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Direct impact and Newton's law of restitution questions
Conservation of momentum gives one equation and two unknowns. The coefficient of restitution supplies the second, and between them they settle any direct collision.
6 original questions · 23 marks · the direct impact and newton's law of restitution notes · Further Mechanics 1
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State Newton's law of restitution and the range of values the coefficient can take, saying what each end means.
Worked answer
The speed of separation equals e times the speed of approach, both measured along the line of centres, with 0 ≤ e ≤ 1. At e = 1 the impact is perfectly elastic and no kinetic energy is lost; at e = 0 the spheres do not separate at all and move off together. B1 for the law, B1 for the range with both ends explained.A sphere of mass 3 kg moving at 6 m/s strikes a stationary sphere of mass 2 kg directly, with e = 0.5. Find both velocities afterwards.
Worked answer
Momentum: 3(6) = 3v₁ + 2v₂, so 3v₁ + 2v₂ = 18. Restitution: v₂ − v₁ = 0.5(6) = 3. Substituting: 3v₁ + 2(v₁ + 3) = 18 gives 5v₁ = 12, so v₁ = 2.4 m/s and v₂ = 5.4 m/s. M1 for conservation of momentum, M1 for the restitution equation, A1 for both velocities.A sphere of mass 3 kg moving at 6 m/s strikes a stationary sphere of mass 2 kg directly. The coefficient of restitution is 0.5, and after the impact the spheres move at 2.4 m/s and 5.4 m/s in the original direction. Find the kinetic energy lost.
Worked answer
Before the impact: ½(3)(6²) = 54 J, the stationary sphere contributing nothing.
After: ½(3)(2.4²) + ½(2)(5.4²) = 8.64 + 29.16 = 37.8 J. Both spheres must appear in this line; omitting the struck one is the usual slip and it inflates the loss.
The loss is 54 − 37.8 = 16.2 J, which is 30% of the original energy. M1 for the kinetic energy before, A1 for 54 J, M1 for the kinetic energy after with both spheres, A1 for a loss of 16.2 J.Explain why momentum is conserved in a collision but kinetic energy usually is not.
Worked answer
During the impact each sphere exerts a force on the other that is equal in size and opposite in direction, for the same length of time, so the impulses cancel and, provided any external impulse is negligible over the brief contact, the total momentum is unchanged. Kinetic energy obeys no such law. Some of it goes into deforming the spheres, and into sound and heat, and does not come back unless the impact is perfectly elastic. B1 for the equal and opposite forces acting for the same time, B1 for the impulses cancelling, B1 for the energy lost to deformation, sound and heat.A sphere of mass 4 kg moving at 5 m/s meets one of mass 6 kg moving at 2 m/s towards it. After the impact the 4 kg sphere moves at 1 m/s in the reverse direction. Find the coefficient of restitution.
Worked answer
Fix a positive direction and keep it for every velocity. Taking the 4 kg sphere's original direction as positive gives u₁ = 5, u₂ = −2 and v₁ = −1. Sign errors here are the most common slip in the whole topic.
Momentum: 4(5) + 6(−2) = 8, so 4(−1) + 6v₂ = 8 and v₂ = 2 m/s.
Speed of approach = 5 − (−2) = 7, and speed of separation = 2 − (−1) = 3.
So e = 3/7 = 0.429, which lies between 0 and 1 as it must. M1 for conservation of momentum, A1 for the second velocity, M1 for the speeds of approach and separation, A1 for e = 0.429. A value outside that range means a sign has slipped.A smooth sphere A of mass 3m moving with speed u strikes directly a smooth sphere B of mass m which is at rest on the same smooth horizontal surface. The coefficient of restitution between the spheres is e. Show that the kinetic energy lost in the impact is 3mu²(1 − e²)/8.
Worked answer
Take the direction of u as positive, and let the velocities after the impact be vA and vB.
Momentum: 3mu = 3mvA + mvB, so 3vA + vB = 3u.
Restitution: vB − vA = eu.
Substituting the second into the first gives 4vA = 3u − eu, so vA = u(3 − e)/4 and vB = 3u(1 + e)/4. Keep the answers as single fractions; the algebra that follows is much worse otherwise.
Kinetic energy before = ½(3m)u² = 3mu²/2.
After = ½(3m)vA² + ½(m)vB² = (3mu²/32)[(3 − e)² + 3(1 + e)²].
Expanding the bracket: (9 − 6e + e²) + (3 + 6e + 3e²) = 12 + 4e². The terms in e cancel, which is the sign that the expansion is right. So the energy after is 3mu²(12 + 4e²)/32 = mu²(9 + 3e²)/8.
Loss = 12mu²/8 − mu²(9 + 3e²)/8 = 3mu²(1 − e²)/8, as required. M1 for the momentum equation, M1 for the restitution equation, A1 for both velocities, M1 for the kinetic energy before and after, A1 for the energy after, M1 for the subtraction, A1 for the printed result.
Two checks. At e = 1 the loss is zero, which is what perfect elasticity means. At e = 0.5 with m = 1 and u = 6, the formula gives 10.125 J, and the velocities 3.75 and 6.75 m/s give the same figure directly.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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