Question 1

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 man 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.

Accepted Answer

(a) +ve (b) -ve (c) -ve (d) + ve (e) - ve. The sign of work depends only on the angle between the force and the displacement — not on how big or "important" the force is. Work done by a force F on a body that undergoes a displacement d is W = F · d = Fd cos θ where θ is the angle between the direction of the force and the direction of displacement. If θ < 90^° (force has a component along the motion), W is positive; if θ 90^° (force opposes the motion), W is negative; if θ = 90^°, W = 0. Go through each case by asking only: which way does this particular force point, and which way does the body actually move? (a) Work done by the man in lifting the bucket The man pulls the rope upward, and the bucket also moves upward (out of the well). The applied force and the displacement point the same way, so θ = 0^° and cos θ = 1. This work is positive. (b) Work done by the gravitational force in the same case Gravity on the bucket always acts vertically downward (it is the bucket's weight, mg, pulling toward the Earth's centre — never confuse this downward force with the man's upward pull). The bucket's displacement is upward. Here θ = 180^° and cos θ = -1. This work is negative. (c) Work done by friction on a body sliding down an inclined plane Kinetic friction always opposes the relative sliding of the surfaces in contact, never assists it. The body slides down the incline, so friction acts up along the incline, directly opposite to the body's displacement. That gives θ = 180^°. This work is negative. (d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity Uniform velocity means zero acceleration, so by Newton's second law the net force on the body is zero. Friction acts backward (opposing the motion); for the net force to vanish, the applied force must act forward, in the same direction as the displacement, to exactly balance friction. So the applied force and displacement are both along the direction of motion: θ = 0^°. This work is positive. (Note: even though the body's kinetic energy is not changing — uniform velocity — the applied force is still doing positive work; that work is exactly canceled by the negative work friction does, consistent with the work–energy theorem giving zero net work.) (e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest A resistive (drag) force always acts opposite to the instantaneous velocity of the body — that is what makes it resistive. At every instant of the pendulum's swing, the air resistance points opposite to the pendulum's motion, so θ = 180^° throughout. This is exactly why the resistive force steadily removes kinetic energy and brings the pendulum to rest. This work is negative. Answer: (a) positive, (b) negative, (c) negative, (d) positive, (e) negative.

Question 2

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 kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, work done by the net force on the body in 10 s, (d) change in kinetic energy of the and interpret your results.

Accepted Answer

(a) 882 J; (b) -247 J; (c) 635 J; (d) 635 J; Work done by the net force on a body equals change in its kinetic energy.. Work is force times the distance actually covered, so before computing any of the four quantities you first need how far the body travels in 10 s — and that means finding the acceleration from the net force, not the applied force alone. Step 1 — Find the friction force. On a horizontal table the normal force N balances the body's weight, so N = mg — this is the point people get wrong: friction depends on the normal force, which comes from weight ( mg ), not from mass alone. Kinetic friction: f = μ k N = μ k mg f = 0.1 × 2 kg × 9.8 m/s^2 = 1.96 N Friction acts opposite to the motion (the applied force pushes the body forward, so friction points backward). Step 2 — Find the net force and acceleration. Newton's second law, F net = ma, with the applied force F = 7 N and friction f = 1.96 N acting in opposite directions: F net = F - f = 7 N - 1.96 N = 5.04 N a = (F net)/m = (5.04 N )/(2 kg ) = 2.52 m/s^2 Step 3 — Find the distance covered in t = 10 s. The body starts from rest ( u = 0 ), so from s = ut + 1/2at^2: s = 0 + 1/2(2.52 m/s^2)(10 s )^2 = 1/2(2.52)(100) m = 126 m (a) Work done by the applied force Work W = F × s (force and displacement point the same way, so no cosine factor is needed): W a = 7 N × 126 m = 882 J (b) Work done by friction Friction acts opposite to the displacement, so its work is negative: W f = -f × s = -1.96 N × 126 m = -246.96 J ≈ -247 J The negative sign is not optional bookkeeping — it says friction removes mechanical energy from the body (converts it to heat), which is why it must be reported as negative, not just as a magnitude. (c) Work done by the net force W net = F net × s = 5.04 N × 126 m = 635.04 J ≈ 635 J Equivalently, W net = W a + W f = 882 J - 246.96 J = 635.04 J — same number, because work adds like any other scalar contribution to the net force. (d) Change in kinetic energy First find the final speed using v = u + at: v = 0 + (2.52 m/s^2)(10 s ) = 25.2 m/s Kinetic energy, KE = 1/2mv^2, with initial KE = 0 since the body starts at rest: Δ KE = 1/2(2 kg )(25.2 m/s )^2 - 0 = 1/2(2)(635.04) J = 635.04 J ≈ 635 J Interpretation. Δ KE comes out equal to W net (both 635 J ) — this is the work–energy theorem: the net work done on a body equals its change in kinetic energy, regardless of how many separate forces (applied force, friction) contribute to that net force. The applied force alone does far more work ( 882 J ) than the body actually gains as kinetic energy, because friction continuously drains part of that work away as heat ( -247 J ); only the leftover, 635 J, shows up as the body's gain in motion energy. Answer: (a) W a = 882 J, (b) W f = -247 J, (c) W net = 635 J, (d) Δ KE = 635 J — and W net = Δ KE, confirming the work–energy theorem.

Question 3

Given in Fig. 5.11 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. Fig. 5.11

