NCERT Solutions · Class 11 Physics Work, Energy and Power

Work is a dot product, so only the force component along the direction of motion does any work.The force is given as \[\vec{F} = -\hat{i} + 2\hat{j} + 3\hat{k} \ \text{N} \] where \(\displaystyle \hat{i}, \hat{j}, \hat{k}\) point along the x-, y- and z-axes.The body moves only along the z-axis, through a distance of $\displaystyle 4$ m, so its displacement vector is \[\vec{d} = 4\hat{k} \ \text{m} \] (the x- and y-components of the displacement are both zero — the body never moves sideways).Work done by a constant force \[W = \vec{F} \cdot \vec{d} \]Taking the dot product component by component: \[W = (-1\ \text{N})(0\ \text{m}) + (2\ \text{N})(0\ \text{m}) + (3\ \text{N})(4\ \text{m}) \]The x- and y-terms vanish because the corresponding displacement components are zero — a common trap here is to add up all three force components regardless of which way the body actually moves, but a force perpendicular to the motion does zero work no matter how large it is. Only the z-component of \(\displaystyle \vec F\) survives: \[W = 3\ \text{N} \times 4\ \text{m} = 12\ \text{N·m} \]Since \(\displaystyle 1\ \text{N·m} = 1\ \text{J}\), \[W = 12\ \text{J} \]The force components and the distance ($\displaystyle 4$ m) are given as exact whole numbers in the problem, so the result is stated as the exact value $\displaystyle 12$ J, with no rounding needed.Answer: The work done is $\displaystyle 12$ J.

Kinetic energy set by a particle's mass and speed does not mean the higher-energy particle moves faster — mass matters just as much, and here the proton is nearly $\displaystyle 2000$ times heavier.Step $\displaystyle 1$: Convert the kinetic energies to joules.The energies are given in keV, so convert using \(\displaystyle 1 \text{ eV} = 1.60\times10^{-19} \text{ J} \):\[KE_e = 10 \text{ keV} = 10 \times 10^3 \times 1.60\times10^{-19}\text{ J} = 1.60\times10^{-15}\text{ J} \] \[KE_p = 100 \text{ keV} = 100 \times 10^3 \times 1.60\times10^{-19}\text{ J} = 1.60\times10^{-14}\text{ J} \]A common slip here is comparing $\displaystyle 10$ and $\displaystyle 100$ directly without converting to the same energy unit as the mass (kg) and using SI throughout — keV cannot be substituted straight into \(\displaystyle KE=\tfrac12 mv^2\).Step $\displaystyle 2$: Get speed from the kinetic energy formula.Kinetic energy: \(\displaystyle KE = \dfrac{1}{2}mv^2\), where \(\displaystyle m\) is the particle's mass and \(\displaystyle v\) its speed. Solving for \(\displaystyle v\):\[v = \sqrt{\dfrac{2\,KE}{m}} \]Electron (\(\displaystyle m_e = 9.11\times10^{-31}\text{ kg}\)):\[v_e = \sqrt{\dfrac{2\times1.60\times10^{-15}\text{ J}}{9.11\times10^{-31}\text{ kg}}} = \sqrt{\dfrac{3.20\times10^{-15}}{9.11\times10^{-31}}}\text{ m/s} = \sqrt{3.5126\times10^{15}}\text{ m/s} \] \[v_e \approx 5.9267\times10^{7}\text{ m/s} \]Proton (\(\displaystyle m_p = 1.67\times10^{-27}\text{ kg}\)):\[v_p = \sqrt{\dfrac{2\times1.60\times10^{-14}\text{ J}}{1.67\times10^{-27}\text{ kg}}} = \sqrt{\dfrac{3.20\times10^{-14}}{1.67\times10^{-27}}}\text{ m/s} = \sqrt{1.9162\times10^{13}}\text{ m/s} \] \[v_p \approx 4.3774\times10^{6}\text{ m/s} \]Since \(\displaystyle v_e > v_p\), the electron is faster, even though the proton carries ten times more kinetic energy — the electron's mass is about $\displaystyle 1835$ times smaller, and speed depends on \(\displaystyle 1/\sqrt{m}\), not on energy alone.Step $\displaystyle 3$: Take the ratio of speeds.Rather than dividing the rounded numbers above, it is cleaner (and avoids compounding rounding error) to combine the two expressions for \(\displaystyle v\) before evaluating:\[\frac{v_e}{v_p} = \sqrt{\frac{KE_e}{KE_p}\cdot\frac{m_p}{m_e}} = \sqrt{\frac{10}{100}\times\frac{1.67\times10^{-27}}{9.11\times10^{-31}}} \]\[= \sqrt{0.100 \times 1833.2} = \sqrt{183.3} \approx 13.5398 \]This matches the direct division of the two speeds found above (\(\displaystyle 5.9267\times10^{7} / 4.3774\times10^{6} = 13.54\)), confirming the numbers are consistent.Step $\displaystyle 4$: Apply significant figures.The input data (the kinetic energies and the masses) carry three significant figures, so the final results are rounded to three significant figures, only at this last step:\[v_e \approx 5.93\times10^{7}\text{ m/s}, \qquad v_p \approx 4.38\times10^{6}\text{ m/s}, \qquad \frac{v_e}{v_p} \approx 13.5 \]Answer: The electron is faster; \(\displaystyle v_e \approx 5.93\times10^{7}\text{ m/s}\), \(\displaystyle v_p \approx 4.38\times10^{6}\text{ m/s}\), and \(\displaystyle v_e : v_p \approx 13.5 : 1\).

