Questions on direct impact and Newton's law of restitution

Question 1

State Newton's law of restitution and the range of values the coefficient can take, saying what each end means.

Accepted 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.

Question 2

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.

Accepted 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.

Question 3

A sphere moving with speed u strikes directly an identical sphere which is at rest. Write down the velocity of each sphere after the impact when e = 1, and when e = 0.

Accepted Answer

With e = 1 the equal spheres exchange velocities: the first stops and the second moves off at u, which satisfies both momentum, mu = 0 + mu, and restitution, separation u = 1 × u. B1. With e = 0 there is no separation, so the two move on together at the common speed u/2, from mu = 2m × u/2. B1.

Question 4

A sphere moving at 7 m/s strikes directly a sphere moving at 2 m/s in the same direction. After the impact the spheres are separating at 2 m/s. Find the coefficient of restitution, and explain why the separation speed could not have been 6 m/s.

Accepted Answer

The speed of approach is the relative speed 7 − 2 = 5 m/s, not 7. M1 for approach as the difference of the velocities. So e = separation/approach = 2/5 = 0.4. A1. A separation speed of 6 m/s would give e = 6/5 = 1.2, and the coefficient cannot exceed 1: the spheres cannot separate faster than they approached, since that would create kinetic energy in the impact. B1 for the range 0 ≤ e ≤ 1 with the reason. The masses are not needed for e, which is fixed by the relative speeds alone; they would be needed to find either velocity on its own.

Question 5

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.

Accepted 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 overstates 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.

Question 6

Explain why momentum is conserved in a collision but kinetic energy usually is not.

Accepted 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.

Question 7

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.

Accepted 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.
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.

Question 8

A sphere A of mass 2 kg moving at 4 m/s catches up and strikes directly a sphere B of mass 4 kg moving at 1 m/s in the same direction. The coefficient of restitution is 0.75. Find the velocity of each sphere after the impact, and the magnitude of the impulse exerted on B.

Accepted Answer

Take the common direction as positive. Momentum: 2(4) + 4(1) = 2v₁ + 4v₂, so v₁ + 2v₂ = 6. Restitution: v₂ − v₁ = 0.75(4 − 1) = 2.25, the approach speed being the difference of the two velocities, not the speed of A alone.
Adding the two equations: 3v₂ = 8.25, so v₂ = 2.75 m/s and v₁ = 0.5 m/s, both in the original direction. A is now slower than B, as the rear sphere must be.
The impulse on B is its change of momentum: 4(2.75 − 1) = 7 N s forwards. As a check, A receives 2(4 − 0.5) = 7 N s backwards, the equal and opposite impulse. M1 for conservation of momentum, M1 for the restitution equation with the relative approach speed, A1 for both velocities, M1 for impulse as change of momentum, A1 for 7 N s.

Question 9

A sphere of mass 2 kg moving at 4 m/s meets directly a sphere of mass 3 kg moving at 1 m/s towards it. The coefficient of restitution is 0.5. Find the velocity of each sphere after the impact.

Accepted Answer

Take the 2 kg sphere's direction as positive, so the velocities before are 4 and −1. Momentum: 2(4) + 3(−1) = 2v₁ + 3v₂, so 2v₁ + 3v₂ = 5. M1. Restitution, with the approach speed 4 − (−1) = 5: v₂ − v₁ = 0.5(5) = 2.5. M1 for the restitution equation with the relative approach speed. Substituting v₂ = v₁ + 2.5: 5v₁ + 7.5 = 5, so v₁ = −0.5 m/s and v₂ = 2 m/s. A1 A1 for the two velocities. Both signs have changed: the 2 kg sphere rebounds at 0.5 m/s and the 3 kg sphere at 2 m/s, each now moving the way the other came. Using an approach speed of 4, or of 3 from subtracting the speeds, misses that the second sphere was closing the gap too; the two velocities have opposite signs, so the relative speed is the sum of the speeds.