Accepted Answer

(a) x a; 0 (c) x < a, x b; - V 1 (b) −∞ < x < ∞; V 1 (d) - b/2 < x < - a / 2, a / 2 < x < b / 2; - V 1. Big idea: kinetic energy can never be negative, so a particle of total energy E is forbidden wherever the graph of V(x) rises above the cross. The total mechanical energy of a particle moving in a conservative one‑dimensional field is E = K + V(x), K = 1/2mv^2 where E is the total energy (in joules, fixed — the horizontal level of the cross), K is the kinetic energy, V(x) is the potential energy at position x (the blue curve), m is the mass and v the speed. Rearranging, K = E - V(x) Since K= 12 mv^2 is a square times a positive mass, K ≥ 0 always. Therefore E - V(x) ≥ 0 ⇒ V(x) ≤ E Rule 1 (forbidden regions): the particle cannot be found at any x where the blue curve lies above the cross. Rule 2 (minimum total energy): for the particle to exist anywhere at all, E must be at least as large as the lowest point of the whole curve, E min = V min (there the particle sits at rest, K=0 ). Graph 1 (top). Reading the picture: for x<a the blue line lies exactly on the x -axis, so V(x)=0 there; at x=a it jumps straight up to the flat level V 0 and stays there for all x a. The cross sits on the V -axis above 0 but clearly below the V 0 tick — measuring against the axis, roughly two‑thirds of the way up, so 0<E<V 0. For x a: K = E - V 0, and since E<V 0 this is negative — impossible. For x<a: K = E - 0 = E 0 — allowed. Forbidden region: x a. Minimum total energy: V min =0, so E min =0 J. Physical context: a sudden step up in potential energy — a ball rolling along a level floor towards a raised platform of height h, with V 0 = mgh. It rolls freely on the lower floor but can never climb onto the platform unless E V 0. Graph 2 (second). Reading the picture: a staircase. A short segment to the left of the origin sits at the lowest level of the entire graph; from x=0 to x=a the line is at V 0; it steps up at a, up again at c, and reaches its top step V 3 between b and d; after d it drops to a level that is still higher than that leftmost segment. The figure prints only the two ordinate labels V 0 and V 3; the lowest step (left of the origin) carries no label, so call it V 1. The cross lies below every part of the curve — below even that lowest step, though still above zero. At every x without exception, K = E - V(x) < 0 Forbidden region: all of them, -∞ < x < ∞; with this E the particle cannot exist anywhere. Minimum total energy: E min = V min = V 1, the lowest step. Physical context: a ball on a flight of stairs, or a charged particle passing through a series of regions each held at a different fixed potential — the energy jumps by a fixed amount at each boundary and is constant in between. Graph 3 (third). Reading the picture: V(x)=V 0 (a flat line above the axis) for x<a and again for x b; between a and b the line plunges below the axis to the flat floor -V 1. The cross sits on the V -axis between 0 and V 0, a little above halfway, so 0<E<V 0. For x<a and x b: K=E-V 0<0 — impossible. For a<x<b: K = E-(-V 1) = E+V 1 0 — allowed, and this is where the particle must be. Forbidden regions: x<a and x b. Minimum total energy: V min =-V 1, so E min = -V 1. Physical context: a finite potential well — a ditch of depth V 1 cut into ground of height V 0. A particle with too little energy to climb out is trapped in the ditch and simply shuttles between x=a and x=b; this is the standard picture of a bound particle. Graph 4 (bottom). Reading the picture: far out on both sides the curve hugs the x -axis ( V → 0 ); coming inwards it rises steeply to the peak value V 0 at x=-a/2 on the left and x=+a/2 on the right; between -a/2 and +a/2 it drops to the flat floor -V 1 below the axis. The dotted horizontal line drawn through the cross cuts the left branch and the right branch, and the two dotted droppers land on the axis exactly where the arrows labelled -b/2 and b/2 point. So V(± b/2) = E, with 0<E<V 0 and b/2 a/2. Comparing curve to cross piece by piece: x b/2: V < E ⇒ K 0 (allowed), a/2 < x < b/2: V E ⇒ K<0 (impossible), x < a/2: V=-V 1 < E ⇒ K = E+V 1 0 (allowed) Forbidden regions: -b/2 < x < -a/2 and a/2 < x < b/2 — the two shoulders of the barrier. Minimum total energy: V min = -V 1, so E min =-V 1. Physical context: an attractive well surrounded by a repulsive barrier that dies away at large distance — the potential energy of an α -particle inside a nucleus (short‑range nuclear attraction inside, Coulomb repulsion outside). Classically a particle inside the well is locked in: it can never cross the barrier region to reach the outside, and one outside can never get in. Answer: In each case the particle is excluded wherever V(x) E, because K=E-V(x) would be negative — Graph 1: forbidden for x a, E min =0; Graph 2: forbidden everywhere (the cross lies below every step), E min =V 1, the lowest step of the staircase; Graph 3: forbidden for x<a and x b, E min =-V 1; Graph 4: forbidden for -b/2<x<-a/2 and a/2<x<b/2, E min =-V 1 — the shapes representing, respectively, a step onto a raised platform, a staircase of potentials, a bound particle in a finite well, and an α -particle held inside a nucleus by a Coulomb barrier.

Question 4

The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx^2 /2, where k is the force constant of the oscillator. For k = 0.5 N m^-1, the graph of V(x) versus x is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.

Accepted Answer

Total energy is fixed, and kinetic energy can never be negative — that single fact pins down the turning points. For a particle undergoing linear simple harmonic motion, the total mechanical energy is conserved and splits into kinetic and potential parts: E = K(x) + V(x) where K(x) is the kinetic energy and V(x) is the potential energy at position x. Since kinetic energy is K = 1/2mv^2, it can never be negative — K(x) ≥ 0 for every physically allowed position. This means the particle can only occupy positions where V(x) ≤ E. Step 1: Write down the potential energy with the given k. V(x) = (kx^2)/2, k = 0.5 N m^-1 V(x) = ((0.5)x^2)/2 = 0.25 x^2 (with (x ) in metres, (V ) in joules) Step 2: Find where V(x) equals the total energy. The particle is given total energy E = 1 J. At a turning point, the particle has momentarily zero velocity, so all of its energy is potential: K = 0 and E = V(x). 1 = 0.25 x^2 x^2 = 1/0.25 = 4 m^2 x = ± 2 m Step 3: Check what happens beyond x = ± 2 m. Take a point just past this, say x = 3 m: V(3) = 0.25 × 3^2 = 2.25 J This exceeds the total energy E = 1 J. If the particle could be at x = 3 m, its kinetic energy would have to be K = E - V(3) = 1 - 2.25 = -1.25 J, which is impossible, since kinetic energy is 12 mv^2 and can never be negative. The same argument holds for any x 2 m: the parabola V(x) = 0.25x^2 rises above E = 1 J there, so the region x 2 m is forbidden to a particle with only 1 J of total energy. Step 4: Interpret this physically. As the particle moves outward from x = 0, its potential energy rises (at the expense of kinetic energy, which falls) until, exactly at x = ± 2 m, all the energy has become potential and the kinetic energy drops to zero — the particle is momentarily at rest. Since it cannot go further (that would demand negative kinetic energy), the restoring force F = -dV/dx = -kx — which always points back toward x = 0 — decelerates it to a stop and then pushes it back the way it came. This is exactly what "turning back" means: x = ± 2 m are the classical turning points of the oscillation. A common slip here is to treat k in V(x) = kx^2/2 as if it needed a factor of 12 applied twice — it doesn't; the 12 is already built into the formula, so with k = 0.5 N m^-1 the coefficient of x^2 is simply k/2 = 0.25, not 0.5 or 0.125. Answer: The particle turns back at x = ± 2 m, since these are the points where V(x) = E = 1 J, forcing kinetic energy to zero; for x 2 m, V(x) 1 J, which would require negative kinetic energy and is therefore forbidden.

Question 5

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 and closer to the earth? (d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), 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?

