Chapter 5: Work, Energy and Power

The angle $$\theta$$ between two vectors is obtained from the scalar (dot) product:

$$\mathbf{F}\cdot\mathbf{d} = |\mathbf{F}|\,|\mathbf{d}|\cos\theta$$

Step 1 — compute the dot product using $$\hat{\mathbf{i}}\cdot\hat{\mathbf{i}} = \hat{\mathbf{j}}\cdot\hat{\mathbf{j}} = \hat{\mathbf{k}}\cdot\hat{\mathbf{k}} = 1$$ and the other cross-products equal to zero:

$$\mathbf{F}\cdot\mathbf{d} = (3)(5) + (4)(4) + (-5)(3) = 15 + 16 - 15 = 16 \text{ unit}$$

Step 2 — compute the magnitudes:

$$|\mathbf{F}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9+16+25} = \sqrt{50} = 5\sqrt{2}$$

$$|\mathbf{d}| = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25+16+9} = \sqrt{50} = 5\sqrt{2}$$

Step 3 — find $$\theta$$:

$$\cos\theta = \dfrac{\mathbf{F}\cdot\mathbf{d}}{|\mathbf{F}|\,|\mathbf{d}|} = \dfrac{16}{(5\sqrt{2})(5\sqrt{2})} = \dfrac{16}{50} = 0.32$$

$$\theta = \cos^{-1}(0.32) \approx 71.3^\circ$$

Step 4 — projection of $$\mathbf{F}$$ on $$\mathbf{d}$$:

$$\mathrm{Proj}_{\mathbf{d}}\,\mathbf{F} = |\mathbf{F}|\cos\theta = \dfrac{\mathbf{F}\cdot\mathbf{d}}{|\mathbf{d}|} = \dfrac{16}{5\sqrt{2}} = \dfrac{16\sqrt{2}}{10} = 1.6\sqrt{2} \approx 2.26 \text{ unit}$$

Given: $$m = 1.00\,\mathrm{g} = 1.00\times 10^{-3}\,\mathrm{kg}$$, height $$h = 1.00\,\mathrm{km} = 10^{3}\,\mathrm{m}$$, final speed $$v = 50.0\,\mathrm{m\,s^{-1}}$$, $$g = 9.8\,\mathrm{m\,s^{-2}}$$.

Work done by gravity. Gravity acts vertically downward through the entire fall of $$h$$, so

$$W_g = mgh = (10^{-3})(9.8)(10^{3}) = 9.8\,\mathrm{J}$$

Work done by the resistive force. The drop starts from rest, so by the work–energy theorem the net work equals the change in kinetic energy:

$$W_{\text{net}} = W_g + W_r = \Delta K = \dfrac{1}{2}mv^2 - 0$$

$$\dfrac{1}{2}mv^2 = \dfrac{1}{2}(10^{-3})(50.0)^2 = \dfrac{1}{2}(10^{-3})(2500) = 1.25\,\mathrm{J}$$

Therefore

$$W_r = \dfrac{1}{2}mv^2 - W_g = 1.25 - 9.8 = -8.55\,\mathrm{J}$$

The negative sign confirms that the resistive force opposes the motion (and dissipates most of the gravitational work as heat in the surrounding air).

(a) Work done by the road on the cycle. The force exerted by the road on the cycle is $$F = 200\,\mathrm{N}$$ directed opposite to the cycle's displacement $$s = 10\,\mathrm{m}$$, so the angle between $$\mathbf{F}$$ and $$\mathbf{s}$$ is $$180^\circ$$:

$$W_{\text{road on cycle}} = Fs\cos 180^\circ = -(200)(10) = -2000\,\mathrm{J} = -2\,\mathrm{kJ}$$

This negative work is what brings the cycle to rest (decreases its kinetic energy).

