of reference. The weightlessness is due to the free fall of the astronaut along with the station, and not due to any balancing of gravity by a centrifugal force.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a box is pushed with a constant force on a rough horizontal surface and the applied force is just equal to the kinetic frictional force, the box moves with a constant velocity.

Reason (R) is also correct: In this situation the applied force is balanced by the frictional force, so the net force on the box is zero. By Newton's first law, the box then continues to move with a uniform velocity.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a car takes a sharp turn at high speed, the passengers do tend to slide outward, away from the centre of the turn.

Reason (R) is also correct: The passengers require a centripetal force to move along the circular path. If the friction between the passengers and the seat is not sufficient to provide this force, their bodies continue to move along the straight line by inertia while the car turns inwards, and so they appear to slide outward.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In uniform circular motion the kinetic energy of the object, $\frac{1}{2}\text{mv}^2$, remains constant, since the centripetal force is perpendicular to the velocity and therefore does no work.

Reason (R) is also correct: The kinetic energy depends only on the mass and the speed of the object. The speed remains constant in uniform circular motion, so the kinetic energy also remains constant.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: A cyclist moving in a circular track does experience a frictional force. On a level circular track it is precisely the friction between the tyres and the road that provides the necessary centripetal force, without which the cyclist could not take the turn at all.

Reason (R) as stated is also not correct in this context: Friction is needed not only to start the motion but also to maintain the circular motion, since it supplies the centripetal force at every instant.

Since the given alternatives do not include an option in which both the assertion and the reason are false, the alternative in which the assertion is false has to be chosen.

Therefore, the assertion is false.

Note: Both the assertion and the reason are false. However, no option for “both are false” is given. Therefore, the options need to be corrected.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: The angle of friction θ_s is the angle which the resultant of the normal reaction and the limiting friction makes with the normal reaction, and it satisfies tan θ_s = μ_s. The angle of repose is the minimum angle of inclination of a plane at which a body placed on it just begins to slide down, and it also satisfies tan θ_s = μ_s. Hence, the two angles are equal.

Reason (R) is also correct: When a body is just at the point of motion, the force of friction acting on it has its maximum value, and this maximum value of static friction is called limiting friction.

However, the Reason merely defines limiting friction. It does not explain why the angle of repose should be equal to the angle of friction.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

The fan continues to rotate due to the inertia of motion. When the current is switched off, no driving force acts on the blades, but by virtue of their inertia they tend to continue rotating. The friction in the bearings and the friction of air gradually oppose this motion and finally bring the fan to rest.

Newton's first law states that a body continues in its state of rest or of uniform motion in a straight line unless it is compelled by an external force to change that state.

This is exactly the property of inertia, by virtue of which a body is unable, by itself, to change its state of rest or of uniform motion in a straight line. Since the first law simply states this property of matter, it is also called the law of inertia.

The second law of motion gives the measure of force. It states that the force acting on a body is the product of its mass and the acceleration produced, F = ma, which enables force to be measured quantitatively. The first law only gives a qualitative definition of force.

Yes. A body acted upon by four external forces can remain at rest provided the vector sum of all the four forces is zero, so that the net force on it is zero. By Newton's first law, the body then continues in its state of rest.

No. A train moving on a circular track is continuously accelerated, since the direction of its velocity changes at every instant. A frame of reference which is accelerated is a non-inertial frame, so the train moving on a circular track is not an inertial frame of reference.

The moment the astronaut is thrown out of the spaceship, no external force acts on him, since he is in interstellar space far from all material objects. Hence, by Newton's first (or second) law, the acceleration of the astronaut is zero and he continues to move with the velocity he had at that instant.

(a) No. When the body moves with a uniform velocity, its acceleration is zero, so by Newton's second law the net force acting on it is zero.

(b) Yes. When the body takes a turn with a uniform speed, the direction of its velocity changes continuously. Hence, it is accelerated, and a force must be acting on it. This is the centripetal force, directed towards the centre of the turn.

Given,

• Mass of the trolley = 300 kg, mass of the sand bag = 25 kg
• Uniform speed, v = 27 km h^-1
• The track is smooth, and the sand leaks out at 0.05 kg s^-1

The trolley is moving uniformly, that is, without acceleration, on a smooth track, so no external force acts on the trolley-bag system. The sand leaks out vertically through a hole in the floor and carries away with it exactly the horizontal velocity it already had, so it exerts no horizontal force on the trolley.

Hence, the speed of the trolley remains 27 km h^-1 even after the entire sand bag is empty.

(i) If the distance travelled is directly proportional to time, the body moves with a uniform velocity. Its acceleration is zero, so the net external force acting on it is zero.

(ii) Yes. A body can remain at rest when external forces act on it, provided there is more than one external force and their vector sum is zero.

(iii) No. If the net force on a body is zero, the body need not be at rest. It may also be moving with a constant velocity in a straight line.

(iv) False. It is quite possible for a body to be moving with a constant velocity while the net external force acting on it is zero.

(v) If the force acts perpendicular to the direction of motion, it has no component along the direction of motion. Hence, there is no change in the speed of the body, but its direction of motion changes.

(vi) To the person sitting in the compartment, the ball appears to go vertically straight up and come straight back, because he and the ball share the same uniform horizontal velocity. To a person standing outside, the ball has a constant horizontal velocity in addition to its vertical motion, so it appears to describe a parabolic path.

(vii) In vector form, the second law of motion is written as

$\vec{\text F} = \text m\vec{\text a} = \frac{\text d\vec{\text p}}{\text{dt}}$

The reading of the balance will be 10 kg-wt.

A spring balance reads the tension in its spring. When the two ends are pulled by forces of 10 kg-wt each in opposite directions, the balance is in equilibrium and the tension in the spring is 10 kg-wt, not the sum of the two forces. The second force merely holds the other end of the balance in position, exactly as a fixed support would.

The increase in length will be 0.1 cm.

When a 1 kg body is suspended from the spring, the upper end is held by the support, which exerts an equal and opposite force on the spring. Thus, the spring is already being stretched by equal and opposite forces of 1 kg-wt at its two ends. When it is laid on a frictionless table and 1 kg bodies are suspended from each end, the situation is exactly the same, since the tension in the spring is again 1 kg-wt. Hence, the increase in length remains 0.1 cm.

He can get away from the frictionless surface by throwing some heavy object, or by blowing air out of his mouth, in the direction opposite to that in which he wants to move.

By the principle of conservation of linear momentum, the momentum given to the object (or to the air) in one direction gives him an equal momentum in the opposite direction, and he starts moving.

He should first fire a bullet in the direction opposite to the line joining A to B. By the principle of conservation of linear momentum he himself recoils, that is, he begins to move towards B along the frictionless ice.

On reaching B, he should fire the second bullet in the direction of the line joining A to B. The recoil now acts in the backward direction and brings him to rest at B.

No, the boat will not start moving.

When the fan throws air on the sail, the air pushes the sail forward, but at the same time the air pushes the fan backward with an equal and opposite force, by Newton's third law. Since the fan is itself a part of the boat, these are internal forces of the boat-fan-air system, and their vector sum is zero. Hence, the boat cannot move. It would move only under a reaction from some external agency.

No, the bubbles will not rise in the water.

The water in the freely falling bottle is in a state of weightlessness. Hence, the pressure in the water does not increase with depth, and no upthrust acts on the bubbles of gas. In the absence of any upthrust, the bubbles do not rise up.

Yes, the cage will appear lighter than before.

The cage is made of wire, so the air inside the cage is not bound with the cage but is in contact with the atmospheric air outside. When the bird starts flying inside the cage, the reaction of the air pushed down by its wings is not transmitted to the cage. Hence, the weight of the bird is no longer experienced and the cage appears lighter.

In a closed glass cage the air inside is bound with the cage, so the reaction of the air pushed down by the wings of the bird is transmitted to the cage.

(i) When the bird flies with a constant velocity, the downward push on the air just balances the weight of the bird. Hence, there is no change in the weight of the cage.

(ii) When the bird flies upward with acceleration, it pushes the air down with a force greater than its weight. Hence, the cage feels heavier than before.

(iii) When the bird flies downward with acceleration, it pushes the air down with a force less than its weight. Hence, the cage feels lighter than before.

An observer whose frame of reference is unaccelerated is an inertial observer, and one whose frame is accelerated is a non-inertial observer.

(a) A child revolving in a "giant wheel" is in circular motion and is therefore accelerated. He is a non-inertial observer.

(b) The driver in a car moving with a constant speed on a straight road has unaccelerated motion. He is an inertial observer.

(c) A pilot in an aircraft which is taking off is accelerated. He is a non-inertial observer.

(d) A cyclist negotiating a sharp turn is accelerated, since the direction of his velocity changes. He is a non-inertial observer.

(e) A guard on a train which is slowing down to a stop is under retardation, that is, he is accelerated. He is a non-inertial observer.

When the second boy pulls the rope, he exerts a force on the first boy through the rope, and by Newton's third law the first boy exerts an equal and opposite force on him.

The surface is frictionless, so no external horizontal force acts on the system and the total linear momentum, which was initially zero, must remain zero. Since the two boys have the same mass, both of them move towards each other with the same speed and meet at the mid-point of the initial distance between them.

Limiting friction is the greatest.

The static friction is a self-adjusting force whose value increases with the applied force up to a maximum, and this maximum value of static friction is called limiting friction. Once the motion begins, the friction falls to the value of kinetic friction. Thus,

$\text{Kinetic friction} \lt \text{Limiting friction}$

There is practically no change in the limiting friction.

By the third law of kinetic friction, the frictional force is reasonably independent of the relative speed of the two surfaces in contact, at least within a certain range of speeds. Hence, moving the wooden block faster does not appreciably alter the frictional force.

Sliding (or kinetic) friction arises when an axle rotates in a sleeve, since the surface of the axle slides over the surface of the sleeve. It is to convert this sliding friction into the much smaller rolling friction that ball bearings are used.

The force of friction is independent of the area of contact of the two surfaces, within wide limits. It depends only on the normal reaction and on the nature of the materials in contact. On the modern theory this is so because, with an increase in the contact area, the adhesive force also increases in the same ratio, so that the adhesive pressure responsible for friction remains constant.

The coefficient of friction depends on

• the nature of the materials of both the surfaces in contact, including their roughness,
• the cleanliness of the surfaces,
• the surface treatment, that is, whether the surfaces are polished or lubricated, and
• the temperature and the humidity.

It does not depend on the area of contact or on the normal reaction.

There is no change in the coefficient of friction.

The coefficient of friction depends only on the nature of the two surfaces in contact and not on the weight of the body. However, since

$\text f_s = \mu_s\text R = \mu_s\text{mg}$

the force of limiting friction becomes doubled when the weight is doubled.

Sand is thrown on the tracks covered with snow in order to increase the force of friction between the tyres of the vehicles and the road. Snow is very smooth and the friction between the tyres and the snow is very small, so the tyres are likely to slip. The rough sand raises the coefficient of friction and the vehicles can move safely.

This is because the coefficient of friction between rubber tyres and the road is much smaller than that between iron wheels and the road. Hence, a smaller force of friction has to be overcome in the case of rubber tyres, and the carts are easier to ply.

When the brakes are on, the wheels cannot rotate freely and can only skid over the road. In this way the rolling friction is converted into sliding friction, which is comparatively much larger. Hence, a much greater force has to be applied and it becomes difficult to move the cycle.

No, a large brake is not more effective than a small one.

The force of friction does not depend on the surface area of contact. It depends only on the normal reaction and on the nature of the materials in contact. Hence, increasing the size of the brake does not by itself increase the braking force.

Wheels are made circular so that, when a vehicle moves, the wheels roll over the road instead of sliding over it. In this way the sliding friction is converted into the much smaller rolling friction, and the effort needed to move the vehicle is greatly reduced.

When a barrel is rolled, only rolling friction has to be overcome, whereas when it is slid, sliding friction has to be overcome. Since rolling friction is much smaller than sliding friction, much less effort is needed and it is easier to roll the barrel than to slide it.

In order to walk, we press the ground backward with our feet, and the horizontal component of the reaction of the ground pushes us forward. This horizontal component is the force of friction.

On ice the friction between our feet and the ice is very small, so the horizontal reaction available is very small and we are unable to push ourselves forward. Hence, it is difficult to walk on ice.

No, we cannot jump off a frictionless horizontal surface.

To jump, we must press the surface with our feet and obtain a reaction from it. A frictionless surface cannot offer the horizontal component of the reaction which is needed to propel the body, so no jump is possible.

The irregular projections on the surfaces of automobile tyres increase the frictional grip between the tyres and the road. This larger friction supplies the centripetal force needed at turns and prevents the wheels from slipping at high speeds and on wet roads.

Lubricants are used in machines to reduce friction between the moving parts. By reducing friction, they reduce the wear and tear of the machine parts, prevent overheating and save a part of the fuel or energy which would otherwise be spent in overcoming friction.

When a lubricant such as oil or grease is introduced between two surfaces in contact, it forms a thin layer between them so that the two surfaces are no longer in direct contact. In this way the sliding friction between the solid surfaces is replaced by the liquid friction between the layers of the lubricant, which is much smaller in magnitude.

Fast moving vehicles are given a streamline shape, that is, sharp in front, in order to reduce the friction due to air considerably. If the shape were blunt, the air would offer a much larger resistance at high speed, and more fuel would be consumed in overcoming it. Aircraft, jet planes and fast cars are all streamlined for this reason.

The angle is 180°.

The force of friction always opposes the relative motion between the surfaces in contact, so it acts in a direction exactly opposite to the instantaneous velocity of the vehicle. When the vehicle moves forth and back, the direction of the frictional force also reverses, and it always remains antiparallel to the instantaneous velocity.

(i) In turning a car : from the force of friction between the tyres and the road.

(ii) In revolving a ball tied to a string in a circle : from the tension in the string.

(iii) In the earth's revolution around the sun : from the gravitational force of attraction exerted on the earth by the sun.

(iv) In the revolution of an electron around the nucleus : from the electrostatic force of attraction between the negatively-charged electron and the positively-charged nucleus.

On a wet road the friction between the tyres and the road is greatly reduced. The centripetal force required to take the turn is supplied by this friction, so when the friction becomes too small to provide the necessary centripetal force, the scooter slips instead of taking the turn.

The centripetal force needed to keep the ball moving in a circle along with the rotating disc can only be provided by the force of friction between the ball and the disc. Since both the ball and the disc are smooth, this friction is absent, so the necessary centripetal force is not provided to the ball and it falls down.

No. The direction of the centripetal force does not depend on the sense of rotation.

The centripetal acceleration, and hence the centripetal force, is always directed radially inwards, towards the centre of the circular path, whether the body revolves clockwise or anticlockwise.

While the stone oscillates, the tension in the string is greatest when it passes through the mean position. At that position the tension has not only to balance the weight of the stone but also to provide the centripetal force required for the circular motion,

$\text T = \text{mg} + \frac{\text{mv}^2}{\text r}$

Since T is much greater than the weight mg of the stone, the string is unable to bear this large tension and breaks.

One newton is that force which, when acting on a body of mass 1 kilogram, produces in it an acceleration of 1 metre per second squared.

$1\ \text N = 1\ \text{kg} \times 1\ \text{m s}^{-2}$

A frame of reference which is accelerated is a non-inertial frame. Examples are a train moving on a circular track, a lift which is accelerating upwards or downwards, a merry-go-round, and a car taking a sharp turn.

$1\ \text N = 10^5\ \text{dyne}$

This follows because 1 N = 1 kg × 1 m s^-2 = 10³ g × 10² cm s^-2 = 10⁵ g cm s^-2 = 10⁵ dyne.

The coefficient of limiting friction is equal to the tangent of the angle of repose,

$\mu_s = \tan \theta_s$

where θ_s is the angle of repose, that is, the minimum angle of inclination of a plane with the horizontal at which a body placed on it just begins to slide down.

Given,

• A body of weight W stretches each of the springs A and B by 5 cm

Each spring is stretched by 5 cm under a tension W, so the force constant of each spring corresponds to an extension of 5 cm per weight W.

When the spring B, together with the weight W, is suspended from the spring A, the tension in the spring B is W (due to the weight hanging from it) and the tension in the spring A is also W (since the springs are weightless). Hence, each spring is stretched by 5 cm.

If now the weight W is removed, the spring B loses its extension of 5 cm and the spring A also loses its extension of 5 cm. The lower end of the spring B therefore rises through the sum of the two contractions,

$5 + 5 = 10\ \text{cm}$

Hence, the increase in each spring is 5 cm, and on removing W the lower end of the spring B rises through 10 cm.

(a) To the observer sitting in the cabin : The cabin and the body both fall freely with the same acceleration g, so there is no relative acceleration between them. Hence, the body appears to remain stationary in air in front of the observer.

(b) To the observer standing on the earth : He is in an inertial frame and sees the body under the action of the gravity force. Hence, the body appears to fall with the acceleration due to gravity g.