Accepted Answer

(a) rocket; (b) For a conservative force work done over a path is minus of change in potential energy. Over a complete orbit, there is no change in potential energy; (c) K.E. increases, but P.E. decreases, and the sum decreases due to dissipation against friction; (d) in the second case.. These four parts are all about which force does work — a force perpendicular to displacement, or on a closed path, does zero work even while things are moving and energy is being transferred. (a) At the rocket's expense, not the atmosphere's. Frictional (drag) force from the air acts on the rocket casing as it moves through the atmosphere. Work done by a force is W = F · d and here the drag force on the rocket points backward, opposite to its velocity, so this work is negative — it removes kinetic energy from the rocket. That lost kinetic energy is exactly what turns into heat at the casing. The atmosphere does not supply the heat; it is only the medium through which the rocket loses its own energy. By Newton's third law the rocket does equal and opposite (positive) work on the air molecules it pushes past, but the heating of the casing specifically is the mechanical energy the rocket itself gives up to friction. So the heat comes out of the rocket's kinetic energy. (b) Because gravity is a conservative force, and a conservative force does zero work around any closed path. The defining property of a conservative force is that the work it does depends only on the start and end positions, not on the path: W conservative = -Δ U = U initial - U final where U is the potential energy (here, gravitational PE, U = -GMm/r, with G the gravitational constant, M the Sun's mass, m the comet's mass, and r the Sun–comet separation). After one complete orbit the comet returns to exactly the same point, so r — and hence U — is unchanged: U initial = U final, which makes W = 0 over the full orbit. This holds even though the force is not perpendicular to the velocity at every instant (it does positive work while the comet falls toward the Sun and negative work of equal size while it climbs back out) — the two exactly cancel because gravity is conservative. A force that is merely non-perpendicular to velocity everywhere would not have this guarantee; it is specifically the closed path plus conservativeness that forces the net work to zero. (c) The satellite is losing total energy, but its potential energy is dropping faster than its total energy — so kinetic energy (and speed) still goes up. For a satellite of mass m in a circular orbit of radius r around the Earth (mass M ), gravity supplies the centripetal force: GMm/r^2 = mv^2/r ⇒ v^2 = GM/r So its kinetic and potential energies are K = 1/2mv^2 = GMm/2r, U = -GMm/r and the total mechanical energy is E = K + U = GMm/2r - GMm/r = -GMm/2r Atmospheric drag is a dissipative (non-conservative) force: it steadily removes mechanical energy, so E becomes more negative and the orbit shrinks — r decreases. Look at what that does to each term separately: U = -GMm/r becomes more negative (drops) as r shrinks, while K = GMm/2r increases as r shrinks. The drop in U is twice as large as the corresponding rise in K; half of the released potential energy pays for the increase in kinetic energy, and the other half (plus whatever margin remains) is what is lost as heat to air resistance. So even though the satellite is continuously losing energy overall, its speed — governed by K, which depends only on the shrinking r — keeps increasing as it spirals inward. This is the mark of the mistake to avoid here: "losing energy" does not mean "slowing down" for an orbiting body, because kinetic and potential energy move in opposite directions as r changes. (d) The work done is greater in the second case, where he pulls the rope — by about 2.9×10^2 J. Case (i) — carrying the mass in his hands: The man's arms exert an upward force (supporting the 15 kg mass against gravity) while he walks a horizontal distance of 2 m. The supporting force is vertical; the displacement is horizontal, so the angle between them is 90^°: W = Fd cos θ = Fd cos 90^° = 0 No work is done on the mass by the man's carrying force, because the mass's height never changes — this is the same reason a table does no work holding up a book, no matter how far you slide the table sideways. Case (ii) — pulling the rope over a pulley: Here the rope runs over a pulley with an identical 15 kg mass hanging from the other end. As the man walks 2 m pulling the rope, the pulley redirects that same length of rope, so the hanging mass is lifted straight up by a height h = 2 m. Using the work–energy relation for lifting against gravity, W = mgh with m = 15 kg, g = 9.8 m/s^2, and h = 2 m: W = (15 kg )(9.8 m/s^2)(2 m ) = 294 J Rounding to two significant figures (matching the precision of the given 15 kg and g = 9.8 m/s^2 ) gives W ≈ 2.9×10^2 J. So walking while carrying the load horizontally does zero work on it, while walking while hauling the rope does real positive work, because in the second case the man's effort actually changes the hanging mass's height — the physical quantity work is sensitive to, not just to how tired his arms get. Answer: (a) the rocket's own kinetic energy; (b) gravity is a conservative force, so its work over any closed path — including one full orbit — is zero; (c) K = GMm/2r rises as the orbit shrinks even while total energy E = -GMm/2r is being drained by drag; (d) case (ii) does more work, about 2.9×10^2 J, versus zero work in case (i).

Question 6

Underline the correct alternative: (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered. (b) Work done by a body against friction always results in a loss of its kinetic/potential energy. The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the 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.

Accepted Answer

(a) decrease; (b) kinetic energy; (c) external force; (d) total linear momentum, and also total energy (if the system of two bodies is isolated).. Underlining the right word here comes down to one distinction each time: which quantity is defined to store energy, which one is dissipated as heat, and which one only cares about internal versus external forces. (a) Conservative force doing positive work → potential energy decreases. For any conservative force, the work it does and the potential energy of the body are tied together by W = -Δ U where W is the work done by the conservative force and Δ U is the change in potential energy. This minus sign is the definition of potential energy — it is built so that energy is conserved as the force does work. If the force does positive work, W 0, then Δ U = -W < 0 so the potential energy decreases. This is exactly what happens when gravity (a conservative force) pulls a falling body down: gravity does positive work on the body, and its gravitational potential energy decreases as it falls. Correct alternative: decreases. (b) Work done against friction → loss of kinetic energy, not potential energy. Potential energy is a bookkeeping device that exists only for conservative forces (gravity, spring force, electrostatic force), because only for those forces is the work path-independent. Friction is dissipative — it converts mechanical energy irreversibly into heat at the rubbing surfaces, so no potential-energy function can be assigned to it. By the work–energy theorem, Δ KE = W net When a body does work against friction, that amount of work is removed from its kinetic energy budget and leaves the body as heat; it is not stored as potential energy anywhere. The aside worth remembering: "loss of energy to friction" always means kinetic energy, because friction has no potential energy associated with it. Correct alternative: kinetic energy. (c) Rate of change of total momentum of a many-particle system → external force only. For a system of particles, Newton's second law generalizes to d p total dt = F ext where p total is the vector sum of the momenta of all the particles and F ext is the sum of forces coming from outside the system. The internal forces — the forces the particles of the system exert on one another — always occur in Newton's-third-law pairs: equal in magnitude, opposite in direction, along the same line of action. When summed over the whole system these pairs cancel exactly, so internal forces can never change the system's total momentum, no matter how complicated the interactions inside the system are (collisions, explosions, chemical reactions). Only a genuinely external force can change p total. Correct alternative: external force. (d) Inelastic collision → total linear momentum is the one that survives unchanged. "Inelastic" is defined precisely by kinetic energy not being conserved — some of it converts into heat, sound, or permanent deformation at the point of impact, so total kinetic energy is ruled out immediately. That converted energy need not stay inside "the system of two bodies" either — heat and sound can radiate away into the surroundings — so even "total energy of the system of two bodies" (as opposed to the universe as a whole) is not guaranteed to stay the same. What is guaranteed is momentum: during a collision the forces the two bodies exert on each other are internal to the two-body system and are Newton's-third-law pairs, exactly as in part (c), while the external forces (like gravity) act for such a short collision time that their impulse is negligible. So the total linear momentum of the two bodies just before impact equals the total linear momentum just after, for any collision — elastic or inelastic. Correct alternative: total linear momentum. Answer: (a) decreases; (b) kinetic energy; (c) external force; (d) total linear momentum.

Question 7

State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. 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 always less than the initial kinetic energy of the system.

