Work, Energy & Power — NCERT Exercises

The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative.

(a) Work done by a person in lifting a bucket out of a well by means of a rope tied to the bucket.

(b) Work done by gravitational force in the above case.

(c) Work done by friction on a body sliding down an inclined plane.

(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.

(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer

(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.

A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of friction = 0.1. Compute the

(a) work done by the applied force in 10 s.

(b) work done by friction in 10 s.

(c) work done by the net force on the body in 10 s.

(d) change in kinetic energy of the body in 10 s. Also, interpret your results. Take g = 10 m s^-2.

Answer

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

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

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

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

The distance moved by the body in 10 s is

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

(a) The work done by the applied force is

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

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

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

(c) The work done by the net force is

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

(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.

Given in Fig. are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

Answer

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 $\left(\text K = \dfrac{1}{2}\text{mv}^2\right)$ 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 $-\dfrac{\text b}{2} \lt \text x \lt -\dfrac{\text a}{2}$ and $\dfrac{\text a}{2} \lt \text x \lt \dfrac{\text b}{2}$. The minimum total energy in the permitted region is − U₁.

The potential energy function for a particle executing linear simple harmonic motion is given by U(x) = $\dfrac{1}{2}$ k x², where k is the force-constant of the oscillator. For k = 0.5 N m^-1, the graph U(x) versus x is shown in the figure. Show that a particle of total energy 1 J moving under this potential energy must 'turn back' when it reaches x = ± 2 m.

Answer

Given,

• Potential energy function, $\text U(\text x) = \dfrac{1}{2}\text{kx}^2$
• 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,

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

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,

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

Substituting the given values,

$\text x = \sqrt{\dfrac{2 \times 1}{0.5}} = \sqrt{4} \\[1em] = \pm 2\ \text m$

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.

Answer the following :

(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?

(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?

(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer to the earth?

(d) In Fig. (a), the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. (b), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater? Take g = 9.8 ms^-2.

Answer

(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

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

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

Underline the correct alternative :

(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unchanged.

(b) The work done by a body against friction always results in a loss of the kinetic energy/potential energy of the body.

(c) The rate of change of total momentum of a many-particles system is proportional to the external force/sum of internal forces on the system.

(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.

Answer

(a) Decreases. A conservative force is the negative gradient of the potential energy, $\text F = -\dfrac{\text{dU}}{\text{dx}}$. 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.

State, with reason, if each of the following statement is true or false :

(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.

(b) The total energy of a system is always conserved, no matter what internal and external forces on the body are present.

(c) The work done in the motion of a body over a closed loop is zero for every force in nature.

(d) In an inelastic collision, the final kinetic energy is less than the initial kinetic energy of the system.

Answer

(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.

Answer carefully, with reasons :

(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e., when they are in contact)?

(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?

(c) What are the answers to (a) and (b) for an inelastic collision?

(d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

Answer

(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.

A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to

1. t^1/2
2. t
3. t^3/2
4. t²

Answer

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.

A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to

1. t^1/2
2. t
3. t^3/2
4. t²

Answer

t^3/2

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

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

Separating the variables,

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

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

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

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

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

Integrating again from s = 0 at t = 0,

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

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.

A body constrained to move along the Z-axis of a coordinate system is subjected to a constant force $\vec{\text F} \space$given by $\vec{\text F} = (-\hat{\text i} + 2\hat{\text j} + 3\hat{\text k})$ N, where $\hat{\text i}, \hat{\text j}, \hat{\text k}$ are unit vectors along X, Y, Z axis of the system respectively. What is work done by this force in moving the body a distance of 4 m along the Z-axis?

Answer

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 $\hat{\text k} \cdot \hat{\text k} = 1$,

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

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

An electron and a proton are detected in a cosmic-ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which one is faster? Obtain the ratio of their speeds. (Electron mass = 9.11 × 10^-31 kg, proton mass = 1.67 × 10^-27 kg, 1 eV = 1.60 × 10^-19 J).

Answer

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

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

Therefore the ratio of the two speeds is

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

Substituting the given values,

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

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 $\text v \propto \sqrt{\dfrac{\text K}{\text m}}$, the electron turns out to be much faster.

A raindrop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed and moves with uniform speed there after. What is the work done by the gravitational force on the drop in the first and the second half of its journey? What is the work done by the resistive force in the entire journey if the speed of the drop on reaching the ground is 10 m s^-1? Take density of water to be 10³ kg m^-3 and g = 9.8 m s^-2.

Answer

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

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

The gravitational force on the drop is

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

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

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

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

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

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,

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

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

A molecule in a gas container hits the wall with speed 200 ms^-1 and at an angle 30° with the normal, and rebounds with the same speed. Is the momentum of the system conserved in the collision? Is the collision elastic or inelastic?

Answer

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 $\left(\dfrac{1}{2}\text{mv}^2\right)$ 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.

A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 minutes. If the tank is 40 m above the ground, and the efficiency of pump is 30%, how much electric power is consumed by the pump? Take density of water to be 10³ kg m^-3 and g = 9.8 m s^-2.

Answer

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

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

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

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

The output power of the pump is

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

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

$\text P_{in} = \text P_{out} \times \dfrac{100}{30} = 13.1 \times \dfrac{100}{30} \\[1em] = 43.6\ \text{kW}$

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

Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed v. If the collision is elastic, which of the following cases is a possible result after collision?

Answer 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

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

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

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

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

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

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

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

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

Only in case (ii) is the final kinetic energy equal to the initial kinetic energy $\dfrac{1}{2}\text{mv}^2$. 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 of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table, as shown. How high does the bob A rise after the collision? Neglect the size of bobs and assume the collision to be elastic.

Answer

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

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

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.

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipates 5% of its initial energy against air resistance? (g = 9.8 m s^-2).

Answer

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,

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

The mass m cancels from both sides, giving

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

Substituting the values,

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

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

A trolley of mass 300 kg carrying a sand bag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s^-1. What is the speed of the trolley after the entire sand bag is empty?

Answer

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.

A body of mass 0.5 kg travels in a straight line with velocity v = a x^3/2, where a = 5 m^-1/2 s^-1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

Answer

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

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

The initial kinetic energy of the body is

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

The final kinetic energy of the body is

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

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,

$\text W = \text K_2 - \text K_1 = 50 - 0 \\[1em] = 50\ \text J$

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

The blades of a windmill sweep out a circle of area A.

(a) If the wind blows at a velocity v perpendicular to the circle, what is the mass of air passing through it in time t?

(b) What is the kinetic energy of the air?

(c) Assume that the windmill converts 25% of the wind's energy into electrical energy, and that A = 30 m², v = 36 km/h and the density of air is 1.2 kg m^-3. What is the electrical power produced?

Answer

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

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

(b) The kinetic energy of this air is

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

(c) Given,

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

Converting the wind speed into SI units,

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

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

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

Therefore the electrical power produced is

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

Substituting the values,

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

Hence, the electrical power produced is 4.5 kW.

A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.

(a) How much work does she do against the gravitational force?

(b) Fat supplies 3.8 × 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up? (g = 9.8 m s^-2)

Answer

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

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

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

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

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

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

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

A family uses 8 kW of power.

(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square metre. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?

(b) Compare this area to that of the roof of a typical house.

Answer

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,

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

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.