(b) Work done by the cycle on the road. By Newton's third law, the cycle exerts an equal and opposite force of $$200\,\mathrm{N}$$ on the road. However, the road does not undergo any displacement, $$s_{\text{road}} = 0$$, so

$$W_{\text{cycle on road}} = (200)(0) = 0$$

Comment. Although the road–cycle pair forms a Newton's third-law action–reaction pair, the works done by these forces on their respective bodies need not be equal and opposite, because the displacements of the two bodies are different. Here, the cycle's kinetic energy is dissipated as heat (in the tyre and the road surface) — not transferred to the road as work.

Let $$v_i = 200\,\mathrm{m\,s^{-1}}$$ be the initial speed and $$v_f$$ the emergent speed. The emergent kinetic energy is 10% of the initial:

$$\dfrac{1}{2}mv_f^2 = 0.10 \times \dfrac{1}{2}mv_i^2$$

The mass $$m$$ and the factor $$\tfrac{1}{2}$$ cancel, giving

$$v_f^2 = 0.10\,v_i^2 \;\Rightarrow\; v_f = v_i\sqrt{0.10} = \dfrac{v_i}{\sqrt{10}}$$

$$v_f = \dfrac{200}{\sqrt{10}} \approx 63.2\,\mathrm{m\,s^{-1}}$$

Notice that the speed is reduced only by a factor of about three (not ten) although the kinetic energy is reduced by a factor of ten — because kinetic energy depends on the square of the speed.

Description of the plot. On a graph of force (N) versus displacement $$x$$ (m):

• The applied force $$F$$ is a horizontal line at $$+100\,\mathrm{N}$$ from $$x=0$$ to $$x=10\,\mathrm{m}$$, then drops linearly from $$+100\,\mathrm{N}$$ at $$x=10\,\mathrm{m}$$ to $$+50\,\mathrm{N}$$ at $$x=20\,\mathrm{m}$$.
• The frictional force $$f$$ is a horizontal line at $$-50\,\mathrm{N}$$ (negative because it opposes the motion) from $$x=0$$ to $$x=20\,\mathrm{m}$$.

Work done by the woman = area under the $$F$$–$$x$$ graph between $$x=0$$ and $$x=20\,\mathrm{m}$$.

From $$0$$ to $$10\,\mathrm{m}$$ (rectangle):

$$W_1 = (100)(10) = 1000\,\mathrm{J}$$

From $$10$$ to $$20\,\mathrm{m}$$ the force decreases linearly from $$100\,\mathrm{N}$$ to $$50\,\mathrm{N}$$ (trapezium):

$$W_2 = \dfrac{1}{2}(100 + 50)(20 - 10) = \dfrac{1}{2}(150)(10) = 750\,\mathrm{J}$$

Total work done by the woman:

$$W_{F} = W_1 + W_2 = 1000 + 750 = 1750\,\mathrm{J}$$

Work done by friction (acts opposite to displacement of $$20\,\mathrm{m}$$):

$$W_f = -(50)(20) = -1000\,\mathrm{J}$$

Initial kinetic energy:

$$K_i = \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}(1)(2)^2 = 2\,\mathrm{J}$$

The retarding force is variable. Work done by it as the block moves from $$x = 0.1\,\mathrm{m}$$ to $$x = 2.01\,\mathrm{m}$$:

$$W_r = \int_{0.1}^{2.01} F_r\,dx = \int_{0.1}^{2.01}\!\left(-\dfrac{k}{x}\right) dx = -k\,[\ln x]_{0.1}^{2.01}$$

$$W_r = -k\,\ln\!\left(\dfrac{2.01}{0.1}\right) = -(0.5)\,\ln(20.1)$$

$$\ln(20.1) \approx 3.00 \;\Rightarrow\; W_r \approx -(0.5)(3.00) = -1.5\,\mathrm{J}$$

Outside the rough patch the retarding force is zero, so the only work on the block from start to finish is $$W_r$$. By the work–energy theorem,

$$K_f = K_i + W_r = 2 - 1.5 = 0.5\,\mathrm{J}$$

Finally,

$$\dfrac{1}{2}mv_f^2 = K_f \;\Rightarrow\; v_f = \sqrt{\dfrac{2K_f}{m}} = \sqrt{\dfrac{2(0.5)}{1}} = 1\,\mathrm{m\,s^{-1}}$$

Take A (lowest point) as the reference for potential energy. Then the heights of B (horizontal level of the support) and C (topmost point) above A are $$L$$ and $$2L$$ respectively.