Accepted Answer

(a) F; (b) F; (c) F; (d) F (true usually but not always, why?). Conservation laws apply to a whole system, never to one body on its own — and "always" is a strong word that a single counter-example destroys. All four statements are false, and each one fails for its own reason. (a) "In an elastic collision of two bodies, the momentum and energy of each body is conserved." — False. The two conservation laws for a collision are statements about the pair of bodies taken together, not about either body alone. Writing m 1, m 2 for the masses, u 1, u 2 for the velocities before impact and v 1, v 2 for the velocities after, conservation of linear momentum for the system says m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 and, because the collision is elastic, conservation of kinetic energy for the system says 1/2m 1u 1^2 + 1/2m 2u 2^2 = 1/2m 1v 1^2 + 1/2m 2v 2^2 Both equations balance a sum over two bodies. Nothing in them forces m 1 u 1 = m 1 v 1. A concrete case settles it. Take an elastic head-on collision of equal masses, m 1 = m 2 = m, with the second body at rest: u 1 = u, u 2 = 0. The two equations above are solved by v 1 = 0, v 2 = u — the bodies exchange velocities. Body 1's momentum goes from mu to 0 and its kinetic energy from 1/2mu^2 to 0; body 2 gains exactly what body 1 lost. Neither body kept its own momentum or its own energy. This is the step people get wrong: an individual body's momentum is constant only when the net force on it is zero. During the impact each body is pushed hard by the other, so a large force acts on it for the contact time and its momentum must change. Momentum and energy are conserved for the system precisely because those impact forces are internal to it and come in equal-and-opposite pairs. (b) "Total energy of a system is always conserved, no matter what internal and external forces on the body are present." — False. The half of this statement about internal forces is fine. If internal friction acts between parts of the system, the mechanical energy it destroys reappears as internal (thermal) energy of those same parts, still inside the system, so the total is unchanged. The claim breaks on external forces. For a system acted on from outside, the energy bookkeeping is Δ E total = W ext where W ext is the work done by external forces on the system and Δ E total is the change in the system's total energy of every kind — kinetic, potential and internal. If an outside agent does work on the system, W ext ≠ 0 and the system's total energy changes by exactly that amount. Take a box at rest on a rough floor as the system. Push it with a horizontal force of 20 N through 3.0 m: W ext = F d = (20 N )(3.0 m ) = 60 J The box ends up with 60 J more energy than it started with, split between kinetic energy and the thermal energy of the rubbed surfaces. Energy was not conserved for the box; 60 J was carried into it from outside. The aside worth keeping: "energy can neither be created nor destroyed" is a statement about an isolated system — one across whose boundary no work and no heat pass — or about the universe as a whole. It is not a licence to say that any system you happen to draw a boundary around holds its energy fixed while something outside is pushing on it. Enlarge the system until nothing external does work on it, and only then is the total constant. (c) "Work done in the motion of a body over a closed loop is zero for every force in nature." — False. Zero work around every closed loop is not a property of forces in general; it is the definition of a conservative force. Gravity and the spring force satisfy it. Friction and viscous drag do not. Drag a block once around a closed loop of total path length L against a constant kinetic friction force f. Friction always opposes the motion, so at every element of the path the work is negative and the contributions add rather than cancel: W friction = -fL For f = 5.0 N and a loop of length L = 4.0 m, W friction = -20 J, not zero, even though the block is back where it began. Air resistance behaves the same way. Since forces exist in nature for which the closed-loop work is non-zero, the word "every" makes the statement false. (d) "In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system." — False, although it is true in almost every collision you will meet. Start from the definition used in this chapter: in every collision the total linear momentum is conserved; the collision is called elastic when the initial and final kinetic energies are equal, and inelastic otherwise. "Otherwise" is a two-way split, and it covers a change in kinetic energy in either direction. Usually the kinetic energy falls, because the collision deforms the bodies and heats them — a lump of putty striking a wall, two cars crumpling together. But a collision can also release energy that was already stored inside the bodies as chemical or elastic potential energy, and that released energy can appear as kinetic energy. A worked counter-example. A trolley of mass m = 2.0 kg carrying a compressed, latched spring moves at u = 1.0 m s^-1 towards an identical stationary trolley. The latch trips on contact and the spring, storing E spring = 9.0 J, pushes the trolleys apart. Momentum is conserved: mu = mv 1 + mv 2 ⇒ v 1 + v 2 = 1.0 m s^-1 Energy released into kinetic form: 1/2mv 1^2 + 1/2mv 2^2 = 1/2mu^2 + E spring = 1.0 J + 9.0 J = 10.0 J Solving the two together gives v 1 = -1.5 m s^-1 and v 2 = 2.5 m s^-1 (the first trolley recoils backwards, the second moves forwards), and you can check the kinetic energy: 1/2(2.0)(1.5)^2 + 1/2(2.0)(2.5)^2 = 2.25 + 6.25 = 10.0 J, against an initial 1.0 J. Kinetic energy is not conserved, so the collision is not elastic — it is inelastic by the definition above — yet the final kinetic energy is ten times the initial. Such collisions are called superelastic or explosive. Total energy is of course still conserved: the extra 9.0 J came out of the spring, not out of nowhere. So the statement holds usually, but not always, and one word decides a true/false question. Read "always" literally every time you meet it in a statement of this kind — a single physically possible counter-example is enough to make the sentence false. Answer: All four statements are false. (a) F — momentum and kinetic energy are conserved for the two-body system, not for each body separately; the bodies exchange them during impact. (b) F — a system's total energy changes by Δ E total = W ext whenever external forces do work on it; conservation holds only for an isolated system. (c) F — zero closed-loop work defines a conservative force; for friction over a loop of length L, W = -fL ≠ 0. (d) F — true usually but not always, since an inelastic collision that releases stored internal energy (a superelastic or explosive collision) can end with more kinetic energy than it started with.

Question 8

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? Fig. 5.13 Fig. 5.12 (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).

