Work, Energy & Power

(a) Positive. The person applies an upward force on the bucket and the bucket is also displaced upwards, so the force and the displacement are in the same direction. Here θ = 0° and W = Fs cos 0° = Fs, which is positive.

(b) Negative. The gravitational force on the bucket acts vertically downwards while the displacement is vertically upwards, so θ = 180° and W = Fs cos 180° = − Fs, which is negative.

(c) Negative. Friction always opposes the relative motion of the body, so the frictional force acts up the incline while the displacement is down the incline. Here θ = 180° and the work done is negative.

(d) Positive. As the body moves with uniform velocity, the applied force is equal and opposite to the force of friction, that is, it acts along the direction of motion. Since the force and the displacement are in the same direction, the work done by the applied force is positive.

(e) Negative. The resistive force of air always acts opposite to the direction of motion of the pendulum bob, so θ = 180° and the work done by this force is negative. It is this negative work that brings the pendulum to rest.

Given,

• Mass of the body, m = 2 kg
• Applied horizontal force, F = 7 N
• Coefficient of friction, μ = 0.1
• Initial velocity, u = 0 (body starts from rest)
• Time, t = 10 s
• g = 10 m s^-2

The force of kinetic friction acting on the body is

$$\begin{aligned}\text f_k = \mu \text R = \mu \text{mg} = 0.1 \times 2 \times 10 \\[1em] = 2\ \text N\end{aligned}$$

The net force acting on the body is

$\text F_{net} = \text F - \text f_k = 7 - 2 = 5\ \text N$

The acceleration produced in the body is

$\text a = \frac{\text F_{net}}{\text m} = \frac{5}{2} = 2.5\ \text{m s}^{-2}$

The distance moved by the body in 10 s is

$$\begin{aligned}\text s = \text{ut} + \frac{1}{2}\text{at}^2 = (0 \times 10) + \frac{1}{2} \times 2.5 \times (10)^2 \\[1em] = 125\ \text m\end{aligned}$$

(a) The work done by the applied force is

$$\begin{aligned}\text W_F = \text F\text s \cos 0^\circ = 7 \times 125 \times 1 \\[1em] = 875\ \text J\end{aligned}$$

(b) The frictional force is opposite to the displacement, so θ = 180°. The work done by friction is

$$\begin{aligned}\text W_f = \text f_k\text s \cos 180^\circ = -2 \times 125 \\[1em] = -250\ \text J\end{aligned}$$

(c) The work done by the net force is

$$\begin{aligned}\text W_{net} = \text F_{net}\text s = 5 \times 125 \\[1em] = 625\ \text J\end{aligned}$$

(d) By the work-kinetic energy theorem, the change in the kinetic energy of the body is equal to the work done by the net force,

$\Delta \text K = \text W_{net} = 625\ \text J$

Hence, the work done by the applied force is 875 J, by friction is −250 J, by the net force is 625 J and the change in kinetic energy is 625 J.

The result shows that the work done by the applied force and that done by friction add up to the work done by the net force (875 − 250 = 625 J), and this net work appears entirely as the gain in the kinetic energy of the body, which verifies the work-kinetic energy theorem.

The total energy of the particle is the sum of its kinetic energy and potential energy,

$\text E = \text K + \text U \quad \Rightarrow \quad \text K = \text E - \text U$

Since the kinetic energy $\left(\text K = \frac{1}{2}\text{mv}^2\right)$ can never be negative, the particle can exist only in those regions where the potential energy U is not greater than the total energy E. Wherever U > E, the kinetic energy would have to be negative, which is not possible, and the particle cannot be found there.

(a) In the region x > a the potential energy is U₀, which is greater than the total energy E. Hence the particle cannot be found in the region x > a. The particle exists between x = 0 and x = a, where the minimum total energy required is zero.

(b) Over the entire X-axis the potential energy is greater than the total energy E. Hence the particle cannot exist anywhere on the X-axis for the given energy.

(c) In the regions x < a and x > b the potential energy is U₀, which is greater than E. Hence the particle cannot be found in the regions x < a and x > b. The particle exists between x = a and x = b, where the minimum total energy is − U₁.

(d) By the same argument, the particle cannot be found in the regions $-\frac{\text b}{2} \lt \text x \lt -\frac{\text a}{2}$ and $\frac{\text a}{2} \lt \text x \lt \frac{\text b}{2}$. The minimum total energy in the permitted region is − U₁.

Given,

• Potential energy function, $\text U(\text x) = \frac{1}{2}\text{kx}^2$
• Force constant of the oscillator, k = 0.5 N m^-1
• Total energy of the particle, E = 1 J

The total energy of the oscillating particle is partly kinetic and partly potential, but by the principle of conservation of mechanical energy the total energy remains constant. Thus,

$\text E = \text K + \text U = \frac{1}{2}\text{mv}^2 + \frac{1}{2}\text{kx}^2$

where m is the mass of the particle and v is its velocity at the displacement x.

The particle turns back at that point where its velocity becomes zero, that is, where the whole of its energy is potential energy. Putting v = 0,

$\text E = 0 + \frac{1}{2}\text{kx}^2 \quad \Rightarrow \quad \text x = \sqrt{\frac{2\text E}{\text k}}$

Substituting the given values,

$$\begin{aligned}\text x = \sqrt{\frac{2 \times 1}{0.5}} = \sqrt{4} \\[1em] = \pm 2\ \text m\end{aligned}$$

Hence, the particle of total energy 1 J must turn back when it reaches x = ± 2 m.

Beyond x = ± 2 m the potential energy would exceed the total energy of 1 J, so the kinetic energy would become negative, which is not possible.

(a) The energy of a rocket in flight, kinetic as well as potential, depends upon its mass. When the casing burns up due to friction, the mass of the rocket decreases and hence its energy decreases. The heat energy required for burning is therefore obtained at the expense of the mass (energy) of the rocket, and not of the atmosphere.

(b) The gravitational force exerted by the sun on the comet is a conservative force. The work done by a conservative force over a closed path is always zero, since the work done depends only on the initial and the final positions of the body and not on the path followed. As the comet returns to the same point after one complete orbit, the initial and the final positions coincide, and hence the work done by the gravitational force over every complete orbit is zero.

(c) As the satellite loses energy due to dissipation against atmospheric resistance, it moves down into orbits closer to the earth and its potential energy decreases. In coming down from a higher altitude the value of g increases, and by the conservation of energy this decrease in potential energy appears mainly as an increase in the kinetic energy of the satellite, the loss against friction being very small. Hence the speed of the satellite increases progressively as it comes closer to the earth, although its total energy decreases slowly.

(d) In Fig. (a) : The man carries the mass on his hands, so the force applied by him is vertically upward while the displacement of 2 m is horizontal. The angle between the force and the displacement is 90°, so the work done is

$\text W = \text{Fs} \cos \theta = \text{Fs} \cos 90^\circ = \textbf{0}$

In Fig. (b) : The rope passes over a pulley, so the man applies the force in the horizontal direction, that is, along his direction of motion. Here θ = 0° and the work done is

$$\begin{aligned}\text W = \text{Fs} \cos 0^\circ = (15 \times 9.8) \times 2 \times 1 \\[1em] = 294\ \text J\end{aligned}$$

Hence, the work done is greater in the second case.

(a) Decreases. A conservative force is the negative gradient of the potential energy, $\text F = -\frac{\text{dU}}{\text{dx}}$. Hence when a conservative force does positive work on a body, the potential energy of the body decreases by an equal amount.

(b) Kinetic energy. Friction always opposes the motion of a body and therefore does negative work on it. By the work-kinetic energy theorem this negative work causes a loss in the kinetic energy of the body.

(c) External force. The total linear momentum of a system of particles can change only when a net external force acts on the system. The internal forces occur in equal and opposite pairs and cannot change the total momentum. Hence the rate of change of the total momentum is proportional to the external force.

(d) Total linear momentum and total energy. In an inelastic collision a part of the kinetic energy of the system is converted into other forms such as heat and sound, so the total kinetic energy changes. However, the total linear momentum and the total energy of the system remain unchanged.

(a) False. In an elastic collision the total momentum and the total kinetic energy of the system are conserved, but the momentum and the energy of each individual body change during the collision. It is only the sum for the whole system that remains constant.

(b) False. The total energy of a system is conserved only if the system is isolated, that is, no external force acts upon it. If external forces act on the system, they can do work on it and thereby change its total energy.

(c) False. The work done over a closed loop is zero only for conservative forces such as the gravitational and the electrostatic force. For non-conservative forces such as friction the work done depends on the path followed, and the net work done over a closed loop is not zero.

(d) True. In an inelastic collision a part of the kinetic energy is converted into other forms such as heat, sound or the potential energy of deformation. Hence the final kinetic energy of the system is less than its initial kinetic energy.

(a) No, the total kinetic energy is not conserved during the contact time. In an elastic collision the total kinetic energy is conserved before and after the collision, but not necessarily during the collision. While the balls are in contact they are deformed, and a part of the kinetic energy is temporarily converted into the elastic potential energy of deformation. This potential energy is fully recovered as kinetic energy when the balls separate, and that is why the collision is elastic overall.

(b) Yes, the total linear momentum is conserved during the collision. During the collision the two balls exert equal and opposite forces on each other, and these are internal forces of the system. Since no external force acts on the system, the total linear momentum remains conserved at every instant, including the short time of contact.

(c) For an inelastic collision : The total kinetic energy is not conserved, either during or after the collision, because a part of it is permanently converted into other forms such as heat, sound and the energy of deformation. The total linear momentum, however, is conserved during as well as after the collision, since momentum conservation holds whenever no external force acts on the system, irrespective of the nature of the collision.

(d) Elastic. If the potential energy depends only on the separation between the centres of the balls, the force during the collision is a conservative force. For a conservative force the total mechanical energy (kinetic + potential) is conserved and no energy is lost as heat or sound. Hence the collision is elastic.

t

The instantaneous power delivered to a body is

$\text P = \vec{\text F} \cdot \vec{\text v} = \text{Fv}$

since the force and the velocity are in the same direction here.

The body starts from rest (u = 0) and moves with a constant acceleration a. Therefore its velocity at time t is

$\text v = \text u + \text{at} = 0 + \text{at} = \text{at}$

Also, by Newton's second law the force is F = ma, which is constant since a is constant. Substituting these,

$\text P = (\text{ma})(\text{at}) = \text{ma}^2\text t$

Since m and a are both constant,

$\text P \propto \text t$

Hence, the power delivered to the body is proportional to t.

t^3/2

The instantaneous power delivered to the body is constant. Now,

$\text P = \frac{\text{dW}}{\text{dt}} = \frac{\text d}{\text{dt}}\left(\frac{1}{2}\text{mv}^2\right) = \text{mv}\frac{\text{dv}}{\text{dt}}$

Separating the variables,

$\text P\text{dt} = \text{mv}\text{dv}$

Integrating, with the body starting from rest (v = 0 at t = 0),

$$\begin{aligned}\int_0^{\text t} \text P\text{dt} = \int_0^{\text v} \text{mv}\text{dv} \\[1em] \text{Pt} = \frac{\text{mv}^2}{2} \quad \Rightarrow \quad \text v = \sqrt{\frac{2\text{Pt}}{\text m}}\end{aligned}$$

Writing $\text v = \frac{\text{ds}}{\text{dt}}$,

$\frac{\text{ds}}{\text{dt}} = \left(\frac{2\text P}{\text m}\right)^{1/2}\text t^{1/2}$

Integrating again from s = 0 at t = 0,

$$\begin{aligned}\int_0^{\text s}\text{ds} = \left(\frac{2\text P}{\text m}\right)^{1/2}\int_0^{\text t}\text t^{1/2}\text{dt} \\[1em] \text s = \left(\frac{2\text P}{\text m}\right)^{1/2} \times \frac{\text t^{3/2}}{3/2} \\[1em] \text s = \frac{2}{3}\left(\frac{2\text P}{\text m}\right)^{1/2}\text t^{3/2}\end{aligned}$$

Since P and m are constant,

$\text s \propto \text t^{3/2}$

Hence, the displacement of the body is proportional to t^3/2.

Given,

• Force, $\vec{\text F} = (-\hat{\text i} + 2\hat{\text j} + 3\hat{\text k})$ N
• The body moves a distance of 4 m along the Z-axis

Since the body is constrained to move along the Z-axis, its displacement is

$\vec{\text s} = 4\hat{\text k}\ \text m$

Work done by a force is the scalar product of the force and the displacement,

$\text W = \vec{\text F} \cdot \vec{\text s}$

Substituting the values,

$\text W = (-\hat{\text i} + 2\hat{\text j} + 3\hat{\text k}) \cdot (4\hat{\text k})$

Using $\hat{\text i} \cdot \hat{\text k} = 0$, $\hat{\text j} \cdot \hat{\text k} = 0$ and $\hat{\text k} \cdot \hat{\text k} = 1$,

$$\begin{aligned}\text W = -4(\hat{\text i} \cdot \hat{\text k}) + 8(\hat{\text j} \cdot \hat{\text k}) + 12(\hat{\text k} \cdot \hat{\text k}) \\[1em] = 0 + 0 + 12 \\[1em] = 12\ \text J\end{aligned}$$

Hence, the work done by the force is 12 J.

Given,

• Kinetic energy of the electron, K_e = 10 keV
• Kinetic energy of the proton, K_p = 100 keV
• Mass of the electron, m_e = 9.11 × 10^-31 kg
• Mass of the proton, m_p = 1.67 × 10^-27 kg

Let v_e and v_p be the speeds of the electron and the proton. The kinetic energy of a body of mass m moving with speed v is

$\text K = \frac{1}{2}\text{mv}^2 \quad \Rightarrow \quad \text v = \sqrt{\frac{2\text K}{\text m}}$

Therefore the ratio of the two speeds is

$\frac{\text v_e}{\text v_p} = \sqrt{\frac{2\text K_e}{\text m_e}} \times \sqrt{\frac{\text m_p}{2\text K_p}} = \sqrt{\frac{\text K_e}{\text K_p} \times \frac{\text m_p}{\text m_e}}$

Substituting the given values,

$$\begin{aligned}\frac{\text v_e}{\text v_p} = \sqrt{\frac{10\ \text{keV}}{100\ \text{keV}} \times \frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}} \\[1em] = \sqrt{0.1 \times 1833.15} \\[1em] = \sqrt{183.3} \\[1em] = 13.5\end{aligned}$$

Hence, the ratio of the speeds is v_e : v_p = 13.5 : 1, that is, the electron moves 13.5 times faster than the proton.

Although the kinetic energy of the proton is ten times that of the electron, the mass of the proton is nearly 1833 times the mass of the electron, and since $\text v \propto \sqrt{\frac{\text K}{\text m}}$, the electron turns out to be much faster.

Given,

• Radius of the raindrop, r = 2 mm = 2 × 10^-3 m
• Total height of fall, h = 500 m
• Speed of the drop on reaching the ground, v = 10 m s^-1
• Density of water, ρ = 10³ kg m^-3
• g = 9.8 m s^-2

Whether the drop falls with decreasing acceleration or with uniform speed, the work done by the gravitational force depends only on the vertical distance fallen. The mass of the drop is

$$\begin{aligned}\text m = \frac{4}{3}\pi \text r^3 \rho = \frac{4}{3} \times 3.14 \times (2 \times 10^{-3})^3 \times 10^3 \\[1em] = 3.35 \times 10^{-5}\ \text{kg}\end{aligned}$$

The gravitational force on the drop is

$$\begin{aligned}\text{mg} = 3.35 \times 10^{-5} \times 9.8 \\[1em] = 3.28 \times 10^{-4}\ \text N\end{aligned}$$

Each half of the journey is of height $\frac{500}{2} = 250$ m, so the work done by the gravitational force in each half is

$$\begin{aligned}\text W = \text{mg} \times \text h = (3.28 \times 10^{-4}) \times 250 \\[1em] = 0.082\ \text J\end{aligned}$$

Hence, the work done by the gravitational force in the first half as well as in the second half of the journey is 0.082 J.

The total work done by the gravitational force in the entire journey is

$\text W_g = 2 \times 0.082 = 0.164\ \text J$

The drop starts from rest, so its initial kinetic energy is zero. Its kinetic energy on reaching the ground is

$$\begin{aligned}\text K = \frac{1}{2}\text{mv}^2 = \frac{1}{2} \times (3.35 \times 10^{-5}) \times (10)^2 \\[1em] = 1.675 \times 10^{-3}\ \text J\end{aligned}$$

Therefore the change in the kinetic energy is

$\Delta \text K = 1.675 \times 10^{-3} - 0 = 1.675 \times 10^{-3}\ \text J$

By the work-kinetic energy theorem, the total work done by all the forces equals the change in kinetic energy. If W_r is the work done by the resistive force,

$$\begin{aligned}\Delta \text K = \text W_g + \text W_r \\[1em] \text W_r = \Delta \text K - \text W_g \\[1em] = (1.675 \times 10^{-3}) - 0.164 \\[1em] = 0.001675 - 0.164 \\[1em] = -0.162\ \text J\end{aligned}$$

Hence, the work done by the resistive force in the entire journey is − 0.162 J.

Given,

• Speed of the molecule before and after the collision, v = 200 m s^-1
• Angle made with the normal, θ = 30°

Conservation of momentum : The momentum of the system (molecule + wall) is conserved in the collision, irrespective of the nature of the collision. During the collision the molecule and the wall exert equal and opposite forces on each other, and these are internal forces of the system. Since no external force acts on the system, the total linear momentum remains conserved. The momentum gained by the wall is exactly equal to the momentum lost by the molecule.

Nature of the collision : The wall is extremely massive compared with the molecule. Hence, even though the wall receives a recoil momentum, the velocity produced in it is negligible, and so the kinetic energy acquired by the wall is negligible. The molecule rebounds with the same speed of 200 m s^-1 with which it strikes the wall, so its kinetic energy $\left(\frac{1}{2}\text{mv}^2\right)$ remains unchanged.

Therefore the total kinetic energy of the system remains conserved in the collision.

Hence, the momentum of the system is conserved and the collision is elastic.

Given,

• Volume of water to be pumped, V = 30 m³
• Time taken, t = 15 min = 15 × 60 = 900 s
• Height through which water is raised, h = 40 m
• Efficiency of the pump, η = 30%
• Density of water, ρ = 10³ kg m^-3
• g = 9.8 m s^-2

The mass of water to be pumped up is

$$\begin{aligned}\text m = \text{volume} \times \text{density} = 30 \times 10^3 \\[1em] = 3 \times 10^4\ \text{kg}\end{aligned}$$

The work required to raise this water through a height of 40 m is equal to the gain in its gravitational potential energy,

$$\begin{aligned}\text W = \text{mgh} = (3 \times 10^4) \times 9.8 \times 40 \\[1em] = 117.6 \times 10^5\ \text J\end{aligned}$$

The output power of the pump is

$$\begin{aligned}\text P_{out} = \frac{\text W}{\text t} = \frac{117.6 \times 10^5}{900} \\[1em] = 1.31 \times 10^4\ \text W = 13.1\ \text{kW}\end{aligned}$$

Since the efficiency of the pump is only 30%, the electric power consumed is

$$\begin{aligned}\text P_{in} = \text P_{out} \times \frac{100}{30} = 13.1 \times \frac{100}{30} \\[1em] = 43.6\ \text{kW}\end{aligned}$$

Hence, the electric power consumed by the pump is 43.6 kW.

Let m be the mass of each ball bearing and v the initial speed of the striking ball. Before the collision, balls 2 and 3 are at rest, so the total kinetic energy of the system is

$\text K_i = \frac{1}{2}\text{mv}^2 + 0 + 0 = \frac{1}{2}\text{mv}^2$

Since the collision is elastic, the total kinetic energy after the collision must also be $\frac{1}{2}\text{mv}^2$. Let us test each case.

Case (i) : Ball 1 comes to rest and balls 2 and 3 move together with speed $\frac{\text v}{2}$

$\text K_f = 0 + \frac{1}{2}(2\text m)\left(\frac{\text v}{2}\right)^2 = \frac{1}{4}\text{mv}^2$

Case (ii) : Balls 1 and 2 come to rest and ball 3 moves with speed v,

$\text K_f = 0 + \frac{1}{2}\text{mv}^2 = \frac{1}{2}\text{mv}^2$

Case (iii) : All the three balls move together with speed $\frac{\text v}{3}$

$\text K_f = \frac{1}{2}(3\text m)\left(\frac{\text v}{3}\right)^2 = \frac{1}{6}\text{mv}^2$

Only in case (ii) is the final kinetic energy equal to the initial kinetic energy $\frac{1}{2}\text{mv}^2$. In cases (i) and (iii) the kinetic energy is less than the initial value, which is not possible for an elastic collision.

Hence, case (ii) is the only possible result of the collision.

The bob A and the bob B are of the same mass and the collision between them is elastic and head-on.

In a one-dimensional elastic collision between two bodies of equal masses, the bodies merely exchange their velocities after the collision. This follows from

$\text v_1 = \left(\frac{\text m_1 - \text m_2}{\text m_1 + \text m_2}\right)\text u_1 + \left(\frac{2\text m_2}{\text m_1 + \text m_2}\right)\text u_2$

Putting m₁ = m₂ = m gives v₁ = u₂ and v₂ = u₁.

Here the bob B is initially at rest, that is, u₂ = 0. Hence after the collision the bob A comes to rest (v₁ = 0) and the bob B moves off with the whole velocity that A had acquired on reaching the lowest point.

Since the bob A is at rest at the lowest position, it has no kinetic energy left to be converted into potential energy.

Hence, the bob A does not rise at all after the collision; it comes to rest at the lowest position originally occupied by B.

Given,

• Length of the pendulum, l = 1.5 m
• Energy dissipated against air resistance = 5% of the initial energy
• g = 9.8 m s^-2

Let m be the mass of the bob. The bob is released from the horizontal position, so its initial height above the lowest point is equal to the length of the pendulum,

$\text h = \text l = 1.5\ \text m$

The potential energy of the bob at the horizontal position is

$\text U = \text{mgh}$

As the bob swings down, this potential energy is converted into kinetic energy. Since 5% of the initial energy is dissipated against air resistance, only 95% of it is available as kinetic energy at the lowest point. Therefore,

$\frac{1}{2}\text{mv}^2 = \text{mgh} \times \frac{95}{100}$

The mass m cancels from both sides, giving

$\text v = \sqrt{2\text{gh} \times \frac{95}{100}}$

Substituting the values,

$$\begin{aligned}\text v = \sqrt{2 \times 9.8 \times 1.5 \times 0.95} \\[1em] = \sqrt{27.93} \\[1em] = 5.3\ \text{m s}^{-1}\end{aligned}$$

Hence, the bob arrives at the lowermost point with a speed of 5.3 m s^-1.

Given,

• Mass of the trolley, M = 300 kg
• Mass of the sand bag, m = 25 kg
• Speed of the trolley, v = 27 km/h
• Rate of leakage of sand = 0.05 kg s^-1
• The track is frictionless

The speed of the trolley remains 27 km/h.

Since the track is frictionless, no external horizontal force acts on the system of the trolley and the sand. The only forces involved between the trolley and the leaking sand are internal forces.

At the instant a bit of sand leaves the hole, it is moving along with the trolley and therefore carries exactly the same horizontal velocity as the trolley has at that moment. The sand simply falls out without pushing the trolley either forward or backward, so no horizontal impulse is given to the trolley.

By the principle of conservation of linear momentum, since there is no net external horizontal force, the horizontal velocity of the trolley cannot change.

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

Given,

• Mass of the body, m = 0.5 kg
• Velocity, v = a x^3/2 where a = 5 m^-1/2 s^-1
• Displacement from x₁ = 0 to x₂ = 2 m

The velocity of the body at a displacement x is

$\text v = \text a\text x^{3/2} = (5)\text x^{3/2}$

At x = 0, the velocity is

$\text v_1 = 5 \times (0)^{3/2} = 0$

At x = 2 m, the velocity is

$$\begin{aligned}\text v_2 = 5 \times (2)^{3/2} = 5 \times 2\sqrt{2} \\[1em] = 10\sqrt{2}\ \text{m s}^{-1}\end{aligned}$$

The initial kinetic energy of the body is

$\text K_1 = \frac{1}{2}\text{mv}_1^2 = \frac{1}{2} \times 0.5 \times (0)^2 = 0$

The final kinetic energy of the body is

$$\begin{aligned}\text K_2 = \frac{1}{2}\text{mv}_2^2 = \frac{1}{2} \times 0.5 \times (10\sqrt{2})^2 \\[1em] = \frac{1}{2} \times 0.5 \times 200 \\[1em] = 50\ \text J\end{aligned}$$

By the work-kinetic energy theorem, the work done by the net force is equal to the change in the kinetic energy of the body,

$$\begin{aligned}\text W = \text K_2 - \text K_1 = 50 - 0 \\[1em] = 50\ \text J\end{aligned}$$

Hence, the work done by the net force during the displacement is 50 J.

Given,

• Area swept by the blades of the windmill = A
• Velocity of the wind perpendicular to the circle = v
• Density of air = ρ

(a) The volume of air passing through the blades per second is A v. Therefore the volume of air passing through the blades in time t is A v t, and the mass of this air is

$$\begin{aligned}\text m = \text{volume} \times \text{density} \\[1em] = \text{A v t}\rho\end{aligned}$$

(b) The kinetic energy of this air is

$$\begin{aligned}\text K = \frac{1}{2}\text{mv}^2 = \frac{1}{2}(\text{A v t}\rho)\text v^2 \\[1em] = \frac{1}{2}\text{A t}\rho\text v^3\end{aligned}$$

(c) Given,

• A = 30 m²
• v = 36 km/h
• ρ = 1.2 kg m^-3

Converting the wind speed into SI units,

$$\begin{aligned}\text v = 36\ \text{km/h} = \frac{36 \times 1000}{60 \times 60} \\[1em] = 10\ \text{m s}^{-1}\end{aligned}$$

The windmill converts 25% of the wind's energy into electrical energy, so the electrical energy produced in time t is

$\text E = \frac{25}{100} \times \frac{1}{2}\text A\text t\rho\text v^3 = \frac{1}{8}\text A\text t\rho\text v^3$

Therefore the electrical power produced is

$\text P = \frac{\text E}{\text t} = \frac{1}{8}\text A\rho\text v^3$

Substituting the values,

$$\begin{aligned}\text P = \frac{1}{8} \times 30 \times 1.2 \times (10)^3 \\[1em] = \frac{1}{8} \times 30 \times 1.2 \times 1000 \\[1em] = 4.5 \times 10^3\ \text W = 4.5\ \text{kW}\end{aligned}$$

Hence, the electrical power produced is 4.5 kW.