Condition at C — string just goes slack. At C the centripetal acceleration is directed downward (toward the centre of the circle). Both gravity $$mg$$ and tension $$T_C$$ point downward, so the centripetal equation is

$$T_C + mg = \dfrac{mv_C^2}{L}$$

The string becomes slack precisely when $$T_C = 0$$, giving

$$mg = \dfrac{mv_C^2}{L} \;\Rightarrow\; v_C^2 = gL$$

(i) Speed at A. Mechanical energy conservation between A and C:

$$\dfrac{1}{2}mv_o^2 = \dfrac{1}{2}mv_C^2 + mg(2L)$$

$$v_o^2 = v_C^2 + 4gL = gL + 4gL = 5gL$$

$$\boxed{\,v_o = \sqrt{5gL}\,}$$

(ii) Speeds at B and C. For B, conservation of energy between A and B (height of B above A is $$L$$):

$$\dfrac{1}{2}mv_o^2 = \dfrac{1}{2}mv_B^2 + mgL$$

$$v_B^2 = v_o^2 - 2gL = 5gL - 2gL = 3gL \;\Rightarrow\; v_B = \sqrt{3gL}$$

And from above, $$v_C = \sqrt{gL}$$.

(iii) Ratio of kinetic energies.

$$\dfrac{K_B}{K_C} = \dfrac{\tfrac{1}{2}mv_B^2}{\tfrac{1}{2}mv_C^2} = \dfrac{v_B^2}{v_C^2} = \dfrac{3gL}{gL} = 3$$

Trajectory after C. At C the string becomes slack, so the bob is no longer constrained. It then moves under gravity alone with an initial horizontal velocity $$v_C = \sqrt{gL}$$ — i.e. it undergoes projectile motion (parabolic path).

Convert the speed to SI units:

$$v = 18.0\,\mathrm{km/h} = \dfrac{18 \times 1000}{3600}\,\mathrm{m\,s^{-1}} = 5\,\mathrm{m\,s^{-1}}$$

The road is smooth, so the only horizontal force doing work on the car is the spring force. At the instant of maximum compression $$x_m$$, the car is momentarily at rest. All the initial kinetic energy is stored in the spring:

$$\dfrac{1}{2}mv^2 = \dfrac{1}{2}k\,x_m^2$$

Solving for $$x_m$$:

$$x_m = v\sqrt{\dfrac{m}{k}} = 5 \times \sqrt{\dfrac{1000}{5.25\times 10^3}}$$

$$x_m = 5 \times \sqrt{0.1905} = 5 \times 0.4365 \approx 2.18\,\mathrm{m}$$

Now the friction force $$f = \mu mg$$ acts opposite to the motion throughout the compression $$x_m$$. The work–energy theorem (from initial state with speed $$v$$ to the moment of maximum compression where speed is zero) becomes:

$$\Delta K = W_{\text{spring}} + W_{\text{friction}}$$

$$0 - \dfrac{1}{2}mv^2 = -\dfrac{1}{2}kx_m^2 - \mu mg\,x_m$$

Rearranging,

$$\dfrac{1}{2}kx_m^2 + \mu mg\,x_m - \dfrac{1}{2}mv^2 = 0$$

Plug in $$m=1000\,\mathrm{kg}$$, $$v=5\,\mathrm{m\,s^{-1}}$$, $$k=5.25\times 10^{3}\,\mathrm{N\,m^{-1}}$$, $$\mu = 0.5$$, $$g=9.8\,\mathrm{m\,s^{-2}}$$:

$$\dfrac{1}{2}(5250)\,x_m^2 + (0.5)(1000)(9.8)\,x_m - \dfrac{1}{2}(1000)(25) = 0$$

$$2625\,x_m^2 + 4900\,x_m - 12500 = 0$$

Using the quadratic formula:

$$x_m = \dfrac{-4900 + \sqrt{4900^2 + 4(2625)(12500)}}{2(2625)} = \dfrac{-4900 + \sqrt{1.5526\times 10^{8}}}{5250}$$

$$x_m = \dfrac{-4900 + 12460}{5250} \approx \dfrac{7560}{5250} \approx 1.44\,\mathrm{m}$$