Accepted Answer

(a) No (b) Yes (c) Linear momentum is conserved during an inelastic collision, kinetic energy is, of course, not conserved even after the collision is over. (d) elastic.. Momentum survives contact; kinetic energy only survives the encounter as a whole. These two conservation laws behave differently during the brief instant the balls are touching, and the reason is Newton's third law. (a) Kinetic energy during contact — NOT conserved instant-by-instant. While the balls are in contact they compress each other, the way a spring compresses. Some of the kinetic energy the balls carried in is converted into elastic potential energy stored in the deformed material of the balls, exactly as compressing a spring turns kinetic energy into spring potential energy (U = 1/2kx^2). At the instant of maximum compression, the balls even share a common velocity, and a large part of the kinetic energy is momentarily stored as potential energy of deformation — it is not present as kinetic energy at all. Because the balls are perfectly elastic, every joule of that stored potential energy is returned as kinetic energy when the balls spring back apart, so the total kinetic energy before the collision equals the total kinetic energy after it. But that is a statement about the two end states, not about every instant of the contact itself. Between those end states, kinetic energy is temporarily traded for potential energy and back again. (b) Linear momentum during contact — IS conserved at every instant. The only forces acting during the collision are the forces the two balls exert on each other. By Newton's third law, ball 1 pushes on ball 2 with a force F 12 exactly equal and opposite to the force F 21 that ball 2 pushes back on ball 1, at every instant of the contact: F 12 = - F 21 These are internal forces to the two-ball system, so they cancel in pairs and contribute nothing to the net force on the system. With no external force acting during the short collision time (gravity and any support force are balanced, and the collision is too brief for friction to matter), Newton's second law for the system, F ext = d p total dt, gives d p total /dt = 0. So the total momentum is constant throughout the contact, not just before and after it — this is why momentum, unlike kinetic energy, is the safer quantity to track during the collision itself. (c) Same questions, for an inelastic collision. - Momentum: conserved during the collision, for exactly the same reason as in (b) — momentum conservation follows only from there being no external force, and that is true whether the collision is elastic or not. Newton's third law does not care what the balls are made of. - Kinetic energy: not conserved — and this time not even between the initial and final states. In an inelastic collision part of the deformation is permanent (the balls are dented, or stick together), so part of the initial kinetic energy is used up in the collision and turns into heat, sound, and permanent deformation of the material instead of being returned as kinetic energy. So kinetic energy is neither conserved during contact nor conserved overall; only the elastic case recovers it all at the end. (d) A potential energy that depends only on separation is the elastic case. If the potential energy of the pair depends only on the distance r between the centres of the two balls, U = U(r), then the interaction force is F = -dU/dr r, which is a conservative force — it depends only on the current separation, not on the balls' history or speed. For a conservative interaction, the total mechanical energy (kinetic + potential) of the two-ball system is conserved throughout: as the balls approach, r decreases, U rises and kinetic energy falls; as they recede back to the same separations they came from, U and kinetic energy retrace their path exactly, with nothing left behind as heat or permanent deformation. That is precisely the definition of an elastic collision: kinetic energy lost during compression is returned in full during separation. So a potential energy that depends only on the separation distance between the centres describes an elastic collision. (Had energy leaked away as heat or sound during contact, U could not be written as a function of r alone — it would also depend on the deformation history, which is exactly what happens in an inelastic collision.) Answer: (a) No — kinetic energy is temporarily converted to elastic potential energy of deformation during contact; it is conserved only comparing before and after, not instant-by-instant during contact. (b) Yes — linear momentum is conserved at every instant during contact, because the balls' mutual forces are internal and equal-and-opposite (Newton's third law), so there is no external force on the pair. (c) For an inelastic collision, momentum is still conserved during contact (same reasoning), but kinetic energy is not conserved even overall — part of it is permanently lost to heat, sound, and deformation. (d) If the potential energy depends only on the separation between centres, the interaction force is conservative, so the collision is elastic.

Question 9

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 (i) t^1/2 (ii) t (iii) t^3/2 (iv) t^2

Accepted Answer

(b) t. Power grows in step with velocity, and velocity grows in step with time — so power is directly proportional to time. Setting up the motion. The body starts from rest, so its initial velocity u = 0, and it moves with constant acceleration a. Newton's second law gives the constant force acting on it, F = ma, where m is the mass of the body. Because F is built only from m and a, and both are constant, the force itself does not change with time. Using the first equation of motion, v = u + at, with u = 0: v = at. The velocity grows linearly with time — this is the key intermediate result, since power depends on velocity, not directly on time. Naming the formula for power. Power delivered by a force is the rate of doing work, and for a force acting along the direction of motion it is P = Fv, where F is the applied force and v is the instantaneous speed. This is the mechanical-power identity: power equals force times velocity, not force times displacement (that would be work) and not force alone. Substituting. P = F v = (ma)(at) = ma^2t. Here m and a are both constants (mass does not change, and acceleration is given as constant), so the entire coefficient ma^2 is a fixed number that does not depend on time. What remains is a single factor of t: P ∝ t. Why not the other powers of t? It is tempting to reach for P ∝ t^3/2 by confusing power with kinetic energy divided by time in a different combination, or to reach for t^2 by forgetting that F here is constant (that behavior would apply only if power itself were held constant, which is a different problem). Checking directly from P = Fv = ma · at avoids both traps: one factor of t comes from v alone, and nothing else in the expression carries time. So among the choices, the power delivered at time t is proportional to t — option (ii). Answer: (ii) — the power delivered is proportional to t (since P = Fv = ma · at = ma^2t, with m and a constant).

Question 10

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

Accepted Answer

(c) t3/2. Constant power does not mean constant force — power fixes how force and speed trade off as the body speeds up. Power delivered by the source is P = Fv where F is the force acting on the body and v is its speed. By Newton's second law, F = mdv/dt, with m the (constant) mass of the body. Since the body starts moving from rest under this source and P is constant, substitute: P = m dv/dt v v dv = P/m dt Step 1: find how speed grows with time. Integrate the left side over speed from 0 to v, and the right side over time from 0 to t (taking the body to start from rest, since only then is the initial power finite while the initial force is unbounded — the standard reading of this problem): ∫ 0^v v dv = P/m∫ 0^t dt v^2/2 = P/m t v = (2P/m)^1/2 t^1/2 So the speed itself grows as v ∝ t^1/2 — this is the step most people skip, jumping straight to x from a wrong assumption of constant acceleration. Under constant power the force is large at first (when v is small, F = P/v is large) and shrinks as the body speeds up — the opposite of constant force, which is why the familiar v = u + at, x ∝ t^2 result does not apply here. Step 2: integrate speed to get displacement. dx/dt = v = (2P/m)^1/2 t^1/2 ∫ 0^x dx = (2P/m)^1/2∫ 0^t t^1/2 dt x = (2P/m)^1/2 · 2/3 t^3/2 All the factors in front of t^3/2 — P and m — are constants of the problem, so they don't affect how x scales with t: x ∝ t^3/2 Answer: The displacement is proportional to t^3/2 (equivalently, the speed grows as v ∝ t^1/2, and x ∝ t^3/2 ).

Question 11

A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by N ˆ 3 ˆ 2 k j i F + − = where,j,i are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?

Accepted Answer

12 J. Work is a dot product, so only the force component along the direction of motion does any work. The force is given as F = - i + 2 j + 3 k N where i, j, k point along the x-, y- and z-axes. The body moves only along the z-axis, through a distance of 4 m, so its displacement vector is d = 4 k m (the x- and y-components of the displacement are both zero — the body never moves sideways). Work done by a constant force W = F · d Taking the dot product component by component: W = (-1 N )(0 m ) + (2 N )(0 m ) + (3 N )(4 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 F survives: W = 3 N × 4 m = 12 N·m Since 1 N·m = 1 J, W = 12 J The force components and the distance (4 m) are given as exact whole numbers in the problem, so the result is stated as the exact value 12 J, with no rounding needed. Answer: The work done is 12 J.

Question 12

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 is faster, the electron or the proton? 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).