Given,

• Mass lifted each time, m = 10 kg
• Number of times the mass is lifted, n = 1000
• Height through which it is lifted each time, h = 0.5 m
• Energy supplied by fat = 3.8 × 10⁷ J kg^-1
• Efficiency of conversion = 20%
• g = 9.8 m s^-2

(a) The work done against the gravitational force in lifting the mass once is mgh. Lifting it 1000 times, the total work done is

$$\begin{aligned}\text W = \text n(\text{mgh}) = 1000 \times (10 \times 9.8 \times 0.5) \\[1em] = 1000 \times 49 \\[1em] = 4.9 \times 10^4\ \text J\end{aligned}$$

Hence, the work done against the gravitational force is 4.9 × 10⁴ J.

(b) Fat supplies 3.8 × 10⁷ J of energy per kilogram, but only 20% of it is converted into mechanical energy. Hence the mechanical energy obtained from 1 kg of fat is

$\frac{20}{100} \times 3.8 \times 10^7 = 7.6 \times 10^6\ \text{J kg}^{-1}$

Therefore the mass of fat used up in performing the work W is

$$\begin{aligned}\text{Fat used} = \frac{\text W}{7.6 \times 10^6} = \frac{4.9 \times 10^4}{7.6 \times 10^6} \\[1em] = 6.4 \times 10^{-3}\ \text{kg}\end{aligned}$$

Hence, the dieter uses up 6.4 × 10^-3 kg of fat.

Given,

• Power used by the family, P = 8 kW = 8000 W
• Rate of incidence of solar energy = 200 W m^-2
• Fraction converted into useful electrical energy = 20%

(a) Let A be the required area. The solar energy incident on this area per second is

$(200\ \text{W m}^{-2}) \times \text A$

Only 20% of this is converted into useful electrical energy, and this must be equal to 8 kW. Therefore,

$$\begin{aligned}\frac{20}{100} \times (200) \times \text A = 8000 \\[1em] 0.2 \times 200 \times \text A = 8000 \\[1em] 40\text A = 8000 \\[1em] \text A = \frac{8000}{40} = 200\ \text m^2\end{aligned}$$

Hence, an area of 200 m² is needed to supply 8 kW.

(b) The roof of a typical house is about 14 m × 14 m, whose area is

$14 \times 14 = 196\ \text m^2$

Hence, the area required (200 m²) is comparable to the area of the roof of a typical house.

Given,

• Mass of the block = m
• Velocity of the block at O = v
• Resistive frictional force, F = ar + br²

The block is brought to rest by the resistive frictional force. Let the total distance covered before it stops be R.

Since the frictional force varies with the distance r, it is a variable force. The work done by a variable force is obtained by integration,

$\text W = \int_0^{\text R} \text F\text{dr} \times \cos 180^\circ = -\int_0^{\text R} (\text{ar} + \text{br}^2)\text{dr}$

the negative sign appearing because the frictional force is opposite to the displacement.

Integrating,

$$\begin{aligned}\text W = -\left[\frac{\text{ar}^2}{2} + \frac{\text{br}^3}{3}\right]_0^{\text R} \\[1em] = -\left(\frac{\text{aR}^2}{2} + \frac{\text{bR}^3}{3}\right)\end{aligned}$$

By the work-kinetic energy theorem, the work done by the net force equals the change in kinetic energy. The block finally comes to rest, so its final kinetic energy is zero,

$\text W = \text K_f - \text K_i = 0 - \frac{1}{2}\text{mv}^2$

Therefore,

$$\begin{aligned}-\left(\frac{\text{aR}^2}{2} + \frac{\text{bR}^3}{3}\right) = -\frac{1}{2}\text{mv}^2 \\[1em] \frac{\text{aR}^2}{2} + \frac{\text{bR}^3}{3} = \frac{\text{mv}^2}{2}\end{aligned}$$

Multiplying throughout by 6,

$3\text{aR}^2 + 2\text{bR}^3 = 3\text{mv}^2$

Hence, the total distance R covered by the block before coming to rest is given by 2bR³ + 3aR² = 3mv².

Given,

• Initial position, $\vec{\text r}_1 = (2\hat{\text i} + 3\hat{\text j})$ m
• Final position, $\vec{\text r}_2 = (4\hat{\text i} + 6\hat{\text j})$ m
• Force, $\vec{\text F} = (3\text x^2\hat{\text i} + 2\text y\hat{\text j})$ N

The force is a variable force, since it depends on the coordinates x and y. The work done by a variable force is

$\text W = \int \vec{\text F} \cdot \text d\vec{\text r}$

Writing $\text d\vec{\text r} = \text{dx}\hat{\text i} + \text{dy}\hat{\text j}$,

$$\begin{aligned}\vec{\text F} \cdot \text d\vec{\text r} = (3\text x^2\hat{\text i} + 2\text y\hat{\text j}) \cdot (\text{dx}\hat{\text i} + \text{dy}\hat{\text j}) \\[1em] = 3\text x^2\text{dx} + 2\text y\text{dy}\end{aligned}$$

From the position vectors, x changes from 2 m to 4 m and y changes from 3 m to 6 m. Therefore,

$\text W = \int_2^4 3\text x^2\text{dx} + \int_3^6 2\text y\text{dy}$

Integrating each term separately,

$$\begin{aligned}\int_2^4 3\text x^2\text{dx} = 3\left[\frac{\text x^3}{3}\right]_2^4 = \left[\text x^3\right]_2^4 \\[1em] = (4)^3 - (2)^3 = 64 - 8 = 56\end{aligned}$$

and

$$\begin{aligned}\int_3^6 2\text y\text{dy} = 2\left[\frac{\text y^2}{2}\right]_3^6 = \left[\text y^2\right]_3^6 \\[1em] = (6)^2 - (3)^2 = 36 - 9 = 27\end{aligned}$$

Adding the two,

$\text W = 56 + 27 = 83\ \text J$

Hence, the work done by the force is 83 J.

Given,

• Length of the simple pendulum = l
• Maximum angular displacement = θ

Let m be the mass of the bob. When the string makes an angle θ with the vertical, the bob is at its extreme position, where its velocity is zero and it possesses only potential energy.

If the bob is raised through a height h above its lowest position, then from the geometry of the figure,

$\text h = \text l - \text l \cos \theta = \text l(1 - \cos \theta)$

The potential energy of the bob at the extreme position is

$\text U = \text{mgh} = \text{mgl}(1 - \cos \theta)$

As the bob swings down towards the lowest position, this potential energy is gradually converted into kinetic energy. At the lowest point the height is zero, so the whole of the potential energy appears as kinetic energy. By the principle of conservation of mechanical energy, the maximum kinetic energy of the bob is

$\text K_{max} = \text{mgl}(1 - \cos \theta)$

Hence, the maximum kinetic energy of the bob is mgl (1 − cos θ), which occurs at the lowest point of the swing.

Given,

• Mass of the bullet = m, moving with horizontal velocity v
• Mass of the block = M, initially at rest
• Length of the string = l

Step 1 : The collision. The bullet gets embedded in the block, so the collision is perfectly inelastic. By the principle of conservation of linear momentum, if V is the common velocity of the bullet and the block just after the impact,

$$\begin{aligned}\text{mv} + \text M(0) = (\text m + \text M)\text V \\[1em] \text V = \frac{\text{mv}}{\text m + \text M}\end{aligned}$$

Step 2 : The rise of the block. After the collision the combined mass swings up on the string. If it rises through a height h before momentarily coming to rest, then by the conservation of mechanical energy,

$$\begin{aligned}\frac{1}{2}(\text m + \text M)\text V^2 = (\text m + \text M)\text{gh} \\[1em] \text h = \frac{\text V^2}{2\text g} = \frac{1}{2\text g}\left(\frac{\text{mv}}{\text m + \text M}\right)^2\end{aligned}$$

Step 3 : The angle of the string. If θ is the maximum angle made by the string with the vertical, then from the geometry,

$\text h = l - l\cos \theta \quad \Rightarrow \quad \cos \theta = 1 - \frac{\text h}{l}$

Substituting the value of h,

$\cos \theta = 1 - \frac{1}{2\text{g}l}\left(\frac{\text{mv}}{\text m + \text M}\right)^2$

Hence, the maximum angle made by the string after the impact is

$\theta = \cos^{-1}\left[1 - \frac{\text m^2\text v^2}{2\text{g}l(\text m + \text M)^2}\right]$

Given,

• Mass of the pendulum bob, m₁ = 10^-2 kg
• Height through which it is raised, h₁ = 5 × 10^-2 m
• Mass picked up at the bottom, m₂ = 10^-3 kg
• g = 10 m s^-2

Step 1 : Velocity of the bob at the bottom of the swing. As the bob falls through the height h₁, its potential energy is converted into kinetic energy. By the conservation of mechanical energy,

$\frac{1}{2}\text m_1\text v^2 = \text m_1\text{gh}_1 \quad \Rightarrow \quad \text v = \sqrt{2\text{gh}_1}$

Substituting the values,

$$\begin{aligned}\text v = \sqrt{2 \times 10 \times 5 \times 10^{-2}} = \sqrt{1} \\[1em] = 1\ \text{m s}^{-1}\end{aligned}$$

Step 2 : Picking up the extra mass. The bob picks up the mass m₂, so the process is a perfectly inelastic collision. By the conservation of linear momentum, if V is the common velocity,

$$\begin{aligned}\text m_1\text v = (\text m_1 + \text m_2)\text V \\[1em] \text V = \frac{\text m_1\text v}{\text m_1 + \text m_2} = \frac{10^{-2} \times 1}{10^{-2} + 10^{-3}}\end{aligned}$$

$$\begin{aligned}\text V = \frac{10^{-2}}{1.1 \times 10^{-2}} = \frac{1}{1.1} \\[1em] = 0.909\ \text{m s}^{-1}\end{aligned}$$

Step 3 : Rise of the combined mass. The kinetic energy of the combined mass is now converted into potential energy as it rises through a height h₂,

$$\begin{aligned}\frac{1}{2}(\text m_1 + \text m_2)\text V^2 = (\text m_1 + \text m_2)\text{gh}_2 \\[1em] \text h_2 = \frac{\text V^2}{2\text g}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text h_2 = \frac{(0.909)^2}{2 \times 10} = \frac{0.826}{20} \\[1em] = 4.13 \times 10^{-2}\ \text m\end{aligned}$$

Hence, the combined mass rises to a height of about 4.13 × 10^-2 m, that is, nearly 4.1 cm.

Given,

• Force constant of the spring, k = 10 N m^-1
• Mass of the attached piece, M = 0.1 kg
• Mass of the striking piece = 0.1 kg
• Velocity of the striking piece, v = 1.0 m s^-1
• Velocity with which it returns back, v′ = 0.6 m s^-1

Maximum displacement of the attached piece : At the collision the whole of the kinetic energy $\left(\frac{1}{2}\text{Mv}^2\right)$ of the striking mass is used in displacing the attached mass, that is, in compressing the spring of force-constant k.

At the position of maximum compression the attached piece is momentarily at rest, so the whole of this energy is stored as the elastic potential energy of the spring. Thus, if the maximum compression of the spring be x, then

$\frac{1}{2}\text{Mv}^2 = \frac{1}{2}\text{kx}^2 \quad \Rightarrow \quad \text x = \sqrt{\frac{\text{Mv}^2}{\text k}}$

Substituting the values,

$$\begin{aligned}\text x = \sqrt{\frac{0.1 \times (1.0)^2}{10}} = \sqrt{\frac{0.1}{10}} \\[1em] = \sqrt{0.01} = 0.1\ \text m\end{aligned}$$

Amplitude of the S.H.M. set in : After the head-on collision the striking piece returns back with the velocity v′, and the attached mass performs S.H.M. during which the spring undergoes extension and compression alternately.

The energy left with the spring-mass system is the difference between the initial kinetic energy of the striking piece and the kinetic energy it carries away on returning. If a is the amplitude of elongation of the spring, then by the conservation of energy,

$\frac{1}{2}\text{Mv}^2 = \frac{1}{2}\text{Mv}'^2 + \frac{1}{2}\text{ka}^2$

$\text a = \sqrt{\frac{\text M(\text v^2 - \text v'^2)}{\text k}}$

Substituting the values,

$\text a = \sqrt{\frac{0.1 \times \left[(1.0)^2 - (0.6)^2\right]}{10}} = \sqrt{\frac{0.1 \times (1 - 0.36)}{10}}$

$$\begin{aligned}= \sqrt{\frac{0.1 \times 0.64}{10}} = \sqrt{0.0064} \\[1em] = 0.08\ \text m\end{aligned}$$

Hence, the maximum displacement of the attached piece is 0.1 m and the amplitude of the S.H.M. set in is 0.08 m.

Given,

• Mass of the block, m = 4.0 kg
• Speed of the block, v = 3.0 m s^-1
• Constant force exerted on the spring, F = 120 N

(a) The block exerts a constant force of 120 N on the spring while compressing it, and the spring exerts the same force of 120 N on the block while returning it to its initial position. This means that the whole of the energy stored in the spring during compression is returned to the block, and no energy is lost.

Hence, the collision is elastic.

(b) The kinetic energy of the block in the beginning is

$$\begin{aligned}\text K = \frac{1}{2}\text{mv}^2 = \frac{1}{2} \times 4.0 \times (3.0)^2 \\[1em] = \frac{1}{2} \times 4.0 \times 9.0 \\[1em] = 18\ \text J\end{aligned}$$

(c) At the position of maximum compression the block momentarily comes to rest, so the whole of its kinetic energy has been used in doing work against the constant force of the spring. If x is the maximum compression,

$$\begin{aligned}\text F \times \text x = \text K \\[1em] 120 \times \text x = 18 \\[1em] \text x = \frac{18}{120} = 0.15\ \text m = 15\ \text{cm}\end{aligned}$$

Hence, the maximum compression of the spring is 0.15 m.

(d) When the spring is compressed through 10 cm = 0.1 m, the potential energy stored in the spring is equal to the work done against the constant force,

$$\begin{aligned}\text U = \text F \times 0.1 = 120 \times 0.1 \\[1em] = 12\ \text J\end{aligned}$$

The kinetic energy left with the block at this instant is

$\text K' = 18 - 12 = 6\ \text J$

Therefore the required ratio is

$\frac{\text K'}{\text U} = \frac{6}{12} = \frac{1}{2}$

Hence, the ratio of the kinetic energy to the potential energy is 1 : 2.

Given,

• Mass of body A = m, mass of body B = 2m
• Mass of body C = m, moving with velocity v₀
• Compression of the spring at time t₀ = x₀

(a) The body C of mass m collides elastically with the body A of mass m. In a one-dimensional elastic collision between two bodies of equal masses, the bodies merely exchange their velocities. Since A is initially at rest, after the collision C comes to rest and A moves with the velocity v₀.

After this, A compresses the spring and pushes B forward. The system of A and B is now free from any external horizontal force, so its linear momentum is conserved. At the time t₀ the two bodies have the same instantaneous velocity, say V. Therefore,

$$\begin{aligned}\text m\text v_0 = (\text m + 2\text m)\text V \\[1em] \text V = \frac{\text m\text v_0}{3\text m} = \frac{\text v_0}{3}\end{aligned}$$

Hence, the common velocity of A and B at time t₀ is $\frac{\text v_0}{3}$.

(b) By the conservation of mechanical energy for the system of A, B and the spring, the initial kinetic energy of A is shared as the kinetic energy of the two bodies and the elastic potential energy of the spring,

$\frac{1}{2}\text m\text v_0^2 = \frac{1}{2}(3\text m)\text V^2 + \frac{1}{2}\text k\text x_0^2$

Substituting $\text V = \frac{\text v_0}{3}$,

$$\begin{aligned}\frac{1}{2}\text m\text v_0^2 = \frac{1}{2}(3\text m)\frac{\text v_0^2}{9} + \frac{1}{2}\text k\text x_0^2 \\[1em] \frac{1}{2}\text m\text v_0^2 = \frac{1}{6}\text m\text v_0^2 + \frac{1}{2}\text k\text x_0^2\end{aligned}$$

Therefore,

$\frac{1}{2}\text k\text x_0^2 = \frac{1}{2}\text m\text v_0^2 - \frac{1}{6}\text m\text v_0^2 = \frac{1}{3}\text m\text v_0^2$

$\text k = \frac{2\text m\text v_0^2}{3\text x_0^2}$

Hence, the spring-constant is $\text k = \frac{2\text m\text v_0^2}{3\text x_0^2}$.

Given,

• Mass of the first body, m₁ = 5 kg moving along the x-axis with v₁ = 2 m s^-1
• Mass of the second body, m₂ = 10 kg moving along the y-axis with $\text v_2 = \sqrt{3}$ m s^-1
• The bodies stick together after the collision

(a) Since the bodies stick together, the collision is perfectly inelastic and the linear momentum is conserved separately along the x- and the y-directions.

The momentum along the x-axis before the collision is

$\text p_x = \text m_1\text v_1 = 5 \times 2 = 10\ \text{kg m s}^{-1}$

The momentum along the y-axis before the collision is

$\text p_y = \text m_2\text v_2 = 10 \times \sqrt{3} = 10\sqrt{3}\ \text{kg m s}^{-1}$

The magnitude of the total momentum after the collision is

$$\begin{aligned}\text p = \sqrt{\text p_x^2 + \text p_y^2} = \sqrt{(10)^2 + (10\sqrt{3})^2} \\[1em] = \sqrt{100 + 300} = \sqrt{400} \\[1em] = 20\ \text{kg m s}^{-1}\end{aligned}$$

The combined mass is m₁ + m₂ = 15 kg, so its velocity is

$$\begin{aligned}\text V = \frac{\text p}{\text m_1 + \text m_2} = \frac{20}{15} \\[1em] = 1.33\ \text{m s}^{-1}\end{aligned}$$

If θ is the angle made by this velocity with the x-axis,

$\tan \theta = \frac{\text p_y}{\text p_x} = \frac{10\sqrt{3}}{10} = \sqrt{3} \quad \Rightarrow \quad \theta = 60^\circ$

Hence, the combined mass moves with a velocity of 1.33 m s^-1 at 60° with the x-axis.

(b) The kinetic energy before the collision is

$$\begin{aligned}\text K_i = \frac{1}{2}\text m_1\text v_1^2 + \frac{1}{2}\text m_2\text v_2^2 \\[1em] = \frac{1}{2} \times 5 \times (2)^2 + \frac{1}{2} \times 10 \times (\sqrt{3})^2 \\[1em] = 10 + 15 = 25\ \text J\end{aligned}$$

The kinetic energy after the collision is

$$\begin{aligned}\text K_f = \frac{1}{2}(\text m_1 + \text m_2)\text V^2 = \frac{1}{2} \times 15 \times (1.33)^2 \\[1em] = \frac{1}{2} \times 15 \times 1.78 \\[1em] = 13.3\ \text J\end{aligned}$$

The loss in kinetic energy appears as heat,

$$\begin{aligned}\Delta \text K = \text K_i - \text K_f = 25 - 13.3 \\[1em] = 11.7\ \text J\end{aligned}$$

Hence, about 11.7 J of heat is liberated in the collision.

Let the body A of mass m₁ move with velocity u₁ and strike the body B of mass m₂, which is initially at rest (u₂ = 0). For a one-dimensional elastic collision, the velocity of B after the collision is

$\text v_2 = \left(\frac{2\text m_1}{\text m_1 + \text m_2}\right)\text u_1$

(a) Greatest speed : Rewriting the expression by dividing the numerator and the denominator by m₁,

$\text v_2 = \frac{2\text u_1}{1 + \frac{\text m_2}{\text m_1}}$

The speed v₂ is greatest when the denominator is least, that is, when $\frac{\text m_2}{\text m_1}$ is as small as possible. In the limit m₂ << m₁, we get v₂ = 2u₁.

Hence, for the greatest speed the mass of B should be much less than the mass of A.

(b) Greatest momentum : The momentum of B after the collision is

$\text p_2 = \text m_2\text v_2 = \frac{2\text m_1\text m_2}{\text m_1 + \text m_2}\text u_1 = \frac{2\text m_1\text u_1}{\frac{\text m_1}{\text m_2} + 1}$

This is greatest when $\frac{\text m_1}{\text m_2}$ is as small as possible, that is, when m₂ is very large. In the limit m₂ >> m₁, the momentum approaches 2m₁u₁.

Hence, for the greatest momentum the mass of B should be much greater than the mass of A.

(c) Greatest kinetic energy : The kinetic energy transferred to B is greatest when the transfer of energy is complete. In a one-dimensional elastic collision the whole of the kinetic energy is transferred from A to B only when the two masses are equal, for then the bodies merely exchange their velocities, A comes to rest and B moves off with the velocity u₁.

Hence, for the greatest kinetic energy the mass of B should be equal to the mass of A.

Given,

• Rest mass of the particle = m₀
• Speed of the particle, $\text v = \frac{\text c}{2}$

Mass : According to the theory of relativity, the mass of a particle moving with a speed v is

$\text m = \frac{\text m_0}{\sqrt{1 - \frac{\text v^2}{\text c^2}}}$

Substituting $\text v = \frac{\text c}{2}$,

$$\begin{aligned}\text m = \frac{\text m_0}{\sqrt{1 - \frac{\text c^2/4}{\text c^2}}} = \frac{\text m_0}{\sqrt{1 - \frac{1}{4}}} = \frac{\text m_0}{\sqrt{\frac{3}{4}}} \\[1em] = \frac{2\text m_0}{\sqrt{3}} = 1.155\text m_0\end{aligned}$$

Momentum : The momentum of the particle is

$$\begin{aligned}\text p = \text{mv} = \frac{2\text m_0}{\sqrt{3}} \times \frac{\text c}{2} \\[1em] = \frac{\text m_0\text c}{\sqrt{3}} = 0.577\text m_0\text c\end{aligned}$$

Total energy : By the mass-energy equivalence relation, the total energy is

$$\begin{aligned}\text E = \text{mc}^2 = \frac{2\text m_0}{\sqrt{3}}\text c^2 \\[1em] = 1.155\text m_0\text c^2\end{aligned}$$

Kinetic energy : The kinetic energy is the difference between the total energy and the rest mass energy,

$$\begin{aligned}\text K = \text{mc}^2 - \text m_0\text c^2 = (\text m - \text m_0)\text c^2 \\[1em] = (1.155\text m_0 - \text m_0)\text c^2 \\[1em] = 0.155\text m_0\text c^2\end{aligned}$$

Hence, the mass is 1.155 m₀, the momentum is 0.577 m₀c, the total energy is 1.155 m₀c² and the kinetic energy is 0.155 m₀c².

Given,

• Mass of the vehicle = m
• Constant power of the engine = P
• The vehicle starts from rest

The power delivered to the vehicle is

$\text P = \text{Fv} = \text{ma}\text v = \text m\frac{\text{dv}}{\text{dt}}\text v$

To express the velocity in terms of the displacement s, we write the acceleration as

$\text a = \frac{\text{dv}}{\text{dt}} = \frac{\text{dv}}{\text{ds}} \times \frac{\text{ds}}{\text{dt}} = \text v\frac{\text{dv}}{\text{ds}}$

Substituting this,

$\text P = \text m\left(\text v\frac{\text{dv}}{\text{ds}}\right)\text v = \text{mv}^2\frac{\text{dv}}{\text{ds}}$

Separating the variables,

$\text P\text{ds} = \text{mv}^2\text{dv}$

Integrating, with v = 0 at s = 0,

$$\begin{aligned}\int_0^{\text s}\text P\text{ds} = \int_0^{\text v}\text{mv}^2\text{dv} \\[1em] \text{Ps} = \text m\left[\frac{\text v^3}{3}\right]_0^{\text v} = \frac{\text{mv}^3}{3}\end{aligned}$$

Therefore,

$\text v^3 = \frac{3\text{Ps}}{\text m} \quad \Rightarrow \quad \text v = \left(\frac{3\text{Ps}}{\text m}\right)^{1/3}$

Hence, the instantaneous velocity of the vehicle as a function of the displacement s is $\text v = \left(\frac{3\text{Ps}}{\text m}\right)^{1/3}$, that is, $\text v \propto \text s^{1/3}$.

Given,

• Velocity of projection of the shell, u = 50 m s^-1
• Angle of projection, θ = 60°
• The shell splits into three equal parts at the highest point

At the highest point of the path, the vertical component of the velocity becomes zero and only the horizontal component remains. Therefore the velocity of the shell at the highest point is

$$\begin{aligned}\text u_x = \text u \cos \theta = 50 \times \cos 60^\circ = 50 \times \frac{1}{2} \\[1em] = 25\ \text{m s}^{-1}\end{aligned}$$

Let the mass of the shell be 3m, so that each of the three equal parts has mass m.

The momentum of the shell just before the explosion, in the horizontal direction, is

$\text p_i = (3\text m)(25) = 75\text m$

After the explosion :

• The first part retraces its path, so it moves backward with a velocity of 25 m s^-1, that is, its momentum is − 25m.
• The second part falls down vertically, so its horizontal momentum is zero.
• Let the third part move with velocity v in the horizontal direction, so its momentum is m v.

By the principle of conservation of linear momentum in the horizontal direction,

$$\begin{aligned}75\text m = (-25\text m) + 0 + \text m\text v \\[1em] \text m\text v = 75\text m + 25\text m = 100\text m \\[1em] \text v = 100\ \text{m s}^{-1}\end{aligned}$$

Hence, the third part moves horizontally with a velocity of 100 m s^-1 in the original direction of motion of the shell.