The compression is smaller than in Example 5.8 (2.18 m) because some kinetic energy has now been dissipated by friction.

The elevator moves up at constant velocity, so its acceleration is zero — i.e. the net force on it is zero. The motor must exert an upward force $$F$$ that balances both the weight and the friction:

$$F = mg + f = (1800)(9.8) + 4000 = 17640 + 4000 = 21640\,\mathrm{N}$$

The power delivered by the motor at speed $$v = 2\,\mathrm{m\,s^{-1}}$$:

$$P = F\,v = (21640)(2) = 43280\,\mathrm{W} \approx 43.3\,\mathrm{kW}$$

Using $$1\,\mathrm{hp} = 746\,\mathrm{W}$$,

$$P = \dfrac{43280}{746} \approx 58.0\,\mathrm{hp}$$

Consider a 1-D elastic collision: a neutron of mass $$m_1$$ with initial speed $$v_{1i}$$ strikes a stationary moderator nucleus of mass $$m_2$$. For an elastic collision (conserving both momentum and kinetic energy), the standard results are

$$v_{1f} = \dfrac{m_1 - m_2}{m_1 + m_2}\,v_{1i}, \qquad v_{2f} = \dfrac{2m_1}{m_1 + m_2}\,v_{1i}$$

The fraction of initial kinetic energy retained by the neutron is

$$\dfrac{K_{1f}}{K_{1i}} = \left(\dfrac{v_{1f}}{v_{1i}}\right)^{\!2} = \left(\dfrac{m_1 - m_2}{m_1 + m_2}\right)^{\!2}$$

The fraction transferred to the moderator nucleus is

$$f = \dfrac{K_{2f}}{K_{1i}} = 1 - \left(\dfrac{m_1 - m_2}{m_1 + m_2}\right)^{\!2} = \dfrac{4\,m_1 m_2}{(m_1 + m_2)^2}$$

For two specific moderators (taking $$m_1 = 1\,\mathrm{u}$$ for the neutron):

Deuterium ($$m_2 = 2\,\mathrm{u}$$):

$$f_D = \dfrac{4(1)(2)}{(1+2)^2} = \dfrac{8}{9} \approx 0.89$$

So about 89% of the neutron's kinetic energy is lost in a single head-on elastic collision with a deuteron.

Carbon ($$m_2 = 12\,\mathrm{u}$$):

$$f_C = \dfrac{4(1)(12)}{(1+12)^2} = \dfrac{48}{169} \approx 0.28$$

About 28% lost per collision with carbon. Note that $$f$$ is maximised when $$m_2 = m_1$$ (giving $$f=1$$, i.e. complete energy transfer). For a very heavy nucleus like uranium ($$m_2 \gg m_1$$) the neutron barely loses any energy — which is why light nuclei (D, C) are used as moderators. After several collisions, the neutron's speed drops from $$\sim 10^7$$ to thermal speeds ($$\sim 10^3$$) at which fission probability is high.