Question 10

A sphere of mass 3 kg moving at 4 m/s strikes directly a sphere of mass 1 kg which is at rest. Find the velocity of each sphere after the impact and the kinetic energy lost, first when the coefficient of restitution is 0 and then when it is 1.

Accepted Answer

With e = 0 the spheres do not separate, so they move on together: 3(4) = (3 + 1)v gives v = 3 m/s for both. M1 for coalescence with the combined mass. Kinetic energy before is ½(3)(4²) = 24 J and after is ½(4)(3²) = 18 J, so 6 J is lost. A1 for 3 m/s and 6 J.
With e = 1: momentum 3(4) = 3v₁ + v₂ and restitution v₂ − v₁ = 4. M1 for the pair of equations. Then 4v₁ = 8, so v₁ = 2 m/s and v₂ = 6 m/s. A1. Kinetic energy after is ½(3)(2²) + ½(1)(6²) = 6 + 18 = 24 J, so none is lost. B1 for showing the energy is unchanged.
Every value of e between gives a loss between 0 and 6 J: 6 J is the most this pair can lose, because momentum must survive and 18 J is the least energy that 12 N s of momentum can be carried with. e = 0 is the only case with one unknown, and e = 1 is the only case where kinetic energy could have been used in place of restitution.

Question 11

A sphere A of mass 5 kg moving at 6 m/s strikes directly a sphere B of mass 3 kg which is at rest. The impulse exerted on B has magnitude 15 N s. Find the velocity of each sphere after the impact, and the coefficient of restitution.

Accepted Answer

The impulse on B is its change of momentum: 3v₂ = 15, so v₂ = 5 m/s. A receives the equal and opposite impulse, so 5v₁ = 5(6) − 15 = 15 and v₁ = 3 m/s. M1 for impulse as change of momentum for one sphere, A1 for both velocities. Momentum checks: 30 before, 15 + 15 after. Then e = separation/approach = (5 − 3)/6 = 1/3. M1 for e from the two relative speeds, A1 for 1/3. The impulse replaces the restitution equation as the second piece of information, and e comes out at the end instead of going in at the start. An impulse of 9 N s would give v₂ = 3 and v₁ = 4.2, with A faster than the sphere in front of it, so that data would describe no possible impact; the 15 N s here leaves 3 ≤ 5, as it must.

Question 12

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.

Accepted Answer

Take the direction of u as positive, and let the velocities after the impact be v_A and v_B.
Momentum: 3mu = 3mv_A + mv_B, so 3v_A + v_B = 3u.
Restitution: v_B − v_A = eu.
Substituting the second into the first gives 4v_A = 3u − eu, so v_A = u(3 − e)/4 and v_B = 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)v_A² + ½(m)v_B² = (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.

Question 13

A smooth sphere A of mass 2m moving with speed u strikes directly a smooth sphere B of mass 3m which is at rest. The coefficient of restitution between the spheres is e. Find the velocity of each sphere after the impact in terms of u and e, and find the range of values of e for which the direction of motion of A is reversed by the impact.

Accepted Answer

Take the direction of u as positive. Momentum: 2mu = 2mv_A + 3mv_B, so 2v_A + 3v_B = 2u. Restitution: v_B − v_A = eu.
Substituting v_B = v_A + eu into the first equation: 5v_A = u(2 − 3e), so v_A = u(2 − 3e)/5 and v_B = 2u(1 + e)/5. v_B is positive for every e, so B always moves off forwards.
A is reversed when v_A < 0, that is when 2 − 3e < 0, so e > 2/3. Since e cannot exceed 1, the range is 2/3 < e ≤ 1. At e = 2/3 exactly, A is brought to rest, and for smaller e it follows B at reduced speed. M1 for conservation of momentum, M1 for the restitution equation, A1 for v_A, A1 for v_B, M1 for the condition v_A < 0, A1 for the range including the upper end.