Given,

• Mass of the sphere, m = 40 kg
• Length of the thread = l
• Angular amplitude = θ₀
• Maximum tension the thread can support = 80 kg f
• g = 9.8 m s^-2 (not given in the question; taken as its standard value)

Converting the given quantities into the absolute unit of force,

$\text{Weight of the sphere} = \text{mg} = 40 \times 9.8 = 392\ \text N$

$\text{Maximum tension} = 80\ \text{kg f} = 80 \times 9.8 = 784\ \text N$

Tension at an angle θ : Let v be the velocity of the sphere when the thread makes an angle θ with the vertical. Taking the extreme position (angular amplitude θ₀) as the position of zero velocity and applying the conservation of mechanical energy between the two positions,

$$\begin{aligned}\frac{1}{2}\text{mv}^2 = \text{mgl}(\cos \theta - \cos \theta_0) \\[1em] \text v^2 = 2\text{gl}(\cos \theta - \cos \theta_0)\end{aligned}$$

At the position θ, the net radial force towards the centre provides the necessary centripetal force,

$\text T - \text{mg}\cos \theta = \frac{\text{mv}^2}{\text l}$

Substituting the value of v²,

$$\begin{aligned}\text T = \text{mg}\cos \theta + \frac{\text m}{\text l}[2\text{gl}(\cos \theta - \cos \theta_0)] \\[1em] = \text{mg}\cos \theta + 2\text{mg}\cos \theta - 2\text{mg}\cos \theta_0 \\[1em] \text T = \text{mg}(3\cos \theta - 2\cos \theta_0)\end{aligned}$$

Substituting mg = 392 N,

$\text T = 392(3\cos \theta - 2\cos \theta_0)\ \text N$

Maximum angular amplitude : The tension is greatest at the lowest point, where θ = 0 and cos θ = 1,

$\text T_{max} = \text{mg}(3 - 2\cos \theta_0)$

The thread can support a maximum tension of 784 N, while the weight of the sphere is 392 N. Therefore,

$784 = 392(3 - 2\cos \theta_0)$

$$\begin{aligned}\frac{784}{392} = 3 - 2\cos \theta_0 \\[1em] 2 = 3 - 2\cos \theta_0\end{aligned}$$

$2\cos \theta_0 = 1 \quad \Rightarrow \quad \cos \theta_0 = \frac{1}{2}$

$\theta_0 = 60^\circ$

Hence, the tension in the thread at an angle θ is 392 (3 cos θ − 2 cos θ₀) N, and the maximum angular amplitude of oscillation is 60°.

Given,

• Mass of the sphere, m = 500 g = 0.5 kg
• Length of the string, l = 1.0 m
• Angular amplitude, θ₀ = 60°
• Angle at which the tension is required, θ = 30°
• g = 10 m s^-2

Tension at an angle θ : Let v be the velocity of the sphere when the thread makes an angle θ with the vertical. Taking the extreme position (angular amplitude θ₀) as the position of zero velocity and applying the conservation of mechanical energy between the two positions,

$$\begin{aligned}\frac{1}{2}\text{mv}^2 = \text{mgl}(\cos \theta - \cos \theta_0) \\[1em] \text v^2 = 2\text{gl}(\cos \theta - \cos \theta_0)\end{aligned}$$

At the position θ, the net radial force towards the centre provides the necessary centripetal force,

$\text T - \text{mg} \cos \theta = \frac{\text{mv}^2}{\text l}$

Substituting the value of v²,

$$\begin{aligned}\text T = \text{mg} \cos \theta + \frac{\text m}{\text l}[2\text{gl}(\cos \theta - \cos \theta_0)] \\[1em] = \text{mg} \cos \theta + 2\text{mg}\cos \theta - 2\text{mg}\cos \theta_0 \\[1em] \text T = \text{mg}(3\cos \theta - 2\cos \theta_0)\end{aligned}$$

Substituting the given values, with cos 30° = $\frac{\sqrt{3}}{2}$ = 0.866 and cos 60° = 0.5,

$$\begin{aligned}\text T = 0.5 \times 10 \times [3(0.866) - 2(0.5)] \\[1em] = 5 \times [2.598 - 1] \\[1em] = 5 \times 1.598 \\[1em] = 7.99\ \text N\end{aligned}$$

Hence, the tension in the string is about 8 N.

Given,

• The man starts from the origin O with constant velocity $\vec{\text v}_1$ along the + Y-axis
• A particle of mass m starts from the point P with uniform velocity $\vec{\text v}_2$ along a circle of radius R

The particle moves along a circle of radius R with a uniform speed v₂, so its angular velocity is

$\omega = \frac{\text v_2}{\text R}$

At the instant t = 0 the particle is at P, where its velocity $\vec{\text v}_2$ is along the + Y-direction. After a time t the radius vector has turned through an angle = $\theta$ = ωt, so the velocity of the particle, which is always along the tangent, is

$\vec{\text v}_2(\text t) = -\text v_2 \sin \omega\text t\hat{\text i} + \text v_2\cos \omega\text t\hat{\text j}$

The velocity of the man is constant,

$\vec{\text v}_1 = \text v_1\hat{\text j}$

The velocity of the particle with respect to the man is

$$\begin{aligned}\vec{\text v}_{rel} = \vec{\text v}_2(\text t) - \vec{\text v}_1 \\[1em] = -\text v_2\sin \omega\text t\hat{\text i} + (\text v_2\cos \omega\text t - \text v_1)\hat{\text j}\end{aligned}$$

Therefore the linear momentum of the particle with respect to the man is

$$\begin{aligned}\vec{\text p}_{rel} = \text m \vec{\text v}_{rel} \\[1em] = \text m\left[-\text v_2\sin\left(\frac{\text v_2\text t}{\text R}\right)\hat{\text i} + \left\lbrace\text v_2\cos\left(\frac{\text v_2\text t}{\text R}\right) - \text v_1\right\rbrace\hat{\text j}\right]\end{aligned}$$

Hence, the linear momentum of the particle with respect to the man varies with time as given above, its magnitude being

$|\vec{\text p}_{rel}| = \text m\sqrt{\text v_1^2 + \text v_2^2 - 2\text v_1\text v_2\cos\left(\frac{\text v_2\text t}{\text R}\right)}$

Given,

• Radius of the frictionless sphere = R
• The particle starts sliding from rest from the top

Let O be the centre of the sphere, A its topmost point and B the point where it touches the ground. Let P be the instantaneous position of the particle of mass m during its sliding, such that ∠POA = θ. Let v be the velocity of the particle at P.

Step 1 : Condition for leaving the surface. At the point P the net radially inward force on the particle is mg cos θ − N, where N is the normal reaction at the surface of the sphere. This supplies the necessary centripetal force, and thus

$\text{mg}\cos \theta - \text N = \frac{\text{mv}^2}{\text R}$

The particle will fly off the surface at that angle θ for which N = 0, that is,

$\text{mg}\cos \theta = \frac{\text{mv}^2}{\text R} \quad \Rightarrow \quad \frac{1}{2}\text{mgR}\cos \theta = \frac{1}{2}\text{mv}^2 \qquad \ldots(\text i)$

Step 2 : Conservation of mechanical energy. The fall in the height of the particle in sliding down from A to P is

$\text{AC} = \text{AO} - \text{CO} = \text R - \text R\cos \theta$

Since the sphere is smooth, the loss in potential energy due to this fall appears as the kinetic energy of the particle at P,

$\text{mg}(\text R - \text R\cos \theta) = \frac{1}{2}\text{mv}^2 \qquad \ldots(\text{ii})$

Step 3 : Solving. Equating (i) and (ii),

$\frac{1}{2}\text{mgR}\cos \theta = \text{mgR}(1 - \cos \theta)$

$$\begin{aligned}\frac{1}{2}\cos \theta = 1 - \cos \theta \\[1em] \frac{3}{2}\cos \theta = 1 \quad \Rightarrow \quad \cos \theta = \frac{2}{3}\end{aligned}$$

Step 4 : Height above the ground. The sphere is fixed on the ground, so the centre O is at a height R above the ground. Hence the height of the point P at which the particle flies off the surface is

$$\begin{aligned}\text{BC} = \text{BO} + \text{OC} = \text R + \text R\cos \theta \\[1em] = \text R + \text R\left(\frac{2}{3}\right)\end{aligned}$$

$= \frac{5\text R}{3}$

Hence, the particle leaves the surface of the sphere at a height $\frac{5\text R}{3}$ above the ground.

Given,

• Mass of the ball, m = 0.5 kg
• Length of the cord (radius), l = 50 cm = 0.5 m
• Speed at the lowest point, v_B = 8 m/s
• g = 10 m/s²

Step 1 : Will the ball complete the circle? For a body to just complete a vertical circle, the minimum speed required at the lowest point is

$$\begin{aligned}\text v_{min} = \sqrt{5\text{gl}} = \sqrt{5 \times 10 \times 0.5} = \sqrt{25} \\[1em] = 5\ \text{m/s}\end{aligned}$$

Since the actual speed at the lowest point (8 m/s) is greater than 5 m/s, the ball is able to complete one round.

Step 2 : Speed at the highest point. Applying the conservation of mechanical energy between the lowest point B and the highest point A, which are separated by a vertical height 2l,

$$\begin{aligned}\frac{1}{2}\text{mv}_B^2 = \frac{1}{2}\text{mv}_A^2 + \text{mg}(2\text l) \\[1em] \text v_A^2 = \text v_B^2 - 4\text{gl}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text v_A^2 = (8)^2 - 4 \times 10 \times 0.5 = 64 - 20 = 44 \\[1em] \text v_A = \sqrt{44} = 6.63\ \text{m/s}\end{aligned}$$

Step 3 : Tension at the highest point. At the highest point the tension and the weight both act towards the centre,

$$\begin{aligned}\text T_A = \frac{\text{mv}_A^2}{\text l} - \text{mg} = \frac{0.5 \times 44}{0.5} - (0.5 \times 10) \\[1em] = 44 - 5 = 39\ \text N\end{aligned}$$

Step 4 : Tension at the lowest point. At the lowest point the tension acts upwards and the weight downwards,

$$\begin{aligned}\text T_B = \frac{\text{mv}_B^2}{\text l} + \text{mg} = \frac{0.5 \times 64}{0.5} + (0.5 \times 10) \\[1em] = 64 + 5 = 69\ \text N\end{aligned}$$

Hence, the ball completes the round, its speed at the highest point is 6.63 m/s, and the tensions in the cord are 39 N at the highest point and 69 N at the lowest point.

Given,

• Mass of the bullet, m = 0.012 kg
• Horizontal speed of the bullet, v = 70 m s^-1
• Mass of the block, M = 0.4 kg
• g = 9.8 m s^-2

Step 1 : Common velocity after the impact. The bullet comes to rest with respect to the block, so the collision is perfectly inelastic. By the conservation of linear momentum,

$$\begin{aligned}\text{mv} = (\text m + \text M)\text V \\[1em] \text V = \frac{\text{mv}}{\text m + \text M} = \frac{0.012 \times 70}{0.012 + 0.4}\end{aligned}$$

$\text V = \frac{0.84}{0.412} = 2.04\ \text{m s}^{-1}$

Step 2 : Height to which the block rises. After the impact the combined mass swings up, and by the conservation of mechanical energy its kinetic energy is converted into potential energy,

$$\begin{aligned}\frac{1}{2}(\text m + \text M)\text V^2 = (\text m + \text M)\text{gh} \\[1em] \text h = \frac{\text V^2}{2\text g} = \frac{(2.04)^2}{2 \times 9.8}\end{aligned}$$

$\text h = \frac{4.16}{19.6} = 0.212\ \text m = 21.2\ \text{cm}$

Step 3 : Heat produced. The heat produced is equal to the loss in the kinetic energy during the collision. The kinetic energy before the collision is

$$\begin{aligned}\text K_i = \frac{1}{2}\text{mv}^2 = \frac{1}{2} \times 0.012 \times (70)^2 \\[1em] = \frac{1}{2} \times 0.012 \times 4900 = 29.4\ \text J\end{aligned}$$

The kinetic energy just after the collision is

$$\begin{aligned}\text K_f = \frac{1}{2}(\text m + \text M)\text V^2 = \frac{1}{2} \times 0.412 \times (2.04)^2 \\[1em] = \frac{1}{2} \times 0.412 \times 4.16 = 0.857\ \text J\end{aligned}$$

Therefore the heat produced is

$$\begin{aligned}\text Q = \text K_i - \text K_f = 29.4 - 0.857 \\[1em] = 28.5\ \text J\end{aligned}$$

Hence, the block rises to a height of about 0.212 m and nearly 28.5 J of heat is produced in the block.

Given,

• Angle of the gradual track, θ₁ = 30°
• Angle of the steep track, θ₂ = 60°
• Common height, h = 10 m
• g = 10 m s^-2

Speeds at the bottom : Both the tracks are frictionless, so for each stone the loss in potential energy is completely converted into kinetic energy. By the conservation of mechanical energy,

$\frac{1}{2}\text{mv}^2 = \text{mgh} \quad \Rightarrow \quad \text v = \sqrt{2\text{gh}}$

This result does not contain the angle of the incline, so both the stones reach the bottom with the same speed,

$$\begin{aligned}\text v = \sqrt{2 \times 10 \times 10} = \sqrt{200} \\[1em] = 14.1\ \text{m s}^{-1}\end{aligned}$$

Hence, the two stones reach the bottom with the same speed of 14.1 m s^-1.

Times taken : The acceleration of a stone sliding down a frictionless incline of angle θ is a = g sin θ, and the length of the incline is $\text l = \frac{\text h}{\sin \theta}$. Starting from rest, using $\text l = 0\times \text t+\frac{1}{2}\text{at}^2$,

$\frac{\text h}{\sin \theta} = \frac{1}{2}(\text g\sin \theta)\text t^2 \quad \Rightarrow \quad \text t = \frac{1}{\sin \theta}\sqrt{\frac{2\text h}{\text g}}$

Here $\sqrt{\frac{2\text h}{\text g}} = \sqrt{\frac{2 \times 10}{10}} = \sqrt{2} = 1.414$ s.

For the gradual track (θ₁ = 30°, sin 30° = 0.5),

$\text t_1 = \frac{1.414}{0.5} = 2.83\ \text s$

For the steep track (θ₂ = 60°, sin 60° = 0.866),

$\text t_2 = \frac{1.414}{0.866} = 1.63\ \text s$

Hence, the stones do not reach the bottom at the same time. The stone on the steep track takes 1.63 s while that on the gradual track takes 2.83 s, since the steeper incline gives a larger acceleration along a shorter path.

Given,

• Mass of the block, m = 1 kg
• Spring constant, k = 100 N m^-1
• Distance moved down the incline, x = 10 cm = 0.1 m
• Angle of the incline, θ = 37°
• g = 10 N/kg, sin 37° = 3/5 = 0.6, cos 37° = 4/5 = 0.8

The block is released from rest with the spring unstretched, and it again comes to rest after moving 10 cm down the incline. Since the block is at rest at both the ends of this displacement, its change in kinetic energy is zero.

Therefore, by the conservation of energy, the loss in the gravitational potential energy of the block is used up in doing work against friction and in storing elastic potential energy in the spring,

$\text{mgx}\sin 37° = \text f\text x + \frac{1}{2}\text{kx}^2$

where the frictional force is f = μ R = μ mg cos 37°. Thus,

$\text{mgx}\sin 37° = \mu\text{mgx}\cos 37° + \frac{1}{2}\text{kx}^2$

Substituting the values,

$1 \times 10 \times 0.1 \times 0.6 = \mu \times 1 \times 10 \times 0.1 \times 0.8 + \frac{1}{2} \times 100 \times (0.1)^2$

$$\begin{aligned}0.6 = 0.8\mu + \frac{1}{2} \times 100 \times 0.01 \\[1em] 0.6 = 0.8\mu + 0.5\end{aligned}$$

$$\begin{aligned}0.8\mu = 0.6 - 0.5 = 0.1 \\[1em] \mu = \frac{0.1}{0.8} = 0.125\end{aligned}$$

Hence, the coefficient of friction between the block and the inclined plane is 0.125.

Given,

• Mass of the bolt, m = 0.3 kg
• Uniform speed of the elevator = 7 m s^-1 (downward)
• Length of the elevator (distance of fall), h = 3 m
• g = 9.8 m s^-2

The elevator moves with a uniform speed, so it is a non-accelerated frame and the relative motion of the bolt with respect to the elevator is exactly the same as it would be in a stationary elevator. With respect to the elevator, the bolt starts from rest and falls freely through 3 m.

The loss in the potential energy of the bolt relative to the floor of the elevator is

$$\begin{aligned}\text U = \text{mgh} = 0.3 \times 9.8 \times 3 \\[1em] = 8.82\ \text J\end{aligned}$$

The bolt does not rebound after hitting the floor, so the whole of this energy is converted into heat,

$\text Q = 8.82\ \text J$

Hence, the heat produced by the impact is 8.82 J.

No, the answer would not be different if the elevator were stationary. Since the elevator moves with a uniform velocity, it is not an accelerated frame; the relative velocity with which the bolt strikes the floor and hence the heat produced remain the same as in a stationary elevator.

Given,

• Mass of the trolley, M = 200 kg
• Speed of the trolley, v = 36 km/h = 10 m s^-1
• Mass of the child, m = 20 kg
• Speed of the child relative to the trolley, u = 4 m s^-1 (opposite to the motion)
• Distance run by the child on the trolley, l = 10 m

Step 1 : Final speed of the trolley. The track is frictionless, so no external horizontal force acts on the system of the trolley and the child. Hence the total linear momentum is conserved.

The initial momentum of the system is

$$\begin{aligned}\text p_i = (\text M + \text m)\text v = (200 + 20) \times 10 \\[1em] = 2200\ \text{kg m s}^{-1}\end{aligned}$$

Let v' be the final speed of the trolley. The child runs with a speed of 4 m s^-1 relative to the trolley, in a direction opposite to its motion, so the velocity of the child with respect to the ground is (v' − 4).

By the conservation of linear momentum,

$$\begin{aligned}2200 = \text M\text v' + \text m(\text v' - 4) \\[1em] 2200 = 200\text v' + 20\text v' - 80\end{aligned}$$

$$\begin{aligned}220\text v' = 2280 \\[1em] \text v' = \frac{2280}{220} = 10.36\ \text{m s}^{-1}\end{aligned}$$

Hence, the final speed of the trolley is 10.36 m s^-1.

Step 2 : Distance moved by the trolley. The child covers 10 m along the trolley with a relative speed of 4 m s^-1, so the time taken is

$\text t = \frac{\text l}{\text u} = \frac{10}{4} = 2.5\ \text s$

During this time the trolley moves with the speed v' = 10.36 m s^-1, so the distance moved by the trolley is

$$\begin{aligned}\text s = \text v'\text t = 10.36 \times 2.5 \\[1em] = 25.9\ \text m\end{aligned}$$

Hence, the trolley moves through about 25.9 m from the time the child begins to run.

Given,

• Radius of the circular loop = R
• Angular frequency of rotation = ω

Let m be the mass of the bead and let the radius vector joining the centre of the loop to the bead make an angle θ with the vertical downward direction. The bead moves in a horizontal circle of radius R sin θ.

Two forces act on the bead : its weight mg vertically downwards and the normal reaction N exerted by the wire along the radius, towards the centre of the loop.

Resolving the normal reaction, the vertical component balances the weight,

$\text N\cos \theta = \text{mg} \qquad \ldots(\text i)$

and the horizontal component provides the centripetal force for the horizontal circle of radius R sin θ,

$\text N\sin \theta = \text m\omega^2(\text R\sin \theta) \qquad \ldots(\text{ii})$

From equation (ii), provided sin θ ≠ 0,

$\text N = \text m\omega^2\text R$

Substituting this in equation (i),

$$\begin{aligned}\text m\omega^2\text R\cos \theta = \text{mg} \\[1em] \cos \theta = \frac{\text g}{\omega^2\text R} \qquad \ldots(\text{iii})\end{aligned}$$

Condition for the bead to stay at the lowermost point : Since cos θ cannot exceed 1, a solution with θ ≠ 0 is possible only when

$\frac{\text g}{\omega^2\text R} \le 1 \quad \Rightarrow \quad \omega \ge \sqrt{\frac{\text g}{\text R}}$

Therefore, when $\omega \le \sqrt{\frac{\text g}{\text R}}$, no such angle exists and the only possible position of the bead is θ = 0, that is, the bead remains at its lowermost point.

Angle for $\omega = \sqrt{\frac{2\text g}{\text R}}$ : Here $\omega^2 = \frac{2\text g}{\text R}$. Substituting in equation (iii),

$\cos \theta = \frac{\text g}{\left(\frac{2\text g}{\text R}\right)\text R} = \frac{\text g}{2\text g} = \frac{1}{2}$

$\theta = 60^\circ$

Hence, the bead remains at its lowermost point for $\omega \le \sqrt{\frac{\text g}{\text R}}$, and for $\omega = \sqrt{\frac{2\text g}{\text R}}$ the radius vector makes an angle of 60° with the vertical downward direction.

Joule

Reason — Work is the product of force and the displacement in the direction of the force, W = F s cos θ. The S.I. absolute unit of work is the joule (J). If a force of 1 newton produces a displacement of 1 metre in the direction of the force, the work done is 1 joule, that is, 1 joule = 1 newton × 1 metre. The watt is the unit of power, the newton is the unit of force and the pascal is the unit of pressure.

work done per unit time

Reason — The rate of doing work by an agent, or a machine, is called the power of the agent, or the machine. Thus,

$\text{power} = \frac{\text{work}}{\text{time}}$

The average power is $\bar{\text P} = \frac{\text W}{\text t}$ and the instantaneous power is $\text P = \frac{\text{dW}}{\text{dt}}$.

Watt

Reason — Since power is work divided by time, its S.I. unit is joule/second (J s^-1), which is called the watt (W). If 1 joule of work is done in 1 second, the power is 1 watt. The joule is the unit of work or energy, the newton is the unit of force and the pascal is the unit of pressure.

Joule

Reason — Energy is the capacity to do work, so energy is measured in terms of the work done by a body. Its unit is therefore the same as that of work, that is, the joule (J) in the S.I. system. The calorie, the erg and the kilowatt-hour are also units of energy, but they belong to other systems of units and are not the S.I. unit.

elastic collision

Reason — An elastic collision is one in which both momentum and total kinetic energy are conserved. The forces involved are conservative in nature and the mechanical energy is not converted into other forms like heat, sound or light. In an inelastic collision the momentum is conserved but the total kinetic energy is not.

1

Reason — The degree of elasticity of a collision is measured by the coefficient of restitution e, defined as the ratio of the relative speed of separation after collision to the relative speed of approach before collision. For a perfectly elastic collision the relative speed of separation is equal to the relative speed of approach, so e = 1. For a perfectly inelastic collision e = 0.

work

Reason — Work is measured by the product of the applied force and the displacement of the body in the direction of the force, that is, W = F s cos θ where F cos θ is the component of the force in the direction of displacement. Power is the rate of doing work, energy is the capacity to do work and momentum is the product of mass and velocity.

kg m s^-1

Reason — Momentum is the product of the mass and the velocity of a body, p = m v. In the S.I. system mass is measured in kilogram and velocity in metre per second, so the unit of momentum is kg m s^-1. The newton is the unit of force, the joule of work and the pascal of pressure.

Kinetic energy

Reason — Kinetic energy is given by $\text K = \frac{1}{2}\text m(\vec{\text v} \cdot \vec{\text v})$. Since mass is a scalar and the scalar product of a vector with itself is a scalar, kinetic energy is a scalar quantity. Force and momentum are vectors, and the potential energy gradient $\left(-\frac{\text{dU}}{\text{dx}}\right)$ is equal to force and is therefore also a vector.

work done is equal to the change in kinetic energy

Reason — The work-kinetic energy theorem states that the work done by the net force acting on a system is equal to the change in its kinetic energy,

$\text W = \Delta \text K = \frac{1}{2}\text{mv}^2 - \frac{1}{2}\text{mu}^2$

The theorem holds for constant as well as variable forces.

kinetic energy

Reason — The energy possessed by a body by virtue of its state of motion is known as kinetic energy. For a body of mass m moving with velocity v, it is given by $\text K = \frac{1}{2}\text{mv}^2$. Potential energy is due to the position or configuration of a body, and is not due to its motion.

50 J

Reason — Given, F = 10 N and s = 5 m, the displacement being in the direction of the force so that θ = 0°.

$$\begin{aligned}\text W = \text F\text s \cos \theta = 10 \times 5 \times \cos 0^\circ \\[1em] = 10 \times 5 \times 1 = 50\ \text J\end{aligned}$$

30 W

Reason — Given, W = 300 J and t = 10 s. Power is the rate of doing work,

$$\begin{aligned}\text P = \frac{\text W}{\text t} = \frac{300}{10} \\[1em] = 30\ \text W\end{aligned}$$

100 J

Reason — Given, F = 50 N and s = 2 m, the displacement being along the direction of the force.

$$\begin{aligned}\text W = \text F\text s = 50 \times 2 \\[1em] = 100\ \text J\end{aligned}$$

9 J

Reason — Given, m = 2 kg and v = 3 m/s. The kinetic energy is

$$\begin{aligned}\text K = \frac{1}{2}\text{mv}^2 = \frac{1}{2} \times 2 \times (3)^2 \\[1em] = \frac{1}{2} \times 2 \times 9 = 9\ \text J\end{aligned}$$

490 J

Reason — Given, m = 10 kg, h = 5 m and g = 9.8 m/s². The gravitational potential energy is

$$\begin{aligned}\text U = \text{mgh} = 10 \times 9.8 \times 5 \\[1em] = 490\ \text J\end{aligned}$$

0.67

Reason — The coefficient of restitution is the ratio of the relative speed of separation after collision to the relative speed of approach before collision,

$$\begin{aligned}\text e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = \frac{4}{6} \\[1em] = 0.67\end{aligned}$$

0 N

Reason — Given, m = 2 kg, r = 1 m, v = 3 m/s and g = 9.8 m/s². At the top of a vertical circle both the tension and the weight act towards the centre, and together they provide the centripetal force,

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

Substituting the values,

$$\begin{aligned}\text T = \frac{2 \times (3)^2}{1} - (2 \times 9.8) \\[1em] = 18 - 19.6 = -1.6\ \text N\end{aligned}$$

A string can only pull and never push, so the tension in it cannot be negative. The negative value shows that the given speed is less than the critical velocity at the top,

$\text v_c = \sqrt{\text{gr}} = \sqrt{9.8 \times 1} = 3.13\ \text{m s}^{-1}$

At a speed below v_c the weight of the particle is already greater than the centripetal force required, so the string slackens and the particle leaves the circular path. The tension in the string is therefore zero.

Note: The printed explanation states that the particle would still maintain its circular motion under gravity alone. That holds only at the critical velocity $\sqrt{\text{gr}} = 3.13$ m s^-1; at the given speed of 3 m s^-1 the particle cannot stay on the circular path.