Accepted Answer

The electron is faster, v e / v p = 13.5. 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 2000 times heavier. Step 1: Convert the kinetic energies to joules. The energies are given in keV, so convert using 1 eV = 1.60×10^-19 J: KE e = 10 keV = 10 × 10^3 × 1.60×10^-19 J = 1.60×10^-15 J KE p = 100 keV = 100 × 10^3 × 1.60×10^-19 J = 1.60×10^-14 J A common slip here is comparing 10 and 100 directly without converting to the same energy unit as the mass (kg) and using SI throughout — keV cannot be substituted straight into KE= 12 mv^2. Step 2: Get speed from the kinetic energy formula. Kinetic energy: KE = 1/2mv^2, where m is the particle's mass and v its speed. Solving for v: v = √ (2 KE)/m Electron ( m e = 9.11×10^-31 kg ): v e = √ (2×1.60×10^-15 J )/(9.11×10^-31 kg ) = √ (3.20×10^-15)/(9.11×10^-31) m/s = √ 3.5126×10^15 m/s v e ≈ 5.9267×10^7 m/s Proton ( m p = 1.67×10^-27 kg ): v p = √ (2×1.60×10^-14 J )/(1.67×10^-27 kg ) = √ (3.20×10^-14)/(1.67×10^-27) m/s = √ 1.9162×10^13 m/s v p ≈ 4.3774×10^6 m/s Since v e v p, the electron is faster, even though the proton carries ten times more kinetic energy — the electron's mass is about 1835 times smaller, and speed depends on 1/√ m, not on energy alone. Step 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 v before evaluating: v e/v p = √ KE e/KE p p/m e = √ 10/100×(1.67×10^-27)/(9.11×10^-31) = √ 0.100 × 1833.2 = √ 183.3 ≈ 13.5398 This matches the direct division of the two speeds found above ( 5.9267×10^7 / 4.3774×10^6 = 13.54 ), confirming the numbers are consistent. Step 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 ≈ 5.93×10^7 m/s, v p ≈ 4.38×10^6 m/s, v e/v p ≈ 13.5 Answer: The electron is faster; v e ≈ 5.93×10^7 m/s, v p ≈ 4.38×10^6 m/s, and v e: v p ≈ 13.5: 1.

Question 13

A rain drop 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 thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s^-1?

Accepted Answer

0.082 J in each half; - 0.163 J. 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 250 m stretch, whether the drop is speeding up in the first half or moving at a steady terminal speed in the second. Step 1 — mass of the drop. The drop is a sphere of radius r = 2 mm = 2 × 10^-3 m. Water has density ρ = 1.0 × 10^3 kg m^-3 (the standard value, since the question itself does not restate it). With the volume of a sphere, V = 4/3π r^3: V = 4/3π (2×10^-3 m )^3 = 4/3π (8×10^-9 m^3) = 3.351×10^-8 m^3 m = ρ V = (1.0×10^3 kg m^-3)(3.351×10^-8 m^3) = 3.351×10^-5 kg Step 2 — weight of the drop. Weight is a force, so it is measured in newtons, not kilograms — the m above is mass; multiplying by g turns it into the force gravity actually exerts. Using g = 9.8 m s^-2: mg = (3.351×10^-5 kg )(9.8 m s^-2) = 3.284×10^-4 N Step 3 — work done by gravity in each half of the fall. The work done by a constant force over a displacement d along the force is W = F d. Gravity pulls straight down with magnitude mg, and the drop falls h/2 = 250 m in each half of the 500 m descent — first from 500 m to 250 m, then from 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 gravity, 1st half = W gravity, 2nd half = mg × h/2 = (3.284×10^-4 N )(250 m ) = 8.21×10^-2 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 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 v = 10 m s^-1, so its final kinetic energy is: KE f = 1/2mv^2 = 1/2(3.351×10^-5 kg )(10 m s^-1)^2 = 1.676×10^-3 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 500 m: gravity and the resistive (viscous) force of the air. W gravity, total + W resistive = KE f - KE i = KE f - 0 The total work by gravity over the entire 500 m fall is just the sum of the two halves already found: W gravity, total = mgh = (3.284×10^-4 N )(500 m ) = 1.642×10^-1 J Solving for the resistive force's work: W resistive = KE f - W gravity, total = 1.676×10^-3 J - 1.642×10^-1 J = -1.625×10^-1 J Rounding once, at the end, to three significant figures: W resistive ≈ -1.63×10^-1 J = -0.163 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 mgh = 0.164 J (a far higher speed); instead nearly all of that is dissipated, and only 1.68×10^-3 J of kinetic energy is left at 10 m s^-1. Answer: Work done by gravity is the same in both halves — 8.21×10^-2 J (≈ 0.082 J) in the first half and 8.21×10^-2 J (≈ 0.082 J) in the second half; the work done by the resistive force over the entire journey is about -1.63×10^-1 J (−0.163 J).

Question 14

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

Accepted Answer

Yes, momentum of the molecule + wall system is conserved. The wall has a recoil momentum such that the momentum of the wall + momentum of the outgoing molecule equals momentum of the incoming molecule, assuming the wall to be stationary initially. However, the recoil momentum produces negligible velocity because of the large mass of the wall. Since kinetic energy is also conserved, the collision is elastic.. 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 y be along the normal to the wall (pointing away from it) and x be along the wall, in the plane containing the molecule's path. The molecule strikes at 30^° 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 v = 200 m s^-1: v x = v sin 30^°, v y = -v cos 30^° ( moving into the wall ) After rebound, same speed v = 200 m s^-1: v x' = v sin 30^°, v y' = +v cos 30^° ( 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, 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 m, the change in momentum has components Δ p x = m(v x' - v x) = 0 Δ p y = m(v y' - v y) = m [v cos 30^° - (-v cos 30^°) ] = 2mv cos 30^° Putting in the number (keeping the mass symbolic, since it is not given): 2v cos 30^° = 2(200 m s^-1)(0.8660) = 346 m s^-1 ( 3 s.f. ) So the molecule's own momentum changes by Δ p = 2mv cos 30^° ≈ 346 m kg m s^-1 (with 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 2mv cos 30^° 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 = 1/2mv^2, and the speed is unchanged — 200 m s^-1 in, 200 m s^-1 out — so 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 2mv cos 30^° ≈ 346 m 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, 200 m s^-1 before and after.

Question 15

A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m^3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

Accepted Answer

43.6 kW. Efficiency tells you the electrical power is bigger than the useful mechanical power, not the other way round — so you divide by 0.30, you don't multiply by it. Step 1 — find the mass of water that must be lifted. The tank holds a volume V = 30 m^3 of water. Using the density of water, ρ = 1000 kg/m^3: m = ρ V = 1000 kg/m^3 × 30 m^3 = 3.0×10^4 kg Step 2 — find the useful work done in lifting it. Lifting a mass m through a height h against gravity requires work W = mgh where g = 9.8 m/s^2 is the acceleration due to gravity and h = 40 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 4 will fix.) W = (3.0×10^4 kg )(9.8 m/s^2)(40 m ) = 1.176×10^7 J Step 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 min = 15×60 s = 900 s P useful = W/t = (1.176×10^7 J )/(900 s ) = 1.3067×10^4 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 30% efficient. Step 4 — use the efficiency to get the electrical power consumed. Efficiency is defined as η = ( useful power output )/( power input (consumed) ) so the power the pump actually consumes is larger than the useful output: P input = P useful η = (1.3067×10^4 W )/0.30 = 4.3556×10^4 W Step 5 — round to the correct number of significant figures. Every given quantity — 30 m^3, 15 min, 40 m, 30 % — carries 2 significant figures, so the final power should also be quoted to 2 significant figures: P input ≈ 4.4×10^4 W = 44 kW Answer: The pump consumes about 4.4×10^4 W ( ≈ 44 kW ) of electric power.

