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Successive impacts and impacts with a wall questions
A wall behaves like a sphere of infinite mass, so only the restitution equation survives. Chain a few impacts together and the question becomes whether the next one happens at all.
7 original questions · 29 marks · the successive impacts and impacts with a wall notes · Further Mechanics 1
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Explain why momentum is not conserved when a ball bounces off the floor, and say what does apply.
Worked 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.A ball is dropped from 3 m onto a floor with e = 0.7. Find the height of the first rebound.
Worked 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.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).
Worked 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. The factor of 2 on all but the first term is where this proof is usually lost.A ball is dropped from 3 m onto a floor with e = 0.7. Find the total distance it travels before coming to rest.
Worked 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. Doubling the first drop as well is the usual slip and gives 11.76.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.
Worked answer
First impact. Momentum: 2(6) = 2vA + vB. Restitution: vB − vA = 0.5(6 − 0) = 3.
Substituting the second into the first: 3vA + 3 = 12, so vA = 3 m/s and vB = 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.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.
Worked 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.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.
Worked answer
A strikes B. Momentum: 3mu = 3mvA + mvB, so 3vA + vB = 3u. Restitution: vB − vA = u/2.
Substituting: 3vA + vA + u/2 = 3u, so 4vA = 5u/2 and vA = 5u/8. Then vB = 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) = mvB′ + 2mvC. Restitution: vC − vB′ = (1/2)(9u/8) = 9u/16.
Substituting: vB′ + 2vB′ + 9u/8 = 9u/8, so 3vB′ = 0 and vB′ = 0. B is brought to rest, and vC = 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.
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