69.6 N

Reason — Given, m = 2 kg, r = 1 m, v = 5 m/s and g = 9.8 m/s². At the lowest point of a vertical circle the tension acts upwards and the weight downwards, and their difference provides the centripetal force,

$$\begin{aligned}\text T = \frac{\text{mv}^2}{\text r} + \text{mg} = \frac{2 \times (5)^2}{1} + (2 \times 9.8) \\[1em] = 50 + 19.6 \\[1em] = 69.6\ \text N\end{aligned}$$

4.4 m/s

Reason — Given, r = 2 m and g = 9.8 m/s². At the highest point of a vertical circle the tension becomes zero for the minimum speed, and the weight alone provides the centripetal force,

$\text{mg} = \frac{\text{mv}_{min}^2}{\text r} \quad \Rightarrow \quad \text v_{min} = \sqrt{\text{gr}}$

Substituting the values,

$$\begin{aligned}\text v_{min} = \sqrt{9.8 \times 2} = \sqrt{19.6} \\[1em] = 4.4\ \text{m s}^{-1}\end{aligned}$$

The minimum speed at the top does not depend on the mass of the particle.

Note: The printed answer key gives option 4 (3.1 m/s), but the textbook's own explanation for this question obtains 4.4 m s^-1, which is option 1. Option 1 has been taken as the correct answer.

100 W

Reason — Given, W = 2000 J and t = 20 s. The power of the machine is

$$\begin{aligned}\text P = \frac{\text W}{\text t} = \frac{2000}{20} \\[1em] = 100\ \text W\end{aligned}$$

Mass and velocity

Reason — The kinetic energy of a body is $\text K = \frac{1}{2}\text{mv}^2$, so it depends on the mass and the velocity of the body. It depends more on the speed than on the mass, since it varies as the square of the speed but only as the first power of the mass.

position

Reason — Potential energy is the energy possessed by a body by virtue of its position or configuration. For instance, a body of mass m raised to a height h above the earth's surface possesses gravitational potential energy U = mgh. The energy due to motion is kinetic energy.

The sum of kinetic and potential energy

Reason — Mechanical energy is associated with the motion and the position of an object, and it is the sum of the kinetic energy and the potential energy of a system. For an isolated system in the presence of conservative forces this sum remains constant throughout the motion.

Potential to kinetic energy and vice-versa

Reason — In a pendulum, the bob at its extreme position is momentarily at rest and possesses only potential energy. As it swings down, this potential energy is gradually converted into kinetic energy, which is maximum at the lowest point. As the bob rises on the other side, the kinetic energy is again converted back into potential energy. Thus the transformation is potential to kinetic energy and vice-versa.

the force is perpendicular to the direction of motion

Reason — The work done by a force is W = F s cos θ. When the force is perpendicular to the direction of motion, θ = 90° and cos 90° = 0, so the work done is zero. For example, when a stone tied to a string is whirled in a circle, the tension is always normal to the displacement and hence does no work on the stone.

increases

Reason — When a car is accelerated its speed increases, and since the kinetic energy $\text K = \frac{1}{2}\text{mv}^2$ varies as the square of the speed, the kinetic energy of the car increases. By the work-kinetic energy theorem, the net work done on the car is positive and equal to this increase.

less than the total kinetic energy before the collision

Reason — In an inelastic collision the total kinetic energy is not conserved, though the momentum is still conserved. A part of the kinetic energy is converted into other forms of energy such as heat, sound and the energy of deformation. Hence the total kinetic energy after the collision is less than that before the collision.

Momentum

Reason — In all types of collisions the total linear momentum of the system remains conserved, because during the collision the colliding bodies exert equal and opposite internal forces on each other and no external force acts on the system. Kinetic energy is conserved only in an elastic collision.

stick together after collision

Reason — When two bodies collide and stick together after the collision, the collision is called a perfectly inelastic collision. The two bodies then move with a common velocity, and the loss of kinetic energy is the maximum possible consistent with the conservation of momentum.

Note: The printed answer key gives option 1 (separate after collision). By the textbook's own definition of a perfectly inelastic collision, the bodies stick together, so option 2 has been taken as the correct answer.

exchange their velocities

Reason — For a one-dimensional elastic collision the final velocities are

$\text v_1 = \left(\frac{\text m_1 - \text m_2}{\text m_1 + \text m_2}\right)\text u_1 + \left(\frac{2\text m_2}{\text m_1 + \text m_2}\right)\text u_2$

Putting m₁ = m₂ = m gives v₁ = u₂ and v₂ = u₁. Thus in a head-on elastic collision of two bodies of equal masses, the bodies merely exchange their velocities after the collision.

is less than the initial kinetic energy

Reason — In an inelastic collision the total kinetic energy is not conserved. Some of it is permanently converted into internal energy forms such as heat, sound and the energy of deformation, so the final kinetic energy is less than the initial kinetic energy of the system.

Momentum only

Reason — In every collision, whether in one dimension or in two, the total linear momentum of the system is conserved, since no external force acts on the system during the collision. The kinetic energy is conserved only if the collision happens to be elastic, so in general only the momentum is conserved.

all of the these

Reason — In an inelastic collision the loss in kinetic energy is converted into other forms of energy. Potential energy changes occur due to the deformation of the bodies, and other forms of energy such as heat and sound are produced. Hence the loss of kinetic energy may be converted into all of these forms.

maintain circular motion

Reason — At the top of a vertical circle the minimum speed is $\text v_c = \sqrt{\text{gl}}$, obtained by putting the tension equal to zero so that the weight alone provides the centripetal force. If the speed at the top falls below this critical velocity, the string slackens and the body leaves the circular path. Hence the minimum speed is required to maintain circular motion.

both gravity and tension

Reason — At the highest point of a vertical circle, the weight mg acts vertically downwards, that is, towards the centre, and the tension T in the string also acts towards the centre. Their sum provides the necessary centripetal force,

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

Hence the centripetal force is provided by both gravity and tension.

maximum

Reason — The gravitational potential energy of a body is U = mgh, which increases with height. The highest point of a vertical circle is at the greatest height above the lowest point, so the potential energy there is maximum. Correspondingly the kinetic energy and the speed are minimum at that point.

maximum

Reason — At the lowest point of a vertical circle the height is minimum, so the potential energy is minimum. Since the total mechanical energy remains constant throughout the motion, the kinetic energy at the lowest point is maximum. This is also why the speed there, $\text v_B = \sqrt{5\text{gl}}$, is the greatest in the whole circle.

Note: The printed answer key gives option 2 (minimum). By the conservation of mechanical energy the kinetic energy is greatest where the potential energy is least, so option 1 has been taken as the correct answer.

constant

Reason — In vertical circular motion the speed and hence the kinetic energy of the particle changes from point to point, and so does the potential energy. However, if only conservative forces act, the sum of the kinetic and the potential energy, that is, the total mechanical energy, remains constant throughout the motion.

greater than the speed at the top

Reason — In going from the top to the bottom of the vertical circle the particle descends through a height 2l, so its potential energy decreases and by the conservation of mechanical energy its kinetic energy increases. Hence the speed at the bottom, $\sqrt{5\text{gl}}$, is greater than the speed at the top, $\sqrt{\text{gl}}$.

It remains constant

Reason — Although the kinetic energy and the potential energy of the particle keep changing at different points of a vertical circle, their sum does not change, provided no non-conservative force like friction acts. Hence the total mechanical energy remains constant.

Speed

Reason — In uniform circular motion the particle moves with a constant speed along the circular path. The velocity changes continuously because its direction, which is always tangential to the circle, keeps changing; the acceleration is centripetal and its direction also changes continuously.

A compressed spring

Reason — Potential energy is the energy possessed by a body by virtue of its position or configuration. When a spring is compressed, work has to be done against the elastic restoring force, and this work is stored in the spring as elastic potential energy, $\text U = \frac{1}{2}\text{kx}^2$. A moving car, flowing water and a running athlete all possess kinetic energy.

halved

Reason — Let the original mass be m and velocity v, so that $\text K = \frac{1}{2}\text{mv}^2$. On doubling the mass and halving the velocity,

$$\begin{aligned}\text K' = \frac{1}{2}(2\text m)\left(\frac{\text v}{2}\right)^2 = \frac{1}{2}(2\text m)\frac{\text v^2}{4} \\[1em] = \frac{1}{2} \times \frac{\text{mv}^2}{2} = \frac{\text K}{2}\end{aligned}$$

Hence the kinetic energy is halved.

zero

Reason — If a body moves with a uniform velocity, its acceleration is zero and hence the resultant force acting on it is zero. Since W = F s cos θ and F = 0, the work done by the resultant force is zero. The same follows from the work-kinetic energy theorem, as the kinetic energy does not change.

equal to the relative speed of approach

Reason — For a one-dimensional elastic collision it is proved that u₁ − u₂ = v₂ − v₁, that is, the relative velocity with which two bodies approach each other before collision is equal to the relative velocity with which they depart from each other after the collision. Correspondingly the coefficient of restitution is e = 1.

move with the same velocity after collision

Reason — In a perfectly inelastic collision the colliding bodies stick together after the impact, so they move with a common velocity. By the conservation of linear momentum this common velocity is

$\text v = \frac{\text m_1\text u_1 + \text m_2\text u_2}{\text m_1 + \text m_2}$

Note: The printed answer key gives option 4 (None of the above). Since the bodies in a perfectly inelastic collision do move with a common velocity, option 1 has been taken as the correct answer.

towards the centre of the circle

Reason — In uniform circular motion the speed is constant but the direction of the velocity changes continuously. The acceleration responsible for this change is the centripetal acceleration $\left(\frac{\text v^2}{\text r}\right)$, which is always directed along the radius towards the centre of the circle.

minimum

Reason — The gravitational potential energy U = mgh increases with the height above the reference level. The lowest point of a vertical circle is at the smallest height, so the potential energy there is minimum, and correspondingly the kinetic energy and the speed are maximum.

momentum

Reason — The impulse of a force is the product of the force and the time for which it acts. By Newton's second law,

$\vec{\text F} = \frac{\text d\vec{\text p}}{\text{dt}} \quad \Rightarrow \quad \vec{\text F}\Delta \text t = \Delta \vec{\text p}$

Hence the impulse imparted to an object is equal to the change in its momentum.

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

Explanation

Assertion (A) is correct: A conservative force is one for which the work done in moving a body from one position to another depends only on the initial and the final positions and not on the path followed. In a closed path the initial and the final positions coincide, so the work done by a conservative force over a closed path is zero.

Reason (R) is also correct: A conservative force depends only on the position of the object, and the work done by it is determined entirely by the end positions and not by the specific path taken between them.

Since the work done depends only on the positions, and in a closed path the two positions are the same, the work done must vanish. The Reason therefore explains the Assertion.

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

If assertion is false and reason is true.

Explanation

Assertion (A) is false: The power delivered by a force is $\text P = \vec{\text F} \cdot \vec{\text v} = \text{Fv}\cos \theta$. When the force acts opposite to the direction of motion, as in the case of a frictional force, θ = 180° and the power becomes negative. Hence the power delivered by a force is not always positive.

Reason (R) is correct: Power is defined as the rate at which work is done, $\text P = \frac{\text W}{\text t}$. Since the work done can itself be positive, negative or zero, the power can also take any of these values.

Therefore, assertion is false and reason is true.

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

Explanation

Assertion (A) is correct: For a conservative force such as gravity or the spring force, the work done in moving an object from one point to another depends only on the initial and the final positions and not on the path taken.

Reason (R) is also correct: Conservative forces have a potential energy associated with them, and the potential energy is a function of position alone. The work done by the force equals the decrease in this potential energy, W = U_i − U_f.

Since the potential energy depends only on position, the difference U_i − U_f depends only on the two end positions, which is exactly why the work done is path-independent. The Reason therefore explains the Assertion.

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 the force and the velocity vectors are perpendicular to each other, the power delivered by the force is zero.

Reason (R) is also correct: The power delivered by a force is the scalar product of the force and the velocity,

$\text P = \vec{\text F} \cdot \vec{\text v} = \text{Fv}\cos \theta$

When the two vectors are perpendicular, θ = 90° and cos 90° = 0, so the scalar product and hence the power is zero. The Reason therefore explains the Assertion.

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 kinetic energy of an object can never be negative. It is positive whenever the object is in motion and becomes zero when the object is at rest.

Reason (R) is also correct: The kinetic energy is $\text K = \frac{1}{2}\text{mv}^2$. The mass m of a body is always positive and v² is the square of the speed, which can never be negative. Hence K ≥ 0 always, even though the velocity vector itself may be negative in a chosen direction. The Reason therefore explains the Assertion.

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 work done by the centripetal force on an object moving in a circular path is zero.

Reason (R) is also correct: The centripetal force always acts along the radius, towards the centre of the circle, while the velocity of the object is always along the tangent to the circle. The two are therefore mutually perpendicular.

Since the work done is W = F s cos θ and here θ = 90°, the work done by the centripetal force is zero. The Reason therefore explains the Assertion.

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 the absence of non-conservative forces, the total mechanical energy of a system, which is the sum of its kinetic and potential energy, remains constant.

Reason (R) is also correct: Non-conservative forces such as friction dissipate the mechanical energy of a system as heat, and this energy is not stored in the form of potential energy which could be recovered.

Because such forces do not store energy in a recoverable potential form, their absence means that no mechanical energy is lost, and hence the total mechanical energy stays constant. The Reason therefore explains the Assertion.

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: If the net work done on an object is positive, its kinetic energy increases.

Reason (R) is also correct: The work-kinetic energy theorem states that the work done by the net force acting on a system is equal to the change in its kinetic energy, W_net = ΔK.

If W_net is positive then ΔK is positive, which means the kinetic energy increases. The Reason therefore explains the Assertion.

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

If assertion is false and reason is true.

Explanation

Assertion (A) is false: The gravitational potential energy of an object is measured with respect to an arbitrarily chosen zero-position. It is zero at the ground only if the ground itself is chosen as the reference level. If some other level is chosen as the reference, the potential energy at the ground will not be zero.

Reason (R) is correct: The value of the gravitational potential energy depends on the chosen reference point, since only the change in potential energy between two positions has a definite physical meaning.

Therefore, assertion is false and reason is true.

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

Explanation

Assertion (A) is correct: An object at rest has no kinetic energy, since $\text K = \frac{1}{2}\text{mv}^2$ and v = 0. It may still possess potential energy, for instance a stone held at a height above the ground.

Reason (R) is also correct: Potential energy is due to the position or configuration of a body, whereas kinetic energy is due to its motion.

Since the two forms of energy arise from entirely different causes, a body which is not moving can still possess potential energy on account of its position. The Reason therefore explains the Assertion.

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 work done by a force can be negative.

Reason (R) is also correct: The work done is W = F s cos θ. When the force has a component opposite to the direction of displacement, θ is more than 90° and cos θ is negative, so the work done is negative. For θ = 180°, W = − F s, which is the minimum work.

Since a force opposing the displacement gives a negative value of cos θ, the work done becomes negative. The Reason therefore explains the Assertion.

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 spring is compressed, work has to be done against the elastic restoring force of the spring, and this work is stored in the spring as elastic potential energy.

Reason (R) is also correct: The potential energy stored in a spring compressed or stretched through a distance x is $\text U = \frac{1}{2}\text{kx}^2$, where k is the spring constant.

This expression is obtained by integrating the work done against the restoring force F = − k x, and it is precisely the energy stored on compression. The Reason therefore explains the Assertion.

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: Power is defined as the rate of doing work by an agent or a machine, so it measures how quickly work is done.

Reason (R) is also correct: The average power is given by $\text P = \frac{\text W}{\text t}$, where W is the work done and t is the time taken.

This formula is exactly the mathematical statement that power is work done per unit time, that is, the rate of doing work. The Reason therefore explains the Assertion.

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: If the forces acting on a moving object are balanced, the net work done on it is zero.

Reason (R) is also correct: Balanced forces mean that the resultant force is zero, so the acceleration is zero and the object moves with a constant velocity.

Since the velocity does not change, the kinetic energy does not change, and by the work-kinetic energy theorem the net work done must be zero. The Reason therefore explains the Assertion.

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

If assertion is false and reason is true.

Explanation

Assertion (A) is false: The mechanical advantage of a machine is not always greater than one. In machines which are used to gain speed or a change of direction rather than a gain of force, the output force is smaller than the input force, and the mechanical advantage is then less than one.

Reason (R) is correct: Mechanical advantage is defined as the ratio of the output force to the input force. This ratio may be greater than, equal to, or less than one, depending on the machine.

Therefore, assertion is false and reason is true.

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

Explanation

Assertion (A) is correct: An elastic collision is one in which both momentum and total kinetic energy are conserved, so the total kinetic energy of the system before and after the collision remains the same.

Reason (R) is also correct: In an elastic collision the forces involved are conservative in nature, and the mechanical energy is not converted into other forms like heat, sound or light.

Since no energy is lost to these other forms, the whole of the kinetic energy is recovered after the collision, which is exactly why it is conserved. The Reason therefore explains the Assertion.

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: Potential energy can be converted into kinetic energy. For example, a body falling freely under gravity loses potential energy and gains an equal amount of kinetic energy.

Reason (R) is also correct: According to the principle of conservation of energy, energy can neither be created nor destroyed, but can only be converted from one form to another.

The conversion of potential energy into kinetic energy is a particular instance of this general principle. The Reason therefore explains the Assertion.

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: Frictional forces always do negative work on a moving object.

Reason (R) is also correct: Friction always opposes the relative motion of a body, so the frictional force acts in a direction opposite to the displacement of the object.

Since the angle between the frictional force and the displacement is 180°, cos 180° = − 1 and the work done W = − f s is negative. The Reason therefore explains the Assertion.

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: The tension in the string is maximum at the lowest point of a vertical circle. In the condition of just crossing the highest point, the tension at the lowest point is T_B = 6 mg, which is the greatest value anywhere on the circle.

Reason (R) is false: At the lowest point the centre of the circle lies vertically above the particle, so the centripetal force is directed upwards, whereas the gravitational force acts downwards. The two are therefore in opposite directions, not in the same direction. This is why the equation of motion at the lowest point is

$\text T - \text{mg} = \frac{\text{mv}^2}{\text l} \quad \Rightarrow \quad \text T = \frac{\text{mv}^2}{\text l} + \text{mg}$

It is precisely because the weight opposes the centripetal direction that the tension must exceed it, and this, together with the speed being greatest at the lowest point, makes the tension maximum there.

Therefore, assertion is true but reason is false.

Note: The printed answer key gives option 1. The Reason as printed states that the gravitational force and the centripetal force act in the same direction at the bottom, which is not correct; the textbook's own explanation for this question describes the centripetal force as directed upwards and the gravitational force as directed downwards.

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

Explanation

Assertion (A) is correct: The speed of an object in a vertical circle is minimum at the top of the circle. In the condition of just crossing the highest point, $\text v_A = \sqrt{\text{gl}}$, which is the least speed anywhere on the circle.

Reason (R) is also correct: The potential energy of the object is maximum at the top of the circle, since the height above the lowest point is greatest there.

By the conservation of mechanical energy, where the potential energy is maximum the kinetic energy and hence the speed must be minimum. The Reason therefore explains the Assertion.

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 work done by the centripetal force on an object moving in a vertical circle is zero.

Reason (R) is also correct: The centripetal force is always directed along the radius, towards the centre, while the displacement of the object is always along the tangent. Hence the centripetal force is always perpendicular to the displacement.

Since W = F s cos 90° = 0, the work done by such a force must vanish. The Reason therefore explains the Assertion.

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 vertical circular motion the total mechanical energy of the system remains constant if only conservative forces act on it, even though the kinetic and the potential energy separately keep changing.

Reason (R) is also correct: Non-conservative forces like friction convert mechanical energy into other forms such as heat, and therefore do not conserve it.

It is precisely the absence of such dissipative forces that keeps the sum of the kinetic and potential energy constant. The Reason therefore explains the Assertion.

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: An object can complete a vertical circle only if its speed at the lowest point is sufficiently high; the minimum required speed there is $\sqrt{5\text{gl}}$.

Reason (R) is also correct: The speed at the bottom must be large enough to ensure that, even after the object has risen through a height 2 l to the top, it still retains the critical speed $\sqrt{\text{gl}}$ needed to provide the centripetal force at the highest point, so that the string does not go slack.

The Reason thus states exactly the condition under which the circle can be completed, and therefore explains the Assertion.

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: At the top of a vertical loop the normal force on a roller coaster car is less than the gravitational force on it.

Reason (R) is also correct: At the top of the loop both the weight mg and the normal reaction N are directed downwards, that is, towards the centre, and together they provide the required centripetal force,

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

Since the weight already supplies a part of the centripetal force, only the remainder has to be supplied by the normal reaction, which therefore turns out to be the smaller of the two for the speeds with which such loops are negotiated. The Reason therefore explains the Assertion.

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: If the object just maintains circular motion, the tension in the string is zero at the top of the vertical circle. This is the condition which defines the critical velocity $\text v_c = \sqrt{\text{gl}}$.

Reason (R) is also correct: At the top of the vertical circle, when the tension has become zero, the weight of the body alone is directed towards the centre and it provides the entire centripetal force,

$\text{mg} = \frac{\text{mv}_c^2}{\text l}$

Since gravity alone is then sufficient, no additional pull from the string is needed and the tension is zero. The Reason therefore explains the Assertion.

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 object in a vertical circle has its maximum kinetic energy at the bottom of the circle, where its speed $\sqrt{5\text{gl}}$ is the greatest.

Reason (R) is also correct: The potential energy is minimum at the bottom of the circle, since the height above the reference level is least there.

By the conservation of mechanical energy the sum of the kinetic and the potential energy is constant, so where the potential energy is minimum the kinetic energy must be maximum. The Reason therefore explains the Assertion.

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

The work done by a force depends on the following three factors :

(i) The magnitude of the applied force (F).

(ii) The magnitude of the displacement produced (s).

(iii) The angle θ between the direction of the force and the direction of the displacement.

These are combined in the relation W = F s cos θ.

The work done by a force is W = F s cos θ.

Maximum work : For a fixed F and a fixed s, the work done is maximum when cos θ is maximum, that is, cos θ = 1 or θ = 0°. Then W_max = F s. This happens when the force and the displacement are in the same direction.

Minimum work : The work done is minimum when cos θ is minimum, that is, cos θ = − 1 or θ = 180°. Then W_min = − F s. This happens when the force and the displacement are in opposite directions.

A force does no work in either of the following two cases :

(i) When the displacement is zero (s = 0), that is, the force is not capable of displacing the point of application. Then W = F × 0 × cos θ = 0.

(ii) When the force is perpendicular to the displacement (θ = 90°), since cos 90° = 0 and hence W = F s cos 90° = 0. For example, the tension in the string of a stone whirled in a circle does no work on the stone.

While moving a body along the horizontal direction, the force of gravity (mg) acts vertically downwards whereas the displacement is horizontal. The angle between the gravitational force and the displacement is therefore 90°, and

$\text W = \text{mg}\text s\cos 90^\circ = 0$

Hence, no work is done against gravity when a body is moved along the horizontal direction.

No, he does not do any work.

To support the load the person exerts a force vertically upwards, against gravity, while he walks with uniform velocity along the horizontal road, so that the displacement is horizontal. The force is thus perpendicular to the direction of displacement, and the work done is

$\text W = \vec{\text F} \cdot \vec{\text s} = \text F\text s\cos 90^\circ = 0$

When the person accelerates on the road, he has to apply an additional force in the horizontal direction, that is, in the direction of motion, in order to produce the acceleration. Since this force and the displacement are in the same direction, work is done by this horizontal force.

However, the force exerted to support the load against gravity is still vertical while the displacement is horizontal, so no work is done by the force exerted against gravity.

No, it does not perform any work.

The satellite revolves around the earth in a circular orbit under the gravitational force exerted by the earth. This gravitational force acts as the centripetal force and is always directed along the radius, towards the centre of the orbit, whereas the displacement of the satellite is always along the tangent to the orbit.

Since the force is always perpendicular to the direction of motion of the satellite, θ = 90° and the work done is zero.

Friction is a non-conservative force because the work done against the frictional force depends upon the path followed between the two positions and not merely on the initial and the final positions. The longer the path between two points, the greater is the work done against friction.

Further, since friction always opposes the motion of a body, work has to be done against it on the return journey as well. Hence the net work done against the frictional force in moving a body through one complete round trip is not zero, which is the characteristic of a non-conservative force.

Kinetic energy can never be negative. The kinetic energy is $\text K = \frac{1}{2}\text{mv}^2$, in which the mass m is always positive and v², being the square of the speed, can never be negative. Hence K ≥ 0 always.

Potential energy can be negative. The potential energy of a system is negative when the force involved between the bodies of the system is one of attraction, for example the gravitational potential energy of a planet-satellite system.

Kilowatt is not the unit of energy. It is the unit of power, being equal to 1000 watt, that is, 1000 joule per second.

The joule, the calorie and the electron-volt are all units of energy. The kilowatt-hour, however, which is the product of power and time, is a unit of energy.

In uniform circular motion.

In uniform circular motion the particle moves with a constant speed, so its kinetic energy $\left(\frac{1}{2}\text{mv}^2\right)$, which depends only on the speed, remains unchanged.

The momentum, however, is a vector quantity $(\vec{\text p} = \text m\vec{\text v})$, and the direction of the velocity keeps changing continuously along the circular path. Hence the momentum changes continuously in direction, though its magnitude remains the same.

The momentum is conserved in both types of collisions, elastic as well as inelastic. During the collision the colliding bodies exert equal and opposite internal forces on each other, and since no external force acts on the system, the total linear momentum remains conserved.

The kinetic energy is conserved only in an elastic collision. In an inelastic collision a part of the kinetic energy is converted into other forms such as heat and sound, so the total kinetic energy is not conserved.

No. In a perfectly inelastic collision the whole of the kinetic energy is not lost.

Only that much kinetic energy is lost as is necessary for the conservation of linear momentum. Since the total momentum of the system must remain the same, the combined body has to move with a common velocity

$\text v = \frac{\text m_1\text u_1 + \text m_2\text u_2}{\text m_1 + \text m_2}$

and therefore it retains some kinetic energy. The kinetic energy would become zero only in the special case when the total momentum of the system happens to be zero.

The tension in the string is greatest at the lowest point of the vertical circle.

At the lowest point the tension acts vertically upwards while the weight acts vertically downwards, so the tension has to balance the weight as well as provide the centripetal force,

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

Moreover the speed of the body is also maximum at the lowest point, which makes the tension there the greatest. In the condition of just crossing the highest point, T_B = 6 mg.

At the top of the circle, two forces act on the object, both directed vertically downwards, that is, towards the centre of the circle :

(i) the weight of the object, mg, and

(ii) the tension T in the string, if any.

Their sum provides the necessary centripetal force at that point,

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

The tension in the string is maximum at the lowest point of the circle.

This is because at the lowest point the tension must not only provide the centripetal force but also balance the weight of the body, and the speed of the body is also greatest there.

The kinetic energy of the object is maximum at the lowest point of the circle.