Let the cue ball, moving with speed $$v_1$$, strike the stationary target ball. After the collision, the cue moves with speed $$v_1'$$ at angle $$\theta_1$$ above the original line and the target moves with speed $$v_2'$$ at angle $$\theta_2$$ below it. Since $$m_1 = m_2 = m$$:

Momentum conservation (resolve along and perpendicular to the original direction):

$$m v_1 = m v_1'\cos\theta_1 + m v_2'\cos\theta_2 \quad\text{(along)}$$

$$0 = m v_1'\sin\theta_1 - m v_2'\sin\theta_2 \quad\text{(perpendicular)}$$

Kinetic energy conservation (elastic collision):

$$\dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_1'^2 + \dfrac{1}{2}mv_2'^2 \;\Rightarrow\; v_1^2 = v_1'^2 + v_2'^2 \quad (*)$$

Now square the two momentum equations (cancelling $$m$$):

$$v_1 = v_1'\cos\theta_1 + v_2'\cos\theta_2$$

$$0 = v_1'\sin\theta_1 - v_2'\sin\theta_2$$

Squaring and adding:

$$v_1^2 = v_1'^2 + v_2'^2 + 2 v_1' v_2'(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2)$$

$$v_1^2 = v_1'^2 + v_2'^2 + 2 v_1' v_2'\cos(\theta_1 + \theta_2)$$

Comparing with $$(*)$$, the cross term must vanish:

$$2 v_1' v_2' \cos(\theta_1 + \theta_2) = 0$$

Assuming both balls are moving after the collision ($$v_1', v_2' \ne 0$$),

$$\cos(\theta_1 + \theta_2) = 0 \;\Rightarrow\; \theta_1 + \theta_2 = 90^\circ$$

That is, for an elastic collision between equal masses (with one initially at rest), the two outgoing velocities are perpendicular. Hence,

$$\theta_1 = 90^\circ - 37^\circ = 53^\circ$$

General principle. The total mechanical energy is $$E = K + V(x)$$ where the kinetic energy $$K = \tfrac{1}{2}mv^{2} \ge 0$$. Hence at every point a classical particle can occupy we must have

$$V(x) \le E.$$

The particle is forbidden in regions where $$V(x) > E$$. The minimum total energy is the global minimum of $$V(x)$$ — any smaller $$E$$ leaves no point at which $$K \ge 0$$.

(a) Single step: $$V = 0$$ for $$x < a$$, $$V = V_{0}$$ for $$x \ge a$$, with the cross marking an $$E < V_{0}$$.

Forbidden region: $$x \ge a$$ (there $$V_{0} > E$$).
Minimum total energy: $$E_{\min} = 0$$ (the global minimum of $$V$$).
Physical context: a body moving on a flat surface that meets a steep vertical step it cannot climb (a ball rolling toward a wall too high to surmount).

(b) Step both ways: $$V = V_{0}$$ everywhere except over a small range, with the cross at $$E < V_{0}$$.

Forbidden region: everywhere on the diagram — the particle cannot exist anywhere because $$V(x) > E$$ at every point.
Minimum total energy: $$E_{\min} = V_{0}$$ (the minimum of $$V$$ on this diagram).
Physical context: a particle which has insufficient energy to penetrate any of the potential plateaus (a body confined outside a closed barrier).

(c) Square well of depth $$V_{1}$$: $$V = V_{0}$$ for $$x < a$$ and $$x > b$$, $$V = -V_{1}$$ for $$a \le x \le b$$, with the cross at $$E$$ in the range $$-V_{1} < E < 0 < V_{0}$$.

Forbidden regions: $$x < a$$ and $$x > b$$ (in both, $$V_{0} > E$$).
Minimum total energy: $$E_{\min} = -V_{1}$$ (the bottom of the well).
Physical context: a particle trapped in a finite potential well — classical analogue of a molecule bound in a potential trough or a ball confined inside a rigid box.

(d) Symmetric pair of barriers: $$V = -V_{1}$$ for $$|x| < a/2$$, $$V = V_{0}$$ for $$a/2 \le |x| \le b/2$$, $$V = -V_{1}$$ outside, with the cross at $$E$$ in the range $$-V_{1} < E < V_{0}$$.

Forbidden regions: the two barrier ranges $$-b/2 < x < -a/2$$ and $$a/2 < x < b/2$$ (there $$V_{0} > E$$).
Minimum total energy: $$E_{\min} = -V_{1}$$ (the floor of each well).
Physical context: two adjacent potential wells separated by a barrier the particle cannot climb — analogous to two bound states of a double-well potential.

