Questions on successive impacts and impacts with a wall

Question 1

Explain why momentum is not conserved when a ball bounces off the floor, and say what does apply.

Accepted Answer

The floor is attached to the Earth, so it is an external body and the ball on its own is not an isolated system: the impulse it receives changes its momentum. What still applies is the law of restitution, which says the rebound speed is e times the approach speed. B1 for the floor being an external body, B1 for restitution still applying.

Question 2

A ball is dropped from 3 m onto a floor with e = 0.7. Find the height of the first rebound.

Accepted Answer

If the drop height is h, the arrival speed is √(2gh), the rebound speed is e√(2gh), and the height reached is e²h. So the rebound reaches 0.49 × 3 = 1.47 m. M1 for a rebound height of e²h, A1 for 0.49 × 3, A1 for 1.47 m. The value of g never has to be worked out.

Question 3

A ball strikes a fixed wall at right angles at 9 m/s and rebounds at 6 m/s. Find the coefficient of restitution, and state the rebound speed there would have been if the impact had been perfectly elastic.

Accepted Answer

Against a fixed wall only restitution applies: e = 6/9 = 2/3. B1. Perfectly elastic means e = 1, so the ball would have rebounded at the full 9 m/s. B1.

Question 4

A ball is dropped from rest at a height of 2.5 m onto a horizontal floor. The coefficient of restitution is 0.8. Find the speed of the ball immediately after its second bounce, taking g = 9.8 m/s².

Accepted Answer

The ball arrives at the floor at √(2 × 9.8 × 2.5) = √49 = 7 m/s. M1 for v² = 2gh. It leaves the first bounce at 0.8 × 7 = 5.6 m/s, rises and returns at 5.6 m/s, and leaves the second bounce at 0.8 × 5.6 = 4.48 m/s. M1 for applying e at each bounce, A1 for 4.48 m/s. Each bounce multiplies the speed by e, so after n bounces the speed is 7(0.8)ⁿ; the heights go as the square of that, with ratio 0.64.

Question 5

A ball is dropped from a height h onto a floor with coefficient of restitution e. Prove that the total distance it travels before coming to rest is h(1 + e²)/(1 − e²).

Accepted Answer

The ball arrives at √(2gh) and leaves at e√(2gh), so it rises to e²h. Repeating gives heights h, e²h, e⁴h, …, a geometric sequence of ratio e².
The first height is travelled once, downwards. Every later height is travelled twice, up and down.
Total = h + 2(e²h + e⁴h + …) = h + 2e²h/(1 − e²), the sum to infinity being valid because e² < 1.
Over a common denominator: h(1 − e² + 2e²)/(1 − e²) = h(1 + e²)/(1 − e²) as required. M1 for the heights forming a geometric sequence of ratio e², A1 for the sum to infinity, M1 for doubling all but the first height, A1 for the printed result.

Question 6

A ball is dropped from 3 m onto a floor with e = 0.7. Find the total distance it travels before coming to rest.

Accepted Answer

The heights form a geometric sequence 3, 1.47, 0.7203, … with ratio e² = 0.49. The first drop is travelled once and every later height twice, so the total is 3 + 2(1.47 + 0.7203 + …) = 3 + 2(1.47/0.51) = 3 + 5.765 = 8.76 m.
The closed form h(1 + e²)/(1 − e²) = 3(1.49)/0.51 gives the same. M1 for the sequence of ratio 0.49, M1 for doubling all but the first drop, A1 for 1.47/0.51, A1 for 8.76 m. The first drop is counted once, not twice; doubling it as well gives 11.76.

Question 7

Sphere A of mass 2 kg moving at 6 m/s strikes stationary sphere B of mass 1 kg with coefficient of restitution 0.5. B then hits a wall at right angles and rebounds with coefficient of restitution 1/3. Show that A and B collide again.

Accepted Answer

First impact. Momentum: 2(6) = 2v_A + v_B. Restitution: v_B − v_A = 0.5(6 − 0) = 3. Substituting the second into the first: 3v_A + 3 = 12, so v_A = 3 m/s and v_B = 6 m/s. Since 3 < 6, A does not pass through B, so the answer is physically possible. At the wall. B approaches at 6 and rebounds at 6 × 1/3 = 2 m/s, now travelling back towards A. After the wall. A is still moving forward at 3 m/s and B is moving backwards at 2 m/s, so they approach each other at 5 m/s and must meet again. M1 for conservation of momentum, M1 for the restitution equation, A1 for 3 m/s and 6 m/s, B1 for B rebounding at 2 m/s, A1 for the approach speed of 5 m/s and the conclusion.

Question 8

A ball strikes a floor at 7 m/s and rebounds to a height of 0.9 m. Find the coefficient of restitution, taking g = 9.8 m/s².

Accepted Answer

The rebound speed satisfies v² = 2(9.8)(0.9) = 17.64, so v = 4.2 m/s. Then e = 4.2/7 = 0.6. M1 for v² = 2gh, A1 for v = 4.2 m/s, A1 for e = 0.6.