A constant force's work depends only on the displacement along its own line of action, not on how fast (or how steadily) the object is moving — so the raindrop's weight does exactly the same work over each $\displaystyle 250$ m stretch, whether the drop is speeding up in the first half or moving at a steady terminal speed in the second.Step $\displaystyle 1$ — mass of the drop.The drop is a sphere of radius \(\displaystyle r = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \). Water has density \(\displaystyle \rho = 1.0 \times 10^{3} \, \text{kg m}^{-3} \) (the standard value, since the question itself does not restate it). With the volume of a sphere, \(\displaystyle V = \dfrac{4}{3}\pi r^{3} \):\[V = \frac{4}{3}\pi (2\times10^{-3}\,\text{m})^{3} = \frac{4}{3}\pi (8\times10^{-9}\,\text{m}^3) = 3.351\times10^{-8}\,\text{m}^3 \]\[m = \rho V = (1.0\times10^{3}\,\text{kg m}^{-3})(3.351\times10^{-8}\,\text{m}^3) = 3.351\times10^{-5}\,\text{kg} \]Step $\displaystyle 2$ — weight of the drop.Weight is a force, so it is measured in newtons, not kilograms — the \(\displaystyle m \) above is mass; multiplying by \(\displaystyle g \) turns it into the force gravity actually exerts. Using \(\displaystyle g = 9.8 \, \text{m s}^{-2} \):\[mg = (3.351\times10^{-5}\,\text{kg})(9.8\,\text{m s}^{-2}) = 3.284\times10^{-4}\,\text{N} \]Step $\displaystyle 3$ — work done by gravity in each half of the fall.The work done by a constant force over a displacement \(\displaystyle d \) along the force is \(\displaystyle W = F d \). Gravity pulls straight down with magnitude \(\displaystyle mg \), and the drop falls \(\displaystyle h/2 = 250 \, \text{m} \) in each half of the $\displaystyle 500$ m descent — first from $\displaystyle 500$ m to $\displaystyle 250$ m, then from $\displaystyle 250$ m down to the ground. Because the force and the displacement are identical in both halves, the work done is identical too, no matter what the speed was doing:\[W_{\text{gravity, 1st half}} = W_{\text{gravity, 2nd half}} = mg \times \frac{h}{2} = (3.284\times10^{-4}\,\text{N})(250\,\text{m}) = 8.21\times10^{-2}\,\text{J} \]It is tempting to think the second half — where the drop moves at constant (terminal) speed with zero acceleration — must involve "less work," but work done by gravity does not know about acceleration; it only multiplies the force by the distance fallen. Equal falls get equal shares of gravitational work.Step $\displaystyle 4$ — work done by the resistive force, from the work–energy theorem over the whole fall.The drop starts at rest (it "falls from a height") and reaches the ground at \(\displaystyle v = 10 \, \text{m s}^{-1} \), so its final kinetic energy is:\[KE_f = \frac{1}{2}mv^2 = \frac{1}{2}(3.351\times10^{-5}\,\text{kg})(10\,\text{m s}^{-1})^2 = 1.676\times10^{-3}\,\text{J} \]The work–energy theorem says the total work done by every force on the drop equals its change in kinetic energy. Two forces act over the full $\displaystyle 500$ m: gravity and the resistive (viscous) force of the air.\[W_{\text{gravity, total}} + W_{\text{resistive}} = KE_f - KE_i = KE_f - 0 \]The total work by gravity over the entire $\displaystyle 500$ m fall is just the sum of the two halves already found:\[W_{\text{gravity, total}} = mgh = (3.284\times10^{-4}\,\text{N})(500\,\text{m}) = 1.642\times10^{-1}\,\text{J} \]Solving for the resistive force's work:\[W_{\text{resistive}} = KE_f - W_{\text{gravity, total}} = 1.676\times10^{-3}\,\text{J} - 1.642\times10^{-1}\,\text{J} = -1.625\times10^{-1}\,\text{J} \]Rounding once, at the end, to three significant figures:\[W_{\text{resistive}} \approx -1.63\times10^{-1}\,\text{J} = -0.163\,\text{J} \]The negative sign says the resistive force takes energy away from the drop, which makes sense — a drag force always opposes the motion it acts against. In fact viscous drag removes almost all of the energy gravity supplies: without air resistance the drop would have arrived with kinetic energy equal to the full \(\displaystyle mgh = 0.164 \, \text{J} \) (a far higher speed); instead nearly all of that is dissipated, and only \(\displaystyle 1.68\times10^{-3}\,\text{J} \) of kinetic energy is left at \(\displaystyle 10 \, \text{m s}^{-1} \).Answer: Work done by gravity is the same in both halves — \(\displaystyle 8.21\times10^{-2}\,\text{J} \) (≈ $\displaystyle 0.082$ J) in the first half and \(\displaystyle 8.21\times10^{-2}\,\text{J} \) (≈ $\displaystyle 0.082$ J) in the second half; the work done by the resistive force over the entire journey is about \(\displaystyle -1.63\times10^{-1}\,\text{J} \) (−$\displaystyle 0.163$ J).