Question 14

A sphere of mass m moving with speed u strikes directly an identical sphere which is at rest, the coefficient of restitution being e. Show that the fraction of the kinetic energy lost in the impact is (1 − e²)/2. Given that exactly one third of the kinetic energy is lost, find e. Explain why no impact between these spheres can lose more than half of the kinetic energy.

Accepted Answer

Momentum: mu = mv₁ + mv₂, so v₁ + v₂ = u. Restitution: v₂ − v₁ = eu. Adding and subtracting: v₁ = u(1 − e)/2 and v₂ = u(1 + e)/2.
Kinetic energy before = ½mu². After = ½m(v₁² + v₂²) = (mu²/8)[(1 − e)² + (1 + e)²] = (mu²/8)(2 + 2e²) = ¼mu²(1 + e²), the terms in e cancelling as they should.
Loss = ½mu² − ¼mu²(1 + e²) = ¼mu²(1 − e²), and dividing by ½mu² gives the fraction (1 − e²)/2, as required.
One third lost means (1 − e²)/2 = 1/3, so e² = 1/3 and e = 1/√3 = 0.577.
The fraction (1 − e²)/2 is largest when e = 0, where it equals ½. Even then the spheres cannot stop: momentum must be conserved, so they coalesce and move on together at u/2, keeping half the energy. At the other extreme, e = 1 gives v₁ = 0 and v₂ = u: the spheres exchange velocities and nothing at all is lost. M1 for conservation of momentum, M1 for the restitution equation, A1 for both velocities, M1 for the kinetic energy before and after, A1 for the fraction (1 − e²)/2, A1 for e = 1/√3, B1 for the maximum of one half at e = 0 with a reason.

Question 15

A sphere of mass m₁ moving with speed u strikes directly a sphere of mass m₂ which is at rest, the coefficient of restitution being e. Show that the kinetic energy lost in the impact is m₁m₂u²(1 − e²)/(2(m₁ + m₂)).

Accepted Answer

Take the direction of u as positive. Momentum: m₁u = m₁v₁ + m₂v₂. Restitution: v₂ − v₁ = eu. M1 for the two equations. Substituting v₂ = v₁ + eu: (m₁ + m₂)v₁ = m₁u − m₂eu, so v₁ = u(m₁ − em₂)/(m₁ + m₂) and v₂ = m₁u(1 + e)/(m₁ + m₂). M1 for solving, A1 for both velocities.
Kinetic energy after = ½m₁v₁² + ½m₂v₂² = [u²/(2(m₁ + m₂)²)][m₁(m₁ − em₂)² + m₂m₁²(1 + e)²]. M1 for the energy after with both spheres over the common denominator.
Expand the bracket: m₁(m₁² − 2em₁m₂ + e²m₂²) + m₁²m₂(1 + 2e + e²) = m₁³ + m₁²m₂ + e²m₁m₂² + e²m₁²m₂, the terms in e cancelling. Grouping, this is m₁²(m₁ + m₂) + e²m₁m₂(m₁ + m₂) = m₁(m₁ + m₂)(m₁ + e²m₂). A1 for the factorised bracket.
So the energy after is m₁u²(m₁ + e²m₂)/(2(m₁ + m₂)), and the loss is ½m₁u² − that = [m₁u²/(2(m₁ + m₂))][(m₁ + m₂) − (m₁ + e²m₂)] = m₁m₂u²(1 − e²)/(2(m₁ + m₂)), as required. A1.
The factor m₁m₂/(m₁ + m₂) is the reduced mass of the pair, and the loss is ½ × reduced mass × (approach speed)² × (1 − e²), which is the form the result takes for any pair of approach speeds. With m₁ = 3m and m₂ = m it gives 3mu²(1 − e²)/8, and with m₁ = m₂ it gives mu²(1 − e²)/4, the two cases derived separately earlier in this bank. The loss is greatest at e = 0 and zero at e = 1, and it is the same whichever sphere is the one at rest, since the expression is symmetric in m₁ and m₂.