No, the angular momentum of the particle does not change.

At any instant the velocity of the particle is directed along the tangent to the circle. When the string is cut, the centripetal force acting on the particle vanishes, so by inertia the particle flies away tangentially in a straight line. Since no external torque acts on it about the axis of revolution, its angular momentum about that axis, mvr, continues to remain constant.

(i) In the spring balance the reading depends on the apparent weight of the body, so when the elevator moves with an upward acceleration the reading of the spring balance increases. In the physical balance the two pans are affected in exactly the same way, so there is no effect on the equilibrium of the physical balance.

(ii) When the lift is accelerated upward, the apparent weight is R = m(g + a), which is greater than the real weight mg. Hence, the apparent weight increases. If the lift goes up with a uniform speed, the acceleration is zero and the apparent weight remains the same as the real weight.

(iii) During the jump the thief and the box both fall freely with the acceleration due to gravity, so there is no reaction between the box and his head. Hence, the weight of the box as experienced by the thief is zero, that is, the box is in the state of weightlessness.

(iv) Given, m = 0.5 kg and u = 10 m s^-1. The wall is perfectly elastic, so the ball rebounds with the same speed, v = −10 m s^-1. The change in momentum is

$$\begin{aligned}\Delta \text p = \text{mv} - \text{mu} = 0.5 \times (-10) - 0.5 \times 10 \\[1em] = -5 - 5 \\[1em] = -10\ \text{kg m s}^{-1}\end{aligned}$$

Hence, the change in momentum of the ball is 10 kg m s^-1, directed opposite to the initial direction of motion.

Given,

• Three particles A, B and C of equal mass m move with equal speeds v along the medians of an equilateral triangle
• After the collision, A comes to rest and B retraces its path with the same speed v

The velocity of C after the collision has to be found.

The three particles move along the medians towards the centroid G, so their three velocities are inclined to one another at 120°. Since the three momenta are equal in magnitude (mv) and are inclined at 120°, their vector sum is zero. Hence, the total linear momentum of the system before the collision is

$\vec{\text p}_{\text{initial}} = 0$

No external force acts on the system during the collision, so by the principle of conservation of linear momentum the total momentum after the collision must also be zero,

$\text m\vec{\text v}_A' + \text m\vec{\text v}_B' + \text m\vec{\text v}_C' = 0$

After the collision A comes to rest, so $\vec{\text v}_A' = 0$, and B retraces its path with the same speed, so $\vec{\text v}_B' = -\vec{\text v}_B$. Therefore,

$$\begin{aligned}\text m(-\vec{\text v}_B) + \text m\vec{\text v}_C' = 0 \\[1em] \vec{\text v}_C' = \vec{\text v}_B\end{aligned}$$

But $\vec{\text v}_B$ was directed towards G, whereas B now moves away from G. Hence, the velocity of C after the collision has the magnitude v and is directed opposite to the direction along which B returns.

Hence, the velocity of C is v, directed opposite to the direction of return of B.

Given,

• Mass of the body, m = 2 kg

The reading of the spring balance is the apparent weight of the body.

When the lift falls downward with the acceleration due to gravity g : Here a = g, so

$$\begin{aligned}\text R = \text m(\text g - \text a) = \text m(\text g - \text g) \\[1em] = 0\end{aligned}$$

Hence, the reading of the balance is zero, that is, the body is in the state of weightlessness.

When the lift goes upward with the same acceleration g : Here the apparent weight is

$$\begin{aligned}\text R = \text m(\text g + \text a) = \text m(\text g + \text g) = 2\text{mg} \\[1em] = 2 \times 2 \times \text g \\[1em] = 4\text g\ \text N = 4\ \text{kg-wt}\end{aligned}$$

Hence, the reading of the balance is 4 kg-wt.

The coefficient of static friction is equal to the tangent of the angle of friction,

$\mu_s = \tan \theta_s$

where θ_s is the angle of friction, that is, the angle which the resultant of the normal reaction and the limiting friction makes with the normal reaction, in the state of limiting friction.

Given,

• Coefficient of friction, $\mu_s = \frac{1}{\sqrt{3}}$

The angle of friction has to be calculated.

The angle of friction θ_s is related to the coefficient of friction by

$\tan \theta_s = \mu_s$

Substituting the given value,

$\tan \theta_s = \frac{1}{\sqrt{3}}$

$$\begin{aligned}\theta_s = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \\[1em] = 30^\circ\end{aligned}$$

Hence, the angle of friction is 30°.

The law of inertia is another name for Newton's first law of motion. It states that a body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled by an external force to change that state.

Inertia is that property of a body by virtue of which it is unable, by itself, to change its state of rest or of uniform motion. It is of two kinds, the inertia of rest and the inertia of motion. For example, a passenger standing in a bus leans backwards when the bus suddenly starts, because of the inertia of rest, and leans forward when the bus suddenly stops, because of the inertia of motion.

Weightlessness is the state in which a body experiences no reaction from its support, so that its apparent weight becomes zero, although the gravity force on it continues to act.

It occurs when the body and its support are both falling freely with the same acceleration, that is, with the acceleration due to gravity. For example, if the rope of a lift breaks, the lift and a man standing in it both fall with the acceleration g. The reaction of the floor on the man is then

$\text R = \text m(\text g - \text g) = 0$

so that the man feels weightless. An astronaut in a satellite orbiting the earth is weightless for the same reason.

Inertia is directly proportional to the mass of a body.

The greater the mass of a body, the greater is the force required to produce a given change in its state of rest or of uniform motion. For example, it is far more difficult to set a loaded truck in motion, or to stop it, than a bicycle. It is for this reason that mass is taken as the quantitative measure of inertia, and mass is often called the measure of inertia of a body.

By Newton's second law of motion, the force acting on a body of mass m producing an acceleration a in it is

$\vec{\text F} = \text m\vec{\text a}$

If no external force acts on the body, then

$\vec{\text F} = 0 \quad \Rightarrow \quad \text m\vec{\text a} = 0$

The mass m of a body can never be zero. Hence,

$\vec{\text a} = 0 \quad \Rightarrow \quad \frac{\text d\vec{\text v}}{\text{dt}} = 0 \quad \Rightarrow \quad \vec{\text v} = \text{constant}$

Thus, in the absence of an external force, the velocity of the body remains constant, that is, the body continues in its state of rest (if v = 0) or of uniform motion in a straight line. This is precisely Newton's first law of motion.

Hence, the first law of motion follows from the second law.

(i) When the car moves with a uniform velocity : There is no acceleration, so no pseudo force acts on the ball. The ball remains suspended vertically in its usual position.

(ii) When the car moves with accelerated motion : In the accelerated frame of the car a pseudo force ma acts on the ball in the backward direction. Hence, the ball moves backward, and the cord makes an angle θ with the vertical given by tan θ = a/g.

(iii) When the car is turning towards the right : The ball needs a centripetal force towards the centre of the turn, which lies on the right. By virtue of its inertia the ball tends to continue in a straight line, so the ball moves towards the left.

Static friction is a self-adjusting force.

When a small horizontal force F is applied to a body lying on a rough surface, the body does not move. This is because the force of static friction f_s comes into play and adjusts itself to be exactly equal and opposite to the applied force, so that f_s = F. As the applied force is gradually increased, the static friction also increases, always remaining equal to the applied force.

This adjustment continues only up to a limit. The maximum value which the static frictional force can take is called the limiting friction, given by

$\text f_s(\text{max}) = \mu_s\text R$

Once the applied force exceeds this limiting value, the friction can no longer increase and the body begins to move. Hence, in general f_s ≤ μ_sR, and it is in this sense that static friction is described as a self-adjusting force.

Rolling friction comes into play when a body such as a wheel, a disc or a cylinder rolls over the surface of another body.

It arises mainly due to the weight of the rolling body, which causes some flattening of that part of the wheel which is in contact with the ground and some denting of the ground itself. The wheel thus has to be continually lifted out of the small depression it creates, and work has to be done in doing so.

The effect is small if both the rim of the wheel and the ground are hard, which is why motor tyres wear out more quickly if they are not kept at the correct pressure. The force of rolling friction is directly proportional to the normal reaction R and inversely proportional to the radius r of the wheel,

$\text f_r = \mu_r\frac{\text R}{\text r}$

The weight of the vehicle and its load causes some flattening of that part of the wheel which is in contact with the ground, and some denting of the ground. Hence, work has to be done in lifting the wheel out of the dent formed on the ground, and this work is done at the cost of the fuel.

If the rubber tyres are kept fully inflated, they are hard and are flattened much less. Less work is therefore done against rolling friction, so less fuel is consumed per unit distance covered and petrol is saved.

In order to start pulling the cart, the horse has to exert a force sufficient to overcome the limiting friction, which is the maximum value of the static friction.

Once the motion has started, the friction opposing the motion is the kinetic friction, which is smaller than the limiting friction. Hence, a smaller force is now sufficient, and the horse has to pull less hard than in the initial stage.

(i) Mud-guards on the wheels of vehicles : The mud particles sticking to the tyres move along a circular path with the wheel. When the centripetal force needed is not available, they fly off along the tangent to the wheel. Mud-guards are fixed so as to intercept the mud thrown off tangentially.

(ii) Banking of the road bed at a circular turning : On a banked road the normal reaction is inclined, and its horizontal component N sin θ provides a part of the centripetal force needed for the turn. Thus, the reliance on friction alone is reduced, and heavy vehicles can negotiate the turn safely at the speed for which the banking angle has been designed, tan θ = v²/rg.

(iii) The outer rail is slightly higher at the turning point : Raising the outer rail banks the track, so that the horizontal component of the normal reaction supplies the necessary centripetal force to the train. The wheel flanges then do not press on the rails, and the wear and tear of the rails and the wheels is reduced.

(iv) On the hills, the roads are made sloping downwards towards the hill : The road is thus banked with its outer edge, which is on the valley side, raised. The horizontal component of the normal reaction then acts towards the centre of the turn and supplies the centripetal force, so that the vehicle does not skid outwards towards the valley.

(v) The wings of an aeroplane are leant inwards while taking a turn : When the aeroplane banks its wings, the lift force becomes inclined. Its horizontal component is directed towards the centre of the turn and provides the necessary centripetal force, while its vertical component balances the weight of the aeroplane.

(vi) In the well of death the rider does not fall : The rider drives on the vertical wall, so the normal reaction of the wall is horizontal and directed towards the centre, supplying the centripetal force. The weight of the rider is balanced by the vertical force of friction between the tyres and the wall. Hence, he does not fall.

(vii) A small smooth ball placed on a smooth rotating circular disc falls down : The centripetal force required to keep the ball moving in a circle with the disc can be provided only by friction between the ball and the disc. Since both are smooth, this friction is absent and the necessary centripetal force is not provided to the ball. Hence, it falls down.

(viii) Water does not fall from an inverted bucket revolved fast in a vertical circle : At the highest point, the weight of the water and the reaction of the bucket together provide the centripetal force needed. If the speed is large enough, the whole of the weight is used up in providing this centripetal force, so there is no force left to make the water fall out of the bucket.

(ix) An electron revolving round the nucleus does not fly away : The electron is acted upon by a centripetal force directed towards the nucleus, which is provided by the electrostatic force of attraction between the negatively-charged electron and the positively-charged nucleus. Hence, it continues to revolve in its orbit.

Cream is lighter than the rest of the milk. When milk is rotated at high speed in a cream separator, every part of it needs a centripetal force to keep moving in a circle. The heavier (denser) milk experiences a larger outward centrifugal tendency and moves away from the axis of rotation, while the lighter cream, being pushed inwards by the heavier milk, collects near the axis. The cream is then drawn off from the centre.

(i) Falling outwards on a revolving merry-go-round : To move in a circle along with the merry-go-round, we need a centripetal force directed towards the centre. When we hold the strings, they pull our body inwards and provide this force. If we do not hold the strings, no such force is available, and by virtue of inertia our body tends to continue in a straight line. In the rotating frame of the merry-go-round this appears as a centrifugal force acting outwards, and we fall outwards.

(ii) The head striking the left wall when the car turns right :

As described by a person standing outside : He is in an inertial frame. To him, the car turns towards the right, while the body of the person sitting inside continues to move along the straight line by virtue of its inertia. Hence, the left wall of the car comes up against the person's head, and no real outward force acts on him.

As described by the person sitting in the car : He is in a rotating, that is, a non-inertial frame. To him, a centrifugal force acts on his body radially outwards, that is, towards the left, and it is this force which pushes his head against the left wall. This centrifugal force is a pseudo force.

(iii) Water not falling from an inverted bucket revolved fast in a vertical circle : At the highest point of the vertical circle, the weight mg of the water and the reaction R of the bucket both act vertically downwards, towards the centre, and together provide the centripetal force,

$\text{mg} + \text R = \frac{\text{mv}^2}{\text r}$

If the bucket is revolved fast enough, the whole of the weight of the water is used up in providing the necessary centripetal force. There is then no force left to make the water fall out, and hence the water does not fall even when the bucket is inverted.

The cord C supports the block and is, in addition, pulled by the force applied through the cord D. Hence, the tension in the cord C is

$\text T_C = \text T + \text{Mg}$

Why C breaks when T is increased steadily : When the force T in the cord D is increased slowly, the block has time to remain in equilibrium at every stage, so the tension in C is always greater than that in D by the weight Mg of the block. Hence, as T is steadily increased, the tension in C reaches the breaking value first, and the cord C breaks even though it is the weaker cord.

Why D breaks on a sudden jerk : When a sudden jerk is given to D, the force acts for a very short interval of time. Owing to its inertia, the block does not move appreciably in this short interval, so the extra tension is not transmitted to the cord C. The whole of the sudden large force is therefore borne by the cord D itself, and hence D breaks rather than C.

The two are related to the coefficient of static friction by μ_r < μ_k < μ_s. It is because rolling friction is so much smaller than sliding friction that vehicles are fitted with wheels and that ball bearings are used in machines.

As the rear wheel rolls forward, the part of the wheel which is in contact with the road tends to slide backwards over the road. The force of friction always opposes this relative motion.

Hence, the force of friction exerted by the earth on the rear wheel acts in the forward direction, along the tangent to the surface of the road at the point of contact. It is this forward frictional force that drives the bicycle. On the front wheel, which is not driven, the frictional force acts in the backward direction.

Let the retarding force F produce a retardation a in the motor car of mass m, so that

$\text a = \frac{\text F}{\text m}$

Since the car finally stops, v = 0. Using the equation of motion $\text v^2 = \text u^2 - 2\text{as}$,

$$\begin{aligned}0 = \text u^2 - 2\text{as} \\[1em] \text s = \frac{\text u^2}{2\text a}\end{aligned}$$

The retarding force, and hence the retardation a, is the same in both the cases. Therefore,

$\text s \propto \text u^2$

If the initial speed is doubled, that is, u' = 2u, then

$\frac{\text s'}{\text s} = \left(\frac{2\text u}{\text u}\right)^2 = 4 \quad \Rightarrow \quad \text s' = 4\text s$

Hence, the car will cover four times the original distance before stopping, that is, three times more distance than before.

Given,

• Three identical blocks A, B and C, each of mass M
• A force F is applied on the frictionless table

Acceleration of the blocks : The three blocks move together as a system of total mass 3M. By Newton's second law,

$\text a = \frac{\text F}{3\text M}$

Net force on the block A : Block A alone has mass M and moves with the same acceleration a. Hence,

$$\begin{aligned}\text F_A = \text{Ma} = \text M \times \frac{\text F}{3\text M} \\[1em] = \frac{\text F}{3}\end{aligned}$$

Force which A applies on B : The blocks B and C together, of mass 2M, are pushed forward only by the contact force exerted on B by A. Hence,

$$\begin{aligned}\text F_{AB} = 2\text M\text a = 2\text M \times \frac{\text F}{3\text M} \\[1em] = \frac{2\text F}{3}\end{aligned}$$

Force which B applies on C : The block C alone, of mass M, is pushed forward by the contact force exerted on it by B. Hence,

$$\begin{aligned}\text F_{BC} = \text{Ma} \\[1em] = \frac{\text F}{3}\end{aligned}$$

Action-reaction pairs : At the contact surface between A and B, A pushes B forward with $\frac{2\text F}{3}$ and B pushes A backward with $\frac{2\text F}{3}$. At the contact surface between B and C, B pushes C forward with $\frac{\text F}{3}$ and C pushes B backward with $\frac{\text F}{3}$.

Given,

• Radius of the chamber, r = 25 m
• Acceleration due to gravity, g = 9.8 m s^-2

Why the motorcyclist does not drop down at the uppermost point : At the uppermost point the motorcyclist is expected to drop down under his own weight, but this does not happen. While driving in vertical loops he experiences a centrifugal force, equal and opposite to the centripetal force. At the uppermost point this centrifugal force $\frac{\text{mv}^2}{\text r}$ is directed vertically upwards, while his weight mg acts vertically downwards. The resultant force $\frac{\text{mv}^2}{\text r} - \text{mg}$ keeps the motorcyclist pressed against the wall of the death-well, and so he does not drop down even though there is no support from below.