Momentum is not conserved for the molecule by itself — it is conserved only when you include the wall it bounces off. The collision itself is elastic, because the molecule's speed comes back unchanged.Set up axes at the point of impact: let \(\displaystyle y\) be along the normal to the wall (pointing away from it) and \(\displaystyle x\) be along the wall, in the plane containing the molecule's path. The molecule strikes at \(\displaystyle 30^\circ\) to the normal and rebounds like light off a mirror — angle of reflection equal to angle of incidence — so its velocity has these components.Before impact, with speed \(\displaystyle v = 200\ \text{m s}^{-1}\): \[v_x = v\sin30^\circ, \qquad v_y = -v\cos30^\circ \quad (\text{moving into the wall}) \]After rebound, same speed \(\displaystyle v = 200\ \text{m s}^{-1}\): \[v_x' = v\sin30^\circ, \qquad v_y' = +v\cos30^\circ \quad (\text{moving away from the wall}) \]The wall can only push the molecule perpendicular to its own surface — a smooth wall exerts no force along itself — so the component of velocity parallel to the wall, \(\displaystyle v_x\), cannot change. Only the normal component reverses. This is the step people miss: they expect "rebounds with the same speed" to mean nothing at all changes, but direction changes even though speed does not.Momentum of the molecule. For a mass \(\displaystyle m\), the change in momentum has components \[\Delta p_x = m(v_x' - v_x) = 0 \] \[\Delta p_y = m(v_y' - v_y) = m\big[v\cos30^\circ - (-v\cos30^\circ)\big] = 2mv\cos30^\circ \]Putting in the number (keeping the mass symbolic, since it is not given): \[2v\cos30^\circ = 2(200\ \text{m s}^{-1})(0.8660) = 346\ \text{m s}^{-1} \ \ (\text{3 s.f.}) \]So the molecule's own momentum changes by \(\displaystyle \Delta p = 2mv\cos30^\circ \approx 346\,m\ \text{kg m s}^{-1}\) (with \(\displaystyle m\) in kg), directed straight away from the wall along the normal. The molecule's momentum is not the same before and after — its direction has flipped along the normal, so as a vector it has clearly changed. Momentum conservation is a statement about an isolated system, not a guarantee that any single object's momentum survives a collision; here the molecule is not isolated, so there is no reason to expect its own momentum to be unchanged.Momentum of the whole system. By Newton's third law, whatever momentum the molecule loses along the normal, the wall gains — the wall (rigidly attached to the container, effectively to the Earth) picks up momentum \(\displaystyle 2mv\cos30^\circ\) directed into it. Because the wall's effective mass is enormous compared to a gas molecule, this shows up as an immeasurably small recoil velocity, but the momentum is real and exactly cancels the molecule's loss. So total momentum of the (molecule + wall) system, before collision equals after collision — momentum conservation holds for the system, just not for the molecule taken alone.Elastic or inelastic. A collision is called elastic when kinetic energy is conserved, which is a separate question from whether momentum of one piece is conserved. The molecule's kinetic energy depends only on its speed, \[KE = \tfrac{1}{2}mv^2, \] and the speed is unchanged — \(\displaystyle 200\ \text{m s}^{-1}\) in, \(\displaystyle 200\ \text{m s}^{-1}\) out — so \(\displaystyle KE_i = KE_f\). Kinetic energy is conserved even though the direction of motion (and hence the momentum vector) is not; a bounce can reverse direction completely and still be elastic, as long as speed is preserved.Answer: The molecule's own momentum is not conserved — it changes by \(\displaystyle 2mv\cos30^\circ \approx 346\,m\ \text{kg m s}^{-1}\) directed away from the wall along the normal — but the momentum of the (molecule + wall) system is conserved, since the wall gains an equal and opposite momentum. The collision is elastic, because the molecule's speed (and hence its kinetic energy) is unchanged, \(\displaystyle 200\ \text{m s}^{-1}\) before and after.

Efficiency tells you the electrical power is bigger than the useful mechanical power, not the other way round — so you divide by $\displaystyle 0.30$, you don't multiply by it.Step $\displaystyle 1$ — find the mass of water that must be lifted.The tank holds a volume \(\displaystyle V = 30\ \text{m}^3 \) of water. Using the density of water, \(\displaystyle \rho = 1000\ \text{kg/m}^3 \):\[m = \rho V = 1000\ \text{kg/m}^3 \times 30\ \text{m}^3 = 3.0\times10^{4}\ \text{kg} \]Step $\displaystyle 2$ — find the useful work done in lifting it.Lifting a mass \(\displaystyle m \) through a height \(\displaystyle h \) against gravity requires work \[W = mgh \] where \(\displaystyle g = 9.8\ \text{m/s}^2 \) is the acceleration due to gravity and \(\displaystyle h = 40\ \text{m} \) is the height of the tank above the pump. (This is potential energy gained, not the pump's own energy use — that distinction is exactly what the efficiency in Step $\displaystyle 4$ will fix.)\[W = (3.0\times10^{4}\ \text{kg})(9.8\ \text{m/s}^2)(40\ \text{m}) = 1.176\times10^{7}\ \text{J} \]Step $\displaystyle 3$ — convert this to a useful (output) power.Power is work done per unit time, and the time must be in seconds, not minutes: \[t = 15\ \text{min} = 15\times60\ \text{s} = 900\ \text{s} \]\[P_{\text{useful}} = \frac{W}{t} = \frac{1.176\times10^{7}\ \text{J}}{900\ \text{s}} = 1.3067\times10^{4}\ \text{W} \]This is the mechanical power actually delivered to the water — it is not the electrical power the pump draws from the mains, because the pump is only $\displaystyle 30$% efficient.Step $\displaystyle 4$ — use the efficiency to get the electrical power consumed.Efficiency is defined as \[\eta = \frac{\text{useful power output}}{\text{power input (consumed)}} \]so the power the pump actually consumes is larger than the useful output:\[P_{\text{input}} = \frac{P_{\text{useful}}}{\eta} = \frac{1.3067\times10^{4}\ \text{W}}{0.30} = 4.3556\times10^{4}\ \text{W} \]Step $\displaystyle 5$ — round to the correct number of significant figures.Every given quantity — \(\displaystyle 30\ \text{m}^3\), \(\displaystyle 15\ \text{min}\), \(\displaystyle 40\ \text{m}\), \(\displaystyle 30\%\) — carries $\displaystyle 2$ significant figures, so the final power should also be quoted to $\displaystyle 2$ significant figures:\[P_{\text{input}} \approx 4.4\times10^{4}\ \text{W} = 44\ \text{kW} \]Answer: The pump consumes about \(\displaystyle 4.4\times10^{4}\ \text{W}\) (\(\displaystyle \approx 44\ \text{kW}\)) of electric power.