The total mechanical energy is

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

Since the kinetic energy must satisfy $$K \ge 0$$, the particle is allowed only in regions where

$$V(x) \le E \;\Rightarrow\; \dfrac{1}{2}kx^2 \le E$$

The turning points are the values of $$x$$ where $$V(x) = E$$ (so that $$K = 0$$, i.e. the particle is momentarily at rest):

$$\dfrac{1}{2}kx^2 = E \;\Rightarrow\; x = \pm\sqrt{\dfrac{2E}{k}}$$

Substituting $$E = 1\,\mathrm{J}$$ and $$k = 0.5\,\mathrm{N\,m^{-1}}$$:

$$x = \pm\sqrt{\dfrac{2 \times 1}{0.5}} = \pm\sqrt{4} = \pm 2\,\mathrm{m}$$

For $$|x| > 2\,\mathrm{m}$$ we would need $$K < 0$$, which is impossible. Hence the particle must reverse its motion (turn back) exactly when it reaches $$x = \pm 2\,\mathrm{m}$$, as required.

The body is constrained to move along the $$z$$-axis, so its displacement is

$$\mathbf{s} = 4\,\hat{\mathbf{k}}\,\mathrm{m}$$

The work done by the constant force $$\mathbf{F}$$ is the scalar product

$$W = \mathbf{F}\cdot\mathbf{s} = (-1)(0) + (2)(0) + (3)(4) = 12\,\mathrm{J}$$

Only the component of the force along the displacement (i.e. along $$\hat{\mathbf{k}}$$) contributes; the $$x$$- and $$y$$-components do no work because the body has no displacement in those directions.

Non-relativistic kinetic energy: $$K = \dfrac{1}{2}mv^2 \;\Rightarrow\; v = \sqrt{\dfrac{2K}{m}}$$. (Even for an electron with $$K = 10\,\mathrm{keV}$$, $$v \sim 6\times 10^7\,\mathrm{m\,s^{-1}} \ll c$$, so the classical formula is acceptable.)

Electron: $$K_e = 10\,\mathrm{keV} = 10^{4} \times 1.6\times 10^{-19} = 1.6\times 10^{-15}\,\mathrm{J}$$

$$v_e = \sqrt{\dfrac{2(1.6\times 10^{-15})}{9.11\times 10^{-31}}} = \sqrt{3.51\times 10^{15}} \approx 5.93\times 10^{7}\,\mathrm{m\,s^{-1}}$$

Proton: $$K_p = 100\,\mathrm{keV} = 10^{5} \times 1.6\times 10^{-19} = 1.6\times 10^{-14}\,\mathrm{J}$$

$$v_p = \sqrt{\dfrac{2(1.6\times 10^{-14})}{1.67\times 10^{-27}}} = \sqrt{1.916\times 10^{13}} \approx 4.38\times 10^{6}\,\mathrm{m\,s^{-1}}$$

The electron is faster — even though it carries less kinetic energy, its mass is about 1836 times smaller. The ratio of speeds is

$$\dfrac{v_e}{v_p} = \sqrt{\dfrac{K_e}{K_p}\cdot \dfrac{m_p}{m_e}} = \sqrt{\dfrac{10}{100}\cdot \dfrac{1.67\times 10^{-27}}{9.11\times 10^{-31}}}$$

$$= \sqrt{0.1 \times 1833.2} = \sqrt{183.3} \approx 13.5$$

So $$v_e : v_p \approx 13.5 : 1$$.