At the lowest point the height above the reference level is minimum, so the potential energy is minimum. By the conservation of mechanical energy, the kinetic energy must therefore be maximum at that point.

When the object just maintains circular motion, its speed at the top has the critical value $\text v_c = \sqrt{\text{gl}}$. At this speed the weight of the body alone provides the entire centripetal force required at the highest point,

$\text{mg} = \frac{\text{mv}_c^2}{\text l}$

Since no additional inward pull is needed from the string, the tension in it becomes zero.

This is most likely to occur at the top of the circle.

At the highest point the speed of the object is minimum and the tension in the string is also minimum. If the speed falls below the critical velocity $\sqrt{\text{gl}}$, the required centripetal force becomes less than the weight, the string slackens and the object leaves the circular path, thereafter performing projectile motion.

The tension in the string must not become negative at any point of the circular path, since a string can only pull and never push.

As the tension is minimum at the highest point, this requires

$\text T_A = \frac{\text{mv}_A^2}{\text l} - \text{mg} \ge 0 \quad \Rightarrow \quad \text v_A \ge \sqrt{\text{gl}}$

Thus the speed at the highest point must be at least the critical velocity $\sqrt{\text{gl}}$, and that at the lowest point at least $\sqrt{5\text{gl}}$.

The speed of the object is lowest at the top of the circle.

In going from the lowest point to the highest point the object rises through a height 2 l, so its potential energy increases and, by the conservation of mechanical energy, its kinetic energy and hence its speed decreases. In the condition of just crossing the top, the speed there is $\sqrt{\text{gl}}$.

The centripetal force required at a point where the string makes an angle θ with the vertical is $\frac{\text{mv}^2}{\text l}$, and it is supplied by the difference

$\text T - \text{mg}\cos \theta = \frac{\text{mv}^2}{\text l}$

Since both the speed v and the radial component of the weight mg cos θ vary with the position of the object, the centripetal force varies from point to point. It is maximum at the lowest point, where the speed is greatest, and minimum at the highest point, where the speed is least.

The minimum speed required at the top of a vertical circle is the critical velocity,

$\text v_c = \sqrt{\text{gl}}$

where g is the acceleration due to gravity and l is the radius of the circle. This is obtained by putting the tension equal to zero at the highest point, so that the weight of the body alone provides the centripetal force.

Conservative force : the gravitational force. The work done by it in moving a body between two positions is independent of the path followed, and the work done over a closed path is zero.

Non-conservative force : the frictional force. The work done against it depends upon the path followed, and the net work done over a closed path is not zero.

The work done by a force is given by W = F s cos θ, where θ is the angle between the direction of the force and the direction of the displacement.

Positive work : When the force applied on a body is in the same direction as the displacement of the body, the angle θ is 0° and

$\text W = \text F\text s\cos 0^\circ = \text{Fs}$

which is positive. For example, when a body falls freely, the work done by gravity is positive.

Negative work : When the force acts opposite to the direction of motion, the angle θ is 180° and

$\text W = \text F\text s\cos 180^\circ = -\text{Fs}$

which is negative. For example, the work done by the frictional force on a moving body is negative.

If mountain roads were to go straight up the slope, the angle θ of the incline would be large. In that case :

(i) The frictional force between the tyres and the road, which is f = μ mg cos θ, would become small, since cos θ decreases as θ increases. Due to this insufficient friction the wheels of the vehicle would slip on the road.

(ii) For a large slope the component of the weight along the incline, mg sin θ, would be large, so more power would have to be used by the engine to climb the road.

By making the roads wind up gradually, the angle θ is kept small. This keeps the friction large enough to prevent slipping and reduces the power required, although the vehicle then has to travel a longer distance.

A system possesses potential energy when its various parts are held a certain distance apart against some force. The work done in holding the parts apart is stored in the system as its potential energy. Since different types of forces exist in nature, the corresponding potential energies are given different names.

Chemical energy : The chemical energy in substances arises due to the chemical bondings between the atoms of the substances. It is the potential energy associated with the forces between the atoms.

Gravitational energy : Gravitational energy results when objects are held at some distance against the gravitational force of attraction between them. For example, a body of mass m raised to a height h above the earth's surface has gravitational potential energy mgh.

Nuclear energy : Nuclear energy is the result of the short-range, strongest nuclear force between the nucleons inside the nucleus.

Hence, chemical, gravitational and nuclear energies are all potential energies corresponding to different types of forces in nature.

Under the given convention, the potential energy of a system of two particles held at a finite distance apart is negative or positive according as the conservative force acting between the particles is one of attraction or of repulsion.

(a) Negative. The electric force between an electron and a positron is one of attraction, since they carry opposite charges. Hence the potential energy of the electron-positron system in a bound state is negative.

(b) Negative. The planet-satellite system remains bound under the gravitational force of attraction between them, and so the potential energy of the system is negative.

(c) Positive. For an electron-electron system the force is one of electrostatic repulsion, since both carry the same negative charge. Hence the potential energy is positive.

When two protons are brought towards each other, the potential energy of the system increases. Both the protons carry a positive charge, so the force between them is one of repulsion. In bringing them nearer, work has to be done against this force of repulsion, and this work is stored in the system in the form of potential energy.

When a proton and an electron are brought nearer, the potential energy decreases. The force between them is one of attraction, since they carry opposite charges. Hence no external work is needed to bring them closer; on the contrary the attractive force itself does positive work, and the potential energy of the system decreases.

(a) Increases. Work has to be done against the elastic restoring force in compressing a spring, and this work is stored in the spring as elastic potential energy.

(b) Increases. Work has to be done against the elastic restoring force in stretching a spring as well, so the potential energy again increases.

(c) Decreases. The force between two dis-similar charges is one of attraction, so in bringing them nearer the attractive force does positive work and the potential energy of the system decreases.

(d) Increases. Work has to be done against the gravitational force of attraction in taking a body away, so the potential energy of the system increases.

(e) Decreases. As an air bubble rises up in water, the water moves down to take its place. The centre of gravity of the system is therefore lowered, and the potential energy of the system decreases.

When the wet piece of cement is thrown on the wall and gets glued to it, it comes to rest, so its kinetic energy is completely lost.

A part of this initial energy is used up in deforming the cement piece, so that it flattens and sticks to the wall. The remaining part is converted into sound and heat at the surface of the wall.

Thus the initial kinetic energy is not destroyed but is transformed into other forms of energy, in accordance with the principle of conservation of energy.

In a thermal station the chain of energy-changes is as follows :

Chemical energy → Heat energy → Kinetic energy of rotation → Electrical energy

(i) The heat energy produced by burning the coal comes from the chemical energy of the coal.

(ii) This heat energy converts water into steam.

(iii) The steam rotates the turbine, so the heat energy is converted into the kinetic energy of rotation.

(iv) This kinetic energy is converted by the generator into electrical energy.

Yes, a body can have energy without momentum. There is an internal energy in a body due to the thermal agitation of its particles, while the vector sum of the momenta of these moving particles may be zero. Thus the body possesses energy but has no net momentum.

No, there can be no momentum without energy. If a body has momentum p = m v, then its velocity v is not zero, and hence it must possess kinetic energy $\frac{1}{2}\text{mv}^2$, which is related to the momentum by $\text K = \frac{\text p^2}{2\text m}$.

The momentum will be conserved. No external force acts on the shot during the explosion; the forces of explosion are internal forces of the system. Hence the total linear momentum of all the fragments together remains the same as that of the shot just before the explosion.

The kinetic energy will increase. During the explosion the chemical potential energy of the explosives in the shot is converted into the kinetic energy of the fragments. Hence the total kinetic energy after the explosion is greater than that before it.

(a) The momentum of a body is p = m v. Since the velocity of the aeroplane is doubled while its mass remains the same, the momentum of the aeroplane becomes double.

The momentum of the system of the plane and the air remains conserved, because with the increase in the momentum of the plane, the momentum of the air increases equally in the opposite direction.

(b) The kinetic energy is $\text K = \frac{1}{2}\text{mv}^2$, which varies as the square of the velocity. On doubling the velocity, the kinetic energy becomes four times.

This additional energy is obtained from the burning of the fuel of the plane. Thus the 'total' energy remains conserved, in accordance with the principle of conservation of energy.

Yes. In an exothermic chemical reaction also, mass is converted into energy, just as it happens in nuclear fission and fusion.

However, in a chemical reaction the mass converted into energy is much smaller, being about a million times less than that in fission and fusion. This is why the change in mass is not detectable in ordinary chemical reactions, and mass is taken to be conserved in them for practical purposes.

(a) No, the total kinetic energy is not conserved during the elastic collision. During the short time of contact the balls are deformed, and in that duration a part of the kinetic energy is converted into the elastic potential energy of deformation. It is recovered fully as kinetic energy only when the balls separate.

(b) Yes, the total linear momentum remains conserved during the collision. During the collision the bodies exert equal and opposite forces on each other and therefore suffer equal and opposite changes in momentum, so that the total momentum of the system does not change at any instant.

(c) In an inelastic collision the linear momentum is conserved during the collision as well as after it, since momentum conservation does not depend on the nature of the collision. The kinetic energy is not conserved, even after the collision is over, because a part of it is permanently converted into heat, sound and the energy of deformation.

(d) The collision is elastic. If the potential energy depends only on the distance between the centres of the balls, the force during the collision is conservative, so the total mechanical energy is conserved and no energy is lost in the collision.

(a) Decreases. A conservative force is the negative gradient of the potential energy, $\text F = -\frac{\text{dU}}{\text{dx}}$. Hence when a conservative force does positive work on a body, the potential energy of the body decreases by an equal amount.

(b) Kinetic energy. Friction always opposes the motion of a body and does negative work on it, and by the work-kinetic energy theorem this causes a loss in the kinetic energy of the body.

(c) External force. The total momentum of a many-particle system can change only when a net external force acts on the system, since the internal forces occur in equal and opposite pairs and cancel out. Hence its rate of change is proportional to the external force.

(d) Total linear momentum and total energy. In an inelastic collision the kinetic energy of the system changes into some other form, but the total linear momentum and the total energy remain unchanged.

The light body has more kinetic energy.

The kinetic energy and the momentum of a body are related as

$\text p = \sqrt{2\text{mK}} \quad \Rightarrow \quad \text K = \frac{\text p^2}{2\text m}$

For a given momentum p, the kinetic energy varies inversely as the mass,

$\text K \propto \frac{1}{\text m}$

Hence, if two bodies of different masses have the same momentum, the kinetic energy of the lighter body will be more.

The heavy body has a greater momentum.

The momentum and the kinetic energy of a body are related as

$\text p = \sqrt{2\text{mK}}$

For a given kinetic energy K, the momentum varies as the square root of the mass,

$\text p \propto \sqrt{\text m}$

Hence, if two bodies of different masses have the same kinetic energy, the momentum of the heavier body will be more.

Given,

• Radius of the chamber, r = 25 m
• g = 9.8 m s^-2

Why the motorcyclist does not drop down : The motorcyclist driving in vertical loops experiences a centrifugal force $\frac{\text{mv}^2}{\text r}$, equal and opposite to the centripetal force. When he is at the uppermost point, this centrifugal force is directed vertically upwards while his weight mg is directed vertically downwards. The resultant force

$\left(\frac{\text{mv}^2}{\text r} - \text{mg}\right)$

keeps the motorcyclist pressed against the wall of the death-well, and hence he does not drop down even though there is no support from below.

Minimum speed : For the minimum speed v_min required to perform the vertical loop, the weight alone provides the centripetal force at the uppermost point,

$\frac{\text{mv}_{min}^2}{\text r} = \text{mg} \quad \Rightarrow \quad \text v_{min} = \sqrt{\text{gr}}$

Substituting the values,

$$\begin{aligned}\text v_{min} = \sqrt{9.8 \times 25} = \sqrt{245} \\[1em] = 15.65\ \text{m s}^{-1}\end{aligned}$$

Hence, the minimum speed required to perform the vertical loop is 15.65 m s^-1.

The energy released per unit mass of the material consumed is much greater in fusion than in fission.

The number of deuterons which undergo fusion in 1 g of heavy hydrogen is much larger than the number of U²³⁵ nuclei which undergo fission in 1 g of uranium, since a deuteron is far lighter than a uranium nucleus. Therefore, although the energy released in a single fusion reaction of light nuclei is much less than that released in a single fission reaction of a heavy nucleus, the energy output per unit mass of the material consumed is much more in the case of the fusion of light nuclei.

Hence, a hydrogen bomb based on nuclear fusion is far more powerful than an atomic bomb based on nuclear fission.

Water is the most effective moderator.

A moderator slows down the neutrons most effectively when the mass of its nuclei is of the same order as the mass of a neutron, because then the neutrons transfer the whole of their kinetic energy to it in a collision. This follows from the fact that in a head-on elastic collision between two bodies of equal masses, the bodies simply exchange their velocities.

The nuclei of lead and zinc are far too heavy, so the neutrons colliding with them would return with almost undiminished speed. The mass of a hydrogen nucleus present in water is of the order of the mass of a neutron.

Hence, out of water, lead and zinc, water is the most effective moderator.

Given,

• Radius of the circle = R
• Kinetic energy, K = a s², where s is the distance covered

Let m be the mass of the particle and v its linear velocity. Then its kinetic energy is

$\text K = \frac{1}{2}\text{mv}^2 = \text a\text s^2 \qquad \ldots(\text i)$

Tangential force : Differentiating equation (i) with respect to time t,

$\frac{1}{2}\text m(2\text v)\frac{\text{dv}}{\text{dt}} = \text a(2\text s)\frac{\text{ds}}{\text{dt}}$

$\text{mv}\frac{\text{dv}}{\text{dt}} = 2\text{as}\frac{\text{ds}}{\text{dt}}$

But $\frac{\text{ds}}{\text{dt}} = \text v$, so on cancelling v from both sides,

$\text m\frac{\text{dv}}{\text{dt}} = 2\text{as}$

The quantity $\text m\frac{\text{dv}}{\text{dt}}$ is the tangential force, so

$\text F_T = 2\text{as}$

Centripetal force : From equation (i), $\text{mv}^2 = 2\text{as}^2$. Therefore

$\text F_C = \frac{\text{mv}^2}{\text R} = \frac{2\text{as}^2}{\text R}$

Resultant force : The tangential and the centripetal forces are mutually perpendicular, so the resultant force is

$$\begin{aligned}\text F = \sqrt{\text F_T^2 + \text F_C^2} = \sqrt{(2\text{as})^2 + \left(\frac{2\text{as}^2}{\text R}\right)^2} \\[1em] = 2\text{as}\sqrt{1 + \frac{\text s^2}{\text R^2}}\end{aligned}$$

Hence, the force acting on the particle is $\text F = 2\text{as}\left(1 + \frac{\text s^2}{\text R^2}\right)^{1/2}$.

(i) It is the result of scalar multiplication of force and displacement vectors.

Work is the scalar product (dot product) of the force and the displacement, $\text W = \vec{\text F} \cdot \vec{\text s} = \text{Fs}\cos \theta$. The scalar product of two vectors is always a scalar, which is why work, though obtained from two vector quantities, is itself a scalar having magnitude but no direction.

(ii) W_A = W_B but P_A > P_B.

Both the persons are of the same weight and carry equal loads of 20 kg to the same height, that is, to the first floor. Since the work done against gravity is W = mgh and m, g and h are the same for both, W_A = W_B.

The power, however, is $\text P = \frac{\text W}{\text t}$, so

$\frac{\text P_A}{\text P_B} = \frac{\text W_A}{\text W_B} \times \frac{\text t_B}{\text t_A} = 1 \times \frac{4}{3} = \frac{4}{3}$

Since $\frac{\text P_A}{\text P_B} \gt 1$, we get P_A > P_B. The person who takes less time generates more power.

(iii) 8 J

Given, $\vec{\text F} = (-\hat{\text i} + 2\hat{\text j} + 4\hat{\text k})$ N and the body moves 4 m along the Y-axis, so $\vec{\text s} = 4\hat{\text j}$ m.

$$\begin{aligned}\text W = \vec{\text F} \cdot \vec{\text s} = (-\hat{\text i} + 2\hat{\text j} + 4\hat{\text k}) \cdot (4\hat{\text j}) \\[1em] = 0 + 8 + 0 = 8\ \text J\end{aligned}$$

(iv) t

The body starts from rest and is subjected to a constant force, so F = ma is constant and v = at. Therefore

$\text P = \text{Fv} = (\text{ma})(\text{at}) = \text{ma}^2\text t$

Since m and a are constant, P ∝ t.

(v) 90° < θ ≤ 180°

The work done is W = F s cos θ. It is negative when cos θ is negative, which happens when the angle θ lies between 90° and 180°. At θ = 90° the work is zero, and for θ less than 90° it is positive.

(i) Frictional force

The gravitational, electrostatic and magnetic forces are conservative forces, for which the work done depends only on the initial and the final positions of the body. Friction is a non-conservative force, since the work done against it depends on the path followed and the net work done over a closed path is not zero.

(ii) Total mechanical energy

For a body falling freely under gravity alone in vacuum, no non-conservative force acts. The kinetic energy goes on increasing and the potential energy goes on decreasing, but their sum, the total mechanical energy, remains constant throughout the fall.

(iii) 250 π² J

Given, m = 5 kg, r = 1 m and n = 300 revolutions per minute.

Converting the frequency into revolutions per second,

$\text n = \frac{300}{60} = 5\ \text{rev s}^{-1}$

The angular velocity is ω = 2πn, so the linear velocity is v = rω = 2πnr. The kinetic energy is

$\text K = \frac{1}{2}\text{mv}^2 = \frac{1}{2}\text m(\text r\omega)^2 = \frac{1}{2}\text{mr}^2 \times 4\pi^2\text n^2$

Substituting the values,

$$\begin{aligned}\text K = \frac{1}{2} \times 5 \times (1)^2 \times 4\pi^2 \times (5)^2 \\[1em] = \frac{1}{2} \times 5 \times 4\pi^2 \times 25 \\[1em] = 250\pi^2\ \text J\end{aligned}$$

(iv) On rising of an air bubble in water.

As an air bubble rises up in water, the water moves down to occupy its place, so the centre of gravity of the system is lowered and the potential energy decreases. In compressing or stretching a spring, work is done against the elastic restoring force, and in moving a body against the gravitational pull work is done against gravity; in all these cases the potential energy increases.

(i) both momentum and kinetic energy remains conserved.

An elastic collision is one in which both the momentum and the total kinetic energy are conserved. The forces involved are conservative in nature, so the mechanical energy is not converted into other forms like heat, sound or light.

(ii) Both (b) and (c).

A collision is an event in which two particles come together and mutually interact for a very brief period. This interaction may or may not involve physical contact, so physical contact of the bodies is not mandatory. Further, in all types of collisions the total momentum of the system remains conserved, since no external force acts on the system.

(iii) m₁ = m₂

For a one-dimensional elastic collision with the target body initially at rest, the velocity of the striking body after the collision is

$\text v_1 = \left(\frac{\text m_1 - \text m_2}{\text m_1 + \text m_2}\right)\text u_1$

For 100% transfer of kinetic energy, the striking body must come to rest, that is, v₁ = 0. This requires m₁ − m₂ = 0, that is, m₁ = m₂. In this case the two bodies simply exchange their velocities.

(iv) 20 m/s in the direction of 10 kg mass

Given, m₁ = 10 kg with u₁ = 50 m/s and m₂ = 5 kg with u₂ = 40 m/s in the opposite direction. Since the bodies stick together, the collision is perfectly inelastic. Taking the direction of motion of the 10 kg mass as positive and applying the conservation of linear momentum,

$\text m_1\text u_1 - \text m_2\text u_2 = (\text m_1 + \text m_2)\text v$

$$\begin{aligned}(10 \times 50) - (5 \times 40) = (10 + 5)\text v \\[1em] 500 - 200 = 15\text v \\[1em] \text v = \frac{300}{15} = 20\ \text{m/s}\end{aligned}$$

Since v is positive, the combined mass moves in the direction of the 10 kg mass with a velocity of 20 m/s.

(v) Only momentum is conserved

The bullet gets embedded in the block, so the collision is perfectly inelastic. In such a collision the total linear momentum is conserved, but a part of the kinetic energy is converted into heat and the energy of deformation, so the kinetic energy is not conserved.

(i) Kinetic energy

In uniform circular motion the particle travels at a constant speed along the circular path. Since the kinetic energy $\left(\frac{1}{2}\text{mv}^2\right)$ depends only on the speed and not on the direction of motion, it remains constant.

(ii) Because the direction of motion changes

Momentum is a vector quantity, $\vec{\text p} = \text m\vec{\text v}$, so it depends on both the magnitude and the direction of the velocity. In uniform circular motion the speed is constant but the direction of the velocity, which is always tangential to the circle, changes continuously. Hence the momentum changes continuously.

(iii) At the top of the loop

At the highest point of the vertical circle the particle is at the greatest height, so its potential energy is maximum. By the conservation of mechanical energy its kinetic energy, and hence its speed, is the lowest there.

(iv) Total mechanical energy

In vertical circular motion the speed, and therefore the kinetic energy and the potential energy, change from point to point. However, in the absence of any energy loss due to non-conservative forces like friction, their sum, that is, the total mechanical energy, remains constant throughout the motion.

(v) It is at its maximum

The gravitational potential energy is U = mgh, which increases with height. At the highest point of a vertical loop the height above the reference level is greatest, so the potential energy is maximum there.

Meaning of work : In the language of Physics, when an applied force happens to displace the point of application in the direction of the force, work is said to have been done. If there is no displacement of the point of application in the direction of the force, no work is done, regardless of the effort involved. Work is a way of transferring energy from one place to another, or transferring it from one form to another.

Expression for the work done by a constant force :

(a) When the force and the displacement are in the same direction : Work is measured by the product of the applied force and the displacement of the body in the direction of the force, that is,

$\text{work} = \text{force} \times \text{displacement in the direction of the force}$

If a force F acting on a body produces a displacement s in the direction of the force, then

$\text W = \text{F s}$

(b) When the force makes an angle θ with the displacement : If the force F makes an angle θ with the direction of the displacement of the body, then the component of the force in the direction of displacement is F cos θ, and the work done is

$\text W = (\text F\cos \theta)\text s = \text F\text s\cos \theta$

Since the body does not move in the vertical direction, no work is done by the component F sin θ.

In vector notation, since both force and displacement are vector quantities, the work done is the scalar product of the force and the displacement,

$\text W = \vec{\text F} \cdot \vec{\text s}$

Therefore work is a scalar quantity.

Angle for maximum work : For a fixed F and s, the work done is maximum when cos θ is maximum, that is, cos θ = 1 or θ = 0°, when both the force and the displacement are in the same direction. Then

$\text W_{max} = \text F\text s$

Angle for minimum work : The work done is minimum when cos θ is minimum, that is, cos θ = − 1 or θ = 180°, when the force acts opposite to the displacement. Then

$\text W_{min} = -\text F\text s$

(a) Kinetic energy : The energy possessed by a body by virtue of its state of motion is known as kinetic energy. Energy possessed by a body by virtue of its translational motion is known as translational kinetic energy.

Examples : A moving bullet can pierce a target, flowing water can turn a turbine, and a moving hammer can drive a nail into wood. In each case the body is able to do work on account of its motion, that is, it possesses kinetic energy.

Expression for kinetic energy : Consider a body of mass m, initially at rest. Let it be subjected to a constant force of magnitude F, causing it to move with velocity v in the direction of the force over a distance s. The work done by the force is

$\text W = \text F \cdot \text s \qquad \ldots(\text i)$

But F = m a, and from the equation of kinematics, since the body starts from rest,

$\text v^2 = 0 + 2\text a\text s \quad \Rightarrow \quad \text a = \frac{\text v^2}{2\text s}$

Substituting these in equation (i),

$\text W = \text m\left(\frac{\text v^2}{2\text s}\right) \times \text s = \frac{1}{2}\text{mv}^2$

This work done is converted into the kinetic energy of the body. Thus the kinetic energy of the body is

$\text K = \frac{1}{2}\text{mv}^2$

The kinetic energy of a moving body is measured by the amount of work which has been done in bringing the body from the rest position to its present state of motion, and vice-versa. Its S.I. unit is the joule (J).

(b) Work-kinetic energy theorem : The work-kinetic energy theorem states that the work done by the net force acting on a system is equal to the change in its kinetic energy. Mathematically,

$\text W = \Delta \text K$

Proof for a constant force : Let a body of mass m be moving with an initial velocity u. When it is subjected to a constant force F, its velocity is changed to v in a displacement s in the direction of the force. The work done by the force is

$\text W = \text F \cdot \text s = \text{mas}$

From the equation of kinematics, v² = u² + 2as, so that

$\text s = \frac{\text v^2 - \text u^2}{2\text a}$

Substituting this value of s,

$\text W = \text{ma} \times \left(\frac{\text v^2 - \text u^2}{2\text a}\right) = \frac{1}{2}\text{mv}^2 - \frac{1}{2}\text{mu}^2$

$\text W = \Delta \text K$

which proves the theorem.

Motion in a vertical circle : When a body tied to the end of a string is revolved in a vertical circle, the speed of the body is different at different points of the circular path. Therefore, the centripetal force on the body and the tension in the string change continuously.

Suppose a small object of mass m tied at one end of a weightless string of length l is being revolved in a vertical circle in the anticlockwise direction. Let O be the centre of the circle, A the highest point and B the lowest point. Suppose that at any instant the object is at the point P of the circle, whose angular displacement from the lowest point B is θ. Let the velocity of the object at the point P be v, directed along the tangent drawn at the point P, and let T be the tension in the string at the point P.

Forces acting on the object : Two forces act upon the object in the position P :

(i) the weight of the object, mg, vertically downwards, and

(ii) the tension T in the string, towards the centre of the circular path.

The weight mg may be resolved into two components : the tangential component mg sin θ, and the radial component mg cos θ which is directed opposite to the tension T.

Expression for the tension : The net radial force on the object towards the centre of the circular path is T − mg cos θ, and this provides the necessary centripetal force for the circular motion of the object. Therefore,

$\text T - \text{mg}\cos \theta = \frac{\text{mv}^2}{\text l}$

where l, the length of the string, is the radius of the circle. Thus the tension in the string at the point P is

$\text T = \frac{\text{mv}^2}{\text l} + \text{mg}\cos \theta \qquad \ldots(\text i)$

Minimum speed at the highest point : As the object goes upwards on the circular path, its velocity v goes on decreasing and its angular displacement θ goes on increasing, that is, cos θ goes on decreasing. Hence, according to equation (i), the tension T in the string goes on decreasing and becomes minimum at the highest point A.