Minimum speed : At the uppermost point the minimum condition is that the whole of the weight is just used up in providing the necessary centripetal force,

$$\begin{aligned}\frac{\text{mv}<em>{min}^2}{\text r} = \text{mg} \\[1em] \text v</em>{min} = \sqrt{\text{rg}}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text v_{min} = \sqrt{25 \times 9.8} = \sqrt{245} \\[1em] = 15.65\ \text{m s}^{-1}\end{aligned}$$

Hence, the minimum speed required to perform the vertical loop is 15.65 m s^-1.

Given,

• Kinetic energy of the particle, K = as²
• Radius of the circle = R

Let m be the mass of the particle and v its linear velocity. Then its kinetic energy is

$\text K = \frac{1}{2}\text{mv}^2 = \text{as}^2 \quad \text{...............(1)}$

Tangential force : Differentiating equation (1) with respect to time t,

$\frac{\text d}{\text{dt}}\left(\frac{1}{2}\text{mv}^2\right) = \frac{\text d}{\text{dt}}(\text{as}^2)$

$\frac{1}{2}\text m \times 2\text v\frac{\text{dv}}{\text{dt}} = \text a \times 2\text s\frac{\text{ds}}{\text{dt}}$

$\text{mv}\frac{\text{dv}}{\text{dt}} = 2\text{as}\frac{\text{ds}}{\text{dt}}$

But $\frac{\text{ds}}{\text{dt}} = \text v$, so dividing both sides by v,

$\text m\frac{\text{dv}}{\text{dt}} = 2\text{as}$

This is the tangential force, which is the product of the mass and the tangential acceleration,

$\text F_T = 2\text{as} \quad \text{...............(2)}$

Centripetal force : From equation (1),

$\text v^2 = \frac{2\text{as}^2}{\text m}$

Therefore, the centripetal force on the particle is

$$\begin{aligned}\text F_C = \frac{\text{mv}^2}{\text R} = \frac{\text m}{\text R} \times \frac{2\text{as}^2}{\text m} \\[1em] = \frac{2\text{as}^2}{\text R} \quad \text{...............(3)}\end{aligned}$$

Resultant force : The tangential force and the centripetal force are mutually perpendicular, so the resultant force on the particle is

$\text F = \sqrt{\text F_T^2 + \text F_C^2} = \sqrt{(2\text{as})^2 + \left(\frac{2\text{as}^2}{\text R}\right)^2}$

$= 2\text{as}\sqrt{1 + \frac{\text s^2}{\text R^2}}$

Hence, the force acting on the particle is $2\text{as}\left(1 + \frac{\text s^2}{\text R^2}\right)^{1/2}$.

Friction arises from the adhesive (molecular) forces between the molecules of the two surfaces in contact. When a surface is polished, its irregularities are removed and the actual area of contact between the two surfaces increases. Beyond a certain limit of polishing, the molecules of the two surfaces come very close to each other, so the intermolecular adhesive forces increase considerably. As a result, the friction increases instead of decreasing.

Given,

• Two balls of the same mass m
• Lengths of the strings, l₁ and l₂, with l₁ > l₂
• The centripetal force is the same in the two cases

The centripetal force required for a ball of mass m moving in a horizontal circle of radius l with a speed v is

$\text F = \frac{\text{mv}^2}{\text l}$

Since the centripetal force is the same for the two balls,

$\frac{\text{mv}_1^2}{\text l_1} = \frac{\text{mv}_2^2}{\text l_2}$

The masses are equal, so cancelling m,

$$\begin{aligned}\frac{\text v_1^2}{\text v_2^2} = \frac{\text l_1}{\text l_2} \\[1em] \frac{\text v_1}{\text v_2} = \sqrt{\frac{\text l_1}{\text l_2}}\end{aligned}$$

It is given that l₁ > l₂, so $\frac{\text l_1}{\text l_2} \gt 1$ and therefore $\frac{\text v_1}{\text v_2} \gt 1$.

Hence, v₁ > v₂.

This statement is wrong.

A particle moving uniformly along a circle is not in equilibrium. Its velocity changes continuously in direction, so it has a centripetal acceleration $\frac{\text v^2}{\text r}$ and hence a net force $\frac{\text{mv}^2}{\text r}$ acting radially inwards. No outward force balances this force.

The centrifugal force is not a real force. It is a pseudo force introduced only by an observer in the rotating (non-inertial) frame, in which the particle appears to be at rest. In an inertial frame no such force exists. Hence, the centripetal and centrifugal forces do not act in the same frame of reference and cannot be treated as a pair of balancing forces.

Given,

• Mass of the bullet = mass of the block = 20 g = 0.02 kg
• Height of the tower = 100 m
• Muzzle velocity of the bullet, u = 100 m s^-1 (upwards)
• Initial velocity of the block = 0
• Acceleration due to gravity, g = 10 m s^-2

Let the bullet and the block meet after a time t, at a height h below the top of the tower. In this time the bullet must cover (100 − h) upwards while the block covers h downwards.

For the bullet, using $\text s = \text{ut} - \frac{1}{2}\text{gt}^2$,

$100 - \text h = 100\text t - \frac{1}{2}\text{gt}^2 \quad \text{...............(1)}$

For the block, starting from rest,

$\text h = \frac{1}{2}\text{gt}^2 \quad \text{...............(2)}$

Adding equations (1) and (2),

$100 = 100\text t \quad \Rightarrow \quad \text t = 1\ \text s$

(i) 1 s

Substituting t = 1 s in equation (2),

$$\begin{aligned}\text h = \frac{1}{2} \times 10 \times (1)^2 \\[1em] = 5\ \text m\end{aligned}$$

So they meet 5 m below the top of the tower, that is, at a height of (100 − 5) = 95 m from the ground.

(ii) 95 m

Using v = u + gt for the block, with u = 0,

$$\begin{aligned}\text v_{block} = 0 + 10 \times 1 \\[1em] = 10\ \text{m s}^{-1}\ (\text{downwards})\end{aligned}$$

(iii) 10 ms^-1

Using v = u − gt for the bullet,

$$\begin{aligned}\text v_{bullet} = 100 - 10 \times 1 \\[1em] = 90\ \text{m s}^{-1}\ (\text{upwards})\end{aligned}$$

(iv) 90 ms^-1

The bullet gets embedded in the block, so by the principle of conservation of linear momentum, taking the upward direction as positive,

$\text{mv}_{bullet} - \text{mv}_{block} = (\text m + \text m)\text v$

Both masses are equal, so

$$\begin{aligned}\text v = \frac{\text v_{bullet} - \text v_{block}}{2} = \frac{90 - 10}{2} \\[1em] = 40\ \text{m s}^{-1}\end{aligned}$$

The maximum height reached by the combined system above the point of impact is obtained from $\text v^2 = \text u^2 - 2\text{gh}$, with the final velocity zero (v = 0),

$$\begin{aligned}\text h = \frac{\text u^2}{2\text g} = \frac{(40)^2}{2 \times 10} = \frac{1600}{20} \\[1em] = 80\ \text m\end{aligned}$$

(v) 80 m

(i) Friction always acts tangential to the surfaces in contact.

Reason — When two bodies come into contact, the contact force can be resolved into a component normal to the surfaces, which is the normal reaction, and a component parallel to the surfaces. It is this parallel component that resists the impending or actual relative motion, and it is called friction. Hence, friction always acts tangential to the surfaces in contact and opposite to the relative motion.

(ii) Static friction is smaller than sliding friction.

Reason — This is the wrong statement. The limiting friction, which is the maximum value of the static friction, is greater than the sliding (kinetic) friction, so that μ_s > μ_k. This is why more force is needed to start a body moving than to keep it moving. The other three statements are correct.

(iii) $\frac{v^2}{2\mu g}$

Reason — When the brakes are applied, the retardation is produced by the frictional force,

$\text{ma} = \mu\text{mg} \quad \Rightarrow \quad \text a = \mu \text g$

Using $\text v^2 = \text u^2 - 2\text{as}$ with the final velocity zero and u = v,

$$\begin{aligned}0 = \text v^2 - 2\mu \text{gs} \\[1em] \text s = \frac{\text v^2}{2\mu \text g}\end{aligned}$$

(iv) 1.5 ms^-2

Reason — The box remains stationary on the floor of the train so long as the force required to accelerate it does not exceed the limiting friction,

$\text{ma} \le \mu_s\text{mg} \quad \Rightarrow \quad \text a \le \mu_s\text g$

Substituting μ_s = 0.15 and g = 9.8 m s^-2,

$$\begin{aligned}\text a_{max} = 0.15 \times 9.8 \\[1em] = 1.47 \text{m s}^{-2} \\[1em] \approx 1.5\ \text{m s}^{-2}\end{aligned}$$

(v) 15 kg

Reason — Let the mass of block C be m_C. The block B hangs from the string, so the tension is T = m_Bg. For block A the normal reaction is

$\text R = (\text m_A + \text m_C)\text g$

To just stop A from slipping, the tension must not exceed the limiting friction,

$\text m_B\text g = \mu(\text m_A + \text m_C)\text g$

Substituting m_A = 10 kg, m_B = 5 kg and μ = 0.2,

$$\begin{aligned}5 = 0.2 \times (10 + \text m_C) \\[1em] 10 + \text m_C = 25 \\[1em] \text m_C = 15\ \text{kg}\end{aligned}$$

(i) Centripetal force

Reason — A body moving along a circular path is continuously accelerated towards the centre, since the direction of its velocity changes at every instant. The force which produces this radially inward acceleration, and which is therefore responsible for the circular motion, is called the centripetal force. Its magnitude is $\frac{\text{mv}^2}{\text r}$.

(ii) $\sqrt{rg\tan\theta}$

Reason — On a banked track of negligible friction, the normal reaction N is inclined at θ to the vertical. Its vertical component balances the weight and its horizontal component supplies the centripetal force,

$\text N\cos \theta = \text{mg} \quad \text{and} \quad \text N\sin \theta = \frac{\text{mv}^2}{\text r}$

Dividing the second equation by the first,

$\tan \theta = \frac{\text v^2}{\text{rg}} \quad \Rightarrow \quad \text v = \sqrt{\text{rg}\tan \theta}$

(iii) 1.73 ms^-1

Reason — On a level circular road the centripetal force is provided entirely by friction, so the maximum safe speed is

$\text v_{max} = \sqrt{\mu \text{rg}}$

Substituting μ = 0.1, r = 3 m and g = 10 m s^-2,

$$\begin{aligned}\text v_{max} = \sqrt{0.1 \times 3 \times 10} = \sqrt{3} \\[1em] = 1.73\ \text{m s}^{-1}\end{aligned}$$

(iv) the inner wheel

Reason — When a car negotiates a flat curved road, the centrifugal effect in the frame of the car acts outwards at the centre of gravity, which is above the road. This forms a couple which tends to overturn the car outwards. As a result, the reaction on the inner wheel is reduced while that on the outer wheel is increased. Hence, it is the inner wheel that leaves the ground first.

(v) All of them

Reason — On a banked road the horizontal component of the normal reaction supplies a part of the centripetal force, so the reliance on friction alone is reduced. This allows higher speed limits on turns, reduces the wear and tear of the tyres and of the moving parts of the vehicle, and makes the turn safer. Hence, all the three benefits listed are correct.

Newton's second law of motion : The rate of change of momentum of a body is directly proportional to the applied force, and this change takes place in the direction of the applied force.

Explanation : Let a force $\vec{\text F} \space$act on a body of mass m and change its momentum. Then, by the law,

$\vec{\text F} \propto \frac{\text d\vec{\text p}}{\text{dt}} \quad \Rightarrow \quad \vec{\text F} = \text k\frac{\text d\vec{\text p}}{\text{dt}}$

where k is a constant of proportionality. The units of force are so defined that the value of k becomes 1. Hence,

$\vec{\text F} = \frac{\text d\vec{\text p}}{\text{dt}}$

Since $\vec{\text p} = \text m\vec{\text v}$and the mass of the body is constant,

$$\begin{aligned}\vec{\text F} = \frac{\text d}{\text{dt}}(\text m\vec{\text v}) = \text m\frac{\text d\vec{\text v}}{\text{dt}} \\[1em] = \text m\vec{\text a}\end{aligned}$$

This is the familiar form of the second law, and it gives the quantitative measure of force. In terms of the rectangular components,

$\text F_x = \text{ma}_x, \quad \text F_y = \text{ma}_y, \quad \text F_z = \text{ma}_z$

The second law is the real law of motion : The second law contains within itself the other two laws, and so it is called the real law.

(a) The first law follows from the second law. If no external force acts on the body, then

$\vec{\text F} = 0 \quad \Rightarrow \quad \text m\vec{\text a} = 0$

Since the mass of a body can never be zero, $\vec{\text a} = 0$, that is, the velocity of the body remains constant. Hence, the body continues in its state of rest or of uniform motion in a straight line, which is Newton's first law.

(b) The third law follows from the second law. Consider two bodies A and B which interact with each other and on which no external force acts. Then the total force on the system is zero,

$$\begin{aligned}\frac{\text d}{\text{dt}}(\vec{\text p}_A + \vec{\text p}_B) = 0 \\[1em] \frac{\text d\vec{\text p}_A}{\text{dt}} = -\frac{\text d\vec{\text p}_B}{\text{dt}}\end{aligned}$$

But $\frac{\text d\vec{\text p}_A}{\text{dt}}$ is the force exerted on A by B, and $\frac{\text d\vec{\text p}_B}{\text{dt}}$ is the force exerted on B by A. Hence,

$\vec{\text F}_{AB} = -\vec{\text F}_{BA}$

that is, action and reaction are equal and opposite, which is Newton's third law.

Hence, the second law is the real law of motion.

Law of conservation of momentum : If no external force acts on a system of bodies, the total linear momentum of the system remains constant, however the bodies may interact among themselves.

Derivation from Newton's third law of motion :

Consider two bodies A and B of masses m₁ and m₂, moving along a straight line with velocities $\vec{\text u}_1$ and $\vec{\text u}_2$ respectively, where u₁ > u₂, so that A overtakes B and they collide. Let the collision last for a time t, and let their velocities after the collision be $\vec{\text v}_1$ and $\vec{\text v}_2$.

During the collision, let A exert a force $\vec{\text F}_{\text{BA}}$ on B, and let B exert a force $\vec{\text F}_{\text{AB}}$ on A. By Newton's third law of motion, these are equal and opposite,

$\vec{\text F}_{\text{AB}} = -\vec{\text F}_{\text{BA}} \quad \text{...............(1)}$

By Newton's second law in its alternative form, each force is the rate of change of momentum of the body on which it acts,

$\vec{\text F}_{\text{AB}} = \frac{\text m_1\vec{\text v}_1 - \text m_1\vec{\text u}_1}{\text t} \quad \text{and} \quad \vec{\text F}_{\text{BA}} = \frac{\text m_2\vec{\text v}_2 - \text m_2\vec{\text u}_2}{\text t}$

Substituting these in equation (1),

$\frac{\text m_1\vec{\text v}_1 - \text m_1\vec{\text u}_1}{\text t} = -\frac{\text m_2\vec{\text v}_2 - \text m_2\vec{\text u}_2}{\text t}$

Multiplying throughout by t,

$\text m_1\vec{\text v}_1 - \text m_1\vec{\text u}_1 = -(\text m_2\vec{\text v}_2 - \text m_2\vec{\text u}_2)$

Rearranging the terms,

$\text m_1\vec{\text u}_1 + \text m_2\vec{\text u}_2 = \text m_1\vec{\text v}_1 + \text m_2\vec{\text v}_2$

That is,

$\text{Total momentum before collision} = \text{Total momentum after collision}$

Hence, in the absence of an external force, the total linear momentum of the system is conserved.

Newton's second law of motion : The rate of change of momentum of a body is directly proportional to the applied force, and this change takes place in the direction of the applied force. Mathematically,

$\vec{\text F} = \frac{\text d\vec{\text p}}{\text{dt}} = \text m\vec{\text a}$

Impulse : When a large force acts on a body for a very short interval of time, the force is called an impulsive force, and the product of the force and the time interval for which it acts is called the impulse of the force,

$\vec{\text J} = \vec{\text F} \times \Delta \text t$

Proof that impulse is equal to the change in momentum :

Let a force $\vec{\text F} \space$act on a body of mass m for a time interval from t₁ to t₂, changing its momentum from $\vec{\text p}_1$ to $\vec{\text p}_2$. By Newton's second law,

$\vec{\text F} = \frac{\text d\vec{\text p}}{\text{dt}}$

$\vec{\text F}\text{dt} = \text d\vec{\text p}$

Integrating both sides over the interval during which the force acts,

$\int_{\text t_1}^{\text t_2} \vec{\text F}\text{dt} = \int_{\vec{\text p}_1}^{\vec{\text p}_2} \text d\vec{\text p}$

Evaluating the integral on the right,

$$\begin{aligned}\int_{\text t_1}^{\text t_2} \vec{\text F}\text{dt} = \left[\vec{\text p}\right]_{\vec{\text p}_1}^{\vec{\text p}_2} \\[1em] = \vec{\text p}_2 - \vec{\text p}_1\end{aligned}$$

The left-hand side is, by definition, the impulse of the force. Hence,

$\vec{\text J} = \vec{\text p}_2 - \vec{\text p}_1 = \text m\vec{\text v} - \text m\vec{\text u}$

If the force is constant during the interval Δt = t₂ − t₁, it can be taken outside the integral and

$\vec{\text J} = \vec{\text F}\Delta \text t = \text m\vec{\text v} - \text m\vec{\text u}$

Hence, the impulse of a force is equal to the change in momentum produced by it. This is known as the impulse-momentum theorem. Graphically, the impulse is the area enclosed by the force-time graph.