In a head-on elastic collision between two equal masses where one is initially at rest, the velocities simply swap — the incoming ball stops dead, and the ball it hit shoots off with the incoming ball's full speed.Call the three identical ball bearings — each of mass \(\displaystyle m \) — A, B and C. A arrives with speed \(\displaystyle V \) and strikes B first; B and C start out touching, both at rest.The key formula. For a $\displaystyle 1$-D elastic collision between mass \(\displaystyle m_1 \) (initial speed \(\displaystyle u_1 \)) and mass \(\displaystyle m_2 \) (initial speed \(\displaystyle u_2 \)), momentum and kinetic energy conservation together give the standard result: \[v_1 = \frac{m_1-m_2}{m_1+m_2}\,u_1 + \frac{2m_2}{m_1+m_2}\,u_2, \qquad v_2 = \frac{2m_1}{m_1+m_2}\,u_1 + \frac{m_2-m_1}{m_1+m_2}\,u_2 \] Here \(\displaystyle v_1, v_2 \) are the final speeds. Setting \(\displaystyle m_1=m_2=m \) and \(\displaystyle u_2=0 \) (the target starts at rest) collapses this to \(\displaystyle v_1=0,\ v_2=u_1 \) — equal masses exchange velocity completely.Step $\displaystyle 1$ — A hits B. Before: A moves at \(\displaystyle V \), B is at rest. Since \(\displaystyle m_A=m_B \), A stops and B takes up the full speed: \[v_A = 0, \qquad v_B = V \]Step $\displaystyle 2$ — B hits C. Because B and C were already touching, the instant B is moving at \(\displaystyle V \) it strikes C, which is still at rest right beside it. This is the same equal-mass, one-at-rest collision again, so B stops and C takes up the speed: \[v_B = 0, \qquad v_C = V \]Net result immediately after the whole event: A and B are both at rest (touching each other where B stopped), and C alone rolls away with speed \(\displaystyle V \) — the exact speed A started with.Check against the two conservation laws (this is the test that tells you which picture in Fig. $\displaystyle 5.14$ can be right):Momentum before: only A is moving, so \(\displaystyle p_i = mV \). Momentum after: \(\displaystyle p_f = m(0) + m(0) + m(V) = mV \). ✓ matches.Kinetic energy before: \(\displaystyle KE_i = \tfrac{1}{2}mV^2 \). Kinetic energy after: \(\displaystyle KE_f = 0 + 0 + \tfrac{1}{2}mV^2 = \tfrac{1}{2}mV^2 \). ✓ matches — this is what makes the collision elastic, as the problem requires.Why the "obvious" alternative fails. A common guess is that B and C share the blow equally, each moving off at \(\displaystyle V/2 \). Check it: momentum \(\displaystyle = m(V/2)+m(V/2) = mV \) — conserved. But kinetic energy \(\displaystyle = 2\times\tfrac12 m(V/2)^2 = \tfrac14 mV^2 \), only half the initial value. Energy that vanished can't reappear in a collision stated to be elastic, so this outcome is impossible here (it would need part of the collision to be inelastic). The same failure — momentum balances but kinetic energy comes up short — rules out any picture where the balls end up moving together or split three ways.So among the results drawn in Fig. $\displaystyle 5.14$, the only one consistent with both conservation laws for an elastic collision is the one showing the first two ball bearings left at rest, touching each other, while the third ball bearing moves off alone with the original speed \(\displaystyle V \).**Answer: the ball bearing that was hit and the one that hit it both end up at rest (touching each other), while the last ball bearing moves off alone with the original speed \(\displaystyle V \) — this is the only picture that conserves both momentum ( \(\displaystyle mV \) ) and kinetic energy ( \(\displaystyle \tfrac12 mV^2 \) ), as an elastic collision must.