At the highest point A, θ = 180° and cos 180° = − 1. If the velocity of the object at A be v_A, then the tension in the string at A is

$$\begin{aligned}\text T_A = \frac{\text{mv}_A^2}{\text l} + \text{mg}\cos 180^\circ \\[1em] \text T_A = \frac{\text{mv}_A^2}{\text l} - \text{mg}\end{aligned}$$

If the velocity of the object at the highest point is decreased, the tension T_A in the string will decrease and become zero at a certain minimum value of the velocity. Let this minimum value of the velocity be v_c. Then, putting T_A = 0 and v_A = v_c in the above equation, we get

$0 = \frac{\text{mv}_c^2}{\text l} - \text{mg}$

$\text v_c = \sqrt{\text{gl}} \qquad \ldots(\text{ii})$

The object will cross over the highest point A, provided the velocity of the object at the highest point be at least $\sqrt{\text{gl}}$. This is called the 'critical velocity' for the motion in a vertical circle. If the velocity of the object is less than this, the string would slack and the object will not remain on the circular path.

Velocity of the object at any point : In going from the point P to the highest point A, the kinetic energy of the object would decrease and the potential energy would increase by the same amount, since the height of A above P is (l + l cos θ). Therefore, by energy-conservation,

$\frac{1}{2}\text{mv}^2 - \frac{1}{2}\text{mv}_A^2 = \text{mg}(\text l + \text l\cos \theta)$

$\text v^2 - \text v_A^2 = 2\text{gl} + 2\text{gl}\cos \theta$

If the object just crosses the highest point A, then $\text v_A = \text v_c = \sqrt{\text{gl}}$, that is, v_A² = gl. Then we have

$$\begin{aligned}\text v^2 - \text{gl} = 2\text{gl} + 2\text{gl}\cos \theta \\[1em] \text v^2 = 3\text{gl} + 2\text{gl}\cos \theta = (3 + 2\cos \theta)\text{gl}\end{aligned}$$

$\text v = \sqrt{(3 + 2\cos \theta)\text{gl}} \qquad \ldots(\text{iii})$

Variation of the tension in the string : Substituting this value of v in equation (i), in the condition of the object just crossing over the highest point, the tension in the string at the point P is given by

$\text T = \frac{\text m}{\text l}\left\lbrace(3 + 2\cos \theta)\text{gl}\right\rbrace + \text{mg}\cos \theta$

$= 3\text{mg} + 2\text{mg}\cos \theta + \text{mg}\cos \theta$

$\text T = 3\text{mg}(1 + \cos \theta) \qquad \ldots(\text{iv})$

Now we consider three positions of the object on the circular path in the condition of just crossing over the highest point A :

(i) At the lowest point B (θ = 0) : From equations (iii) and (iv),

$\text v_B = \sqrt{(3 + 2\cos 0^\circ)\text{gl}} = \sqrt{5\text{gl}}$

$\text T_B = 3\text{mg}(1 + \cos 0^\circ) = 6\text{mg}$

(ii) When the string is horizontal, that is, at the point C (θ = 90°) :

$\text v_C = \sqrt{(3 + 2\cos 90^\circ)\text{gl}} = \sqrt{3\text{gl}}$

$\text T_C = 3\text{mg}(1 + \cos 90^\circ) = 3\text{mg}$

(iii) At the highest point A (θ = 180°) :

$\text v_A = \sqrt{(3 + 2\cos 180^\circ)\text{gl}} = \sqrt{\text{gl}}$

$\text T_A = 3\text{mg}(1 + \cos 180^\circ) = 0$

Hence, the minimum speed at the highest point is $\sqrt{\text{gl}}$ and the corresponding minimum speed at the lowest point is $\sqrt{5\text{gl}}$. The tension is maximum at the lowest point, where T_B = 6 mg, falls to 3 mg when the string is horizontal, and becomes zero at the highest point. Thus, if on revolving an object in a vertical circle the tension in the string at the highest point be zero, then the tension at the lowest point would be six times the weight of the object.

Work : When an applied force happens to displace the point of application in the direction of the force, work is said to have been done. Work is measured by the product of the applied force and the displacement of the body in the direction of the force,

$\text W = \text F\text s\cos \theta = \vec{\text F} \cdot \vec{\text s}$

S.I. unit of work : The S.I. absolute unit of work is the joule (J). If a force of 1 newton produces a displacement of 1 metre in the direction of the force, then the work done is 1 joule, that is,

$1\ \text{joule} = 1\ \text{newton} \times 1\ \text{metre}$

In the C.G.S. system the unit of work is the erg, and 1 joule = 10⁷ erg.

Positive work : When the applied force has a component in the direction of the displacement, that is, when θ is less than 90°, cos θ is positive and the work done is positive. It is maximum when θ = 0°.

Examples : (i) The work done by a person in lifting a bucket out of a well. (ii) The work done by gravity on a freely falling body.

Negative work : When the applied force has a component opposite to the direction of the displacement, that is, when θ is more than 90°, cos θ is negative and the work done is negative. It is minimum when θ = 180°.

Examples : (i) The work done by friction on a body sliding down an inclined plane. (ii) The work done by gravity when a body is lifted upwards.

Zero work : The work done is zero in two cases : when the displacement is zero, and when the force is perpendicular to the displacement (θ = 90°), since cos 90° = 0.

Examples : (i) The work done by a person carrying a load on his head while walking on a horizontal road. (ii) The work done by the tension in the string of a stone whirled in a circle.

Work done by a variable force : When the force acting on a body is not constant, the total displacement is divided into a large number of infinitesimally small displacements Δx, over each of which the force F(x) may be regarded as constant. The small work done over one such displacement is F(x) Δx, and the total work done in moving the body from x₁ to x₂ is the sum of all such terms,

$\text W = \sum_{\text x_1}^{\text x_2}\text F(\text x)\Delta \text x$

In the language of integral calculus, this becomes

$\text W = \int_{\text x_1}^{\text x_2}\text F(\text x)\text{dx}$

Numerically this quantity is equal to the area between the force curve and the x-axis between the values x₁ and x₂. Hence, the work done by a variable force is equal to the area between the force curve and the displacement axis.

Conservative force : A force is said to be conservative if the work done by the force, or against the force, in moving a body from one position to another, depends only on the initial and the final positions and not upon the path followed between the two positions.

Examples : the gravitational force, the electrostatic force, the magnetic force and the elastic force of a spring.

Properties of conservative forces :

(i) The work done by a conservative force in moving a body between two positions is independent of the path followed.

(ii) The work done by a conservative force over a closed path is zero. If a particle goes from a position A to a position B along one path and returns to A along another path, then

$\text W_{A \to B} + \text W_{B \to A} = 0$

(iii) A potential energy can be associated with a conservative force, and the force is the negative gradient of the potential energy, $\text F = -\frac{\text{dU}}{\text{ds}}$.

(iv) In the presence of conservative forces alone, the total mechanical energy of a system remains constant.

A central force, that is, a force acting upon a particle which is always directed towards or away from a fixed point and whose magnitude depends only on the distance of the particle from that point, is a conservative force.

Non-conservative force : A force is said to be non-conservative if the work done by the force, or against the force, in moving a body from one position to another, depends upon the path followed between the two positions.

Examples : the frictional force, the viscous force and the induction force acting on an electron in a betatron.

Properties of non-conservative forces :

(i) The work done depends upon the path followed. The longer the path between two points, the greater is the work done against the force.

(ii) The net work done in moving a body through one complete round trip is not zero. This is because a frictional force always opposes the motion of a body, so work has to be done against it on the return journey as well.

(iii) No potential energy can be associated with such a force, since the mechanical energy is dissipated, usually as heat, and is not recoverable.

(a) Different forms of energy : Energy is the capacity to do work. It exists in various forms, each with unique characteristics and applications. Some of the primary forms are :

(i) Mechanical energy : It is associated with the motion and the position of objects. Mechanical energy is the sum of the kinetic energy and the potential energy of a system. A moving vehicle and a stretched spring both possess mechanical energy.

(ii) Thermal energy : It is the energy possessed by a body on account of the thermal agitation of its particles. When we rub our hands together, mechanical energy is converted into thermal energy.

(iii) Chemical energy : It is the potential energy which arises due to the chemical bondings between the atoms of a substance. The burning of coal and the working of a cell are examples of the release of chemical energy.

(iv) Electrical energy : It is the energy associated with the flow of electric charge. An electric heater converts electrical energy into heat energy.

(v) Nuclear energy : It is the result of the short-range, strongest nuclear force between the nucleons inside the nucleus. It is released in the processes of nuclear fission and fusion.

(vi) Sound energy : It is the energy carried by the vibrations of a medium. A loudspeaker converts electrical energy into sound energy.

(vii) Light energy : It is the energy carried by radiation. An electric bulb converts electrical energy into light energy.

(b) Mass-energy equivalence : Till the beginning of the 20th century it was believed that matter and energy are two different entities and that, like energy, the mass of the universe is also conserved. Einstein for the first time established that mass and energy are two different forms of the same entity and they are inter-convertible. He gave the mass-energy equivalence relation

$\text E = \text{mc}^2$

where E is the energy obtained when a mass m is completely converted into energy and c is the velocity of light in free space (= 3 × 10⁸ m/s).

Significance in physics :

(i) Since c² is an enormously large number, the relation indicates that an enormous amount of energy is obtained from the complete conversion of a very small amount of mass.

(ii) It explains the release of energy in nuclear fusion, in which a small amount of mass is converted into a large amount of energy when hydrogen nuclei fuse to form helium within the sun.

(iii) It also explains the reverse process. In particle accelerators, particles can be accelerated to high energies, and when they collide the energy from the collision can create new particles, effectively converting kinetic energy into mass. An example is the process of pair production, in which a photon of energy hν converts into an electron and a positron in the presence of a nucleus,

$\text h\nu + \text N \rightarrow \text e^- + \text e^+ + \text N$

Collision : A collision is an event where two particles come together and mutually interact for a very brief period. This interaction, which may or may not involve physical contact, leads to measurable changes in their relative motions. Notably, during a collision, no external force acts on the system, and hence the total momentum of the system remains conserved in all types of collisions.

Key characteristics of collisions : there is a mutual force exerted between the particles; the interaction lasts for a finite, often very brief, period; the paths or trajectories of the particles are altered; and both the energy and the momentum of the individual particles change.

Elastic collision : An elastic collision is one in which both momentum and total kinetic energy are conserved. The final particles are identical to the initial particles. Collisions between atomic, nuclear and fundamental particles are usually elastic.

Characteristics of an elastic collision :

(i) The total linear momentum before and after the collision remains the same.

(ii) The total angular momentum remains unchanged.

(iii) The mechanical energy and the total energy of the system are conserved.

(iv) Mechanical energy is not converted into other forms like heat, sound or light.

(v) The forces involved are conservative in nature.

(vi) The particles after the collision are the same as those before the collision.

Inelastic collision : An inelastic collision is one in which the total kinetic energy is not conserved, though the momentum is still conserved. When two bodies collide and stick together after the collision, some kinetic energy is lost, and this type of collision is known as a perfectly inelastic collision. For instance, the collision between two automobiles on a road is inelastic.

Characteristics of an inelastic collision :

(i) The momentum remains conserved.

(ii) The angular momentum is conserved.

(iii) The total energy of the system is conserved.

(iv) The total kinetic energy is not conserved.

(v) Potential energy changes and other forms of energy, such as heat and sound in macroscopic cases or light and gamma rays in microscopic cases, are produced.

In practice, there is no collision which is perfectly elastic or perfectly inelastic.

Elastic collision in two dimensions (oblique collision) : An oblique collision in two dimensions is a type of collision where the objects involved do not move along the same line before or after the collision. In such collisions the objects typically have velocities that form an angle with each other, rather than being perfectly head-on.

Suppose an elastic body of mass m₁ moving with velocity $\vec{\text u}_1$ along a straight line, say the X-axis, collides with another elastic body of mass m₂, initially at rest. After the collision the bodies m₁ and m₂ move with velocities $\vec{\text v}_1$ and $\vec{\text v}_2$ in the X-Y plane, making angles θ₁ and θ₂ respectively with the X-axis.

By the law of conservation of linear momentum,

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

Taking the scalar components along the X- and the Y-directions and applying the law of conservation of linear momentum separately for the two directions, we have

Along the X-axis :

$\text m_1\text u_1 = \text m_1\text v_1\cos \theta_1 + \text m_2\text v_2\cos \theta_2 \qquad \ldots(\text i)$

Along the Y-axis : Since there is no momentum along the Y-direction before the collision,

$0 = \text m_1\text v_1\sin \theta_1 - \text m_2\text v_2\sin \theta_2 \qquad \ldots(\text{ii})$

Since the collision is elastic, the kinetic energy is also conserved. That is,

$$\begin{aligned}\frac{1}{2}\text m_1\text u_1^2 = \frac{1}{2}\text m_1\text v_1^2 + \frac{1}{2}\text m_2\text v_2^2 \\[1em] \text m_1\text u_1^2 = \text m_1\text v_1^2 + \text m_2\text v_2^2 \qquad \ldots(\text{iii})\end{aligned}$$

In equations (i), (ii) and (iii) we may assume that m₁, m₂ and u₁ are known. But the motion after the collision involves four unknown quantities v₁, v₂, θ₁ and θ₂, which cannot all be determined from three equations. At least one of them should be given. For example, for a given value of θ₁, the quantities θ₂, v₁ and v₂ can be computed from the three equations.

Glancing collision : For such a collision the object of mass m₁ has θ₁ ≈ 0° and the object of mass m₂, initially at rest, then moves at θ₂ ≈ 90°. From equation (i),

$\text m_1\text u_1 = \text m_1\text v_1\cos 0^\circ + \text m_2\text v_2\cos 90^\circ \quad \Rightarrow \quad \text u_1 = \text v_1$

that is, the striking body continues with practically the same velocity.

Inelastic collision : An inelastic collision is one in which the total kinetic energy is not conserved, though the momentum is still conserved. Some of the kinetic energy is converted into other forms of energy such as heat, sound and the energy of deformation.

When two bodies collide and stick together after the collision, the collision is called a perfectly inelastic collision. In this case the two bodies move with a common velocity after the impact.

Common velocity : Let a body of mass m₁ moving with velocity u₁ collide with a body of mass m₂ moving with velocity u₂, and let them stick together and move with a common velocity v. By the conservation of linear momentum,

$$\begin{aligned}\text m_1\text u_1 + \text m_2\text u_2 = (\text m_1 + \text m_2)\text v \\[1em] \text v = \frac{\text m_1\text u_1 + \text m_2\text u_2}{\text m_1 + \text m_2}\end{aligned}$$

Loss of kinetic energy : Let the second body be initially at rest (u₂ = 0). The kinetic energy of the system before the collision is $\text K_1 = \frac{1}{2}\text m_1\text u_1^2$ and that after the collision is $\text K_2 = \frac{1}{2}(\text m_1 + \text m_2)\text v^2$. Substituting the value of v,

$\frac{\text K_2}{\text K_1} = \frac{(\text m_1 + \text m_2)(\text m_1\text u_1)^2}{(\text m_1 + \text m_2)^2\text m_1\text u_1^2} = \frac{\text m_1}{\text m_1 + \text m_2}$

Since m₁ is less than (m₁ + m₂), we see that K₂ < K₁. Thus the inelastic collision results in a loss of kinetic energy.

Properties of an inelastic collision :

(i) The total linear momentum remains conserved.

(ii) The angular momentum is conserved.

(iii) The total energy of the system is conserved.

(iv) The total kinetic energy is not conserved.

(v) Potential energy changes and other forms of energy, such as heat and sound, are produced.

It is not necessary that in an inelastic collision there is always a loss of kinetic energy. If, in a collision, the potential energy of a body is released, then the kinetic energy would increase.

Potential energy : The energy possessed by a body by virtue of its position or configuration is called potential energy. The potential energy of a body is measured by the amount of work which has been done in bringing the body from its zero-position to the present position, or by the amount of work which the body can do in going from its present position to the zero-position.

Potential energy of a block attached to a massless spring :

Suppose one end of a massless spring is fixed to a rigid wall at a point O and the other end is attached to a block placed on a smooth horizontal surface along the X-axis. Initially, when the spring is unstretched, the block is at a point A, where x = 0. This is the equilibrium position.

When the block is displaced from its equilibrium position A to a point B, where AB = x, then an elastic restoring force F is set up in the spring which is proportional to the displacement x and directed opposite to the displacement, that is,

$\text F \propto -\text x, \quad \text{or} \quad \text F = -\text{kx}$

where k is the force constant of the spring, or the spring constant. The negative sign signifies that the force F acts opposite to the displacement x.

Thus, to stretch the spring through a distance x, we must exert on it a force F′, equal and opposite to the restoring force F, that is,

$\text F' = -\text F = \text{kx}$

Now, suppose the block is displaced further from B through an infinitesimally small distance dx. The work done in stretching the spring through dx by the applied force F′ is

$\text{dW} = \text F'\text{dx} = \text{kx}\text{dx}$

the variable force F′ being taken constant over the infinitesimally small distance dx. Therefore, the total work done in stretching the spring through the distance x is obtained by integration,

$$\begin{aligned}\text W = \int_0^{\text x}\text{kx}\text{dx} = \text k\left[\frac{\text x^2}{2}\right]_0^{\text x} \\[1em] = \frac{1}{2}\text{kx}^2\end{aligned}$$

This work is stored in the spring in the form of elastic potential energy. Hence the potential energy of the spring is

$\text U = \frac{1}{2}\text{kx}^2$

The same expression is obtained when the spring is compressed through a distance x, since the expression involves x².

Mechanical energy : The mechanical energy of a body is the sum of its kinetic energy and potential energy. It is associated with the motion and the position of the body.

Principle of conservation of mechanical energy : For an isolated system or a body in the presence of conservative forces, the sum of the kinetic and the potential energy at any point remains constant throughout the motion. It does not depend upon time.

Conservation of mechanical energy for a freely falling body :

Suppose a body of mass m is held at a point A at a height h above the ground and is allowed to fall freely under gravity. Let g be the acceleration due to gravity.

At the point A : In the beginning the body is at rest at A, so that its kinetic energy is zero. It has only potential energy mgh. Thus the total energy of the body at A is

$\text E_A = \text{K.E.} + \text{P.E.} = 0 + \text{mgh} = \text{mgh} \qquad \ldots(\text i)$

At the point B : Now, suppose during the free fall, when the body is passing across a point B at a distance x vertically below A, its velocity is v. Then from the equation v² = u² + 2as, with u = 0 and s = x,

$\text v^2 = (0)^2 + 2\text{gx} = 2\text{gx}$

Therefore the kinetic energy of the body at B is

$\text K = \frac{1}{2}\text{mv}^2 = \frac{1}{2}\text m(2\text{gx}) = \text{mgx}$

The height of the body above the ground at B is (h − x), so its potential energy at B is mg (h − x). Thus the total energy at B is

$$\begin{aligned}\text E_B = \text{mgx} + \text{mg}(\text h - \text x) \\[1em] = \text{mgx} + \text{mgh} - \text{mgx} = \text{mgh} \qquad \ldots(\text{ii})\end{aligned}$$

At the point C (the ground) : Here the height is zero, so the potential energy is zero. The velocity on reaching the ground is given by v² = 2gh, so the kinetic energy is

$\text K = \frac{1}{2}\text m(2\text{gh}) = \text{mgh}$

Thus the total energy at C is

$\text E_C = \text{mgh} + 0 = \text{mgh} \qquad \ldots(\text{iii})$

From equations (i), (ii) and (iii),

$\text E_A = \text E_B = \text E_C = \text{mgh}$

Hence, the mechanical energy of a body falling freely under gravity remains constant at every point of its motion.

To prove : The negative potential energy gradient is equal to the force.

Consider a particle placed in the field of a conservative force F. Let an external force F_ext be applied on it, so that the particle experiences a net force

$\text F_{net} = \text F_{ext} + \text F \qquad \ldots(\text i)$

Work done by an external force can change either the kinetic energy or the potential energy of a body, or both. Thus, if the particle is displaced from a position A to a position B through a small displacement Δs,

$\text F_{ext}\Delta \text s = \Delta \text K + \Delta \text U \qquad \ldots(\text{ii})$

where ΔK is the change in the kinetic energy and ΔU is the change in the potential energy.

If the particle moves through the small displacement Δs, the work-kinetic energy theorem shows that

$\text F_{net}\Delta \text s = \Delta \text K \qquad \ldots(\text{iii})$

From equations (i) and (iii),

$$\begin{aligned}(\text F_{ext} + \text F)\Delta \text s = \Delta \text K \\[1em] \text F_{ext}\Delta \text s = \Delta \text K + (-\text F\Delta \text s) \qquad \ldots(\text{iv})\end{aligned}$$

Comparing equations (ii) and (iv), we get

$$\begin{aligned}\Delta \text K + \Delta \text U = \Delta \text K + (-\text F\Delta \text s) \\[1em] \Delta \text U = -\text F\Delta \text s\end{aligned}$$

In the limit Δs → 0, this gives

$\frac{\text{dU}}{\text{ds}} = -\text F \quad \text{or} \quad \text F = -\frac{\text{dU}}{\text{ds}}$

Hence, the force is equal to the negative potential energy gradient, that is, the negative of the rate of change of potential energy with distance. The negative sign shows that the conservative force always acts in the direction in which the potential energy decreases.

Note: The question writes the relation as $\text F = -\frac{\text{dv}}{\text{ds}}$; here the symbol in the numerator stands for the potential energy U, as used in the textbook.

To prove : In a one-dimensional elastic collision, the relative velocity of approach before impact is equal to the relative velocity of separation after impact.

Let us consider two elastic bodies A and B of respective masses m₁ and m₂ moving along the same straight line with velocities u₁ and u₂ respectively, where u₁ > u₂, so that they collide head-on. Let v₁ and v₂ be their velocities after the collision. Since the collision is head-on, all the velocities are along the same straight line.

By the law of conservation of linear momentum,

$$\begin{aligned}\text m_1\text u_1 + \text m_2\text u_2 = \text m_1\text v_1 + \text m_2\text v_2 \\[1em] \text m_1(\text u_1 - \text v_1) = \text m_2(\text v_2 - \text u_2) \qquad \ldots(\text i)\end{aligned}$$

Since the collision is elastic, the kinetic energy is also conserved,

$\frac{1}{2}\text m_1\text u_1^2 + \frac{1}{2}\text m_2\text u_2^2 = \frac{1}{2}\text m_1\text v_1^2 + \frac{1}{2}\text m_2\text v_2^2$

$\text m_1(\text u_1^2 - \text v_1^2) = \text m_2(\text v_2^2 - \text u_2^2) \qquad \ldots(\text{ii})$

Dividing equation (ii) by equation (i),

$\frac{\text u_1^2 - \text v_1^2}{\text u_1 - \text v_1} = \frac{\text v_2^2 - \text u_2^2}{\text v_2 - \text u_2}$

Factorising the numerators as the difference of two squares,

$\frac{(\text u_1 - \text v_1)(\text u_1 + \text v_1)}{\text u_1 - \text v_1} = \frac{(\text v_2 - \text u_2)(\text v_2 + \text u_2)}{\text v_2 - \text u_2}$

$\text u_1 + \text v_1 = \text v_2 + \text u_2$

Rearranging the terms,

$\text u_1 - \text u_2 = \text v_2 - \text v_1$

Here (u₁ − u₂) is the relative velocity with which the bodies A and B approach each other before the collision, and (v₂ − v₁) is the relative velocity with which they separate from each other after the collision.

Hence, the relative velocity with which two bodies approach each other before collision is equal to the relative velocity with which they depart from each other after the collision.

Let two bodies A and B of masses m₁ and m₂ move along the same straight line with velocities u₁ and u₂ respectively, where u₁ > u₂, and collide elastically. Let v₁ and v₂ be their velocities after the collision.

By the law of conservation of linear momentum,

$\text m_1(\text u_1 - \text v_1) = \text m_2(\text v_2 - \text u_2) \qquad \ldots(\text i)$

Since the collision is elastic, the kinetic energy is conserved,

$\text m_1(\text u_1^2 - \text v_1^2) = \text m_2(\text v_2^2 - \text u_2^2) \qquad \ldots(\text{ii})$

Dividing equation (ii) by equation (i),

$\text u_1 + \text v_1 = \text v_2 + \text u_2 \quad \Rightarrow \quad \text u_1 - \text u_2 = \text v_2 - \text v_1 \qquad \ldots(\text{iii})$

Velocity of A after the collision : From equation (iii), v₂ = u₁ − u₂ + v₁. Putting this value in equation (i),

$$\begin{aligned}\text m_1(\text u_1 - \text v_1) = \text m_2(\text u_1 + \text v_1 - 2\text u_2) \\[1em] -\text v_1(\text m_1 + \text m_2) = \text u_1(\text m_2 - \text m_1) - 2\text m_2\text u_2\end{aligned}$$

$\text v_1 = \left(\frac{\text m_1 - \text m_2}{\text m_1 + \text m_2}\right)\text u_1 + \left(\frac{2\text m_2}{\text m_1 + \text m_2}\right)\text u_2 \qquad \ldots(\text{iv})$

Velocity of B after the collision : Similarly, putting v₁ = v₂ + u₂ − u₁ from equation (iii) in equation (i),

$\text v_2 = \left(\frac{2\text m_1}{\text m_1 + \text m_2}\right)\text u_1 + \left(\frac{\text m_2 - \text m_1}{\text m_1 + \text m_2}\right)\text u_2 \qquad \ldots(\text v)$

Special cases :

(a) Both the colliding bodies have the same mass. Putting m₁ = m₂ = m in equations (iv) and (v),

$\text v_1 = \text u_2 \quad \text{and} \quad \text v_2 = \text u_1$

That is, in a one-dimensional elastic collision of two bodies of equal masses, the bodies merely exchange their velocities after the collision.

(b) One of the bodies is initially at rest. Putting u₂ = 0 in equations (iv) and (v),

$\text v_1 = \left(\frac{\text m_1 - \text m_2}{\text m_1 + \text m_2}\right)\text u_1 \quad \text{and} \quad \text v_2 = \left(\frac{2\text m_1}{\text m_1 + \text m_2}\right)\text u_1$

(c) A light body collides with a heavy body at rest. Here m₁ << m₂ and u₂ = 0, so m₁ may be neglected in comparison with m₂,

$\text v_1 = \left(\frac{-\text m_2}{\text m_2}\right)\text u_1 = -\text u_1 \quad \text{and} \quad \text v_2 = \left(\frac{2\text m_1}{\text m_2}\right)\text u_1 \approx 0$

That is, the light body rebounds with almost its original speed while the heavy body practically remains at rest.