Static friction : When a body lying on a surface is acted upon by a small force which is not sufficient to move it, the frictional force which comes into play is called static friction. It is a self-adjusting force, always equal and opposite to the applied force, so that

$\text f_s \le \mu_s\text R$

Its maximum value, which comes into play when the body is just about to slide, is called the limiting friction.

Kinetic friction : Once the body actually begins to slide over the surface, the frictional force acting on it is called kinetic friction. It acts in a direction opposite to the relative motion of the surfaces, and is given by

$\text f_k = \mu_k\text R$

Kinetic friction is smaller than limiting friction, so that μ_k < μ_s. It is reasonably independent of the relative speed of the surfaces and of the apparent area of contact.

Coefficient of friction : By the first law of limiting friction, the limiting frictional force f_s is directly proportional to the normal reaction R,

$\text f_s \propto \text R \quad \Rightarrow \quad \text f_s = \mu_s\text R$

The constant of proportionality μ_s is called the coefficient of static friction. It is the ratio of the limiting friction to the normal reaction,

$\mu_s = \frac{\text f_s}{\text R}$

It has no unit and no dimensions, and it depends on the nature of the two surfaces in contact.

Angle of friction : In the state of limiting friction, the resultant P of the normal reaction R and the limiting friction f_s makes an angle with the normal reaction. This angle is called the angle of friction, and is denoted by θ_s.

Relation between them : Resolving the resultant P, the component along the surface is the limiting friction f_s and the component normal to the surface is the normal reaction R. Hence, from the geometry of the figure,

$\tan \theta_s = \frac{\text f_s}{\text R}$

But f_s = μ_sR, so

$$\begin{aligned}\tan \theta_s = \frac{\mu_s\text R}{\text R} \\[1em] = \mu_s\end{aligned}$$

Hence, the coefficient of static friction is equal to the tangent of the angle of friction.

Centripetal force : When a body moves along a circular path with a uniform speed, the direction of its velocity changes continuously. Hence, the body is accelerated, and by Newton's second law a force must act on it. The force which is required to keep a body moving in a circular path, and which is always directed along the radius towards the centre of the circular path, is called the centripetal force.

It is always perpendicular to the instantaneous velocity of the body, so it changes only the direction of the velocity and not its magnitude. It is not a new kind of force but is supplied in different situations by different agencies, such as the tension in a string, the force of friction between the tyres and the road, or the gravitational force of attraction.

(i) Acceleration of the particle towards the centre :

Let a particle of mass m move along a circle of radius r with a uniform speed v, and let it move from a point A to a nearby point B in a small time interval Δt, the radius turning through a small angle Δθ. The velocities at A and B have the same magnitude v but different directions.

The change in velocity is $\Delta \vec{\text v} = \vec{\text v}_B - \vec{\text v}_A$. For a small angle Δθ, the magnitude of this change is

$\Delta \text v = \text v\Delta \theta$

Hence, the magnitude of the acceleration is

$$\begin{aligned}\text a_c = \lim_{\Delta \text t \to 0} \frac{\Delta \text v}{\Delta \text t} = \text v \lim_{\Delta \text t \to 0} \frac{\Delta \theta}{\Delta \text t} \\[1em] = \text v\omega\end{aligned}$$

where ω is the angular speed. Since v = rω,

$$\begin{aligned}\text a_c = \text v \times \frac{\text v}{\text r} \\[1em] = \frac{\text v^2}{\text r}\end{aligned}$$

As Δθ tends to zero, the direction of $\Delta \vec{\text v} \space$becomes perpendicular to the velocity, that is, it points along the radius towards the centre. Hence, the acceleration of the particle is

$\text a_c = \frac{\text v^2}{\text r} = \text r\omega^2 \quad (\text{directed towards the centre})$

(ii) Centripetal force acting on the particle :

By Newton's second law of motion, the force is the product of the mass and the acceleration,

$$\begin{aligned}\text F_c = \text{ma}_c = \frac{\text{mv}^2}{\text r} \\[1em] = \text{mr}\omega^2\end{aligned}$$

Hence, the centripetal acceleration is $\frac{\text v^2}{\text r}$ and the centripetal force is $\frac{\text{mv}^2}{\text r}$, both directed towards the centre of the circular path.

Inertial frame of reference : A frame of reference in which Newton's first law of motion is valid is called an inertial frame of reference. In such a frame a body continues in its state of rest or of uniform motion in a straight line so long as no external force acts on it. An inertial frame is one which is either at rest or is moving with a constant velocity, that is, it is unaccelerated.

Examples :

• A frame fixed to the earth, for ordinary terrestrial experiments, since the acceleration of the earth is very small.
• A car moving with a constant speed on a straight road. A passenger in it finds that a ball placed on the seat stays at rest, exactly as Newton's first law requires.
• A spaceship moving with a constant velocity in interstellar space, far from all material objects.

Non-inertial frame of reference : A frame of reference in which Newton's first law of motion is not valid is called a non-inertial frame of reference. Such a frame is accelerated. In a non-inertial frame a body may appear to be accelerated even when no real force acts on it, and to apply Newton's laws in such a frame one has to introduce a pseudo force (or inertial force), which is not due to any physical interaction.

Examples :

• A lift which is accelerating upwards or downwards. To a person inside, his apparent weight changes although the gravity force on him is unaltered.
• A car taking a sharp turn, or a train moving on a circular track, since the direction of the velocity changes continuously.
• A merry-go-round or any rotating platform. A person standing on it feels an outward centrifugal force, which is a pseudo force.
• An aircraft which is taking off.

Frictional force : When the surface of one body slides, or tends to slide, over the surface of another body, each body exerts on the other a force parallel to the surfaces in contact which opposes this relative motion. This force is called the frictional force. The frictional force on each body is opposite to the direction of its motion relative to the other.

Cause of friction : On the modern theory, friction between two surfaces is due to the strong atomic or molecular forces at the points of actual contact. The area of actual contact is much smaller than the apparent area of contact, so the adhesive pressure at the points of contact becomes very large and the molecules there form cold welds. To move one body over the other, these cold welds have to be broken, and the external force required for this purpose is a measure of the friction between the two surfaces.

This theory also explains why friction is independent of the apparent area of contact, since with an increase in the contact area the adhesive force increases in the same ratio and the adhesive pressure remains constant. It further explains why friction increases when two surfaces are made smooth beyond a certain limit, since the molecules then come closer and the adhesive force increases.

Limiting friction : As the applied force F is gradually increased, the force of static friction f_s also increases, always remaining equal to F. But after a certain limit f_s cannot increase any further. The maximum value of the static frictional force, which comes into play when a body just begins to slide over the surface of another body, is called the limiting frictional force. It is equal to the force required to start the motion.

Laws of limiting friction :

(i) First law : The limiting friction f_s is directly proportional to the normal reaction R,

$\text f_s \propto \text R \quad \Rightarrow \quad \text f_s = \mu_s\text R$

where μ_s is the coefficient of static friction. In general, f_s ≤ μ_sR.

(ii) Second law : The direction of the limiting frictional force is always opposite to the direction in which one body tends to move relative to the other body.

(iii) Third law : The limiting frictional force is independent of the apparent area of contact between the two surfaces. It depends only on the normal force and on the nature of the material in contact, and not on the contact area.

(iv) Fourth law : The value of the coefficient of limiting friction depends on the nature of the surfaces in contact, such as their roughness, their material composition, and any surface treatment or lubricant present.

Let m be the combined mass of the rider and the bicycle, v the speed of the bicycle and r the radius of the circular turn.

Why the rider leans inwards : The centripetal force $\frac{\text{mv}^2}{\text r}$ necessary to take the turn is provided by the friction between the tyres and the road. Hence, when the bicycle is turned, a frictional force F acting towards the centre of the turn is exerted at the point A where the tyres touch the road.

Now imagine two equal and opposite forces F₁ and F₂, each equal and parallel to F, acting at the centre of gravity G of the rider and the bicycle. Since they are equal and opposite and act at the same point, their resultant is zero and their imaginary existence does not affect the rider. The force F acting at A can now be replaced by the force F₁ acting at G, together with the anticlockwise couple formed by F and F₂.

If the rider were to remain straight while turning, his weight mg acting vertically downwards at G and the normal reaction R (= mg) acting vertically upwards at A would cancel each other, and the anticlockwise couple would overturn the bicycle outwards. If, however, the rider leans inwards towards the centre of the turn, then mg and R form a clockwise couple which balances the anticlockwise couple. Hence, he moves round the turn without any risk of overturning.

If θ is the angle through which the rider leans from the vertical, then equating the moments of the two couples,

$$\begin{aligned}\text{mg} \times \text{GA}\sin \theta = \text F \times \text{GA}\cos \theta \\[1em] \tan \theta = \frac{\text F}{\text{mg}}\end{aligned}$$

Substituting $\text F = \frac{\text{mv}^2}{\text r}$,

$$\begin{aligned}\tan \theta = \frac{\text{mv}^2/\text r}{\text{mg}} \\[1em] = \frac{\text v^2}{\text{rg}}\end{aligned}$$

Why he decreases his velocity and takes a path of large radius : Since the centripetal force $\frac{\text{mv}^2}{\text r}$ is provided by the frictional force, its value must not exceed the limiting frictional force μ_sR, otherwise the bicycle would slip. Hence,

$\frac{\text{mv}^2}{\text r} \le \mu_s\text{mg} \quad \Rightarrow \quad \text v \le \sqrt{\mu_s\text{rg}}$

The rider therefore slows down, so that $\frac{\text{mv}^2}{\text r}$ is reduced, and follows a path of larger radius r, which reduces the required centripetal force still further. During the rainy season the frictional force decreases appreciably and may fail to provide the necessary centripetal force, which is why bicycle-riders usually slip on the roads while taking a turn in the rains.

Centrifugal force : Centrifugal force is an apparent force which appears to act radially outwards on a body when the body is observed from a rotating frame of reference. Its magnitude is

$\text F = \frac{\text{mv}^2}{\text r}$

where m is the mass of the body, r the radius of the circular path and v its speed.

Explanation : Consider a body of mass m placed on a smooth circular table which is rotating with a uniform angular velocity about a nail passing through its centre, the body being tied to the nail by a string.

A man standing on the earth, looking at the rotating table, sees the body rotating around the nail and finds a centripetal force $\frac{\text{mv}^2}{\text r}$ acting on it. This force is provided by the tension in the string, and is a real force.

To another man who is standing on the table itself, the body appears to be at rest, because its position does not change with respect to him. From his point of view there should be no net force acting on the body. But the body is actually acted upon by an inward force $\frac{\text{mv}^2}{\text r}$. Hence, according to the man standing on the table, a force of magnitude $\frac{\text{mv}^2}{\text r}$ must also be acting outwards, so that the net force on the body is zero. This apparent outward force is called the centrifugal force.

Why it is called a pseudo force :

• It is not a real force, since it does not arise from any physical interaction between two bodies. There is no body which exerts it, and it has no reaction, so it does not obey Newton's third law.
• It appears only in a non-inertial (rotating) frame of reference. To an observer in an inertial frame no such force exists; to him the body simply moves along the tangent by inertia when the string is cut.
• It arises from the inertia of the body, which tends to maintain its straight line motion while the frame of reference accelerates towards the centre.

For these reasons the centrifugal force is called a pseudo force, or a fictitious force. Its effects are nevertheless clearly observed by an observer in the rotating frame, as when a person standing on a merry-go-round falls outwards if he does not hold its strings.

When a vehicle negotiates a flat curved road, the necessary centripetal force is provided entirely by the frictional force between the tyres and the road. The maximum safe speed is then

$\text v_{max} = \sqrt{\mu_s\text{rg}}$

Frictional forces, however, are never reliable. They keep varying from time to time, because roads become slippery during the rainy season and tyres lose their tread and become smooth over time due to wear and tear. In both these situations the coefficient of friction gets reduced, and the maximum safe speed becomes quite low, especially at sharp turns.

Banking refers to the process in which the road surface at the bend is tilted inwards, that is, the outer side of the road is raised above its inner side.

On a banked road the normal reaction N is no longer vertical. Its horizontal component N sin θ is directed towards the centre of the curve and provides a part of the necessary centripetal force, while its vertical component N cos θ balances the weight. For a road banked at the angle θ with friction neglected,

$\text N\sin \theta = \frac{\text{mv}^2}{\text r} \quad \text{and} \quad \text N\cos \theta = \text{mg}$

Dividing,

$\tan \theta = \frac{\text v^2}{\text{rg}} \quad \Rightarrow \quad \theta = \tan^{-1}\left(\frac{\text v^2}{\text{rg}}\right)$

Hence, roads are banked at sharp turns because :

• the banking creates a component of the normal force directed towards the centre of the curve, which provides additional centripetal force;
• this reduces the reliance on friction alone, making the turn safer, especially at higher speeds;
• it protects the moving parts of the vehicle from unwanted wear and tear due to friction;
• it thus enhances safety, comfort, efficiency and longevity, contributing to better road transportation.

When a train negotiates a curve on the railway line, it requires a centripetal force directed towards the centre of the curve. If the track were perfectly level, this force could be supplied only by the sideways push of the rails on the flanges of the wheels. This would cause severe wear and tear of the rails and of the wheel flanges, and there would also be a danger of the train overturning outwards.

To avoid this, the outer rail is raised slightly above the inner rail, so that the track is banked at an angle θ with the horizontal. The normal reaction N of the rails on the train is then perpendicular to the inclined track, and can be resolved into two components.

The vertical component balances the weight of the train,

$\text N\cos \theta = \text{mg}$

and the horizontal component provides the necessary centripetal force,

$\text N\sin \theta = \frac{\text{mv}^2}{\text r}$

Dividing the second equation by the first,

$\tan \theta = \frac{\text v^2}{\text{rg}}$

If h is the height by which the outer rail is raised and d the distance between the rails, then for a small angle θ,

$\tan \theta \approx \frac{\text h}{\text d} \quad \Rightarrow \quad \text h = \frac{\text{dv}^2}{\text{rg}}$

Hence, the outer rail is raised so that the horizontal component of the normal reaction supplies the centripetal force needed by the train, the wheel flanges do not press against the rails, and the train can negotiate the turn safely without wear and tear or risk of overturning.

Concurrent forces : Forces whose lines of action pass through a single common point are called concurrent forces. Since all the forces act at the same point, they can be added by the laws of vector addition.

Condition of equilibrium : A body acted upon by concurrent forces is said to be in equilibrium when it either remains at rest or continues to move with a uniform velocity, that is, when its acceleration is zero.

Proof :

Let three concurrent forces $\vec{\text F}_1$, $\vec{\text F}_2$ and $\vec{\text F}_3$ act at a point on a body of mass m. Being concurrent, they can be added vectorially, so the net force acting on the body is

$\vec{\text F} = \vec{\text F}_1 + \vec{\text F}_2 + \vec{\text F}_3$

By Newton's second law of motion, this net force produces an acceleration $\vec{\text a} \space$in the body,

$\vec{\text F}_1 + \vec{\text F}_2 + \vec{\text F}_3 = \text m\vec{\text a}$

If the body is in equilibrium, its acceleration is zero, that is, $\vec{\text a} = 0$. Therefore,

$$\begin{aligned}\vec{\text F}_1 + \vec{\text F}_2 + \vec{\text F}_3 = \text m \times 0 \\[1em] = 0\end{aligned}$$

Conversely, if $\vec{\text F}_1 + \vec{\text F}_2 + \vec{\text F}_3 = 0$, then

$\text m\vec{\text a} = 0$

The mass m of a body can never be zero, so $\vec{\text a} = 0$. Hence, the body has no acceleration and is in equilibrium.

Hence, a body acted upon by three concurrent forces is in equilibrium if and only if $\vec{\text F}_1 + \vec{\text F}_2 + \vec{\text F}_3 = 0$.