In an elastic collision between two equal masses where one starts at rest, the moving object stops dead and hands over its entire velocity to the one it hits. That single fact settles this problem before you even need the release angle.Step $\displaystyle 1$ — Speed of A just before impact. Let bob A have mass \(\displaystyle m\) and let the pendulum have length \(\displaystyle L\). Released from rest at \(\displaystyle 30^{o}\) to the vertical, A falls through a height \[h_{0}=L(1-\cos 30^{o}) \] Using conservation of mechanical energy (loss in PE = gain in KE) for the swing down to the lowest point, where it strikes B: \[mgh_{0}=\tfrac12 mu^{2}\ \Rightarrow\ u=\sqrt{2gh_{0}} \] Here \(\displaystyle u\) is A's speed at the bottom, an instant before collision. You will see in Step $\displaystyle 3$ that its actual value never has to be computed — the length \(\displaystyle L\) is not even given in the problem, which is itself a hint that it will cancel out.Step $\displaystyle 2$ — Set up the elastic collision. At the lowest point A (moving at speed \(\displaystyle u\)) collides elastically with B (mass \(\displaystyle m\), the same as A, at rest on the table). Let \(\displaystyle v_{A}\) and \(\displaystyle v_{B}\) be their speeds right after collision, both measured along the same line (the collision is head-on).Conservation of momentum: \[mu=mv_{A}+mv_{B}\ \Rightarrow\ u=v_{A}+v_{B} \qquad (1) \]Conservation of kinetic energy (this is what "elastic" means): \[\tfrac12 mu^{2}=\tfrac12 mv_{A}^{2}+\tfrac12 mv_{B}^{2}\ \Rightarrow\ u^{2}=v_{A}^{2}+v_{B}^{2} \qquad (2) \]Step $\displaystyle 3$ — Solve ($\displaystyle 1$) and ($\displaystyle 2$) together. Square equation ($\displaystyle 1$): \[u^{2}=v_{A}^{2}+2v_{A}v_{B}+v_{B}^{2} \] Subtract equation ($\displaystyle 2$) from this: \[0=2v_{A}v_{B}\ \Rightarrow\ v_{A}v_{B}=0 \] So either \(\displaystyle v_{A}=0\) or \(\displaystyle v_{B}=0\). The root \(\displaystyle v_{B}=0\) just says "nothing happened" (A sails through untouched), which is not a real collision — mass B was struck and cannot stay at rest. So the physical solution is \[v_{A}=0,\qquad v_{B}=u \] This is the general equal-mass elastic-collision result: the incoming ball stops, and the target ball leaves with the incoming ball's original speed. It is worth naming because it is the step most students try to "compute" instead of deriving — for equal masses it is not approximate, it is exact.Step $\displaystyle 4$ — Convert A's post-collision speed into a rise in height. Bob A leaves the collision point with speed \(\displaystyle v_{A}=0\). Applying energy conservation to A's motion after the collision (KE at the bottom converts back to PE at the top of its rise, height \(\displaystyle h\)): \[\tfrac12 mv_{A}^{2}=mgh\ \Rightarrow\ h=\dfrac{v_{A}^{2}}{2g}=\dfrac{0}{2g}=0 \]Because \(\displaystyle v_A=0\) exactly, this is one of the rare NCERT numbers that is exact rather than a rounded measurement — there is no significant-figure question here. Notice also that the \(\displaystyle 30^{o}\) release angle, the length \(\displaystyle L\), and the value of \(\displaystyle u\) all cancelled out along the way: for any release angle, an elastic head-on collision between equal masses always leaves the striking bob at rest. That is the physical insight the "$\displaystyle 30$°" in the question is there to tempt you into using unnecessarily.So bob A does not swing back up at all. It comes to rest exactly at the bottom of its arc (hanging vertically, in contact with B), while bob B moves off along the table with the speed A used to have.Answer: Bob A rises to a height of $\displaystyle 0$ — it stops completely at the lowest point of the swing, because for an elastic collision between equal masses the moving bob transfers all its velocity (and hence all its kinetic energy) to the bob it strikes.

The bob loses height as it swings, and only $\displaystyle 95$% of that lost gravitational PE shows up as kinetic energy at the bottom — the rest is eaten by air resistance.Setting up the height dropThe bob starts with the string held horizontal, so the bob is level with the pivot. At the lowest point of the swing, it hangs straight down. The vertical distance it has fallen is exactly the length of the pendulum: \[h = l = 1.5\ \text{m} \]This is the one geometric fact the problem needs — no angle or arc length is involved, only the drop in height between the start and the bottom of the swing.Applying energy conservation with a loss termBy the law of conservation of energy, the gravitational potential energy given up as the bob falls either becomes kinetic energy or is dissipated as heat/sound by air resistance: \[mgh = \left(\tfrac{1}{2}mv^2\right)_{\text{gained}} + E_{\text{dissipated}} \] Here \(\displaystyle m\) is the bob's mass, \(\displaystyle g\) is the acceleration due to gravity (\(\displaystyle 9.8\ \text{m s}^{-2}\), not to be confused with the gravitational constant \(\displaystyle G\)), and \(\displaystyle v\) is the speed at the lowest point.Since $\displaystyle 5$% of the initial energy \(\displaystyle mgh\) is dissipated, the fraction converted to kinetic energy is $\displaystyle 95$%: \[\tfrac{1}{2}mv^2 = 0.95\, mgh \]The mass \(\displaystyle m\) appears on both sides and cancels — this is why the problem never states the bob's mass; the answer cannot depend on it.Solving for v\[v^2 = 2(0.95)gh = 1.9\,gh \] \[v = \sqrt{1.9\,g\,h} \]Substituting \(\displaystyle g = 9.8\ \text{m s}^{-2}\) and \(\displaystyle h = 1.5\ \text{m}\): \[v = \sqrt{1.9 \times 9.8\ \text{m s}^{-2} \times 1.5\ \text{m}} = \sqrt{27.93\ \text{m}^2\text{s}^{-2}} \] \[v = 5.285\ \text{m s}^{-1} \]RoundingThe length \(\displaystyle l = 1.5\ \text{m}\) carries only two significant figures, so that limits the precision of the final answer. Rounding \(\displaystyle 5.285\ \text{m s}^{-1}\) to two significant figures: \[v \approx 5.3\ \text{m s}^{-1} \]This is a speed (a magnitude only) — at the lowest point of the swing the bob's velocity points horizontally, tangent to its circular path, but the question asks only for how fast it is moving, not the direction, so no vector needs to be quoted.Answer: The bob arrives at the lowest point with speed \(\displaystyle v \approx 5.3\ \text{m s}^{-1}\) (obtained from \(\displaystyle v=\sqrt{1.9gl}\), unrounded value \(\displaystyle 5.285\ \text{m s}^{-1}\)).