Question 9

Sphere A of mass 1 kg moving at 6 m/s strikes directly a stationary sphere B of mass 3 kg, the coefficient of restitution being 0.8. B then strikes a wall at right angles and rebounds with coefficient of restitution 0.5. Determine whether A and B collide again.

Accepted Answer

Take A's original direction as positive. Momentum: 1(6) = v_A + 3v_B. Restitution: v_B − v_A = 0.8(6) = 4.8. M1 M1 for the two equations. Adding: 4v_B = 10.8, so v_B = 2.7 m/s and v_A = 2.7 − 4.8 = −2.1 m/s: the lighter sphere bounces back. A1 for both, with the sign.
At the wall B rebounds at 0.5 × 2.7 = 1.35 m/s, now moving back towards where A was. B1.
Whether they meet again is a comparison of signed velocities. A is retreating from the wall at 2.1 m/s and B is following at only 1.35 m/s, so the gap between them grows and they do not collide again. A1 for the comparison and the conclusion.
Had A been moving forwards, or at rest, any rebound of B would have produced a second collision.

Question 10

A ball is dropped from rest at a height of 1.25 m onto a horizontal floor, with coefficient of restitution 0.6. Find the time between the first bounce and the second, taking g = 9.8 m/s².

Accepted Answer

The ball reaches the floor at √(2 × 9.8 × 1.25) = √24.5 = 4.95 m/s. M1 for v² = 2gh, A1 for 4.95 m/s. It rebounds at 0.6 × 4.95 = 2.97 m/s. A ball projected upwards at speed v returns to the same level after 2v/g, so the time between the bounces is 2 × 2.97/9.8 = 0.606 s. M1 for the flight time 2v/g with the rebound speed, A1 for 0.606 s. The interval is measured from bounce to bounce, up and down, so the 2 is needed; halving it gives the time to the top of the rebound alone. Later intervals are each 0.6 times the one before, since the rebound speed is.

Question 11

A ball of mass 0.4 kg moving horizontally at 10 m/s strikes a fixed vertical wall at right angles. The coefficient of restitution is 0.6. Find the magnitude of the impulse the wall exerts on the ball, and the kinetic energy lost in the impact.

Accepted Answer

The ball rebounds at 0.6 × 10 = 6 m/s. Taking the direction away from the wall as positive, u = −10 and v = 6, so the impulse is 0.4(6) − 0.4(−10) = 2.4 + 4 = 6.4 N s away from the wall. M1 for the change in momentum with the rebound counted as a change of sign, A1 for 6.4 N s. Kinetic energy before is ½(0.4)(10²) = 20 J and after is ½(0.4)(6²) = 7.2 J, so 12.8 J is lost. M1 for both kinetic energies, A1 for 12.8 J. The wall receives an equal and opposite impulse, which is why the ball's momentum is not conserved: 6.4 N s has gone into the wall and the Earth it is fixed to. Subtracting the speeds instead of adding them gives 1.6 N s, the impulse for slowing the ball from 10 to 6 without turning it round.

Question 12

Three smooth spheres A, B and C, of masses 3m, m and 2m, lie at rest in that order in a straight line on a smooth horizontal table. A is projected directly towards B with speed u. The coefficient of restitution is 1/2 at every impact. Find the velocity of each sphere after A has struck B and B has struck C, and determine whether A strikes B a second time.

Accepted Answer

A strikes B. Momentum: 3mu = 3mv_A + mv_B, so 3v_A + v_B = 3u. Restitution: v_B − v_A = u/2. Substituting: 3v_A + v_A + u/2 = 3u, so 4v_A = 5u/2 and v_A = 5u/8. Then v_B = 5u/8 + 4u/8 = 9u/8. Check 5u/8 < 9u/8, so A does not overtake B. B strikes C. B arrives at 9u/8 and C is at rest. Momentum: m(9u/8) = mv_B′ + 2mv_C. Restitution: v_C − v_B′ = (1/2)(9u/8) = 9u/16. Substituting: v_B′ + 2v_B′ + 9u/8 = 9u/8, so 3v_B′ = 0 and v_B′ = 0. B is brought to rest, and v_C = 9u/16. Check: momentum 9mu/8 before, 0 + 2m(9u/16) = 9mu/8 after. Conclusion. A is still moving forward at 5u/8 and B is now at rest, so A closes on B at 5u/8 and does strike B again. C moves away at 9u/16, so the question is whether B can catch it. Work that third impact through and B leaves it at 45u/64 while C is still travelling at 9u/16 = 36u/64, so B does catch C and a fourth impact follows. Judging it by B's speed while B is still at rest is the trap: the line above has already said B is about to be struck. M1 for momentum at the first impact, M1 for restitution there, A1 for 5u/8, A1 for 9u/8, M1 for momentum and restitution at the second impact, A1 for B brought to rest, A1 for 9u/16, A1 for the conclusion that A strikes B again. Use the restitution equation for the impact in front of it, never for the pair A and C, which never touch. Feeding u rather than 9u/8 into the second restitution equation is the error that ends this question.