Mass of the drop. With radius $$r = 2\,\mathrm{mm} = 2\times 10^{-3}\,\mathrm{m}$$ and water density $$\rho = 1000\,\mathrm{kg\,m^{-3}}$$,

$$m = \dfrac{4}{3}\pi r^3 \rho = \dfrac{4}{3}\pi (2\times 10^{-3})^3(1000) = \dfrac{32\pi}{3}\times 10^{-6}\,\mathrm{kg} \approx 3.35\times 10^{-5}\,\mathrm{kg}$$

Work done by gravity. Gravity acts vertically downward, in the direction of motion, so over any vertical drop $$h$$:

$$W_g = mgh$$

First half: $$h_1 = 250\,\mathrm{m}$$,

$$W_{g,1} = (3.35\times 10^{-5})(9.8)(250) \approx 0.082\,\mathrm{J}$$

Second half: $$h_2 = 250\,\mathrm{m}$$,

$$W_{g,2} \approx 0.082\,\mathrm{J}$$

(The two are equal because gravity is independent of speed.)

Work done by the resistive force. Take initial speed to be zero (drop starts from rest) and final speed $$v = 10\,\mathrm{m\,s^{-1}}$$. By the work–energy theorem over the entire 500 m journey,

$$W_g + W_r = \Delta K = \dfrac{1}{2}mv^2 - 0$$

$$\dfrac{1}{2}mv^2 = \dfrac{1}{2}(3.35\times 10^{-5})(10)^2 \approx 1.675\times 10^{-3}\,\mathrm{J}$$

Total work by gravity = $$0.082 + 0.082 = 0.164\,\mathrm{J}$$. Hence

$$W_r = \Delta K - W_g = 1.675\times 10^{-3} - 0.164 \approx -0.162\,\mathrm{J}$$

The negative sign confirms that the resistive force opposes the motion (and dissipates most of the gravitational work as heat).

Momentum. Resolve the molecule's velocity into a component normal to the wall and a component parallel to it:

$$v_\perp = v\cos 30^\circ, \qquad v_\parallel = v\sin 30^\circ$$

After rebound (same speed, mirror-reflected about the normal), the parallel component is unchanged while the normal component reverses sign. Hence the molecule's momentum does change — by $$\Delta p = 2 m v\cos 30^\circ$$ directed away from the wall — so momentum of the molecule alone is not conserved.

However, by Newton's third law the wall (and hence the container, which is effectively part of the molecule + wall system) receives an equal and opposite impulse. So for the closed system (molecule + wall) the total linear momentum is conserved. In this sense, momentum is conserved in the collision.

Energy. The molecule rebounds with the same speed, so its kinetic energy is unchanged:

$$K_f = \dfrac{1}{2}mv^2 = K_i$$

No kinetic energy is dissipated. The collision is therefore elastic.

Mass of water to be lifted (water density $$\rho = 1000\,\mathrm{kg\,m^{-3}}$$):

$$m = \rho V = (1000)(30) = 30000\,\mathrm{kg}$$

Useful work done (in lifting the water through $$h = 40\,\mathrm{m}$$ against gravity):

$$W_{\text{useful}} = mgh = (30000)(9.8)(40) = 1.176\times 10^{7}\,\mathrm{J}$$

Time taken: $$t = 15\,\mathrm{min} = 900\,\mathrm{s}$$. Useful (mechanical) power:

$$P_{\text{useful}} = \dfrac{W_{\text{useful}}}{t} = \dfrac{1.176\times 10^{7}}{900} \approx 1.307\times 10^{4}\,\mathrm{W}$$

The efficiency is $$\eta = 30\% = 0.30$$. The electric power consumed is

$$P_{\text{electric}} = \dfrac{P_{\text{useful}}}{\eta} = \dfrac{1.307\times 10^{4}}{0.30} \approx 4.36\times 10^{4}\,\mathrm{W}$$

So the pump consumes about $$43.6\,\mathrm{kW}$$ of electric power.