In terms of the rectangular components, this condition means that the sums of the components of all the forces along each of the three axes must separately be zero,

$\Sigma \text F_x = 0, \quad \Sigma \text F_y = 0, \quad \Sigma \text F_z = 0$

Geometrically, three concurrent forces in equilibrium can be represented in magnitude and direction by the three sides of a triangle taken in the same order.

Impulse-momentum theorem : The impulse of a force acting on a body is equal to the change in momentum produced in the body.

$\vec{\text J} = \vec{\text F} \times \Delta \text t = \text m\vec{\text v} - \text m\vec{\text u}$

Explanation : When a large force acts on a body for a very short interval of time, the force is called an impulsive force. Such a force cannot be measured conveniently at every instant, and neither can the very short time interval always be measured. What can be measured is the total effect of the force, which is the change in momentum it produces. This total effect is called the impulse.

Proof : By Newton's second law of motion in its alternative form,

$\vec{\text F} = \frac{\text d\vec{\text p}}{\text{dt}} \quad \Rightarrow \quad \vec{\text F}\text{dt} = \text d\vec{\text p}$

Integrating both sides over the interval from t₁ to t₂, during which the momentum changes from $\vec{\text p}_1$ to $\vec{\text p}_2$,

$\int_{\text t_1}^{\text t_2} \vec{\text F}\text{dt} = \int_{\vec{\text p}_1}^{\vec{\text p}_2} \text d\vec{\text p}$

$\int_{\text t_1}^{\text t_2} \vec{\text F}\text{dt} = \left[\vec{\text p}\right]_{\vec{\text p}_1}^{\vec{\text p}_2} = \vec{\text p}_2 - \vec{\text p}_1$

The left-hand side is the impulse of the force. If the force is constant during the interval, it can be taken outside the integral, giving

$\vec{\text J} = \vec{\text F}(\text t_2 - \text t_1) = \vec{\text p}_2 - \vec{\text p}_1$

Hence proved.

Unit and dimensions : The SI unit of impulse is the newton second (N s), which is the same as kg m s^-1, and its dimensions are MLT^-1, the same as those of momentum.

Applications : The theorem explains why a cricketer moves his hands backward while holding a catch, and why vehicles are provided with shock absorbers. In each case the change in momentum is fixed, so by increasing the time interval Δt the force experienced is reduced. Graphically, the impulse is equal to the area enclosed by the force-time graph.

Friction is a necessary evil. This means that friction is harmful in many ways, and yet life as we know it would be impossible without it.

Friction as an evil :

• Friction is a non-conservative force which always opposes the motion of one body over another, and it dissipates energy into non-recoverable forms such as heat and noise.
• The excessive heat produced in machines due to friction may cause damage to them.
• Friction causes wear and tear of the parts of machinery which are in mutual contact.
• A considerable part of the fuel consumed in machines, engines and vehicles is spent merely in overcoming friction, so the efficiency is reduced.

Friction as a necessity :

• If friction were totally absent, we could not walk on a road, since we walk only by pressing the ground backward and using the horizontal component of the reaction.
• Vehicles could not run on the ground, and if somehow set in motion, they could not be stopped, because the brakes would not function.
• Nuts and bolts could not hold the parts of machinery together, and nails could not be fixed in a wall.
• Writing on paper would not be possible, adhesives would not work, a match-stick could not be lighted by rubbing, and the threads of a rope could not be held together.

Conclusion : Friction cannot be totally eliminated; it can only be reduced, by polishing the surfaces, by lubrication with oil or grease, by the proper selection of materials, by the use of ball bearings, and by streamlining. Since its harmful effects can never be entirely removed and yet our day-to-day life would be impossible without it, friction is rightly described as a necessary evil.

The magnitude of each is $\frac{\text{mv}^2}{\text r}$, but they do not act on the same body in the same frame of reference and therefore cannot be treated as a pair of balancing forces.

Examples of centripetal force :

• The tension in a string when a stone tied to it is whirled in a circle.
• The force of friction between the tyres and the road when a car takes a turn.
• The gravitational force of attraction of the sun on the earth, which keeps the earth in its orbit.
• The electrostatic force of attraction between the nucleus and an electron revolving around it.

Examples of centrifugal force :

• A person standing on a rotating merry-go-round falls outwards if he does not hold its strings.
• A passenger sitting in a car which suddenly turns to the right is pushed against the left-hand wall of the car.
• The separation of cream from milk in a centrifuge, in which the heavier milk particles move towards the circumference and the lighter cream particles collect near the centre.

Given,

• Mass of the ship, m = 3 × 10⁷ kg
• Initial velocity, u = 0
• Force applied, F = 5 × 10⁴ N
• Distance covered, s = 3 m

The speed attained by the ship has to be calculated.

By Newton's second law of motion, the acceleration produced is

$\text a = \frac{\text F}{\text m}$

Substituting the values,

$$\begin{aligned}\text a = \frac{5 \times 10^4}{3 \times 10^7} \\[1em] = 1.667 \times 10^{-3}\ \text{m s}^{-2}\end{aligned}$$

Using the equation of motion $\text v^2 = \text u^2 + 2\text{as}$ with u = 0,

$$\begin{aligned}\text v^2 = 2 \times 1.667 \times 10^{-3} \times 3 \\[1em] = 10^{-2}\end{aligned}$$

$$\begin{aligned}\text v = \sqrt{10^{-2}} \\[1em] = 0.1\ \text{m s}^{-1}\end{aligned}$$

Hence, the speed attained by the ship is 0.1 m s^-1.

Given,

• Mass on the table, m = 8 kg
• Hanging masses, m₁ = 2 kg and m₂ = 4 kg
• The table and the pulleys are frictionless and the strings are massless
• Acceleration due to gravity, g = 9.8 m s^-2

Since m₂ is heavier than m₁, the 4 kg block moves down, the 8 kg block moves towards the right and the 2 kg block moves up, all with the same acceleration a.

For the 2 kg block, moving up,

$\text T_1 - \text m_1\text g = \text m_1\text a \quad \text{...............(1)}$

For the 8 kg block on the smooth table,

$\text T_2 - \text T_1 = \text{ma} \quad \text{...............(2)}$

For the 4 kg block, moving down,

$\text m_2\text g - \text T_2 = \text m_2\text a \quad \text{...............(3)}$

Adding equations (1), (2) and (3),

$$\begin{aligned}(\text m_2 - \text m_1)\text g = (\text m_1 + \text m + \text m_2)\text a \\[1em] \text a = \frac{(\text m_2 - \text m_1)\text g}{\text m_1 + \text m + \text m_2}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text a = \frac{(4 - 2) \times 9.8}{2 + 8 + 4} = \frac{2 \times 9.8}{14} \\[1em] = \frac{19.6}{14} \\[1em] = 1.4\ \text{m s}^{-2}\end{aligned}$$

From equation (1),

$$\begin{aligned}\text T_1 = \text m_1(\text g + \text a) = 2 \times (9.8 + 1.4) \\[1em] = 2 \times 11.2 \\[1em] = 22.4\ \text N\end{aligned}$$

From equation (3),

$$\begin{aligned}\text T_2 = \text m_2(\text g - \text a) = 4 \times (9.8 - 1.4) \\[1em] = 4 \times 8.4 \\[1em] = 33.6\ \text N\end{aligned}$$

Hence, the acceleration of the system is 1.4 m s^-2, and the tensions in the two strings are 22.4 N and 33.6 N.

Given,

• Maximum tension the strings can bear, T = 72 kg-weight
• Mass of the person, m = 96 kg
• Acceleration due to gravity, g = 9.8 m s^-2

The minimum acceleration with which the person can descend has to be calculated.

While descending with an acceleration a, the net downward force on the person is

$\text{mg} - \text T = \text{ma}$

Expressing the maximum tension in newton, T = 72g N, and substituting,

$$\begin{aligned}\text{mg} - 72\text g = \text{ma} \\[1em] \text a = \frac{(\text m - 72)\text g}{\text m}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text a = \frac{(96 - 72) \times 9.8}{96} = \frac{24 \times 9.8}{96} \\[1em] = \frac{235.2}{96} \\[1em] = 2.45\ \text{m s}^{-2}\end{aligned}$$

If the acceleration were smaller than this, the tension in the strings would exceed 72 kg-weight and the strings would break.

Hence, the person must descend with a minimum acceleration of 2.45 m s^-2.

Given,

• Mass of the lift, m = 400 kg
• Acceleration due to gravity, g = 9.8 m s^-2

The tension in the wire has to be calculated in each case.

(a) At rest : The acceleration is zero, so

$$\begin{aligned}\text T = \text{mg} = 400 \times 9.8 \\[1em] = 3920\ \text N\end{aligned}$$

(b) Moving upward with a constant velocity of 1.0 m s^-1 : The acceleration is again zero, so

$$\begin{aligned}\text T = \text{mg} \\[1em] = 3920\ \text N\end{aligned}$$

(c) Moving upward with an acceleration of 2.0 m s^-2 : The net upward force is T − mg, so

$$\begin{aligned}\text T = \text m(\text g + \text a) = 400 \times (9.8 + 2.0) \\[1em] = 400 \times 11.8 \\[1em] = 4720\ \text N\end{aligned}$$

(d) Moving downward with an acceleration of 2.0 m s^-2 : The net downward force is mg − T, so

$$\begin{aligned}\text T = \text m(\text g - \text a) = 400 \times (9.8 - 2.0) \\[1em] = 400 \times 7.8 \\[1em] = 3120\ \text N\end{aligned}$$

Hence, the tensions in the wire are 3920 N, 3920 N, 4720 N and 3120 N respectively.

Given,

• Mass of the body, m = 10 kg
• Acceleration of the lift, a = 2 m s^-2
• Acceleration due to gravity, g = 10 m s^-2

The reading of the spring balance, which is the apparent weight of the body, has to be calculated.

(i) Lift ascending with an acceleration of 2 m s^-2 :

$$\begin{aligned}\text R = \text m(\text g + \text a) = 10 \times (10 + 2) \\[1em] = 120\ \text N = \frac{120}{10}\ \text{kg-wt} = 12\ \text{kg-wt}\end{aligned}$$

(ii) Lift descending with the same acceleration :

$$\begin{aligned}\text R = \text m(\text g - \text a) = 10 \times (10 - 2) \\[1em] = 80\ \text N = 8\ \text{kg-wt}\end{aligned}$$

(iii) Lift descending with a constant velocity of 2 m s^-1 : Here the acceleration is zero, so

$$\begin{aligned}\text R = \text{mg} = 10 \times 10 \\[1em] = 100\ \text N = 10\ \text{kg-wt}\end{aligned}$$

Hence, the readings of the balance are 12 kg-wt, 8 kg-wt and 10 kg-wt respectively.

Given,

• Initial speed, u = 10 m s^-1
• Final speed, v = 0
• Distance covered, s = 50 m
• Acceleration due to gravity, g = 10 m s^-2

The coefficient of friction has to be calculated.

The body is brought to rest by the frictional force alone. By Newton's second law,

$\text{ma} = \mu\text{mg} \quad \Rightarrow \quad \text a = \mu \text g$

Using the equation of motion $\text v^2 = \text u^2 - 2\text{as}$ with v = 0,

$$\begin{aligned}0 = \text u^2 - 2\mu \text{gs} \\[1em] \mu = \frac{\text u^2}{2\text{gs}}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\mu = \frac{(10)^2}{2 \times 10 \times 50} = \frac{100}{1000} \\[1em] = 0.1\end{aligned}$$

Hence, the coefficient of friction is 0.1.

Given,

• Distance between the rails, d = 1 m
• Radius of the turn, r = 400 m
• Speed of the train, v = 20 m s^-1
• Acceleration due to gravity, g = 10 m s^-2

The height by which the outer rail should be raised has to be calculated.

For the train to cross the turn without friction, the track must be banked at an angle θ such that the horizontal component of the normal reaction provides the necessary centripetal force,

$\tan \theta = \frac{\text v^2}{\text{rg}}$

If h is the height by which the outer rail is raised above the inner rail, then for a small angle of banking

$\tan \theta = \frac{\text h}{\text d}$

Comparing the two expressions,

$\frac{\text h}{\text d} = \frac{\text v^2}{\text{rg}} \quad \Rightarrow \quad \text h = \frac{\text{dv}^2}{\text{rg}}$

Substituting the values,

$$\begin{aligned}\text h = \frac{1 \times (20)^2}{400 \times 10} = \frac{400}{4000} \\[1em] = 0.1\ \text m = 10\ \text{cm}\end{aligned}$$

Hence, the outer rail should be raised through 10 cm relative to the inner rail.

Given,

• Radius of the turn, r = 500 m
• Width of the road, d = 10 m
• Height of the outer edge above the inner edge, h = 0.15 m
• Acceleration due to gravity, g = 10 m s^-2

The speed for which this banking is ideal has to be calculated.

The angle of banking θ is given by

$$\begin{aligned}\tan \theta = \frac{\text h}{\text d} = \frac{0.15}{10} \\[1em] = 0.015\end{aligned}$$

For the ideal speed, the whole of the centripetal force is provided by the horizontal component of the normal reaction, so that

$\tan \theta = \frac{\text v^2}{\text{rg}} \quad \Rightarrow \quad \text v = \sqrt{\text{rg}\tan \theta}$

Substituting the values,

$$\begin{aligned}\text v = \sqrt{500 \times 10 \times 0.015} \\[1em] = \sqrt{75} \\[1em] = 8.66\ \text{m s}^{-1}\end{aligned}$$

Hence, the banking is ideal for a speed of 8.66 m s^-1.

Given,

• Mass of the first body, m₁ = 0.5 kg, with acceleration a₁ = 16 m s^-2
• Acceleration produced in the second body, a₂ = 4.0 m s^-2
• The same force acts in both the cases

The acceleration produced when both the bodies are fastened together has to be calculated.

By Newton's second law of motion, the force is

$$\begin{aligned}\text F = \text m_1\text a_1 = 0.5 \times 16 \\[1em] = 8\ \text N\end{aligned}$$

The same force produces an acceleration of 4.0 m s^-2 in the second body, so its mass is

$$\begin{aligned}\text m_2 = \frac{\text F}{\text a_2} = \frac{8}{4.0} \\[1em] = 2\ \text{kg}\end{aligned}$$

When the two bodies are fastened together, the total mass is

$$\begin{aligned}\text m = \text m_1 + \text m_2 = 0.5 + 2 \\[1em] = 2.5\ \text{kg}\end{aligned}$$

Hence, the acceleration produced by the same force is

$$\begin{aligned}\text a = \frac{\text F}{\text m} = \frac{8}{2.5} \\[1em] = 3.2\ \text{m s}^{-2}\end{aligned}$$

Hence, the acceleration produced is 3.2 m s^-2.

Given,

• Applied force, F = 1.5 N at 30° with the horizontal, in a vertical plane
• Masses of the carriages, m₁ = 40 g = 0.04 kg and m₂ = 20 g = 0.02 kg
• The table is frictionless

Only the horizontal component of the applied force produces the motion, since the vertical component is balanced by the normal reaction of the table. The horizontal component is

$$\begin{aligned}\text F_x = \text F\cos \theta = 1.5 \times \cos 30^\circ \\[1em] = 1.5 \times \frac{\sqrt{3}}{2} \\[1em] = 1.299\ \text N\end{aligned}$$

(i) and (ii) The two carriages are connected by a string, so they move together as a system of total mass

$$\begin{aligned}\text m = \text m_1 + \text m_2 = 0.04 + 0.02 \\[1em] = 0.06\ \text{kg}\end{aligned}$$

Hence, the acceleration of each carriage is

$$\begin{aligned}\text a = \frac{\text F_x}{\text m} = \frac{1.299}{0.06} \\[1em] = 21.6\ \text{m s}^{-2}\end{aligned}$$

(iii) The 20 g carriage is pulled forward only by the tension in the string joining the two carriages. Hence,

$$\begin{aligned}\text T = \text m_2\text a = 0.02 \times 21.6 \\[1em] = 0.43\ \text N\end{aligned}$$

Hence, the acceleration of each carriage is 21.6 m s^-2 and the tension in the string is 0.43 N.

The forces acting on the two carriages separately are shown in the figure.

Given,

• Masses hanging from the two ends of the cord, m₁ = 1 kg and m₂ = 5 kg
• The pulley is massless and frictionless

The reading of the spring balance is the total downward pull on the pulley, which is twice the tension in the cord.

If the weights were held at rest, the tension in the cord on either side would be equal to the corresponding weight and the balance would read (1 + 5) = 6 kg-wt. But the weights move under the action of gravity, so the tension has to be calculated for the accelerated motion.