The speed does not change — the leaking sand carries no horizontal push, so it cannot slow the trolley down.Take the trolley and the sand inside it together as one system moving on a frictionless track. Because the track is frictionless, there is no external horizontal force on this system, so its total horizontal momentum is conserved — this is Newton's first law applied to the system as a whole (no net external force means no change in the total momentum).Now look at how the sand actually leaves. It leaks out through a hole in the floor of the trolley — that is, it falls vertically relative to the trolley. At the instant a grain of sand leaves, it is still moving horizontally at exactly the trolley's own speed \(\displaystyle v\); the trolley exerts no horizontal force on it as it drops out, and by Newton's third law the sand exerts no horizontal reaction force back on the trolley either.This is the detail that trips people up: leaking mass only changes a body's speed (as in rocket propulsion) when the ejected mass leaves with some horizontal velocity different from the body's own velocity, because then it carries away (or gives back) horizontal momentum unevenly with respect to the remaining mass. Here that relative horizontal velocity is zero — the sand's horizontal motion when it exits is identical to the trolley's — so every grain that leaves carries away exactly its own share of momentum at the trolley's current speed, and none in excess or deficit.To see this formally, let \(\displaystyle M(t)\) be the combined mass of the trolley and remaining sand at time \(\displaystyle t\), and \(\displaystyle v(t)\) its speed. The rate of change of the system's momentum is \[\frac{d}{dt}\big(M(t)\,v(t)\big) = M\frac{dv}{dt} + v\frac{dM}{dt}. \] The term \(\displaystyle v\,\dfrac{dM}{dt}\) is exactly the momentum carried away by the leaking sand, because that sand leaves at the same speed \(\displaystyle v\) as the trolley (no horizontal ejection velocity relative to the trolley, unlike a rocket exhaust). Since there is no external horizontal force, \[\frac{d}{dt}\big(M(t)v(t)\big) = v\frac{dM}{dt} \quad\Rightarrow\quad M\frac{dv}{dt} = 0 \quad\Rightarrow\quad \frac{dv}{dt} = 0. \]So \(\displaystyle v\) never changes while sand is leaking, no matter how much mass is lost or how fast it leaks — the rate \(\displaystyle 0.05\ \text{kg s}^{-1}\) and the sandbag's mass of \(\displaystyle 25\ \text{kg}\) (which only fix how long, \(\displaystyle 500\ \text{s}\), the leaking lasts) never enter the answer at all. Once the bag is empty, the trolley is simply moving at the same uniform speed it always had.\[v_{\text{final}} = v_{\text{initial}} = 27\ \text{km h}^{-1} = 27 \times \frac{5}{18}\ \text{m s}^{-1} = 7.5\ \text{m s}^{-1}. \]Answer: The trolley's speed stays unchanged at $\displaystyle 27$ km/h ($\displaystyle 7.5$ m/s) — leaking sand exerts no horizontal force on the trolley since it falls out with the trolley's own horizontal velocity, so momentum conservation on the frictionless track requires no change in speed.

The net force does work equal to the change in kinetic energy — you don't need to find the force at all.This is the work–energy theorem: for any net force acting on a body, \[W_{\text{net}} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \] where \(\displaystyle m\) is the mass, \(\displaystyle v_i\) is the speed at the start of the displacement, and \(\displaystyle v_f\) is the speed at the end. It holds no matter how the force varies along the path, so you never have to integrate \(\displaystyle F\,dx\) — just evaluate the velocity at the two positions.Step $\displaystyle 1$: Write down what's given. \[v = a\,x^{3/2}, \qquad a = 5\ \text{m}^{-1/2}\text{s}^{-1}, \qquad m = 0.5\ \text{kg} \] The displacement is from \(\displaystyle x = 0\) to \(\displaystyle x = 2\ \text{m}\).Check the units of \(\displaystyle a\) so the velocity comes out right: \(\displaystyle x^{3/2}\) carries units of \(\displaystyle \text{m}^{3/2}\), so \[[a][x^{3/2}] = (\text{m}^{-1/2}\text{s}^{-1})(\text{m}^{3/2}) = \text{m}\,\text{s}^{-1} \] which is indeed a speed. This confirms \(\displaystyle a\)'s odd-looking units are exactly what's needed to make \(\displaystyle v = ax^{3/2}\) dimensionally consistent — not a typo.Step $\displaystyle 2$: Find the speed at each endpoint.At \(\displaystyle x = 0\): \[v_i = a(0)^{3/2} = 0 \]At \(\displaystyle x = 2\ \text{m}\): \[v_f = a(2)^{3/2} = 5 \times 2^{3/2}\ \text{m/s} \] Since \(\displaystyle 2^{3/2} = 2\sqrt{2} = 2.828\), it's easier to keep \(\displaystyle v_f^2\) in exact form rather than round \(\displaystyle v_f\) itself: \[v_f^2 = a^2 x^3 = (5)^2 (2)^3 = 25 \times 8 = 200\ \text{m}^2/\text{s}^2 \] (Squaring \(\displaystyle a^2x^3\) directly avoids the mid-calculation rounding of an irrational square root — round only the final answer.)Step $\displaystyle 3$: Apply the work–energy theorem. \[W_{\text{net}} = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2 = \frac{1}{2}(0.5\ \text{kg})(200\ \text{m}^2/\text{s}^2) - 0 \] \[W_{\text{net}} = \frac{1}{2}(0.5)(200)\ \text{J} = 50\ \text{J} \]The mass ($\displaystyle 0.5$ kg) is given to one significant figure and \(\displaystyle a\) is given to one significant figure, so the result is stated to the same precision: $\displaystyle 50$ J (no further decimal places are justified).A common mistake here is to try to find \(\displaystyle F(x)\) from \(\displaystyle v(x)\) using \(\displaystyle F = ma = m\,\dfrac{dv}{dt}\) and then integrate \(\displaystyle F\,dx\) — that route works too, but it's far more algebra for the same answer, because the work–energy theorem already packages that integral into just the initial and final speeds.Answer: $\displaystyle 50$ J

Picture the wind as a moving cylinder of air — in time \(\displaystyle t\), a column of length \(\displaystyle vt\) and cross-section \(\displaystyle A\) sweeps through the blades.(a) Mass of air passing through in time \(\displaystyle t\)In time \(\displaystyle t\), every particle of air in front of the blades moves a distance \(\displaystyle vt\) (distance = speed × time). So the air that has passed through the circle of area \(\displaystyle A\) forms a cylinder of length \(\displaystyle vt\) and cross-sectional area \(\displaystyle A\).Volume of this cylinder: \[V = A \times (vt) = Avt \]Using density \(\displaystyle \rho = \dfrac{\text{mass}}{\text{volume}}\), the mass of air in this volume is\[m = \rho V = \rho A v t \]Here \(\displaystyle \rho\) is the density of air, \(\displaystyle A\) is the swept area, \(\displaystyle v\) is the wind speed, and \(\displaystyle t\) is the time elapsed. Every quantity on the right is something you are given or can measure, so this is the mass that has flowed through the blades.(b) Kinetic energy of this airThe law here is the definition of kinetic energy, \(\displaystyle KE = \dfrac{1}{2}mv^2\), where \(\displaystyle m\) is the mass of the moving object and \(\displaystyle v\) its speed. Substituting the mass found in part (a):\[KE = \frac{1}{2}mv^2 = \frac{1}{2}(\rho A v t)v^2 \]\[KE = \frac{1}{2}\rho A v^3 t \]Notice the wind speed enters as \(\displaystyle v^3\), not \(\displaystyle v^2\) — one factor of \(\displaystyle v\) comes from how much air arrives per second, and two more come from the \(\displaystyle v^2\) in the kinetic-energy formula. This is why wind power is so sensitive to wind speed.(c) Electrical power producedPower is energy delivered per unit time, \(\displaystyle P = \dfrac{\text{energy}}{\text{time}}\). Dividing the kinetic energy from part (b) by \(\displaystyle t\) gives the power available in the wind:\[P_{\text{wind}} = \frac{KE}{t} = \frac{1}{2}\rho A v^3 \]Only $\displaystyle 25$% of this is converted to electrical energy, so the electrical power is\[P_{\text{elec}} = 0.25 \times \frac{1}{2}\rho A v^3 = \frac{1}{8}\rho A v^3 \]Before substituting, convert the wind speed to SI units — this is the step it is easy to slip on, since mixing km/h with metres and seconds elsewhere would give a wrong answer by a factor of \(\displaystyle 3.6\): \[v = 36 \text{ km/h} = 36 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 10 \text{ m/s} \]Now substitute \(\displaystyle \rho = 1.2\ \text{kg m}^{-3}\), \(\displaystyle A = 30\ \text{m}^2\), and \(\displaystyle v = 10\ \text{m/s}\):\[P_{\text{elec}} = \frac{1}{8}(1.2\ \text{kg m}^{-3})(30\ \text{m}^2)(10\ \text{m/s})^3 \]\[P_{\text{elec}} = \frac{1}{8}(1.2)(30)(1000)\ \text{kg m}^{-1}\text{s}^{-3} \cdot \text{m}^2 \cdot \text{m}^3\text{s}^{-3} \]Working through the arithmetic: \(\displaystyle 1.2 \times 30 = 36\), then \(\displaystyle 36 \times 1000 = 36000\), and \(\displaystyle 36000/8 = 4500\).\[P_{\text{elec}} = 4500\ \text{W} \]The given data (\(\displaystyle A = 30\ \text{m}^2\), \(\displaystyle v = 36\ \text{km/h}\), \(\displaystyle \rho = 1.2\ \text{kg m}^{-3}\)) each carry two significant figures, so the final answer should be rounded to two significant figures:\[P_{\text{elec}} = 4.5 \times 10^{3}\ \text{W} = 4.5\ \text{kW} \]Answer: (a) \(\displaystyle m = \rho A v t\); (b) \(\displaystyle KE = \dfrac{1}{2}\rho A v^3 t\); (c) \(\displaystyle P_{\text{elec}} = 4.5\ \text{kW}\) (\(\displaystyle 4500\ \text{W}\)).