Question 16

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 (Fig. 5.14) is a possible result after collision? Fig. 5.14

Accepted Answer

(b). 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 m — A, B and C. A arrives with speed V and strikes B first; B and C start out touching, both at rest. The key formula. For a 1-D elastic collision between mass m 1 (initial speed u 1 ) and mass m 2 (initial speed u 2 ), momentum and kinetic energy conservation together give the standard result: v 1 = m 1-m 2/(m 1+m 2) u 1 + 2m 2/(m 1+m 2) u 2, v 2 = 2m 1/(m 1+m 2) u 1 + m 2-m 1/(m 1+m 2) u 2 Here v 1, v 2 are the final speeds. Setting m 1=m 2=m and u 2=0 (the target starts at rest) collapses this to v 1=0, v 2=u 1 — equal masses exchange velocity completely. Step 1 — A hits B. Before: A moves at V, B is at rest. Since m A=m B, A stops and B takes up the full speed: v A = 0, v B = V Step 2 — B hits C. Because B and C were already touching, the instant B is moving at 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, 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 V — the exact speed A started with. Check against the two conservation laws (this is the test that tells you which picture in Fig. 5.14 can be right): Momentum before: only A is moving, so p i = mV. Momentum after: p f = m(0) + m(0) + m(V) = mV. ✓ matches. Kinetic energy before: KE i = 1/2mV^2. Kinetic energy after: KE f = 0 + 0 + 1/2mV^2 = 1/2mV^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 V/2. Check it: momentum = m(V/2)+m(V/2) = mV — conserved. But kinetic energy = 2× 12 m(V/2)^2 = 14 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. 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 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 V — this is the only picture that conserves both momentum ( mV ) and kinetic energy ( 12 mV^2 ), as an elastic collision must.

Question 17

The bob A of a pendulum released from 30^o to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Accepted Answer

It transfers its entire momentum to the ball on the table, and does not rise at all.. 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 1 — Speed of A just before impact. Let bob A have mass m and let the pendulum have length L. Released from rest at 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= 12 mu^2 ⇒ u=√ 2gh 0 Here u is A's speed at the bottom, an instant before collision. You will see in Step 3 that its actual value never has to be computed — the length L is not even given in the problem, which is itself a hint that it will cancel out. Step 2 — Set up the elastic collision. At the lowest point A (moving at speed u ) collides elastically with B (mass m, the same as A, at rest on the table). Let v A and 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 ⇒ u=v A+v B (1) Conservation of kinetic energy (this is what "elastic" means): 12 mu^2= 12 mv A^2+ 12 mv B^2 ⇒ u^2=v A^2+v B^2 (2) Step 3 — Solve (1) and (2) together. Square equation (1): u^2=v A^2+2v Av B+v B^2 Subtract equation (2) from this: 0=2v Av B ⇒ v Av B=0 So either v A=0 or v B=0. The root 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, 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 4 — Convert A's post-collision speed into a rise in height. Bob A leaves the collision point with speed 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 h ): 12 mv A^2=mgh ⇒ h=(v A^2)/2g=0/2g=0 Because 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 30^o release angle, the length L, and the value of 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 "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 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.

Question 18

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 dissipated 5% of its initial energy against air resistance?

Accepted Answer

5.3 m s^-1 Reprint 2026-27 ANSWERS 165. The bob loses height as it swings, and only 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 drop The 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 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 term By 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 = (1/2mv^2) gained + E dissipated Here m is the bob's mass, g is the acceleration due to gravity ( 9.8 m s^-2, not to be confused with the gravitational constant G ), and v is the speed at the lowest point. Since 5% of the initial energy mgh is dissipated, the fraction converted to kinetic energy is 95%: 1/2mv^2 = 0.95 mgh The mass 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 = √ 1.9 g h Substituting g = 9.8 m s^-2 and h = 1.5 m: v = √ 1.9 × 9.8 m s^-2 × 1.5 m = √ 27.93 m^2 s^-2 v = 5.285 m s^-1 Rounding The length l = 1.5 m carries only two significant figures, so that limits the precision of the final answer. Rounding 5.285 m s^-1 to two significant figures: v ≈ 5.3 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 v ≈ 5.3 m s^-1 (obtained from v=√ 1.9gl, unrounded value 5.285 m s^-1 ).

Question 19

A trolley of mass 300 kg carrying a sandbag 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?

Accepted Answer

27 km h^-1 (no change in speed). 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 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 M(t) be the combined mass of the trolley and remaining sand at time t, and v(t) its speed. The rate of change of the system's momentum is d/dt (M(t) v(t) ) = Mdv/dt + vdM/dt. The term v dM/dt is exactly the momentum carried away by the leaking sand, because that sand leaves at the same speed v as the trolley (no horizontal ejection velocity relative to the trolley, unlike a rocket exhaust). Since there is no external horizontal force, d/dt (M(t)v(t) ) = vdM/dt ⇒ Mdv/dt = 0 ⇒ dv/dt = 0. So v never changes while sand is leaking, no matter how much mass is lost or how fast it leaks — the rate 0.05 kg s^-1 and the sandbag's mass of 25 kg (which only fix how long, 500 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 final = v initial = 27 km h^-1 = 27 × 5/18 m s^-1 = 7.5 m s^-1. Answer: The trolley's speed stays unchanged at 27 km/h (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.

Question 20

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?

Accepted Answer

50 J. 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 net = Δ KE = 1/2mv f^2 - 1/2mv i^2 where m is the mass, v i is the speed at the start of the displacement, and 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 F dx — just evaluate the velocity at the two positions. Step 1: Write down what's given. v = a x^3/2, a = 5 m^-1/2 s^-1, m = 0.5 kg The displacement is from x = 0 to x = 2 m. Check the units of a so the velocity comes out right: x^3/2 carries units of m^3/2, so [a][x^3/2] = ( m^-1/2 s^-1)( m^3/2) = m s^-1 which is indeed a speed. This confirms a 's odd-looking units are exactly what's needed to make v = ax^3/2 dimensionally consistent — not a typo. Step 2: Find the speed at each endpoint. At x = 0: v i = a(0)^3/2 = 0 At x = 2 m: v f = a(2)^3/2 = 5 × 2^3/2 m/s Since 2^3/2 = 2√ 2 = 2.828, it's easier to keep v f^2 in exact form rather than round v f itself: v f^2 = a^2 x^3 = (5)^2 (2)^3 = 25 × 8 = 200 m^2/ s^2 (Squaring a^2x^3 directly avoids the mid-calculation rounding of an irrational square root — round only the final answer.) Step 3: Apply the work–energy theorem. W net = 1/2m v f^2 - 1/2m v i^2 = 1/2(0.5 kg )(200 m^2/ s^2) - 0 W net = 1/2(0.5)(200) J = 50 J The mass (0.5 kg) is given to one significant figure and a is given to one significant figure, so the result is stated to the same precision: 50 J (no further decimal places are justified). A common mistake here is to try to find F(x) from v(x) using F = ma = m dv/dt and then integrate 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: 50 J

Question 21

The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the 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 m^2, v = 36 km/h and the density of air is 1.2 kg m^-3. What is the electrical power produced?