The acceleration of the system is

$$\begin{aligned}\text a = \frac{(\text m_2 - \text m_1)\text g}{\text m_1 + \text m_2} = \frac{(5 - 1)\text g}{1 + 5} \\[1em] = \frac{4\text g}{6} = \frac{2\text g}{3}\end{aligned}$$

The tension in the cord is

$$\begin{aligned}\text T = \frac{2\text m_1\text m_2\text g}{\text m_1 + \text m_2} = \frac{2 \times 1 \times 5 \times \text g}{6} \\[1em] = \frac{10\text g}{6} = \frac{5\text g}{3}\end{aligned}$$

Hence, the reading of the spring balance is

$$\begin{aligned}2\text T = 2 \times \frac{5\text g}{3} = \frac{10\text g}{3} \\[1em] = 3.33\ \text{kg-wt}\end{aligned}$$

Hence, the spring balance will read less than 6 kg-wt, its reading being 3.33 kg-wt during the motion of the weights.

Given,

• Mass of the person, m = 70 kg
• Maximum tension the rope can bear, T = 100 kg-weight
• Acceleration due to gravity, g = 9.8 m s^-2

The maximum acceleration with which the helicopter should rise has to be calculated.

While the person is being lifted with an acceleration a, the net upward force on him is

$\text T - \text{mg} = \text{ma}$

Expressing the maximum tension in newton, T = 100g N, and substituting,

$$\begin{aligned}100\text g - \text{mg} = \text{ma} \\[1em] \text a = \frac{(100 - \text m)\text g}{\text m}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text a = \frac{(100 - 70) \times 9.8}{70} = \frac{30 \times 9.8}{70} \\[1em] = \frac{294}{70} \\[1em] = 4.2\ \text{m s}^{-2}\end{aligned}$$

Hence, the helicopter should rise with a maximum acceleration of 4.2 m s^-2 so that the rope may not break.

Given,

• Mass of the man, m = 60 kg
• Upward acceleration of the lift, a = 4.9 m s^-2
• Acceleration due to gravity, g = 9.8 m s^-2

Lift going upwards with an acceleration of 4.9 m s^-2 : The apparent weight is

$$\begin{aligned}\text R = \text m(\text g + \text a) = 60 \times (9.8 + 4.9) \\[1em] = 60 \times 14.7 \\[1em] = 882\ \text N = \frac{882}{9.8}\ \text{kg-wt} = 90\ \text{kg-wt}\end{aligned}$$

Lift rising with a uniform velocity of 4.9 m s^-1 : The acceleration is zero, so

$$\begin{aligned}\text R = \text{mg} = 60 \times 9.8 \\[1em] = 588\ \text N = 60\ \text{kg-wt}\end{aligned}$$

If the rope of the lift is broken : The lift and the man both fall freely with the acceleration due to gravity, so a = g and

$$\begin{aligned}\text R = \text m(\text g - \text g) \\[1em] = 0\end{aligned}$$

Hence, the apparent weight of the man is 90 kg-wt in the first case, 60 kg-wt in the second case, and zero if the rope is broken, when he is in the state of weightlessness.

Given,

• Mass of the machine gun, M = 5 kg
• Mass of each bullet, m = 50 g = 0.05 kg
• Number of bullets fired, n = 30 per minute, that is, in t = 60 s
• Speed of each bullet, v = 400 m s^-1

The force required to keep the machine gun in position has to be calculated.

The momentum carried away by the bullets in one minute is

$$\begin{aligned}\text{nmv} = 30 \times 0.05 \times 400 \\[1em] = 600\ \text{kg m s}^{-1}\end{aligned}$$

By Newton's second law of motion, the force is equal to the rate of change of momentum,

$$\begin{aligned}\text F = \frac{\text{nmv}}{\text t} = \frac{600}{60} \\[1em] = 10\ \text N\end{aligned}$$

By Newton's third law of motion, an equal and opposite force acts on the gun, and this must be balanced to keep the gun in position.

Hence, a force of 10 N must be exerted to keep the machine gun in position.

Given,

• Mass of the body, m = 10 g = 0.01 kg
• Force, F = 10 N
• Time for which the force acts, t = 3 μs = 3 × 10^-6 s

The impulse and the change in velocity have to be calculated.

The impulse of a constant force is the product of the force and the time interval for which it acts,

$\text J = \text F \times \text t$

Substituting the values,

$$\begin{aligned}\text J = 10 \times 3 \times 10^{-6} \\[1em] = 3 \times 10^{-5}\ \text{N s}\end{aligned}$$

By the impulse-momentum theorem, the impulse is equal to the change in momentum,

$\text J = \text m\Delta \text v \quad \Rightarrow \quad \Delta \text v = \frac{\text J}{\text m}$

Substituting the values,

$$\begin{aligned}\Delta \text v = \frac{3 \times 10^{-5}}{0.01} \\[1em] = 3 \times 10^{-3}\ \text{m s}^{-1}\end{aligned}$$

Hence, the impulse is 3 × 10^-5 N s and the change in velocity of the body is 3 × 10^-3 m s^-1.

Given,

• Mass of the hammer, m = 1.5 kg
• Initial speed, u = 10 m s^-1
• Final speed, v = 0
• Duration of the impact, t = 0.005 s

The average impact force and the distance of penetration have to be calculated.

By Newton's second law of motion, the average force is the rate of change of momentum,

$\text F = \frac{\text m(\text u - \text v)}{\text t}$

Substituting the values,

$$\begin{aligned}\text F = \frac{1.5 \times (10 - 0)}{0.005} = \frac{15}{0.005} \\[1em] = 3000\ \text N\end{aligned}$$

The retardation produced is

$$\begin{aligned}\text a = \frac{\text u - \text v}{\text t} = \frac{10}{0.005} \\[1em] = 2000\ \text{m s}^{-2}\end{aligned}$$

Using the equation of motion $\text v^2 = \text u^2 - 2\text{as}$ with v = 0, the distance of penetration is

$$\begin{aligned}\text s = \frac{\text u^2}{2\text a} = \frac{(10)^2}{2 \times 2000} = \frac{100}{4000} \\[1em] = 0.025\ \text m = 2.5\ \text{cm}\end{aligned}$$

Hence, the average impact force is 3000 N and the nail penetrates through 2.5 cm into the block.

Given,

• Mass of each bullet, m = 2.0 g = 0.002 kg
• Speed of each bullet before striking, u = 11 m s^-1
• Speed of each bullet after striking, v = −11 m s^-1 (it returns with the same speed)
• Number of bullets fired, n = 3 per second

The average force exerted on the disc has to be calculated.

The change in momentum of one bullet is

$\Delta \text p = \text{mv} - \text{mu} = \text m(\text v - \text u)$

$$\begin{aligned}= 0.002 \times (-11 - 11) \\[1em] = 0.002 \times (-22) \\[1em] = -0.044\ \text{kg m s}^{-1}\end{aligned}$$

By Newton's third law of motion, an equal and opposite change of momentum, of magnitude 0.044 kg m s^-1, is imparted to the disc by each bullet. Since 3 bullets strike the disc in one second, the average force is the total rate of change of momentum,

$$\begin{aligned}\text F = \text n \times |\Delta \text p| = 3 \times 0.044 \\[1em] = 0.132\ \text N\end{aligned}$$

Hence, the average force exerted on the disc is 0.132 N.

Given,

• Mass of the first body, m₁ = 5 kg, moving along +x with v₁ = 4 m s^-1
• Mass of the second body, m₂ = 10 kg, moving along +y with v₂ = 2 m s^-1
• They collide and stick together

The final velocity has to be calculated.

The momentum of the first body, along the +x direction, is

$$\begin{aligned}\text p_1 = \text m_1\text v_1 = 5 \times 4 \\[1em] = 20\ \text{kg m s}^{-1}\end{aligned}$$

The momentum of the second body, along the +y direction, is

$$\begin{aligned}\text p_2 = \text m_2\text v_2 = 10 \times 2 \\[1em] = 20\ \text{kg m s}^{-1}\end{aligned}$$

The two momenta are mutually perpendicular, so the magnitude of their resultant is

$$\begin{aligned}\text p = \sqrt{\text p_1^2 + \text p_2^2} = \sqrt{(20)^2 + (20)^2} \\[1em] = \sqrt{800} = 20\sqrt{2} \\[1em] = 28.28\ \text{kg m s}^{-1}\end{aligned}$$

No external force acts on the system, so by the principle of conservation of linear momentum this is also the momentum of the combined body after the collision. The combined mass is

$$\begin{aligned}\text m = \text m_1 + \text m_2 = 5 + 10 \\[1em] = 15\ \text{kg}\end{aligned}$$

Therefore, the final velocity is

$$\begin{aligned}\text v = \frac{\text p}{\text m} = \frac{28.28}{15} \\[1em] = 1.88\ \text{m s}^{-1}\end{aligned}$$

If θ is the angle made by the resultant with the +x direction, then

$\tan \theta = \frac{\text p_2}{\text p_1} = \frac{20}{20} = 1 \quad \Rightarrow \quad \theta = 45^\circ$

Hence, the final velocity is 1.88 m s^-1, directed at an angle of 45° with the +x direction.

Given,

• Weight of the body, W = 200 N
• The rope carrying T₁ makes 30° and the rope carrying T₂ makes 45° with the horizontal, as shown

The tensions T₁ and T₂ have to be calculated.

The point at which the ropes meet is in equilibrium under the two tensions and the weight of the body. Resolving the tensions into horizontal and vertical components,

Vertical components :

$\text T_1\sin 30^\circ + \text T_2\sin 45^\circ = 200 \quad \text{...............(1)}$

Horizontal components :

$\text T_1\cos 30^\circ = \text T_2\cos 45^\circ \quad \text{...............(2)}$

From equation (2),

$$\begin{aligned}\text T_2 = \frac{\text T_1\cos 30^\circ}{\cos 45^\circ} = \frac{\text T_1 \times 0.866}{0.7071} \\[1em] = 1.2247\text T_1\end{aligned}$$

Substituting this in equation (1),

$$\begin{aligned}\text T_1 \times 0.5 + 1.2247\text T_1 \times 0.7071 = 200 \\[1em] 0.5\text T_1 + 0.866\text T_1 = 200 \\[1em] 1.366\text T_1 = 200 \\[1em] \text T_1 = \frac{200}{1.366} \\[1em] = 146.4\ \text N\end{aligned}$$

Therefore,

$$\begin{aligned}\text T_2 = 1.2247 \times 146.4 \\[1em] = 179.3\ \text N\end{aligned}$$

Hence, the tensions in the two ropes are 146.4 N and 179.3 N.

Given,

• Mass of the block, m = 1.5 kg
• Applied horizontal force, F = 1.2 kg-f = 1.2 × 9.8 = 11.76 N
• Coefficient of friction, μ = 0.3
• Acceleration due to gravity, g = 9.8 m s^-2

The acceleration produced has to be calculated in terms of g.

The block rests on a horizontal surface, so the normal reaction is

$$\begin{aligned}\text R = \text{mg} = 1.5 \times 9.8 \\[1em] = 14.7\ \text N\end{aligned}$$

The force of friction opposing the motion is

$$\begin{aligned}\text f = \mu \text R = 0.3 \times 14.7 \\[1em] = 4.41\ \text N\end{aligned}$$

Hence, the net force acting on the block is

$$\begin{aligned}\text F_{net} = \text F - \text f = 11.76 - 4.41 \\[1em] = 7.35\ \text N\end{aligned}$$

By Newton's second law of motion, the acceleration produced is

$$\begin{aligned}\text a = \frac{\text F_{net}}{\text m} = \frac{7.35}{1.5} \\[1em] = 4.9\ \text{m s}^{-2}\end{aligned}$$

Since g = 9.8 m s^-2,

$\text a = \frac{9.8}{2} = \frac{\text g}{2}$

Hence, the acceleration produced in the block is $\frac{\text g}{2}$.

Given,

• Mass of each block, m = 3.0 kg
• Applied horizontal force, F = 20 N
• Acceleration produced in each block, a = 0.50 m s^-2
• The frictional force f is the same on the two blocks

The tension in the string and the frictional force have to be calculated.

For the two blocks taken together : The total mass is 2m = 6.0 kg and the total frictional force is 2f. By Newton's second law,

$\text F - 2\text f = 2\text{ma}$

Substituting the values,

$$\begin{aligned}20 - 2\text f = 6.0 \times 0.50 \\[1em] 20 - 2\text f = 3 \\[1em] 2\text f = 17 \\[1em] \text f = 8.5\ \text N\end{aligned}$$

For the rear block, which is pulled only by the string : The forces on it are the tension T in the forward direction and the frictional force f in the backward direction. By Newton's second law,

$\text T - \text f = \text{ma}$

Substituting the values,

$$\begin{aligned}\text T - 8.5 = 3.0 \times 0.50 \\[1em] \text T = 1.5 + 8.5 \\[1em] = 10\ \text N\end{aligned}$$

Hence, the tension in the string is 10 N and the frictional force on each block is 8.5 N.

Given,

• Length of the inclined plane, l = 4.0 m
• Height of the upper end, h = 1.0 m
• The block just begins to slide down

The coefficient of static friction has to be calculated.

The block just begins to slide down, so the angle of inclination is equal to the angle of repose, and

$\mu_s = \tan \theta$

From the geometry of the inclined plane,

$$\begin{aligned}\sin \theta = \frac{\text h}{\text l} = \frac{1.0}{4.0} \\[1em] = 0.25\end{aligned}$$

The base of the inclined plane is

$$\begin{aligned}\text b = \sqrt{\text l^2 - \text h^2} = \sqrt{(4.0)^2 - (1.0)^2} \\[1em] = \sqrt{16 - 1} = \sqrt{15} \\[1em] = 3.873\ \text m\end{aligned}$$

Therefore,

$$\begin{aligned}\mu_s = \tan \theta = \frac{\text h}{\text b} = \frac{1.0}{3.873} \\[1em] = 0.258\end{aligned}$$

Hence, the coefficient of static friction is 0.258.

Given,

• The body is in limiting equilibrium at an inclination of 30°
• The inclination is then increased to 45°
• Acceleration due to gravity, g = 9.8 m s^-2

The acceleration with which the body slides down at 45° has to be calculated.

At 30° the body is in limiting equilibrium, so the angle of repose is 30° and

$$\begin{aligned}\mu = \tan 30^\circ = \frac{1}{\sqrt{3}} \\[1em] = 0.5774\end{aligned}$$

When the plane is inclined at θ = 45°, the body slides down. Resolving the gravity force into components parallel and perpendicular to the plane, the normal reaction is R = mg cos θ and the kinetic frictional force is μmg cos θ. By Newton's second law,

$\text{mg}\sin \theta - \mu \text{mg}\cos \theta = \text{ma}$

$\text a = \text g(\sin \theta - \mu\cos \theta)$

Substituting the values, with sin 45° = cos 45° = 0.7071,

$$\begin{aligned}\text a = 9.8 \times (0.7071 - 0.5774 \times 0.7071) \\[1em] = 9.8 \times 0.7071 \times (1 - 0.5774) \\[1em] = 6.93 \times 0.4226 \\[1em] = 2.93\ \text{m s}^{-2}\end{aligned}$$

Hence, the body slides down with an acceleration of 2.93 m s^-2.

Given,

• Mass of the body, m = 1 kg
• Rate of revolution, n = 3 rev/s
• Radius of the circle, r = 10 cm = 0.1 m

The linear velocity, the acceleration and the tension have to be calculated.

In one revolution the body covers the circumference of the circle, so the linear velocity is

$\text v = 2\pi \text{rn}$

Substituting the values,

$$\begin{aligned}\text v = 2 \times 3.14 \times 0.1 \times 3 \\[1em] = 1.88\ \text{m s}^{-1}\end{aligned}$$

The acceleration of the body is the centripetal acceleration,

$$\begin{aligned}\text a = \frac{\text v^2}{\text r} = \frac{(1.88)^2}{0.1} = \frac{3.534}{0.1} \\[1em] = 35.3\ \text{m s}^{-2}\end{aligned}$$

The necessary centripetal force is provided by the tension in the string,

$$\begin{aligned}\text T = \text{ma} = 1 \times 35.3 \\[1em] = 35.3\ \text N\end{aligned}$$

Hence, the linear velocity is 1.88 m s^-1, the acceleration is 35.3 m s^-2 and the tension in the string is 35.3 N.

If the string breaks : The centripetal force vanishes instantly, so no force acts on the body towards the centre. By Newton's first law of motion the body then flies off along the tangent to the circular path at the point where the string breaks.

Given,

• Maximum tension the string can bear, T = 3.16 kg-wt = 3.16 × 9.8 = 30.97 N
• Mass of the stone, m = 49 g = 0.049 kg
• Length of the string, r = 1 m
• $\pi = \frac{22}{7}$

The maximum number of revolutions per second has to be calculated.