Lifting the mass slowly means all the work she does becomes gravitational potential energy — no kinetic energy is left over, so \(\displaystyle W = \Delta PE = mgh \) for one lift.Part (a): Work done against gravityEach time she raises the $\displaystyle 10$ kg mass by $\displaystyle 0.5$ m, the work done against gravity equals the gain in potential energy, \[W_1 = mgh \] where \(\displaystyle m\) is the mass, \(\displaystyle g\) is the acceleration due to gravity, and \(\displaystyle h\) is the height raised. This is the same law that gives a body's PE near the Earth's surface — note that \(\displaystyle g\) (acceleration, m/s²) here is not to be confused with \(\displaystyle G\), the universal gravitational constant.Substituting \(\displaystyle m = 10\text{ kg}\), \(\displaystyle g = 9.8\text{ m/s}^2\), \(\displaystyle h = 0.5\text{ m}\): \[W_1 = 10 \times 9.8 \times 0.5 = 49\text{ J} \]She repeats this \(\displaystyle n = 1000\) times, so the total work done against the gravitational force is \[W = nW_1 = 1000 \times 49\text{ J} = 49000\text{ J} = 4.9 \times 10^{4}\text{ J} \]Note that this is the work done lifting the mass each time — the problem tells us the PE gained on each lift is lost (dissipated as heat in muscles/joints) when she lowers it back down, so this energy is not recovered. It has to be supplied afresh by her body on every single lift.Part (b): Fat used upThe $\displaystyle 20$% efficiency means only one-fifth of the chemical energy released from fat actually turns into the mechanical work \(\displaystyle W\) found above; the rest is lost as heat. So the total energy her body must draw from fat is \[E = \frac{W}{\text{efficiency}} = \frac{4.9 \times 10^{4}\text{ J}}{0.20} \] \[E = 2.45 \times 10^{5}\text{ J} \]Each kilogram of fat supplies \(\displaystyle 3.8 \times 10^{7}\text{ J}\) of energy, so the mass of fat used is this required energy divided by the energy content per kilogram: \[m_{\text{fat}} = \frac{E}{3.8 \times 10^{7}\text{ J/kg}} = \frac{2.45 \times 10^{5}\text{ J}}{3.8 \times 10^{7}\text{ J/kg}} \] \[m_{\text{fat}} = 0.0064473\ldots\text{ kg} \]Rounding to three significant figures (matching the three-figure data given for the efficiency and energy content), \[m_{\text{fat}} \approx 6.45 \times 10^{-3}\text{ kg} \]That is only about $\displaystyle 6.45$ grams of fat — a reminder of how much mechanical work the body must do to burn even a small amount of stored fat, precisely because most of the chemical energy is wasted as heat rather than turned into useful work.Answer: (a) \(\displaystyle 4.9 \times 10^{4}\text{ J}\) of work is done against gravity. (b) About \(\displaystyle 6.45 \times 10^{-3}\text{ kg}\) (≈ $\displaystyle 6.45$ g) of fat is used up.

Power in equals power out — the panel's electrical output can never exceed the sunlight falling on it times the conversion efficiency.Step $\displaystyle 1$ — Write down what "$\displaystyle 8$ kW" means as power, in watts. The family's power requirement is \[P_{\text{required}} = 8\ \text{kW} = 8 \times 10^{3}\ \text{W} = 8000\ \text{W} \]Step $\displaystyle 2$ — Find how much useful (electrical) power each square metre of sunlit roof actually delivers. Sunlight arrives at the rate \(\displaystyle I = 200\ \text{W/m}^2 \) (this is an intensity — power per unit area — not a power by itself). Only $\displaystyle 20$% of that gets converted to electricity, so the useful power density is \[P_{\text{useful per m}^2} = \eta \, I = 0.20 \times 200\ \text{W/m}^2 = 40\ \text{W/m}^2 \] This is the step people skip: you cannot divide $\displaystyle 8000$ W by $\displaystyle 200$ W/m² directly, because $\displaystyle 200$ W/m² is the raw solar rate, not the electrical rate the family can actually draw on.Step $\displaystyle 3$ — Divide the required power by the useful power per square metre to get the area. Since power scales with area, \(\displaystyle P_{\text{required}} = P_{\text{useful per m}^2} \times A \), so \[A = \frac{P_{\text{required}}}{P_{\text{useful per m}^2}} = \frac{8000\ \text{W}}{40\ \text{W/m}^2} = 200\ \text{m}^2 \]Both numbers going into this division ($\displaystyle 0.20$ and $\displaystyle 200$ W/m²) carry two significant figures, so the area is quoted the same way: \[A = 200\ \text{m}^2 = 2.0 \times 10^{2}\ \text{m}^2 \]Step $\displaystyle 4$ — Compare this to an actual rooftop. A single-storey house of modest size — say \(\displaystyle 10\ \text{m} \times 10\ \text{m}\) in plan — has a roof of around \(\displaystyle 100\text{–}150\ \text{m}^2\) once you allow for the slope. The \(\displaystyle 200\ \text{m}^2\) worked out above is therefore larger than the whole roof of such a house — roughly one-and-a-half to two times it. In other words, covering only the rooftop of a typical home with $\displaystyle 20$%-efficient panels would fall short of meeting the family's full $\displaystyle 8$ kW demand; you would need a roof close to the size of a fairly large house (about \(\displaystyle 14\ \text{m} \times 14\ \text{m}\)), or extra ground-mounted panels besides the roof, to collect the full requirement from sunlight alone.Answer: (a) \(\displaystyle A = 200\ \text{m}^2\) (i.e., \(\displaystyle 2.0 \times 10^{2}\ \text{m}^2\)); (b) this is comparable to — in fact somewhat larger than — the entire roof of a typical single-storey house, so the roof of an average home is not quite big enough to supply the family's full $\displaystyle 8$ kW this way.