Let each ball have mass $$m$$. Initial momentum and initial kinetic energy of the system:

$$p_i = mV, \qquad K_i = \dfrac{1}{2}mV^2$$

Check each option (only option (ii) is reproduced here, since options (i) and (iii) fail):

Option (i): ball 1 stops; balls 2 and 3 move together with speed $$V/2$$.

$$p_f = 0 + 2m\cdot \dfrac{V}{2} = mV \;\checkmark, \quad K_f = \dfrac{1}{2}(2m)\left(\dfrac{V}{2}\right)^{\!2} = \dfrac{mV^2}{4}$$

Since $$K_f = K_i/2 \ne K_i$$, kinetic energy is not conserved — not consistent with an elastic collision.

Option (ii): balls 1 and 2 remain at rest; ball 3 moves off with speed $$V$$.

$$p_f = mV \;\checkmark, \quad K_f = \dfrac{1}{2}mV^2 = K_i \;\checkmark$$

Both momentum and kinetic energy are conserved — this is consistent with an elastic collision.

Option (iii): all three balls move together with speed $$V/3$$.

$$p_f = (3m)\left(\dfrac{V}{3}\right) = mV \;\checkmark, \quad K_f = \dfrac{1}{2}(3m)\left(\dfrac{V}{3}\right)^{\!2} = \dfrac{mV^2}{6}$$

$$K_f \ne K_i$$, so kinetic energy is not conserved — not elastic.

Hence only option (ii) is a possible result of the elastic collision: the incoming ball 1 stops, ball 2 remains at rest, and ball 3 moves off with the initial speed $$V$$ — the momentum and energy are 'passed through' the middle ball.

For an elastic one-dimensional collision between two equal masses, the standard result (target initially at rest) is

$$v_A' = \dfrac{m_A - m_B}{m_A + m_B}\,v_A = 0\,, \qquad v_B' = \dfrac{2m_A}{m_A + m_B}\,v_A = v_A$$

That is, the incoming ball A comes to rest and ball B moves off with the velocity that A had just before the collision. Since A has zero velocity immediately after the collision, it has zero kinetic energy and so cannot rise to any height — it simply remains at its lowest point.

Hence bob A rises to a height of $$0$$ after the collision.

(All of A's kinetic energy is transferred to B.)

Take the lowest point of the swing as the reference for potential energy. Released from the horizontal position, the bob falls through a height equal to the length of the pendulum, $$h = L = 1.5\,\mathrm{m}$$. The initial mechanical energy (entirely potential) is

$$E_i = mgh = mgL$$

5% of this is dissipated against air resistance; 95% is available as kinetic energy at the bottom:

$$\dfrac{1}{2}mv^2 = 0.95\,mgL$$

The mass cancels:

$$v = \sqrt{2 \times 0.95\,gL} = \sqrt{1.9 \times 9.8 \times 1.5}$$

$$v = \sqrt{27.93} \approx 5.28\,\mathrm{m\,s^{-1}}$$

Consider the trolley + sand as the system. As the sand leaks out, it falls vertically with respect to the trolley — it carries away exactly the horizontal velocity the trolley had at that instant. The leak does not exert any horizontal thrust on the trolley.

Since the track is frictionless and no external horizontal force acts on the trolley, the trolley's velocity is unchanged by the leaking process. (You can also see this from momentum conservation: the horizontal momentum of the falling sand equals the horizontal momentum it carried while on the trolley, so the trolley's momentum is unchanged.)

Therefore the speed of the trolley remains

$$v = 27\,\mathrm{km/h}$$

By the work–energy theorem, the work done by the net force equals the change in kinetic energy:

$$W_{\text{net}} = \Delta K = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2$$

Compute kinetic energy at each end using $$v = a x^{3/2}$$, i.e. $$v^2 = a^2 x^3$$:

At $$x = 0$$: $$v_i = 0 \Rightarrow K_i = 0$$.

At $$x = 2\,\mathrm{m}$$:

$$v_f^2 = (5)^2 (2)^3 = 25 \times 8 = 200\,\mathrm{m^2\,s^{-2}}$$

$$K_f = \dfrac{1}{2}(0.5)(200) = 50\,\mathrm{J}$$

Therefore

$$W_{\text{net}} = K_f - K_i = 50 - 0 = 50\,\mathrm{J}$$