Question 13

A ball is dropped from rest at a height h onto a horizontal floor, the coefficient of restitution being e, where 0 < e < 1. Prove that the total time from release until the ball stops bouncing is √(2h/g) × (1 + e)/(1 − e). Find this time for h = 1.25 m and e = 0.6, taking g = 9.8 m/s².

Accepted Answer

The first fall takes t₀ = √(2h/g), from h = ½gt₀², and the ball arrives at speed gt₀. M1 for the first fall, with the arrival speed in terms of t₀.
After the nth bounce the rebound speed is eⁿ × gt₀, and a ball projected upwards at speed v is back at the floor after 2v/g, so the nth flight lasts 2eⁿt₀. A1 for the flight times.
The total time is t₀ + 2t₀(e + e² + e³ + …). The bracket is a geometric series with first term e and ratio e, and 0 < e < 1 so its sum to infinity is e/(1 − e). M1 for summing the series, A1 for e/(1 − e).
Hence T = t₀[1 + 2e/(1 − e)] = t₀(1 − e + 2e)/(1 − e) = √(2h/g) × (1 + e)/(1 − e), as required. A1.
With h = 1.25 and g = 9.8, t₀ = √(2.5/9.8) = 0.505 s, so T = 0.505 × 1.6/0.4 = 2.02 s. A1.
The first fall is one way and every later flight is a round trip, which is where the factor 2 comes from, exactly as in the distance formula h(1 + e²)/(1 − e²); the ratio here is e, not e², because flight time is proportional to the rebound speed while height is proportional to its square. Infinitely many bounces fit into a finite time because the series converges.

Question 14

A sphere A of mass m moving with speed u strikes directly a stationary sphere B of mass km, the coefficient of restitution being e. B then strikes a wall at right angles to its motion and rebounds, the coefficient of restitution at the wall being f, where f > 0. Show that A and B collide again if and only if f(1 + e) > ke − 1, and deduce that when k ≤ 1 a further collision always occurs.

Accepted Answer

Take A's original direction as positive. Momentum: mu = mv_A + kmv_B, so v_A + kv_B = u. Restitution: v_B − v_A = eu. M1 for the two equations. Substituting, (1 + k)v_B = u(1 + e), so v_B = u(1 + e)/(1 + k) and v_A = u(1 − ke)/(1 + k). A1 for both.
v_B is positive, so B reaches the wall and rebounds at fv_B = fu(1 + e)/(1 + k) towards A. B1.
A and B meet again exactly when they are closing on each other: A's velocity towards the wall plus B's velocity back from it must be positive, v_A + fv_B > 0. M1 for the closing condition with signed velocities. That is u(1 − ke)/(1 + k) + fu(1 + e)/(1 + k) > 0, and since u and 1 + k are positive this is 1 − ke + f(1 + e) > 0, that is f(1 + e) > ke − 1. A1. Every step is reversible, so the condition is necessary as well as sufficient.
If k ≤ 1 then ke − 1 ≤ e − 1 ≤ 0, while f(1 + e) > 0 because f > 0. So the inequality holds and a further collision always occurs. A1.
A second collision can fail only when B is heavier than A, when A can be sent backwards faster than B returns. For k = 3, e = 0.8 and f = 0.5 the left side is 0.9 and the right is 1.4, so those spheres never meet again; for k = ½, e = ½ and f = 1/3 the right side is negative and they must. Using the closing condition v_A > fv_B, with the rebound treated as positive, tests the wrong thing: it asks whether A outruns B, not whether they approach.

Question 15

A ball is dropped from rest at a height of 3 m onto a horizontal floor, the coefficient of restitution being 0.7. Find the number of the bounce after which the rebound height first falls below 3 cm. Explain how the ball can bounce infinitely many times and yet come to rest in a finite time.

Accepted Answer

Each bounce multiplies the rebound height by e² = 0.49, so after n bounces the height is 3(0.49)ⁿ m. M1 for the height after n bounces. The condition 3(0.49)ⁿ < 0.03 is (0.49)ⁿ < 0.01, and taking logarithms, with the inequality reversed because log 0.49 is negative, n > log 0.01/log 0.49 = 6.46. M1 for solving the inequality. So the seventh bounce is the first whose rebound is below 3 cm: the sixth reaches 3(0.49)⁶ = 4.15 cm and the seventh 2.03 cm. A1 for n = 7 with the check.
The bounces never stop in the model, since e²ⁿ is never zero. But the flight time after each bounce is proportional to the rebound speed, which is multiplied by e = 0.7 each time, so the flight times form a geometric series with ratio 0.7. B1 for the flight times forming a geometric sequence with ratio e. A geometric series with ratio less than 1 has a finite sum to infinity, so the infinitely many flights are over after a finite total time, after which the ball is at rest on the floor. B1 for the convergent series giving a finite total time.
The modelling assumption underneath is that e stays at 0.7 however small the bounce, which fails once the rebound is comparable with the deformation of the ball; in practice the ball settles a little sooner than the series predicts, not later.