The necessary centripetal force is provided by the tension in the string. If n is the number of revolutions per second, the angular speed is ω = 2πn, and

$$\begin{aligned}\text T = \text{mr}\omega^2 = \text{mr}(2\pi \text n)^2 \\[1em] = 4\pi^2\text{mrn}^2\end{aligned}$$

Therefore,

$\text n = \frac{1}{2\pi}\sqrt{\frac{\text T}{\text{mr}}}$

Substituting the values,

$$\begin{aligned}\frac{\text T}{\text{mr}} = \frac{30.97}{0.049 \times 1} \\[1em] = 632.0\ \text s^{-2}\end{aligned}$$

$$\begin{aligned}\sqrt{\frac{\text T}{\text{mr}}} = \sqrt{632.0} \\[1em] = 25.14\ \text s^{-1}\end{aligned}$$

$$\begin{aligned}\text n = \frac{25.14}{2 \times \frac{22}{7}} = \frac{25.14 \times 7}{44} \\[1em] = \frac{175.98}{44} \\[1em] = 4\ \text{rev s}^{-1}\end{aligned}$$

Hence, the stone can be made to complete a maximum of 4 revolutions per second without breaking the string.

Given,

• Speed of the motor-cyclist, v = 0.6 km/min = $\frac{600}{60}$ = 10 m s^-1
• Radius of curvature, r = 50 m
• Acceleration due to gravity, g = 10 m s^-2

The centripetal acceleration and the slope with the vertical have to be calculated.

The centripetal acceleration is

$$\begin{aligned}\text a_c = \frac{\text v^2}{\text r} = \frac{(10)^2}{50} = \frac{100}{50} \\[1em] = 2.0\ \text{m s}^{-2}\end{aligned}$$

While taking the turn the motor-cyclist leans inwards through an angle θ from the vertical, given by

$\tan \theta = \frac{\text v^2}{\text{rg}}$

Substituting the values,

$$\begin{aligned}\tan \theta = \frac{(10)^2}{50 \times 10} = \frac{100}{500} \\[1em] = 0.2\end{aligned}$$

$\theta = \tan^{-1}(0.2)$

Hence, the centripetal acceleration is 2.0 m s^-2 and the motor-cyclist leans through an angle of tan^-1(0.2) with the vertical.

Given,

• Radius of the curve, r = 100 m
• Coefficient of friction, μ = 0.4
• Acceleration due to gravity, g = 10 m s^-2

The maximum speed at which the car can turn has to be calculated.

On a level road the necessary centripetal force is provided entirely by the force of friction between the tyres and the road. For the car not to skid,

$\frac{\text{mv}^2}{\text r} \le \mu \text{mg}$

Hence, the maximum speed is

$\text v_{max} = \sqrt{\mu \text{rg}}$

Substituting the values,

$$\begin{aligned}\text v_{max} = \sqrt{0.4 \times 100 \times 10} \\[1em] = \sqrt{400} \\[1em] = 20\ \text{m s}^{-1}\end{aligned}$$

Hence, the maximum speed at which the car can turn round the curve is 20 m s^-1.

Given,

• Mass of the car, m = 1000 kg
• Speed of the car, v = 30 m s^-1
• Radius of the circular path, r = 200 m
• Coefficient of friction, μ = 0.8
• Acceleration due to gravity, g = 9.8 m s^-2

The frictional force required and the maximum speed have to be calculated.

The frictional force must supply the whole of the necessary centripetal force,

$\text F = \frac{\text{mv}^2}{\text r}$

Substituting the values,

$$\begin{aligned}\text F = \frac{1000 \times (30)^2}{200} = \frac{1000 \times 900}{200} \\[1em] = 4500\ \text N\end{aligned}$$

For the car not to skid, this force must not exceed the limiting friction μmg. Hence, the maximum speed is

$\text v_{max} = \sqrt{\mu \text{rg}}$

Substituting the values,

$$\begin{aligned}\text v_{max} = \sqrt{0.8 \times 200 \times 9.8} \\[1em] = \sqrt{1568} \\[1em] = 39.6\ \text{m s}^{-1}\end{aligned}$$

Hence, a frictional force of 4500 N is needed, and the maximum speed of the car can be 39.6 m s^-1.

Given,

• Masses of the two blocks, 20 kg and 10 kg, placed in contact
• Applied horizontal force, F = 5.0 N
• The table is smooth

The two blocks are in contact, so they move together with a common acceleration. By Newton's second law,

$$\begin{aligned}\text a = \frac{\text F}{20 + 10} = \frac{5.0}{30} \\[1em] = \frac{1}{6}\ \text{m s}^{-2}\end{aligned}$$

(i) When the force is applied on the 20 kg block : The 10 kg block is accelerated only by the contact force N exerted on it by the 20 kg block. Hence,

$$\begin{aligned}\text N = 10 \times \frac{1}{6} \\[1em] = \frac{10}{6}\ \text N\end{aligned}$$

Hence, the 20 kg block presses the 10 kg block with a force of $\frac{10}{6}$ N towards the right.

(ii) When the force is applied on the other side of the 10 kg block : Now the 20 kg block is accelerated only by the contact force N' exerted on it by the 10 kg block. Hence,

$$\begin{aligned}\text N' = 20 \times \frac{1}{6} \\[1em] = \frac{20}{6}\ \text N\end{aligned}$$

By Newton's third law of motion the 20 kg block presses the 10 kg block with an equal force.

Hence, the 20 kg block will press the 10 kg block with a force of $\frac{20}{6}$ N towards the right.

Given,

• Masses of the three blocks, m, 3m and 5m
• Tension in the first string, P = 16 N
• The horizontal surface is frictionless

The force F pulls all the three blocks, the tension P pulls the blocks 3m and 5m, and the tension Q pulls only the block 5m.

(i) Considering the blocks 3m and 5m, which are pulled by the tension P,

$\text P = (3\text m + 5\text m)\text a = 8\text{ma}$

Substituting P = 16 N,

$$\begin{aligned}16 = 8\text{ma} \\[1em] \text a = \frac{2}{\text m}\ \text{m s}^{-2}\end{aligned}$$

(ii) The block 5m is pulled only by the tension Q, so

$$\begin{aligned}\text Q = 5\text{ma} = 5\text m \times \frac{2}{\text m} \\[1em] = 10\ \text N\end{aligned}$$

(iii) The force F pulls all the three blocks, of total mass 9m, so

$$\begin{aligned}\text F = (\text m + 3\text m + 5\text m)\text a = 9\text m \times \frac{2}{\text m} \\[1em] = 18\ \text N\end{aligned}$$

Hence, the acceleration of the blocks is $\frac{2}{\text m}$ m s^-2, the tension Q in the second string is 10 N and the force F is 18 N.

Given,

• Mass of the upper block, m₁ = 2 kg
• Mass of the lower block, m₂ = 4 kg
• Upward acceleration required, a = 2 m s^-2
• Acceleration due to gravity, g = 9.8 m s^-2

(i) Tension in the upper string in the rest position : The upper string supports the whole of the two blocks, so

$$\begin{aligned}\text T = (\text m_1 + \text m_2)\text g = (2 + 4) \times 9.8 \\[1em] = 6 \times 9.8 \\[1em] = 58.8\ \text N\end{aligned}$$

(ii) Force F required for an upward acceleration of 2 m s^-2 : The net upward force on the two blocks must be (m₁ + m₂)a, so

$$\begin{aligned}\text F - (\text m_1 + \text m_2)\text g = (\text m_1 + \text m_2)\text a \\[1em] \text F = (\text m_1 + \text m_2)(\text g + \text a)\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text F = 6 \times (9.8 + 2) \\[1em] = 6 \times 11.8 \\[1em] = 70.8\ \text N\end{aligned}$$

(iii) Tension in the string between the two blocks : This string carries only the lower block of mass m₂, which also moves upwards with the acceleration a. Hence,

$$\begin{aligned}\text T' = \text m_2(\text g + \text a) = 4 \times 11.8 \\[1em] = 47.2\ \text N\end{aligned}$$

Hence, the tension in the upper string in the rest position is 58.8 N, the force F required is 70.8 N, and the tension in the string between the two blocks is then 47.2 N.

Given,

• Mass of the proton, m = 1.67 × 10^-27 kg, moving left with u₁ = 7.0 × 10⁶ m s^-1
• Mass of the neutron is nearly equal to that of the proton, moving right with u₂ = 4.0 × 10⁶ m s^-1
• They combine to form a deutron of mass 2m

The velocity of the deutron has to be calculated.

Taking the leftward direction as positive, the total linear momentum before the combination is

$\text p = \text{mu}_1 + \text m(-\text u_2) = \text m(\text u_1 - \text u_2)$

Substituting the values,

$$\begin{aligned}\text p = \text m(7.0 \times 10^6 - 4.0 \times 10^6) \\[1em] = \text m \times 3.0 \times 10^6\ \text{kg m s}^{-1}\end{aligned}$$

No external force acts on the system, so by the principle of conservation of linear momentum this momentum is carried by the deutron of mass 2m,

$$\begin{aligned}2\text m\text v = \text m \times 3.0 \times 10^6 \\[1em] \text v = \frac{3.0 \times 10^6}{2} \\[1em] = 1.5 \times 10^6\ \text{m s}^{-1}\end{aligned}$$

The positive sign shows that the deutron moves in the direction in which the proton was moving.

Hence, the velocity of the deutron is 1.5 × 10⁶ m s^-1, directed towards the left.

Given,

• Momentum of the ball, p = 15 kg m s^-1
• Angle with the normal to the wall, θ = 60°, both before and after reflection
• The speed is unchanged

The impulse has to be calculated.

Resolving the momentum of the ball into components parallel and normal to the wall :

The components parallel to the wall are each p sin θ and they are in the same direction, so the change in momentum along the wall is

$\text p\sin \theta - \text p\sin \theta = 0$

The components normal to the wall are just reversed. Hence, the change in the normal momentum is

$$\begin{aligned}\Delta \text p = \text p\cos \theta - (-\text p\cos \theta) \\[1em] = 2\text p\cos \theta\end{aligned}$$

By the impulse-momentum theorem, this change in momentum is the impulse imparted to the ball,

$\text J = 2\text p\cos \theta$

Substituting the values,

$$\begin{aligned}\text J = 2 \times 15 \times \cos 60^\circ \\[1em] = 30 \times \frac{1}{2} \\[1em] = 15\ \text{kg m s}^{-1}\end{aligned}$$

Hence, the impulse is 15 kg m s^-1, directed normal to the wall.

Given,

• Mass of the shell, m₁ = 0.02 kg
• Mass of the gun, m₂ = 100 kg
• Muzzle speed of the shell, v₁ = 80 m s^-1

The recoil speed of the gun has to be calculated.

Before firing, both the gun and the shell are at rest, so the total linear momentum of the system is zero. No external force acts on the system, so by the principle of conservation of linear momentum,

$$\begin{aligned}\text m_1\text v_1 + \text m_2\text v_2 = 0 \\[1em] \text v_2 = -\frac{\text m_1\text v_1}{\text m_2}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text v_2 = -\frac{0.02 \times 80}{100} = -\frac{1.6}{100} \\[1em] = -0.016\ \text{m s}^{-1}\end{aligned}$$

The negative sign shows that the gun moves opposite to the direction of the shell.

Hence, the recoil speed of the gun is 0.016 m s^-1.

Given,

• m₁ = 2 kg with v₁ = 6 m/s, directed horizontally
• m₂ = 3 kg with v₂ = 4 m/s, directed vertically downwards

The magnitude and the direction of the combined momentum have to be determined.

The momentum of the first sphere is

$$\begin{aligned}\text p_1 = \text m_1\text v_1 = 2 \times 6 \\[1em] = 12\ \text{kg m s}^{-1}\ (\text{horizontal})\end{aligned}$$

The momentum of the second sphere is

$$\begin{aligned}\text p_2 = \text m_2\text v_2 = 3 \times 4 \\[1em] = 12\ \text{kg m s}^{-1}\ (\text{vertically downwards})\end{aligned}$$

The two momenta are mutually perpendicular, so by the parallelogram law of vector addition the magnitude of their resultant is

$$\begin{aligned}\text p = \sqrt{\text p_1^2 + \text p_2^2} = \sqrt{(12)^2 + (12)^2} \\[1em] = \sqrt{144 + 144} = \sqrt{288} \\[1em] = 12\sqrt{2} = 17\ \text{kg m s}^{-1}\ (\text{nearly})\end{aligned}$$

If θ is the angle made by the resultant with the horizontal, then

$\tan \theta = \frac{\text p_2}{\text p_1} = \frac{12}{12} = 1 \quad \Rightarrow \quad \theta = 45^\circ$

Hence, the combined momentum is nearly 17 kg m s^-1, directed at an angle of 45° with the horizontal.

Given,

• Initial speed of the car, u = 7 m s^-1
• Final speed, v = 0
• Shortest stopping distance, s = 10 m
• Acceleration due to gravity, g = 9.8 m s^-2

Let m be the mass of the car and a the retardation produced by the total frictional force. Using the equation of motion $\text v^2 = \text u^2 - 2\text{as}$ with v = 0,

$$\begin{aligned}0 = \text u^2 - 2\text{as} \\[1em] \text a = \frac{\text u^2}{2\text s}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text a = \frac{(7)^2}{2 \times 10} = \frac{49}{20} \\[1em] = 2.45\ \text{m s}^{-2}\end{aligned}$$

By Newton's second law of motion, the total frictional force opposing the motion is

$\text f = \text{ma} = 2.45\text m$

The weight of the car is

$\text W = \text{mg} = 9.8\text m$

Therefore,

$$\begin{aligned}\frac{\text f}{\text W} = \frac{2.45\text m}{9.8\text m} = \frac{2.45}{9.8} \\[1em] = \frac{1}{4}\end{aligned}$$

$\text f = \frac{1}{4}\text W$

Hence, the total frictional force opposing the motion is one-fourth of the weight of the car.

Given,

• Speed of the train, v = 72 km/h = $72 \times \frac{5}{18}$ = 20 m s^-1
• Radius of the curve, r = 400 m
• Weight of the block, mg = 5 kg-wt, so m = 5 kg
• Acceleration due to gravity, g = 10 m s^-2

The reading of the spring balance has to be calculated.

The forces acting on the suspended block are its weight mg, acting vertically downwards, and the spring-tension T along the string. Their resultant must provide the necessary centripetal force $\frac{\text{mv}^2}{\text r}$, which is horizontal and directed towards the centre of the curve.

Since the weight is vertical and the centripetal force is horizontal, these two are mutually perpendicular, and

$\text T^2 = (\text{mg})^2 + \left(\frac{\text{mv}^2}{\text r}\right)^2$

Substituting the values,

$\text{mg} = 5 \times 10 = 50\ \text N$

$$\begin{aligned}\frac{\text{mv}^2}{\text r} = \frac{5 \times (20)^2}{400} = \frac{5 \times 400}{400} \\[1em] = 5\ \text N\end{aligned}$$

Therefore,

$$\begin{aligned}\text T^2 = (50)^2 + (5)^2 \\[1em] = 2500 + 25 = 2525\end{aligned}$$

$$\begin{aligned}\text T = \sqrt{2525} \\[1em] = 50.25\ \text N = \frac{50.25}{10}\ \text{kg-wt} = 5.025\ \text{kg-wt}\end{aligned}$$

Hence, the reading of the balance is 5.025 kg-wt.

Given,

• Length of the inclined plane, l = 6.4 m
• Angle of inclination, θ = 30°
• Coefficient of kinetic friction, μ_k = 0.2
• Initial velocity, u = 0
• Acceleration due to gravity, g = 9.8 m s^-2

The time taken by the block to reach the bottom has to be calculated.

Resolving the gravity force into components parallel and perpendicular to the plane, the normal reaction is R = mg cos θ and the kinetic frictional force is μ_kmg cos θ, acting up the plane. By Newton's second law,

$$\begin{aligned}\text{mg}\sin \theta - \text{f}_k = \text{ma} \\[1em] \text{mg}\sin \theta - \mu_k\text{mg}\cos \theta = \text{ma} \\[1em] \text a = \text g(\sin \theta - \mu_k\cos \theta)\end{aligned}$$

Substituting the values, with sin 30° = 0.5 and cos 30° = 0.866,

$$\begin{aligned}\text a = 9.8 \times (0.5 - 0.2 \times 0.866) \\[1em] = 9.8 \times (0.5 - 0.1732) \\[1em] = 9.8 \times 0.3268 \\[1em] = 3.2\ \text{m s}^{-2}\end{aligned}$$

Using $\text s = \text{ut} + \frac{1}{2}\text{at}^2$ with u = 0 and s = l,

$\text l = \frac{1}{2}\text{at}^2 \quad \Rightarrow \quad \text t = \sqrt{\frac{2\text l}{\text a}}$

Substituting the values,

$$\begin{aligned}\text t = \sqrt{\frac{2 \times 6.4}{3.2}} = \sqrt{\frac{12.8}{3.2}} = \sqrt{4} \\[1em] = 2\ \text s\end{aligned}$$

Hence, the block takes 2 s to reach the bottom of the plane.