(d) A heavy body collides with a light body at rest. Here m₁ >> m₂ and u₂ = 0, so m₂ may be neglected in comparison with m₁,

$\text v_1 = \left(\frac{\text m_1}{\text m_1}\right)\text u_1 = \text u_1 \quad \text{and} \quad \text v_2 = \left(\frac{2\text m_1}{\text m_1}\right)\text u_1 = 2\text u_1$

That is, the heavy body continues to move with almost the same velocity while the light body moves off with twice the velocity of the heavy body.

Given,

• Minimum speed at the top of the circle, $\text v = \sqrt{\text{gr}}$
• Radius of the circle = r, mass of the body = m

Speed when the string is horizontal : Let A be the highest point, C the point where the string is horizontal and B the lowest point. The point C is at a vertical depth r below the highest point A.

Applying the principle of conservation of mechanical energy between A and C, taking the level of C as the reference,

$$\begin{aligned}\frac{1}{2}\text{mv}_C^2 = \frac{1}{2}\text{mv}_A^2 + \text{mgr} \\[1em] \text v_C^2 = \text v_A^2 + 2\text{gr}\end{aligned}$$

Since the body just completes the circle, $\text v_A = \sqrt{\text{gr}}$, that is, v_A² = gr. Therefore,

$$\begin{aligned}\text v_C^2 = \text{gr} + 2\text{gr} = 3\text{gr} \\[1em] \text v_C = \sqrt{3\text{gr}}\end{aligned}$$

Tension when the string is horizontal : At the point C the string is horizontal, so the tension T_C acts horizontally towards the centre while the weight mg acts vertically downwards, that is, perpendicular to the string. Hence the weight has no component along the string, and the tension alone provides the centripetal force,

$$\begin{aligned}\text T_C = \frac{\text{mv}_C^2}{\text r} = \frac{\text m(3\text{gr})}{\text r} \\[1em] = 3\text{mg}\end{aligned}$$

Hence, when the string becomes horizontal the speed of the body is $\sqrt{3\text{gr}}$ and the tension in the string is 3 mg.

Given,

• Mass of the man, m₁ = 55 kg
• Mass of the weight, m₂ = 20 kg
• Distance moved, s = 20 m
• g = 10 m s^-2

The total mass carried up is

$\text m = \text m_1 + \text m_2 = 55 + 20 = 75\ \text{kg}$

(i) On a horizontal road : The force of gravity acts vertically downwards while the displacement is horizontal, so the angle between them is 90°,

$\text W = \text{mg}\text s\cos 90^\circ = 0$

Hence, the work done against gravity on the horizontal road is zero.

(ii) On an incline of 1 in 5 : An incline of 1 in 5 means that the body rises 1 m for every 5 m travelled along the incline. Hence the vertical height gained is

$\text h = \frac{\text s}{5} = \frac{20}{5} = 4\ \text m$

The work done against gravity is

$$\begin{aligned}\text W = \text{mgh} = 75 \times 10 \times 4 \\[1em] = 3000\ \text J\end{aligned}$$

Hence, the work done against gravity on the incline is 3000 J.

Given,

• Mass of the weight, m = 25 kg
• Vertical height, h = 2.0 m
• Angle of the inclined plane, θ = 30°
• g = 9.8 m s^-2

(i) Lifted vertically : The work done against gravity is

$$\begin{aligned}\text W = \text{mgh} = 25 \times 9.8 \times 2.0 \\[1em] = 490\ \text J\end{aligned}$$

(ii) Pushed up the inclined plane : The length of the inclined plane needed to reach the same height is

$\text l = \frac{\text h}{\sin \theta} = \frac{2.0}{\sin 30^\circ} = \frac{2.0}{0.5} = 4\ \text m$

The force required to push the weight up the smooth incline is the component of the weight along the incline,

$\text F = \text{mg}\sin \theta = 25 \times 9.8 \times 0.5 = 122.5\ \text N$

Therefore the work done is

$$\begin{aligned}\text W = \text F \times \text l = 122.5 \times 4 \\[1em] = 490\ \text J\end{aligned}$$

Hence, the work done is 490 J in both the cases.

The work done against gravity depends only on the vertical height through which the body is raised and not on the path followed, since the gravitational force is a conservative force.

Given,

• Force applied, F = 20 kg f
• Angle with the ground, θ = 60°
• Distance moved, s = 10 m
• Time taken, t = 1 min = 60 s
• g = 10 m s^-2

Converting the force into the absolute unit,

$\text F = 20\ \text{kg f} = 20 \times 10 = 200\ \text N$

The work done by the force is

$$\begin{aligned}\text W = \text F\text s\cos \theta = 200 \times 10 \times \cos 60^\circ \\[1em] = 200 \times 10 \times 0.5 \\[1em] = 1000\ \text J\end{aligned}$$

The power dissipated is

$$\begin{aligned}\text P = \frac{\text W}{\text t} = \frac{1000}{60} \\[1em] = 16.7\ \text W\end{aligned}$$

Hence, the power dissipated is 16.7 W.

Given,

• Volume of blood forced per minute, V = 4 × 10^-3 m³
• Pressure head, h = 0.13 m
• Density of blood, ρ = 1.03 × 10³ kg m^-3
• g = 9.8 m s^-2, 1 H.P. = 746 W

The pressure exerted by the heart is

$$\begin{aligned}\text P = \text h\rho\text g = 0.13 \times (1.03 \times 10^3) \times 9.8 \\[1em] = 1312.2\ \text{N m}^{-2}\end{aligned}$$

The work done in forcing a volume V of blood against this pressure is

$$\begin{aligned}\text W = \text{Pressure} \times \text{Volume} = 1312.2 \times (4 \times 10^{-3}) \\[1em] = 5.25\ \text J\end{aligned}$$

This work is done in one minute, that is, in 60 s. Hence the power of the heart is

$$\begin{aligned}\text{Power} = \frac{\text W}{\text t} = \frac{5.25}{60} \\[1em] = 0.0875\ \text W\end{aligned}$$

Converting into horsepower,

$$\begin{aligned}\text{H.P.} = \frac{0.0875}{746} \\[1em] = 1.17 \times 10^{-4}\ \text{H.P.}\end{aligned}$$

Hence, the horsepower of the heart is 1.17 × 10^-4 H.P.

Given,

• Mass of the body, m = 2.0 kg, initially at rest
• First force, F₁ = 30 N acting over s₁ = 3.0 m
• Second force, F₂ = 15 N acting over s₂ = 2.0 m

Both the forces act in the same direction as the displacement, so the work done by each is positive.

The work done by the first force is

$\text W_1 = \text F_1\text s_1 = 30 \times 3.0 = 90\ \text J$

The work done by the second force is

$\text W_2 = \text F_2\text s_2 = 15 \times 2.0 = 30\ \text J$

The total work done on the body is

$\text W = \text W_1 + \text W_2 = 90 + 30 = 120\ \text J$

By the work-kinetic energy theorem, the work done by the net force is equal to the change in the kinetic energy of the body. Since the body starts from rest, its initial kinetic energy is zero, so

$\text K_f - 0 = 120\ \text J$

Hence, the final kinetic energy of the body is 120 J.

Given,

• Mass of the ball, m = 0.2 kg
• Length of the thread, l = 1 m
• Angle with the vertical, θ = 30°
• g = 10 m s^-2

The height through which the ball is raised is

$$\begin{aligned}\text h = \text l(1 - \cos \theta) = 1 \times (1 - \cos 30^\circ) \\[1em] = 1 \times (1 - 0.866) \\[1em] = 0.134\ \text m\end{aligned}$$

The work done against gravity is equal to the gain in the potential energy of the ball,

$$\begin{aligned}\text W = \text{mgh} = 0.2 \times 10 \times 0.134 \\[1em] = 0.268\ \text J\end{aligned}$$

Hence, the work done against gravity is 0.268 J.

When the ball is released, this potential energy is converted into kinetic energy at the lowest point. By the conservation of mechanical energy,

$\frac{1}{2}\text{mv}^2 = \text{mgh} \quad \Rightarrow \quad \text v = \sqrt{2\text{gh}}$

Substituting the values,

$$\begin{aligned}\text v = \sqrt{2 \times 10 \times 0.134} = \sqrt{2.68} \\[1em] = 1.637\ \text{m s}^{-1}\end{aligned}$$

Hence, the velocity of the ball at the lowest point is 1.637 m s^-1.

Given,

• Mass of the bullet, m = 20 g = 0.02 kg
• Speed of the bullet, u = 150 m s^-1
• Distance pierced into the block, s = 10 cm = 0.1 m
• Final velocity, v = 0

The kinetic energy of the bullet before striking the block is

$$\begin{aligned}\text K = \frac{1}{2}\text{mu}^2 = \frac{1}{2} \times 0.02 \times (150)^2 \\[1em] = \frac{1}{2} \times 0.02 \times 22500 \\[1em] = 225\ \text J\end{aligned}$$

The bullet comes to rest inside the block, so the whole of this kinetic energy is used up in doing work against the force of resistance offered by the block. By the work-kinetic energy theorem,

$\text F \times \text s = \text K$

Substituting the values,

$$\begin{aligned}\text F \times 0.1 = 225 \\[1em] \text F = \frac{225}{0.1} = 2250\ \text N\end{aligned}$$

Hence, the average force of resistance offered by the block is 2250 N.

Given,

• Mass of each block, m = 5 kg
• Speed of each block, u = 2 m s^-1, moving towards each other
• The blocks stick together and come to rest
• The horizontal surface is frictionless

The total kinetic energy of the system before the collision is

$$\begin{aligned}\text K_i = \frac{1}{2}\text{mu}^2 + \frac{1}{2}\text{mu}^2 = 2 \times \frac{1}{2} \times 5 \times (2)^2 \\[1em] = 20\ \text J\end{aligned}$$

Since the two blocks stick together and come to rest, the final kinetic energy of the system is

$\text K_f = 0$

(i) Work done by external forces : The surface is frictionless and horizontal, so the weight of the blocks and the normal reaction of the surface are both perpendicular to the displacement and do no work. No external horizontal force acts on the system.

$\text W_{ext} = 0$

(ii) Work done by internal forces : By the work-kinetic energy theorem applied to the system, the total work done is equal to the change in its kinetic energy,

$$\begin{aligned}\text W_{ext} + \text W_{int} = \text K_f - \text K_i \\[1em] 0 + \text W_{int} = 0 - 20\end{aligned}$$

$\text W_{int} = -20\ \text J$

Hence, the work done by the external forces is zero and that done by the internal forces is − 20 J, that is, 20 J of kinetic energy is lost in the collision and appears as heat and the energy of deformation.

Note: The printed answer gives (i) 20 J and (ii) zero. Since no external horizontal force acts on the system on a frictionless surface, the whole of the change in kinetic energy must be accounted for by the internal forces of the collision, as worked out above.

Given,

• Height from which the ball is dropped, h = 60 m
• Energy lost on striking the ground = 25% of its energy

The ball is dropped from rest, so on reaching the ground the whole of its potential energy mgh has been converted into kinetic energy. Its energy on striking the ground is therefore

$\text E = \text{mgh} = \text m\text g \times 60$

On striking the ground the ball loses 25% of its energy, so the energy retained by it is

$\text E' = \frac{75}{100} \times \text E = 0.75 \times \text m\text g \times 60$

If the ball rebounds to a height h′, this retained energy is again converted into potential energy,

$\text{mgh}' = 0.75 \times \text{mg} \times 60$

The mass m and g cancel from both sides, giving

$$\begin{aligned}\text h' = 0.75 \times 60 \\[1em] = 45\ \text m\end{aligned}$$

Hence, the ball rebounds to a height of 45 m.

Given,

• Mass of the body, m = 10 kg
• Height from which it is dropped, h = 20 m
• g = 9.8 m s^-2

Potential energy before dropping : Taking the ground as the reference level,

$$\begin{aligned}\text U = \text{mgh} = 10 \times 9.8 \times 20 \\[1em] = 1960\ \text J\end{aligned}$$

Kinetic energy at a height of 8.0 m : The body has fallen through a distance of

$20 - 8.0 = 12\ \text m$

The loss in potential energy over this fall appears as kinetic energy,

$$\begin{aligned}\text K = \text{mg} \times 12 = 10 \times 9.8 \times 12 \\[1em] = 1176\ \text J\end{aligned}$$

Kinetic energy on hitting the ground : At the ground the height is zero, so the potential energy is zero and the whole of the initial potential energy has been converted into kinetic energy,

$$\begin{aligned}\text K = \text{mgh} = 10 \times 9.8 \times 20 \\[1em] = 1960\ \text J\end{aligned}$$

Hence, the potential energy before dropping is 1960 J, the kinetic energy at a height of 8.0 m is 1176 J and the kinetic energy on hitting the ground is 1960 J.

Given,

• Mass of the block, m = 2 kg
• Height of fall, h = 40 cm = 0.4 m
• Force constant of the spring, k = 1960 N/m
• g = 9.8 m s^-2

Let x be the maximum compression of the spring. The block falls through the height h and then compresses the spring through a further distance x, so the total fall of the block is (h + x).

At the position of maximum compression the block is momentarily at rest, so its kinetic energy is zero. Hence, by the conservation of energy, the whole of the loss in the gravitational potential energy of the block is stored as the elastic potential energy of the spring,

$\text{mg}(\text h + \text x) = \frac{1}{2}\text{kx}^2$

Substituting the values,

$$\begin{aligned}2 \times 9.8 \times (0.4 + \text x) = \frac{1}{2} \times 1960 \times \text x^2 \\[1em] 19.6(0.4 + \text x) = 980\text x^2\end{aligned}$$

$$\begin{aligned}7.84 + 19.6\text x = 980\text x^2 \\[1em] 980\text x^2 - 19.6\text x - 7.84 = 0\end{aligned}$$

Dividing throughout by 19.6,

$50\text x^2 - \text x - 0.4 = 0$

Solving this quadratic equation,

$$\begin{aligned}\text x = \frac{1 \pm \sqrt{(-1)^2 - 4(50)(-0.4)}}{2 \times 50} = \frac{1 \pm \sqrt{1 + 80}}{100} \\[1em] = \frac{1 \pm 9}{100}\end{aligned}$$

Rejecting the negative root, since the compression cannot be negative,

$\text x = \frac{1 + 9}{100} = 0.1\ \text m = 10\ \text{cm}$

Hence, the maximum compression of the spring is 10 cm.

Given,

• Height to which the ball rises on either side, h = 10 cm = 0.1 m
• g = 9.8 m/s²

Let m be the mass of the ball and v its speed at the lowest point A. The cup is smooth, so no energy is lost against friction.

At the highest point of its swing the ball is momentarily at rest, so it possesses only potential energy mgh. At the lowest point A the height is zero, so it possesses only kinetic energy $\frac{1}{2}\text{mv}^2$.

By the principle of conservation of mechanical energy,

$\frac{1}{2}\text{mv}^2 = \text{mgh} \quad \Rightarrow \quad \text v = \sqrt{2\text{gh}}$

Substituting the values,

$$\begin{aligned}\text v = \sqrt{2 \times 9.8 \times 0.1} = \sqrt{1.96} \\[1em] = 1.4\ \text{m/s}\end{aligned}$$

Hence, the speed of the ball at the lowest point A is 1.4 m/s.

Given,

• Vertical height of the curved track, h = 40 cm = 0.4 m
• Angle of the inclined plane, θ = 30°
• g = 9.8 m s^-2

Speed at the bottom of the curved track : The track is smooth, so the whole of the loss in potential energy appears as kinetic energy. By the conservation of mechanical energy,

$\frac{1}{2}\text{mv}^2 = \text{mgh} \quad \Rightarrow \quad \text v = \sqrt{2\text{gh}}$

Substituting the values,

$$\begin{aligned}\text v = \sqrt{2 \times 9.8 \times 0.4} = \sqrt{7.84} \\[1em] = 2.8\ \text{m/s}\end{aligned}$$

Distance covered along the inclined plane : The inclined plane is also smooth, so the body rises along it until the whole of its kinetic energy is again converted into potential energy. It therefore rises to the same vertical height h = 0.4 m.

If d is the distance travelled along the incline,

$\text h = \text d\sin \theta \quad \Rightarrow \quad \text d = \frac{\text h}{\sin \theta}$

Substituting the values,

$$\begin{aligned}\text d = \frac{0.4}{\sin 30^\circ} = \frac{0.4}{0.5} \\[1em] = 0.8\ \text m = 80\ \text{cm}\end{aligned}$$

Hence, the speed of the body at the bottom of the curved track is 2.8 m/s and it goes 80 cm along the inclined plane.

Given,

• Speed of the automobile, v = 72 km h^-1
• Angle of the incline, θ = 30°
• g = 9.8 m s^-2

Converting the speed into SI units,

$\text v = 72\ \text{km h}^{-1} = \frac{72 \times 1000}{60 \times 60} = 20\ \text{m s}^{-1}$

The incline is smooth and the engine is switched off, so the whole of the kinetic energy of the automobile is converted into potential energy as it goes up the incline and comes to rest. If h is the vertical height risen,

$\frac{1}{2}\text{mv}^2 = \text{mgh} \quad \Rightarrow \quad \text h = \frac{\text v^2}{2\text g}$

Substituting the values,

$$\begin{aligned}\text h = \frac{(20)^2}{2 \times 9.8} = \frac{400}{19.6} \\[1em] = 20.41\ \text m\end{aligned}$$

If d is the distance travelled along the incline, then h = d sin θ, so

$$\begin{aligned}\text d = \frac{\text h}{\sin \theta} = \frac{20.41}{\sin 30^\circ} = \frac{20.41}{0.5} \\[1em] = 40.8\ \text m\end{aligned}$$

Hence, the automobile goes 40.8 m up the incline before stopping.

Given,

• Length of the pendulum, l = 1 m
• Mass of the bob, m = 0.1 kg
• The bob is released from the horizontal position
• g = 9.8 m s^-2

Since the bob is released from the horizontal position, its initial height above the lowest point is equal to the length of the pendulum, that is, 1 m. Its initial potential energy is therefore

$\text U = \text{mgl} = 0.1 \times 9.8 \times 1 = 0.98\ \text J$

When the thread makes an angle θ with the vertical, the height of the bob above the lowest point is l (1 − cos θ), so its potential energy at that position is mgl (1 − cos θ). By the conservation of mechanical energy, its kinetic energy is

$\text K = \text{mgl} - \text{mgl}(1 - \cos \theta) = \text{mgl}\cos \theta$

(i) When θ = 0° :

$$\begin{aligned}\text K = \text{mgl}\cos 0^\circ = 0.98 \times 1 \\[1em] = 0.98\ \text J\end{aligned}$$

(ii) When θ = 30° :

$$\begin{aligned}\text K = \text{mgl}\cos 30^\circ = 0.98 \times \frac{\sqrt{3}}{2} \\[1em] = 0.49\sqrt{3}\ \text J = 0.849\ \text J\end{aligned}$$

Hence, the kinetic energy of the bob is 0.98 J at 0° and $0.49\sqrt{3}$ J, that is, about 0.849 J, at 30° with the vertical.

Given,

• Mass of the block, M = 5 kg
• Mass of the bullet, m = 20 g = 0.02 kg
• Height to which the block is raised, h = 15 cm = 0.15 m
• g = 9.8 m s^-2

Step 1 : Common velocity just after the impact. The block with the bullet embedded in it rises through the height h, so by the conservation of mechanical energy,

$$\begin{aligned}\frac{1}{2}(\text M + \text m)\text V^2 = (\text M + \text m)\text{gh} \\[1em] \text V = \sqrt{2\text{gh}}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text V = \sqrt{2 \times 9.8 \times 0.15} = \sqrt{2.94} \\[1em] = 1.715\ \text{m s}^{-1}\end{aligned}$$

Step 2 : Initial velocity of the bullet. The bullet is embedded in the block, so the collision is perfectly inelastic. By the conservation of linear momentum,

$\text{mu} = (\text M + \text m)\text V$

Substituting the values,

$$\begin{aligned}0.02 \times \text u = (5 + 0.02) \times 1.715 \\[1em] 0.02\text u = 5.02 \times 1.715 = 8.61\end{aligned}$$

$\text u = \frac{8.61}{0.02} = 430.5\ \text{m s}^{-1}$

Hence, the initial velocity of the bullet is about 430 m/s.

Given,

• Mass of the bullet, m = 0.02 kg
• Speed of the bullet, u = 200 m s^-1
• Mass of the block, M = 2 kg
• Length of the thread, l = 1 m
• g = 9.8 m s^-2

Step 1 : Common velocity after the impact. The bullet is embedded in the block, so the collision is perfectly inelastic. By the conservation of linear momentum,

$$\begin{aligned}\text{mu} = (\text M + \text m)\text V \\[1em] \text V = \frac{0.02 \times 200}{2 + 0.02} = \frac{4}{2.02}\end{aligned}$$

$\text V = 1.98\ \text{m s}^{-1}$

Step 2 : Height to which the block rises. By the conservation of mechanical energy,

$$\begin{aligned}\text h = \frac{\text V^2}{2\text g} = \frac{(1.98)^2}{2 \times 9.8} = \frac{3.92}{19.6} \\[1em] = 0.2\ \text m\end{aligned}$$

Step 3 : Maximum inclination of the thread. If θ is the maximum angle made by the thread with the vertical, then

$\text h = \text l(1 - \cos \theta) \quad \Rightarrow \quad \cos \theta = 1 - \frac{\text h}{\text l}$

Substituting the values,

$$\begin{aligned}\cos \theta = 1 - \frac{0.2}{1} = 0.8 \\[1em] \theta = \cos^{-1}(0.8) = 37^\circ\end{aligned}$$

Hence, the maximum inclination of the thread with the vertical is 37°.

Given,

• Mass of the body, m = 1 kg
• Time of fall, t = 5 s
• Initial velocity, u = 0 (falls freely)
• g = 10 m s^-2

The velocity of the body after 5 s of free fall is

$$\begin{aligned}\text v = \text u + \text{gt} = 0 + 10 \times 5 \\[1em] = 50\ \text{m s}^{-1}\end{aligned}$$

Momentum :

$$\begin{aligned}\text p = \text{mv} = 1 \times 50 \\[1em] = 50\ \text{kg m s}^{-1}\end{aligned}$$

Kinetic energy :

$$\begin{aligned}\text K = \frac{1}{2}\text{mv}^2 = \frac{1}{2} \times 1 \times (50)^2 \\[1em] = \frac{1}{2} \times 2500 \\[1em] = 1250\ \text J\end{aligned}$$

Hence, the momentum of the body is 50 kg m s^-1 and its kinetic energy is 1250 J.

Given,

• Mass of the first body, m₁ = 1 kg
• Mass of the second body, m₂ = 4 kg
• The two bodies have equal linear momenta, p₁ = p₂ = p

The kinetic energy of a body in terms of its momentum is

$\text K = \frac{\text p^2}{2\text m}$

For a given momentum, the kinetic energy varies inversely as the mass, that is, $\text K \propto \frac{1}{\text m}$. Therefore

$\frac{\text K_1}{\text K_2} = \frac{\text m_2}{\text m_1} = \frac{4}{1}$

Hence, the ratio of the kinetic energies is K₁ : K₂ = 4 : 1.

Given,

• Mass of the first body, m₁ = 1 kg
• Mass of the second body, m₂ = 4 kg
• The two bodies have equal kinetic energies, K₁ = K₂ = K

The momentum of a body in terms of its kinetic energy is

$\text p = \sqrt{2\text{mK}}$

For a given kinetic energy, the momentum varies as the square root of the mass, that is, $\text p \propto \sqrt{\text m}$. Therefore

$\frac{\text p_1}{\text p_2} = \sqrt{\frac{\text m_1}{\text m_2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

Hence, the ratio of the linear momenta is p₁ : p₂ = 1 : 2.

Given,

• Mass of the neutron, m₁ = 1.67 × 10^-27 kg
• Speed of the neutron, u₁ = 1.0 × 10⁸ m s^-1
• Mass of the deuteron, m₂ = 3.34 × 10^-27 kg, initially at rest

The neutron sticks to the deuteron, so the collision is perfectly inelastic and the two move together as a single particle, the triton. By the conservation of linear momentum,

$\text m_1\text u_1 = (\text m_1 + \text m_2)\text v$

Substituting the values,

$(1.67 \times 10^{-27}) \times (1.0 \times 10^8) = (1.67 \times 10^{-27} + 3.34 \times 10^{-27})\text v$

$1.67 \times 10^{-19} = (5.01 \times 10^{-27})\text v$

$$\begin{aligned}\text v = \frac{1.67 \times 10^{-19}}{5.01 \times 10^{-27}} \\[1em] = 3.33 \times 10^{7}\ \text{m s}^{-1}\end{aligned}$$

Hence, the speed of the composite particle triton is 3.33 × 10⁷ m s^-1.

Given,

• Mass of the ball A, m₁ = 2.4 kg
• The ball B is initially at rest, u₂ = 0
• After the collision, $\text v_1 = \frac{\text u_1}{5}$

For a one-dimensional elastic collision with the target body initially at rest, the velocity of the striking body after the collision is

$\text v_1 = \left(\frac{\text m_1 - \text m_2}{\text m_1 + \text m_2}\right)\text u_1$

Since the ball A continues in the same direction with one-fifth of its original speed,

$\frac{\text u_1}{5} = \left(\frac{\text m_1 - \text m_2}{\text m_1 + \text m_2}\right)\text u_1$

Cancelling u₁ from both sides,

$$\begin{aligned}\frac{1}{5} = \frac{\text m_1 - \text m_2}{\text m_1 + \text m_2} \\[1em] \text m_1 + \text m_2 = 5\text m_1 - 5\text m_2\end{aligned}$$

$6\text m_2 = 4\text m_1 \quad \Rightarrow \quad \text m_2 = \frac{2}{3}\text m_1$

Substituting m₁ = 2.4 kg,

$$\begin{aligned}\text m_2 = \frac{2}{3} \times 2.4 \\[1em] = 1.6\ \text{kg}\end{aligned}$$

Hence, the mass of the ball B is 1.6 kg.