Accepted Answer

(a) m Avt ρ = (b) 3 /2 K Av t ρ = (c) P = 4.5 kW. Picture the wind as a moving cylinder of air — in time t, a column of length vt and cross-section A sweeps through the blades. (a) Mass of air passing through in time t In time t, every particle of air in front of the blades moves a distance vt (distance = speed × time). So the air that has passed through the circle of area A forms a cylinder of length vt and cross-sectional area A. Volume of this cylinder: V = A × (vt) = Avt Using density ρ = ( mass )/( volume ), the mass of air in this volume is m = ρ V = ρ A v t Here ρ is the density of air, A is the swept area, v is the wind speed, and 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 air The law here is the definition of kinetic energy, KE = 1/2mv^2, where m is the mass of the moving object and v its speed. Substituting the mass found in part (a): KE = 1/2mv^2 = 1/2(ρ A v t)v^2 KE = 1/2ρ A v^3 t Notice the wind speed enters as v^3, not v^2 — one factor of v comes from how much air arrives per second, and two more come from the v^2 in the kinetic-energy formula. This is why wind power is so sensitive to wind speed. (c) Electrical power produced Power is energy delivered per unit time, P = ( energy )/( time ). Dividing the kinetic energy from part (b) by t gives the power available in the wind: P wind = KE/t = 1/2ρ A v^3 Only 25% of this is converted to electrical energy, so the electrical power is P elec = 0.25 × 1/2ρ A v^3 = 1/8ρ 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 3.6: v = 36 km/h = 36 × (1000 m )/(3600 s ) = 10 m/s Now substitute ρ = 1.2 kg m^-3, A = 30 m^2, and v = 10 m/s: P elec = 1/8(1.2 kg m^-3)(30 m^2)(10 m/s )^3 P elec = 1/8(1.2)(30)(1000) kg m^-1 s^-3 · m^2 · m^3 s^-3 Working through the arithmetic: 1.2 × 30 = 36, then 36 × 1000 = 36000, and 36000/8 = 4500. P elec = 4500 W The given data ( A = 30 m^2, v = 36 km/h, ρ = 1.2 kg m^-3 ) each carry two significant figures, so the final answer should be rounded to two significant figures: P elec = 4.5 × 10^3 W = 4.5 kW Answer: (a) m = ρ A v t; (b) KE = 1/2ρ A v^3 t; (c) P elec = 4.5 kW ( 4500 W ).

Question 22

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 × 10^7 J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Accepted Answer

(a) 49,000 J (b) 6.45 10^-3 kg. Lifting the mass slowly means all the work she does becomes gravitational potential energy — no kinetic energy is left over, so W = Δ PE = mgh for one lift. Part (a): Work done against gravity Each time she raises the 10 kg mass by 0.5 m, the work done against gravity equals the gain in potential energy, W 1 = mgh where m is the mass, g is the acceleration due to gravity, and h is the height raised. This is the same law that gives a body's PE near the Earth's surface — note that g (acceleration, m/s²) here is not to be confused with G, the universal gravitational constant. Substituting m = 10 kg, g = 9.8 m/s^2, h = 0.5 m: W 1 = 10 × 9.8 × 0.5 = 49 J She repeats this n = 1000 times, so the total work done against the gravitational force is W = nW 1 = 1000 × 49 J = 49000 J = 4.9 × 10^4 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 up The 20% efficiency means only one-fifth of the chemical energy released from fat actually turns into the mechanical work W found above; the rest is lost as heat. So the total energy her body must draw from fat is E = W/( efficiency ) = (4.9 × 10^4 J )/0.20 E = 2.45 × 10^5 J Each kilogram of fat supplies 3.8 × 10^7 J of energy, so the mass of fat used is this required energy divided by the energy content per kilogram: m fat = E/(3.8 × 10^7 J/kg ) = (2.45 × 10^5 J )/(3.8 × 10^7 J/kg ) m fat = 0.0064473… kg Rounding to three significant figures (matching the three-figure data given for the efficiency and energy content), m fat ≈ 6.45 × 10^-3 kg That is only about 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) 4.9 × 10^4 J of work is done against gravity. (b) About 6.45 × 10^-3 kg (≈ 6.45 g) of fat is used up.

Question 23

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 meter. 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. Fig. 5.15

Accepted Answer

(a) 200 m^2 (b) comparable to the roof of a large house of dimension 14m × 14m. Chapter 6 6.1 The geometrical centre of each. No, the CM may lie outside the body, as in case of a ring, a hollow sphere, a hollow cylinder, a hollow cube etc. 6.2 Located on the line joining H and C1 nuclei at a distance of 1.24 Å from the H end. 6.3 The speed of the CM of the (trolley + child) system remains unchanged (equal to v) because no external force acts on the system. The forces involved in running on the trolley are internal to this system. 6.6 l z = xp y - yp x, l x = yp z - zp y, l y = zp x - xp z. Power in equals power out — the panel's electrical output can never exceed the sunlight falling on it times the conversion efficiency. Step 1 — Write down what "8 kW" means as power, in watts. The family's power requirement is P required = 8 kW = 8 × 10^3 W = 8000 W Step 2 — Find how much useful (electrical) power each square metre of sunlit roof actually delivers. Sunlight arrives at the rate I = 200 W/m^2 (this is an intensity — power per unit area — not a power by itself). Only 20% of that gets converted to electricity, so the useful power density is P useful per m^2 = η I = 0.20 × 200 W/m^2 = 40 W/m^2 This is the step people skip: you cannot divide 8000 W by 200 W/m² directly, because 200 W/m² is the raw solar rate, not the electrical rate the family can actually draw on. Step 3 — Divide the required power by the useful power per square metre to get the area. Since power scales with area, P required = P useful per m^2 × A, so A = P required P useful per m^2 = (8000 W )/(40 W/m^2) = 200 m^2 Both numbers going into this division (0.20 and 200 W/m²) carry two significant figures, so the area is quoted the same way: A = 200 m^2 = 2.0 × 10^2 m^2 Step 4 — Compare this to an actual rooftop. A single-storey house of modest size — say 10 m × 10 m in plan — has a roof of around 100 – 150 m^2 once you allow for the slope. The 200 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 20%-efficient panels would fall short of meeting the family's full 8 kW demand; you would need a roof close to the size of a fairly large house (about 14 m × 14 m ), or extra ground-mounted panels besides the roof, to collect the full requirement from sunlight alone. Answer: (a) A = 200 m^2 (i.e., 2.0 × 10^2 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 8 kW this way.

NCERT Solutions · Class 11 Physics Work, Energy and Power

Exercises 5.1–5.10 (part 1 of 2)

Exercise 5.1

Exercise 5.2

Exercise 5.3

Exercise 5.4

Exercise 5.5

Exercise 5.6

Exercise 5.7

Exercise 5.8

Exercise 5.9

Exercise 5.10

Exercises 5.11–5.23 (part 2 of 2)

Exercise 5.11

Exercise 5.12

Exercise 5.13

Exercise 5.14

Exercise 5.15

Exercise 5.16

Exercise 5.17

Exercise 5.18

Exercise 5.19

Exercise 5.20

Exercise 5.21

Exercise 5.22

Exercise 5.23