Given,

• Mass of the body = m, released from rest at the top
• Length of the inclined plane = L, height of the top above the bottom = h
• Frictional force = f_k

Let θ be the angle of inclination of the plane, so that

$\sin \theta = \frac{\text h}{\text L}$

Resolving the gravity force into components, the component along the plane, directed down the plane, is mg sin θ, and the frictional force f_k acts up the plane, opposing the motion. Hence, the net force along the plane is

$\text F = \text{mg}\sin \theta - \text f_k$

By Newton's second law of motion, the acceleration of the body down the plane is

$\text a = \frac{\text{mg}\sin \theta - \text f_k}{\text m}$

Substituting $\sin \theta = \frac{\text h}{\text L}$

$\text a = \frac{\text{mg}\frac{\text h}{\text L} - \text f_k}{\text m} = \frac{\text{mgh} - \text f_k\text L}{\text{mL}}$

The body starts from rest and covers the whole length L of the plane. Using the equation of motion $\text v^2 = \text u^2 + 2\text{as}$ with u = 0 and s = L,

$$\begin{aligned}\text v^2 = 2 \times \frac{\text{mgh} - \text f_k\text L}{\text{mL}} \times \text L \\[1em] = \frac{2}{\text m}(\text{mgh} - \text f_k\text L)\end{aligned}$$

Taking the square root,

$\text v = \sqrt{\frac{2}{\text m}(\text{mgh} - \text f_k\text L)}$

Hence proved.

Given,

• The minimum force required to move the body up the plane is three times the minimum force required to prevent it from sliding down
• Coefficient of friction, $\mu = \frac{1}{2\sqrt{3}}$

The angle of inclination of the plane has to be calculated.

Let θ be the angle of inclination and m the mass of the body. The normal reaction is R = mg cos θ, so the limiting frictional force is μmg cos θ.

Force required to move the body up the plane : Here the friction acts down the plane, so

$\text F_1 = \text{mg}\sin \theta + \mu \text{mg}\cos \theta$

Force required to prevent the body from sliding down : Here the friction acts up the plane, so

$\text F_2 = \text{mg}\sin \theta - \mu \text{mg}\cos \theta$

It is given that F₁ = 3F₂. Therefore,

$\text{mg}\sin \theta + \mu \text{mg}\cos \theta = 3(\text{mg}\sin \theta - \mu \text{mg}\cos \theta)$

Dividing throughout by mg,

$$\begin{aligned}\sin \theta + \mu\cos \theta = 3\sin \theta - 3\mu\cos \theta \\[1em] 4\mu\cos \theta = 2\sin \theta \\[1em] \tan \theta = 2\mu\end{aligned}$$

Substituting the given value of μ,

$$\begin{aligned}\tan \theta = 2 \times \frac{1}{2\sqrt{3}} \\[1em] = \frac{1}{\sqrt{3}}\end{aligned}$$

$$\begin{aligned}\theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \\[1em] = 30^\circ\end{aligned}$$

Hence, the angle of inclination of the plane is 30°.

Given,

• Coefficient of friction, μ = 0.5
• The normal reaction is twice the resultant downward force along the incline

The angle between the inclined plane and the horizontal has to be calculated.

Let θ be the angle of inclination and m the mass of the body. The normal reaction is

$\text R = \text{mg}\cos \theta$

The body is sliding down, so the frictional force μmg cos θ acts up the plane. Hence, the resultant downward force along the incline is

$\text F = \text{mg}\sin \theta - \mu \text{mg}\cos \theta$

It is given that R = 2F. Therefore,

$\text{mg}\cos \theta = 2(\text{mg}\sin \theta - \mu \text{mg}\cos \theta)$

Dividing throughout by mg,

$\cos \theta = 2\sin \theta - 2\mu\cos \theta$

Substituting μ = 0.5,

$$\begin{aligned}\cos \theta = 2\sin \theta - \cos \theta \\[1em] 2\cos \theta = 2\sin \theta \\[1em] \tan \theta = 1\end{aligned}$$

$$\begin{aligned}\theta = \tan^{-1}(1) \\[1em] = 45^\circ\end{aligned}$$

Hence, the angle between the inclined plane and the horizontal is 45°.

$\frac{2ma}{g+a}$

Reason — Let F be the upthrust on the balloon, which remains unchanged.

While descending with an acceleration a : The net downward force is mg − F, so

$\text{mg} - \text F = \text{ma} \quad \Rightarrow \quad \text F = \text m(\text g - \text a) \quad \text{...............(1)}$

After a mass m' is removed, ascending with an acceleration a : The remaining mass is (m − m'), and the net upward force is F − (m − m')g, so

$$\begin{aligned}\text F - (\text m - \text m')\text g = (\text m - \text m')\text a \\[1em] \text F = (\text m - \text m')(\text g + \text a) \quad \text{...............(2)}\end{aligned}$$

Equating (1) and (2),

$$\begin{aligned}\text m(\text g - \text a) = (\text m - \text m')(\text g + \text a) \\[1em] \text m - \text m' = \frac{\text m(\text g - \text a)}{\text g + \text a}\end{aligned}$$

$$\begin{aligned}\text m' = \text m - \frac{\text m(\text g - \text a)}{\text g + \text a} = \frac{\text m(\text g + \text a) - \text m(\text g - \text a)}{\text g + \text a} \\[1em] = \frac{2\text{ma}}{\text g + \text a}\end{aligned}$$

0.7 m/s

Reason — Given, m = 20 g = 0.02 kg, u = 1 m/s, s = 20 cm = 0.2 m and F = 2.5 × 10^-2 N.

The retardation produced is

$$\begin{aligned}\text a = \frac{\text F}{\text m} = \frac{2.5 \times 10^{-2}}{0.02} \\[1em] = 1.25\ \text{m s}^{-2}\end{aligned}$$

Using $\text v^2 = \text u^2 - 2\text{as}$,

$$\begin{aligned}\text v^2 = (1)^2 - 2 \times 1.25 \times 0.2 \\[1em] = 1 - 0.5 = 0.5\end{aligned}$$

$$\begin{aligned}\text v = \sqrt{0.5} = \frac{1}{\sqrt{2}} \\[1em] = 0.7\ \text{m s}^{-1}\end{aligned}$$

6.5 m/s and 6.3 m/s

Reason — Let the horizontal direction towards the right be the x-axis and the vertical direction upwards be the y-axis, the origin being the point of collision.

Momenta before the collision :

The particle of mass M moves with a speed of 10 m/s, directed at 30° below the horizontal. Resolving it into components,

$\text u_{1x} = 10\cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 8.66\ \text{m s}^{-1}$

$\text u_{1y} = -10\sin 30^\circ = -10 \times \frac{1}{2} = -5\ \text{m s}^{-1}$

The particle of mass 2M moves with a speed of 5 m/s, directed at 45° above the horizontal. Resolving it into components,

$\text u_{2x} = 5\cos 45^\circ = 5 \times \frac{1}{\sqrt{2}} = 3.54\ \text{m s}^{-1}$

$\text u_{2y} = 5\sin 45^\circ = 3.54\ \text{m s}^{-1}$

Hence, the components of the total momentum before the collision are

$$\begin{aligned}\text p_x = \text M(8.66) + 2\text M(3.54) \\[1em] = 8.66\text M + 7.07\text M = 15.73\text M\end{aligned}$$

$$\begin{aligned}\text p_y = \text M(-5) + 2\text M(3.54) \\[1em] = -5\text M + 7.07\text M = 2.07\text M\end{aligned}$$

Momenta after the collision :

The particle of mass 2M moves with the speed v₁ at 30° above the horizontal, and the particle of mass M moves with the speed v₂ at 45° below the horizontal. Hence, the components of the total momentum after the collision are

$$\begin{aligned}\text p_x' = 2\text M\text v_1\cos 30^\circ + \text M\text v_2\cos 45^\circ \\[1em] = 1.732\text M\text v_1 + 0.707\text M\text v_2\end{aligned}$$

$$\begin{aligned}\text p_y' = 2\text M\text v_1\sin 30^\circ - \text M\text v_2\sin 45^\circ \\[1em] = \text M\text v_1 - 0.707\text M\text v_2\end{aligned}$$

Applying the principle of conservation of linear momentum :

No external force acts on the system during the collision, so the momentum is conserved separately along each axis. Cancelling M throughout,

$1.732\text v_1 + 0.707\text v_2 = 15.73 \quad \text{...............(1)}$

$\text v_1 - 0.707\text v_2 = 2.07 \quad \text{...............(2)}$

Adding equations (1) and (2), the terms in v₂ cancel,

$$\begin{aligned}2.732\text v_1 = 17.80 \\[1em] \text v_1 = \frac{17.80}{2.732} \\[1em] = 6.52\ \text{m s}^{-1}\end{aligned}$$

Substituting this value of v₁ in equation (2),

$$\begin{aligned}6.52 - 0.707\text v_2 = 2.07 \\[1em] 0.707\text v_2 = 6.52 - 2.07 = 4.45 \\[1em] \text v_2 = \frac{4.45}{0.707} \\[1em] = 6.29\ \text{m s}^{-1}\end{aligned}$$

Hence, v₁ is nearly 6.5 m s^-1 and v₂ is nearly 6.3 m s^-1.

$2\sqrt{2}v$

Reason — By the principle of conservation of linear momentum along the original direction of motion,

$\text M(2\text v) + (2\text M)\text v = \text M \times 0 + 2 \times \text{Mv}'\cos 45^\circ$

where v' is the speed of each of the two particles of mass M moving at 45° to the original direction. Substituting $\cos 45^\circ = \frac{1}{\sqrt{2}}$,

$$\begin{aligned}2\text{Mv} + 2\text{Mv} = 2\text{Mv}' \times \frac{1}{\sqrt{2}} \\[1em] 4\text{Mv} = \sqrt{2}\text{Mv}'\end{aligned}$$

$$\begin{aligned}\text v' = \frac{4\text v}{\sqrt{2}} \\[1em] = 2\sqrt{2}\text v\end{aligned}$$

The components perpendicular to the original direction are equal and opposite, so they cancel out.

100 N

Reason — The mass is in equilibrium under three forces : its weight mg vertically downwards, the applied horizontal force F, and the tension T along the rope, which makes 45° with the vertical.

Resolving the tension into components,

$\text T\cos 45^\circ = \text{mg} \quad \text{and} \quad \text T\sin 45^\circ = \text F$

Dividing the second equation by the first,

$\tan 45^\circ = \frac{\text F}{\text{mg}} \quad \Rightarrow \quad \text F = \text{mg}\tan 45^\circ$

Substituting m = 10 kg, g = 10 m s^-2 and tan 45° = 1,

$$\begin{aligned}\text F = 10 \times 10 \times 1 \\[1em] = 100\ \text N\end{aligned}$$

0.036 s

Reason — The ball bounces back with the same speed, so the change in momentum is

$\Delta \text p = \text{mv} - (-\text{mv}) = 2\text{mv}$

Substituting m = 0.15 kg and v = 12 m s^-1,

$$\begin{aligned}\Delta \text p = 2 \times 0.15 \times 12 \\[1em] = 3.6\ \text{kg m s}^{-1}\end{aligned}$$

By the impulse-momentum theorem, F × t = Δp, so

$$\begin{aligned}\text t = \frac{\Delta \text p}{\text F} = \frac{3.6}{100} \\[1em] = 0.036\ \text s\end{aligned}$$

$\frac{2}{3}$

Reason — For two masses connected over a frictionless pulley, the acceleration of the system is

$\text a = \frac{(\text M_2 - \text M_1)\text g}{\text M_1 + \text M_2}$

When M₂ = 2M₁ :

$$\begin{aligned}\text a_1 = \frac{(2\text M_1 - \text M_1)\text g}{\text M_1 + 2\text M_1} = \frac{\text M_1\text g}{3\text M_1} \\[1em] = \frac{\text g}{3}\end{aligned}$$

When M₂ = 3M₁ :

$$\begin{aligned}\text a_2 = \frac{(3\text M_1 - \text M_1)\text g}{\text M_1 + 3\text M_1} = \frac{2\text M_1\text g}{4\text M_1} \\[1em] = \frac{\text g}{2}\end{aligned}$$

Therefore,

$$\begin{aligned}\frac{\text a_1}{\text a_2} = \frac{\text g/3}{\text g/2} \\[1em] = \frac{2}{3}\end{aligned}$$

3360 N

Reason — Let a be the acceleration of the block M in the horizontal direction, towards the right, and let T be the tension in the string.

Constraint relation : The string is inextensible, so the sum of the lengths of its horizontal and vertical portions is constant. The mass m₂ rises through 2 m s^-2, so the vertical portion shortens at this rate and the horizontal portion lengthens at the same rate. Hence, m₁ moves backward relative to M with an acceleration of 2 m s^-2, and its acceleration with respect to the ground is

$\text a_1 = \text a - 2$

The mass m₂ is pressed against the vertical face of M, so it shares the horizontal acceleration a of the block, while it also rises with an acceleration of 2 m s^-2.

Vertical motion of m₂ : The forces on it are the tension T upwards and its weight m₂g downwards. By Newton's second law,

$\text T - \text m_2\text g = \text m_2 \times 2$

Substituting the values,

$$\begin{aligned}\text T = 20 \times 10 + 20 \times 2 \\[1em] = 200 + 40 \\[1em] = 240\ \text N\end{aligned}$$

Horizontal motion of m₁ : The upper surface of M is frictionless, so the only horizontal force on m₁ is the tension T, which pulls it towards the pulley. By Newton's second law,

$\text T = \text m_1(\text a - 2)$

Substituting the values,

$$\begin{aligned}240 = 10(\text a - 2) \\[1em] \text a - 2 = 24 \\[1em] \text a = 26\ \text{m s}^{-2}\end{aligned}$$

Motion of the whole system : The floor is frictionless and the tensions and the normal contact forces are all internal to the system. Hence, F is the only external force in the horizontal direction, and it must produce the horizontal accelerations of all the three masses,

$\text F = \text{Ma} + \text m_1(\text a - 2) + \text m_2\text a$

Substituting the values,

$$\begin{aligned}\text F = 100 \times 26 + 10 \times 24 + 20 \times 26 \\[1em] = 2600 + 240 + 520 \\[1em] = 3360\ \text N\end{aligned}$$

Rope will break while climbing upward

Reason — Given, m = 50 kg, g = 10 m s^-2 and the rope can withstand a tension of 350 N.

While climbing down with an acceleration of 4 m s^-2 :

$$\begin{aligned}\text T = \text m(\text g - \text a) = 50 \times (10 - 4) \\[1em] = 300\ \text N\end{aligned}$$

This is less than 350 N, so the rope does not break.

While climbing up with an acceleration of 5 m s^-2 :

$$\begin{aligned}\text T = \text m(\text g + \text a) = 50 \times (10 + 5) \\[1em] = 750\ \text N\end{aligned}$$

This exceeds 350 N, so the rope breaks while the monkey is climbing upward.

$2\sqrt{2}v$

Reason — Let the masses of the three fragments be 2k, 2k and k, so that the total mass is m = 5k.

The two fragments of equal mass 2k fly off along mutually perpendicular directions with the same speed v, so each has a momentum 2kv, and these two momenta are perpendicular. The magnitude of their resultant is

$\text p = \sqrt{(2\text{kv})^2 + (2\text{kv})^2} = 2\text{kv}\sqrt{2}$

The shell was initially at rest, so the total momentum must remain zero. Hence, the third fragment of mass k must carry an equal and opposite momentum,

$$\begin{aligned}\text k\text v' = 2\sqrt{2}\text{kv} \\[1em] \text v' = 2\sqrt{2}\text v\end{aligned}$$

5

Reason — No external horizontal force acts on the system, so by the principle of conservation of linear momentum,

$\text m_1\text v_1 = (\text m_1 + \text m_2)\text v$

Substituting m₁ = 1000 kg, v₁ = 6 m s^-1 and m₂ = 200 kg,

$$\begin{aligned}1000 \times 6 = (1000 + 200)\text v \\[1em] \text v = \frac{6000}{1200} \\[1em] = 5\ \text{m s}^{-1}\end{aligned}$$

6 N

Reason — The two blocks slide together over the frictionless surface, so they have a common acceleration. By Newton's second law,

$$\begin{aligned}\text a = \frac{\text F}{\text m_A + \text m_B} = \frac{10}{2 + 3} \\[1em] = 2\ \text{m s}^{-2}\end{aligned}$$

The block B is accelerated only by the contact force (F_BA) exerted on it by the block A. Hence,

$$\begin{aligned}\text F_{BA} = \text m_B\text a = 3 \times 2 \\[1em] = 6\ \text N\end{aligned}$$

The curve in which the acceleration remains zero for an initial interval and thereafter increases linearly with time.

Reason — The block is initially at rest on the ground, so the force of static friction opposes the applied force. As the applied force F increases linearly with time, the static friction, being self-adjusting, increases with it and the block does not move. Hence, the acceleration remains zero during this period.

Once the applied force exceeds the limiting friction, the block begins to move. The friction then becomes the constant kinetic friction f_k, and the acceleration is

$\text a = \frac{\text F - \text f_k}{\text m}$

Since F increases linearly with time, the acceleration also increases linearly with time thereafter.