Given,

• Mass of the first body, m₁ = 30 kg with u₁ = 18.0 m s^-1
• Mass of the second body, m₂ = 90 kg with u₂ = − 14.4 m s^-1 (opposite direction)

For a one-dimensional elastic collision, the velocities after the collision are

$\text v_1 = \left(\frac{\text m_1 - \text m_2}{\text m_1 + \text m_2}\right)\text u_1 + \left(\frac{2\text m_2}{\text m_1 + \text m_2}\right)\text u_2$

$\text v_2 = \left(\frac{2\text m_1}{\text m_1 + \text m_2}\right)\text u_1 + \left(\frac{\text m_2 - \text m_1}{\text m_1 + \text m_2}\right)\text u_2$

Here m₁ + m₂ = 30 + 90 = 120 kg.

Velocity of the 30 kg mass :

$$\begin{aligned}\text v_1 = \left(\frac{30 - 90}{120}\right)(18.0) + \left(\frac{2 \times 90}{120}\right)(-14.4) \\[1em] = \left(\frac{-60}{120}\right)(18.0) + \left(\frac{180}{120}\right)(-14.4)\end{aligned}$$

$$\begin{aligned}= (-0.5)(18.0) + (1.5)(-14.4) \\[1em] = -9 - 21.6 = -30.6\ \text{m s}^{-1}\end{aligned}$$

Velocity of the 90 kg mass :

$$\begin{aligned}\text v_2 = \left(\frac{2 \times 30}{120}\right)(18.0) + \left(\frac{90 - 30}{120}\right)(-14.4) \\[1em] = (0.5)(18.0) + (0.5)(-14.4)\end{aligned}$$

$= 9 - 7.2 = 1.8\ \text{m s}^{-1}$

Hence, after the collision the 30 kg mass moves with a velocity of 30.6 m s^-1 in the reversed direction and the 90 kg mass moves with a velocity of 1.8 m s^-1 in the original direction of the 30 kg mass.

Given,

• Mass of the ball A, m₁ = 10 kg with u₁ = 8.0 m s^-1
• Mass of the ball B, m₂ = 20 kg, initially at rest
• Angle made by A after the collision, θ₁ = 30°
• Angle made by B after the collision, θ₂ = 45°

This is an oblique collision in two dimensions. Let v₁ and v₂ be the final speeds of A and B. Applying the law of conservation of linear momentum separately along the two directions :

Along the initial direction of motion (X-axis) :

$\text m_1\text u_1 = \text m_1\text v_1\cos \theta_1 + \text m_2\text v_2\cos \theta_2$

$$\begin{aligned}10 \times 8.0 = 10\text v_1\cos 30^\circ + 20\text v_2\cos 45^\circ \\[1em] 80 = 8.66\text v_1 + 14.14\text v_2 \qquad \ldots(\text i)\end{aligned}$$

Perpendicular to the initial direction (Y-axis) : There is no momentum along this direction before the collision, so

$0 = \text m_1\text v_1\sin \theta_1 - \text m_2\text v_2\sin \theta_2$

$$\begin{aligned}10\text v_1\sin 30^\circ = 20\text v_2\sin 45^\circ \\[1em] 5\text v_1 = 14.14\text v_2\end{aligned}$$

$\text v_1 = 2.828\text v_2 \qquad \ldots(\text{ii})$

Solving : Substituting equation (ii) in equation (i),

$$\begin{aligned}80 = 8.66(2.828\text v_2) + 14.14\text v_2 \\[1em] 80 = 24.49\text v_2 + 14.14\text v_2\end{aligned}$$

$80 = 38.63\text v_2 \quad \Rightarrow \quad \text v_2 = 2.07\ \text{m s}^{-1}$

From equation (ii),

$\text v_1 = 2.828 \times 2.07 = 5.85\ \text{m s}^{-1}$

Hence, the final speeds are 5.85 m s^-1 for the ball A and 2.07 m s^-1 for the ball B.

Given,

• Mass of water, m = 5 kg
• Radius of the vertical circular path, r = 1.8 m
• g = 10 m/s²

Minimum speed at the top : The water does not fall out of the bucket provided the centrifugal force on it at the highest point is at least equal to its weight, that is,

$\frac{\text{mv}_A^2}{\text r} \ge \text{mg} \quad \Rightarrow \quad \text v_A \ge \sqrt{\text{gr}}$

Therefore the minimum speed at the top is

$$\begin{aligned}\text v_A = \sqrt{\text{gr}} = \sqrt{10 \times 1.8} = \sqrt{18} \\[1em] = 4.2\ \text{m s}^{-1}\end{aligned}$$

Normal contact force at the lowest point : Applying the conservation of mechanical energy between the highest point A and the lowest point B, which are separated by a vertical height 2r,

$$\begin{aligned}\text v_B^2 = \text v_A^2 + 4\text{gr} = 18 + (4 \times 10 \times 1.8) \\[1em] = 18 + 72 = 90\ \text m^2\text s^{-2}\end{aligned}$$

At the lowest point the normal contact force N exerted by the bucket acts upwards and the weight mg acts downwards, their difference providing the centripetal force,

$$\begin{aligned}\text N - \text{mg} = \frac{\text{mv}_B^2}{\text r} \\[1em] \text N = \frac{\text{mv}_B^2}{\text r} + \text{mg}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text N = \frac{5 \times 90}{1.8} + (5 \times 10) \\[1em] = 250 + 50 = 300\ \text N\end{aligned}$$

Hence, the minimum speed at the top of the path is 4.2 m s^-1 and the normal contact force at the lowest point is 300 N.

Note: The printed answer for the contact force is 294 N, which is obtained by taking g = 9.8 m s^-2. The value given in the question, g = 10 m s^-2, has been used here.

Given,

• Mass of the ball, m = 1.0 kg
• Length of the string (radius), l = 5.0 m
• g = 9.8 m/s²

(i) Speed at the highest point for zero tension : At the highest point of the vertical circle, both the tension T and the weight mg act towards the centre,

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

Putting T = 0, the weight alone provides the centripetal force,

$\text{mg} = \frac{\text{mv}_A^2}{\text l} \quad \Rightarrow \quad \text v_A = \sqrt{\text{gl}}$

Substituting the values,

$$\begin{aligned}\text v_A = \sqrt{9.8 \times 5.0} = \sqrt{49} \\[1em] = 7\ \text{m/s}\end{aligned}$$

(ii) Speed at the lowest point : Applying the conservation of mechanical energy between the highest point A and the lowest point B, separated by a vertical height 2 l,

$$\begin{aligned}\frac{1}{2}\text{mv}_B^2 = \frac{1}{2}\text{mv}_A^2 + \text{mg}(2\text l) \\[1em] \text v_B^2 = \text v_A^2 + 4\text{gl} = \text{gl} + 4\text{gl} = 5\text{gl}\end{aligned}$$

$$\begin{aligned}\text v_B = \sqrt{5\text{gl}} = \sqrt{5 \times 9.8 \times 5.0} = \sqrt{245} \\[1em] = 7\sqrt{5}\ \text{m/s} = 15.65\ \text{m/s}\end{aligned}$$

Hence, the tension is zero when the speed at the highest point is 7 m/s, and the corresponding speed at the lowest point is $7\sqrt{5}$ m/s.

Given,

• Mass of the weight, m = 10 kg
• Height through which it is raised, h = 2.0 m
• Acceleration, a = 1.5 m s^-2 (upwards)
• g = 10 m s^-2

Since the weight is raised with an upward acceleration a, the applied force must not only balance the weight but also produce the acceleration,

$\text F = \text{mg} + \text{ma} = \text m(\text g + \text a)$

Substituting the values,

$$\begin{aligned}\text F = 10 \times (10 + 1.5) = 10 \times 11.5 \\[1em] = 115\ \text N\end{aligned}$$

The force and the displacement are both in the upward direction, so the work done is

$$\begin{aligned}\text W = \text F \times \text h = 115 \times 2.0 \\[1em] = 230\ \text J\end{aligned}$$

Hence, the work done is 230 J.

Given,

• Load pulled, m = 500 metric ton = 500 × 10³ = 5 × 10⁵ kg
• Gradient of the inclined plane = 1 in 100
• Speed, v = 10 m s^-1
• Frictional force, f = 2000 N
• g = 9.8 m s^-2

A rise of 1 in 100 means that for every 100 m travelled along the plane the load rises 1 m, so

$\sin \theta = \frac{1}{100}$

The component of the weight acting down the incline is

$$\begin{aligned}\text{mg}\sin \theta = (5 \times 10^5) \times 9.8 \times \frac{1}{100} \\[1em] = 49000\ \text N\end{aligned}$$

The engine must overcome this component of the weight as well as the frictional force, so the total force exerted by the engine is

$$\begin{aligned}\text F = \text{mg}\sin \theta + \text f = 49000 + 2000 \\[1em] = 51000\ \text N\end{aligned}$$

The power of the engine is

$$\begin{aligned}\text P = \text F \times \text v = 51000 \times 10 \\[1em] = 5.1 \times 10^5\ \text W = 510\ \text{kW}\end{aligned}$$

Hence, the power of the engine is 510 kW.

Given,

• Mass of the car, m = 1500 kg
• Constant speed, v = 5.0 m s^-1
• Frictional force, f = 500 N
• Gradient of the inclined plane = 1 in 10
• g = 9.8 m s^-2

Power on the level road : On a level road the engine has only to overcome the frictional force, since the car moves with a constant speed. Therefore,

$$\begin{aligned}\text P = \text f \times \text v = 500 \times 5.0 \\[1em] = 2500\ \text W = 2.5\ \text{kW}\end{aligned}$$

Extra power on the inclined plane : A gradient of 1 in 10 means

$\sin \theta = \frac{1}{10}$

To maintain the same speed up the incline, the engine must exert an additional force equal to the component of the weight along the incline,

$$\begin{aligned}\text F' = \text{mg}\sin \theta = 1500 \times 9.8 \times \frac{1}{10} \\[1em] = 1470\ \text N\end{aligned}$$

The extra power required is therefore

$$\begin{aligned}\text P' = \text F' \times \text v = 1470 \times 5.0 \\[1em] = 7350\ \text W = 7.35\ \text{kW}\end{aligned}$$

Hence, the power expended on the level road is 2.5 kW and the extra power required on the inclined plane is 7.35 kW.

Given,

• Mass of the bullet, m = 20 g = 0.02 kg
• Mass of the block, M = 980 g = 0.98 kg
• Force constant of the spring, k = 100 N/m
• Compression of the spring, x = 10 cm = 0.1 m

(a) Velocity of the block after the collision : After the collision the block, with the bullet embedded in it, compresses the spring and comes momentarily to rest. Hence the whole of its kinetic energy is stored as the elastic potential energy of the spring,

$\frac{1}{2}(\text M + \text m)\text V^2 = \frac{1}{2}\text{kx}^2$

Here M + m = 0.98 + 0.02 = 1 kg. Substituting the values,

$$\begin{aligned}\frac{1}{2} \times 1 \times \text V^2 = \frac{1}{2} \times 100 \times (0.1)^2 \\[1em] \text V^2 = 100 \times 0.01 = 1\end{aligned}$$

$\text V = 1.0\ \text{m/s}$

(b) Velocity of the bullet : The bullet is embedded in the block, so the collision is perfectly inelastic. By the conservation of linear momentum,

$$\begin{aligned}\text m\text v = (\text M + \text m)\text V \\[1em] 0.02 \times \text v = 1 \times 1.0\end{aligned}$$

$\text v = \frac{1.0}{0.02} = 50\ \text{m/s}$

(c) Loss in the kinetic energy : The kinetic energy of the bullet before the collision is

$$\begin{aligned}\text K_i = \frac{1}{2}\text m\text v^2 = \frac{1}{2} \times 0.02 \times (50)^2 \\[1em] = \frac{1}{2} \times 0.02 \times 2500 = 25\ \text J\end{aligned}$$

The kinetic energy of the combined mass just after the collision is

$$\begin{aligned}\text K_f = \frac{1}{2}(\text M + \text m)\text V^2 = \frac{1}{2} \times 1 \times (1.0)^2 \\[1em] = 0.5\ \text J\end{aligned}$$

Therefore the loss in kinetic energy is

$$\begin{aligned}\Delta \text K = \text K_i - \text K_f = 25 - 0.5 \\[1em] = 24.5\ \text J\end{aligned}$$

Hence, the velocity of the block after the collision is 1.0 m/s, the velocity of the bullet is 50 m/s and the loss in kinetic energy is 24.5 J.

Given,

• Mass of the sphere, m = 10 kg
• Speed at the point A, v_A = 10 m/s
• Heights : A at 20 m, B at 0 m, C at 10 m and D at 3 m
• g = 10 m/s²

(a) Total energy at the point A : The total energy is the sum of the kinetic and the potential energy,

$$\begin{aligned}\text E_A = \frac{1}{2}\text{mv}_A^2 + \text{mgh}_A \\[1em] = \frac{1}{2} \times 10 \times (10)^2 + (10 \times 10 \times 20)\end{aligned}$$

$= 500 + 2000 = 2500\ \text J$

Since the surface is smooth, this total energy remains the same at every point.

(b) At the point B : The height of B above the reference level is zero, so its potential energy is

$\text U_B = 0$

Therefore its kinetic energy is

$\text K_B = \text E_A - \text U_B = 2500 - 0 = 2500\ \text J$

(c) At the point C : The height of C is 10 m, so

$\text U_C = \text{mgh}_C = 10 \times 10 \times 10 = 1000\ \text J$

$\text K_C = 2500 - 1000 = 1500\ \text J$

(d) At the point D : The height of D is 3 m, so

$\text U_D = \text{mgh}_D = 10 \times 10 \times 3 = 300\ \text J$

$\text K_D = 2500 - 300 = 2200\ \text J$

(e) Since the sphere still possesses a kinetic energy of 2200 J at the point D, it will go beyond D.

Hence, the total energy at A is 2500 J; at B the kinetic energy is 2500 J and the potential energy is zero; the kinetic energy is 1500 J at C and 2200 J at D; and the sphere goes beyond D.

Given,

• The momentum of the body is increased by 50%

Let the initial momentum be p and the initial kinetic energy be K. The kinetic energy in terms of the momentum is

$\text K = \frac{\text p^2}{2\text m}$

Since the mass remains the same, K ∝ p².

The new momentum after an increase of 50% is

$\text p' = \text p + \frac{50}{100}\text p = 1.5\text p$

Therefore the new kinetic energy is

$$\begin{aligned}\text K' = \frac{(\text p')^2}{2\text m} = \frac{(1.5\text p)^2}{2\text m} = 2.25 \times \frac{\text p^2}{2\text m} \\[1em] = 2.25\text K\end{aligned}$$

The percentage increase in the kinetic energy is

$$\begin{aligned}\frac{\text K' - \text K}{\text K} \times 100 = \frac{2.25\text K - \text K}{\text K} \times 100 \\[1em] = 1.25 \times 100 = 125%\end{aligned}$$

Hence, the kinetic energy of the body increases by 125%.

Given,

• Mass of the smaller fragment, m₁ = 0.4 kg
• Mass of the bigger fragment, m₂ = 4 kg
• Kinetic energy of the bigger fragment, K₂ = 100 J

The bomb is initially stationary, so its momentum is zero. By the conservation of linear momentum, the two fragments must fly apart with equal and opposite momenta, that is, their magnitudes are equal,

$\text p_1 = \text p_2 = \text p$

The kinetic energy in terms of the momentum is

$\text K = \frac{\text p^2}{2\text m}$

so that for the same momentum, $\text K \propto \frac{1}{\text m}$. Therefore

$\frac{\text K_1}{\text K_2} = \frac{\text m_2}{\text m_1} = \frac{4}{0.4} = 10$

Hence the kinetic energy of the smaller fragment is

$$\begin{aligned}\text K_1 = 10 \times \text K_2 = 10 \times 100 \\[1em] = 1000\ \text J\end{aligned}$$

Hence, the kinetic energy of the smaller fragment is 1000 J.

Given,

• Mass of each ball, m = 0.1 kg
• Speed of the moving ball, u = 10 m s^-1
• After the collision both balls move symmetrically, each making θ = 30° with the original direction

Let v be the new speed of each ball. Since the two balls move symmetrically, their momentum components perpendicular to the original direction are equal and opposite and therefore cancel out.

Applying the conservation of linear momentum along the original direction of motion,

$\text{mu} = \text{mv}\cos \theta + \text{mv}\cos \theta = 2\text{mv}\cos \theta$

Cancelling m from both sides,

$\text u = 2\text v\cos 30^\circ$

Substituting the values,

$$\begin{aligned}10 = 2 \times \text v \times 0.866 = 1.732\text v \\[1em] \text v = \frac{10}{1.732} = 5.77\ \text{m s}^{-1}\end{aligned}$$

Testing whether the collision is elastic : The kinetic energy before the collision is

$\text K_i = \frac{1}{2}\text{mu}^2 = \frac{1}{2} \times 0.1 \times (10)^2 = 5\ \text J$

The kinetic energy after the collision is

$$\begin{aligned}\text K_f = 2 \times \frac{1}{2}\text{mv}^2 = 0.1 \times (5.77)^2 \\[1em] = 0.1 \times 33.3 = 3.33\ \text J\end{aligned}$$

Since K_f is less than K_i, the kinetic energy is not conserved.

Hence, the new speed of each ball is 5.77 m s^-1 and the collision is not elastic.

Given,

• Mass of the box, m = 25 kg
• Coefficient of sliding friction, μ = 0.20
• Distance moved, s = 20 m
• g = 9.8 m s^-2

Since the box is pulled at a constant speed, its acceleration is zero, so the applied force is exactly equal to the force of sliding friction,

$\text F = \text f = \mu\text R = \mu\text{mg}$

Substituting the values,

$$\begin{aligned}\text F = 0.20 \times 25 \times 9.8 \\[1em] = 49\ \text N\end{aligned}$$

The applied force is horizontal and the displacement is also horizontal, so the work done is

$$\begin{aligned}\text W = \text F \times \text s = 49 \times 20 \\[1em] = 980\ \text J\end{aligned}$$

Hence, the work done in pulling the box through 20 m is 980 J.

Given,

• Mass of the man, m = 60 kg
• Height climbed, h = 10 m
• Time taken, t = half a minute = 30 s
• g = 9.8 m s^-2

The work done by the man against gravity is

$$\begin{aligned}\text W = \text{mgh} = 60 \times 9.8 \times 10 \\[1em] = 5880\ \text J\end{aligned}$$

The power of the man is the rate of doing this work,

$$\begin{aligned}\text P = \frac{\text W}{\text t} = \frac{5880}{30} \\[1em] = 196\ \text W\end{aligned}$$

Hence, the power of the man is 196 W.

Given,

• Power of the pump-motor, P = 2 kW = 2000 W
• Height to which water is raised, h = 10 m
• Time, t = 1 min = 60 s
• g = 10 m/s²

The work done by the pump-motor in one minute is

$$\begin{aligned}\text W = \text P \times \text t = 2000 \times 60 \\[1em] = 1.2 \times 10^5\ \text J\end{aligned}$$

This work is used in raising a mass m of water through a height of 10 m, so it is equal to the gain in the gravitational potential energy of the water,

$\text W = \text{mgh} \quad \Rightarrow \quad \text m = \frac{\text W}{\text{gh}}$

Substituting the values,

$$\begin{aligned}\text m = \frac{1.2 \times 10^5}{10 \times 10} = \frac{1.2 \times 10^5}{100} \\[1em] = 1200\ \text{kg}\end{aligned}$$

Hence, the pump-motor can raise 1200 kg of water per minute to a height of 10 m.

Given,

• Work done in stretching the spring, W = 4 J
• Extension of the spring, x = 10 cm = 0.1 m

The work done in stretching a spring through a distance x is stored in it as elastic potential energy,

$\text W = \frac{1}{2}\text{kx}^2$

where k is the spring constant. Rearranging,

$\text k = \frac{2\text W}{\text x^2}$

Substituting the values,

$$\begin{aligned}\text k = \frac{2 \times 4}{(0.1)^2} = \frac{8}{0.01} \\[1em] = 800\ \text{N m}^{-1}\end{aligned}$$

Hence, the spring-constant is 800 N m^-1.

Given,

• Mass of the solid, m = 2 kg
• Velocity of the solid, v = 10 m/s
• Compression produced in the spring, x = 25 cm = 0.25 m

The spring is ideal and weightless, so the whole of the kinetic energy of the solid is stored in the spring as elastic potential energy at the position of maximum compression,

$\frac{1}{2}\text{mv}^2 = \frac{1}{2}\text{kx}^2$

The kinetic energy of the solid is

$\frac{1}{2}\text{mv}^2 = \frac{1}{2} \times 2 \times (10)^2 = 100\ \text J$

Therefore,

$$\begin{aligned}100 = \frac{1}{2} \times \text k \times (0.25)^2 \\[1em] 100 = \frac{1}{2} \times \text k \times 0.0625\end{aligned}$$

$$\begin{aligned}\text k = \frac{200}{0.0625} \\[1em] = 3200\ \text{N/m}\end{aligned}$$

Hence, the force-constant of the spring is 3200 N/m.

Given,

• Mass of the moving body, m₁ = 20 kg with u₁ = 10 m s^-1
• Mass of the stationary body, m₂ = 5 kg with u₂ = 0
• The masses stick together after the collision

(a) Since the masses stick together, the collision is perfectly inelastic. By the conservation of linear momentum,

$$\begin{aligned}\text m_1\text u_1 = (\text m_1 + \text m_2)\text v \\[1em] 20 \times 10 = (20 + 5)\text v\end{aligned}$$

$\text v = \frac{200}{25} = 8\ \text{m s}^{-1}$

(b) The kinetic energy of the system before the collision is

$$\begin{aligned}\text K_i = \frac{1}{2}\text m_1\text u_1^2 = \frac{1}{2} \times 20 \times (10)^2 \\[1em] = 1000\ \text J\end{aligned}$$

(c) The kinetic energy of the system after the collision is

$$\begin{aligned}\text K_f = \frac{1}{2}(\text m_1 + \text m_2)\text v^2 = \frac{1}{2} \times 25 \times (8)^2 \\[1em] = \frac{1}{2} \times 25 \times 64 = 800\ \text J\end{aligned}$$

(d) The loss in kinetic energy is

$\Delta \text K = 1000 - 800 = 200\ \text J$

This energy has not been destroyed. In this perfectly inelastic collision it has been converted into heat, along with sound and the energy of deformation of the bodies.

Hence, the combined mass moves with a speed of 8 m s^-1, the kinetic energy is 1000 J before and 800 J after the collision, and the remaining 200 J is changed into heat.

Given,

• Mass of the first body, m₁ = 2 kg with u₁ = 3 m s^-1
• Mass of the second body, m₂ = 1 kg with u₂ = 4 m s^-1 in the opposite direction
• The two bodies stick together after the collision

Taking the direction of motion of the first body as positive, u₂ = − 4 m s^-1. Since the bodies stick together, the collision is perfectly inelastic. By the conservation of linear momentum,

$\text m_1\text u_1 + \text m_2\text u_2 = (\text m_1 + \text m_2)\text v$

Substituting the values,

$$\begin{aligned}(2 \times 3) + (1 \times -4) = (2 + 1)\text v \\[1em] 6 - 4 = 3\text v\end{aligned}$$

$\text v = \frac{2}{3}\ \text{m s}^{-1}$

Since v is positive, the sticked bodies move in the original direction of the 2 kg body.

Hence, the velocity of the sticked bodies is $\frac{2}{3}$ m s^-1, that is, about 0.67 m s^-1, in the direction of motion of the 2 kg body.

Given,

• Mass of the proton, m₁ = 1.67 × 10^-27 kg
• Mass of the α-particle, m₂ = 6.68 × 10^-27 kg, initially at rest
• Speed of the α-particle after the collision, v₂ = 8 × 10⁵ m s^-1

Here m₁ + m₂ = (1.67 + 6.68) × 10^-27 = 8.35 × 10^-27 kg.

Speed of the proton before the collision : For a one-dimensional elastic collision with the target initially at rest, the velocity of the target after the collision is

$\text v_2 = \left(\frac{2\text m_1}{\text m_1 + \text m_2}\right)\text u_1$

Substituting the values,

$$\begin{aligned}8 \times 10^5 = \left(\frac{2 \times 1.67 \times 10^{-27}}{8.35 \times 10^{-27}}\right)\text u_1 \\[1em] 8 \times 10^5 = \left(\frac{3.34}{8.35}\right)\text u_1 = 0.4\text u_1\end{aligned}$$

$\text u_1 = \frac{8 \times 10^5}{0.4} = 2 \times 10^6\ \text{m s}^{-1}$

Speed of the proton after the collision :

$\text v_1 = \left(\frac{\text m_1 - \text m_2}{\text m_1 + \text m_2}\right)\text u_1$

Substituting the values,

$\text v_1 = \left(\frac{1.67 - 6.68}{8.35}\right)(2 \times 10^6) = \left(\frac{-5.01}{8.35}\right)(2 \times 10^6)$

$= (-0.6)(2 \times 10^6) = -1.2 \times 10^6\ \text{m s}^{-1}$

The negative sign shows that the proton rebounds, that is, it moves back in the opposite direction.

Hence, the speed of the proton before the collision is 2 × 10⁶ m s^-1 and after the collision it is 1.2 × 10⁶ m s^-1 in the reversed direction.

Given,

• Mass of the first particle, m₁ = 1.8 × 10^-27 kg with u₁ = 10⁶ m s^-1
• Mass of the second particle, m₂ = 3.6 × 10^-27 kg, initially at rest

Here m₁ + m₂ = (1.8 + 3.6) × 10^-27 = 5.4 × 10^-27 kg.

Speed of the first particle after the collision : For a one-dimensional elastic collision with the target initially at rest,

$\text v_1 = \left(\frac{\text m_1 - \text m_2}{\text m_1 + \text m_2}\right)\text u_1$

Substituting the values,

$$\begin{aligned}\text v_1 = \left(\frac{1.8 - 3.6}{5.4}\right)(10^6) = \left(\frac{-1.8}{5.4}\right)(10^6) \\[1em] = -\frac{1}{3} \times 10^6\ \text{m s}^{-1}\end{aligned}$$

The negative sign shows that the first particle rebounds.

Speed of the second particle after the collision :

$\text v_2 = \left(\frac{2\text m_1}{\text m_1 + \text m_2}\right)\text u_1$

Substituting the values,

$$\begin{aligned}\text v_2 = \left(\frac{2 \times 1.8}{5.4}\right)(10^6) = \left(\frac{3.6}{5.4}\right)(10^6) \\[1em] = \frac{2}{3} \times 10^6\ \text{m s}^{-1}\end{aligned}$$

Hence, the first particle rebounds with a speed of $\frac{1}{3} \times 10^6$ m s^-1 while the second particle moves forward with a speed of $\frac{2}{3} \times 10^6$ m s^-1.