Questions on oblique impact and impact with a smooth surface

Question 1

State what happens to each component of a ball's velocity when it strikes a smooth plane surface, and why.

Accepted Answer

The component parallel to the surface is unchanged, because a smooth surface exerts no force along itself. The component perpendicular to the surface is reversed and multiplied by e, exactly as in a direct impact, because the impulse acts along that direction only. B1 B1 for the two components, each with its reason.

Question 2

A ball moving at 12 m/s strikes a smooth wall at 40° to the wall, with e = 0.6. Find the speed immediately afterwards.

Accepted Answer

Parallel component = 12cos40° = 9.19 m/s, unchanged. Perpendicular component = 12sin40° = 7.71 m/s, becoming 0.6 × 7.71 = 4.63 m/s the other way. Speed = √(9.19² + 4.63²) = √105.9 = 10.3 m/s. M1 for resolving and applying e to the perpendicular component, A1 for 4.63 m/s, A1 for the resultant speed.

Question 3

A ball moving with velocity (4i − 3j) m/s strikes a smooth floor that lies along the direction of i. The coefficient of restitution is 0.5. Write down the velocity of the ball immediately after the impact.

Accepted Answer

The i component is parallel to the floor and is unchanged. The j component is perpendicular to it, so it is reversed and multiplied by e: −3 becomes −0.5(−3) = 1.5. The velocity is (4i + 1.5j) m/s. M1 for keeping i and multiplying j by −e, A1 for the vector.

Question 4

A ball of mass 0.3 kg moving with velocity (6i − 5j) m/s strikes a smooth vertical wall lying along the direction of i. The coefficient of restitution is 0.4. Find the impulse exerted by the wall on the ball, giving its magnitude and direction.

Accepted Answer

The wall lies along i, so the i component is unchanged and the j component is reversed and multiplied by e: −5 becomes −0.4(−5) = 2, and the velocity afterwards is (6i + 2j) m/s. M1 for the velocity after the impact. The impulse is the change in momentum, m(v − u) = 0.3((6i + 2j) − (6i − 5j)) = 0.3(7j) = 2.1j N s. M1 for impulse as m(v − u). So the impulse has magnitude 2.1 N s and acts perpendicular to the wall, in the direction of j, away from the wall. A1 for the magnitude with the direction. The i components cancel in the subtraction, which is the algebraic form of the wall being smooth: it can push only along its normal. Subtracting the other way round gives −2.1j, the impulse the ball exerts on the wall; taking the change in the j component as 2 − 5 instead of 2 − (−5) gives 0.9 N s and forgets that the ball's direction has been reversed.

Question 5

A ball moving at 12 m/s strikes a smooth wall at 40° to the wall, with e = 0.6. Find the angle its path makes with the wall afterwards, and the percentage of kinetic energy lost.

Accepted Answer

The components afterwards are 9.19 m/s along the wall and 0.6 × 12sin40° = 4.63 m/s away from it, so the angle is arctan(4.63/9.19) = 26.7°, shallower than the 40° it arrived at.
For the energy, work with squared speeds and let the mass cancel: 9.19² + 4.63² = 105.9 remains out of 12² = 144. The fraction lost is 1 − 105.9/144 = 0.264, that is 26.4%. M1 A1 for the angle, M1 A1 for the percentage lost.

Question 6

A ball strikes a smooth wall at 45° to the wall and leaves at 30° to it. Find the coefficient of restitution.

Accepted Answer

Let the parallel component be p, unchanged by the impact. Before, the perpendicular component is p tan45°; after, it is p tan30°. Restitution applies to that component alone, so e = p tan30°/(p tan45°) = tan30°/tan45° = 0.577 to three decimal places. M1 for the parallel component unchanged, M1 for restitution applied to the perpendicular components, A1 for tan30°/tan45°, A1 for 0.577.

Question 7

Explain why the angle to a smooth surface always decreases in an oblique impact unless e = 1.

Accepted Answer

The tangent of the angle to the surface is the perpendicular component divided by the parallel one. The impact multiplies the numerator by e and leaves the denominator alone, so the tangent is multiplied by e. For e < 1 that makes the angle smaller. M1 for the tangent as the ratio of the two components, A1 for the conclusion. Only e = 1 leaves it unchanged, the mirror case.

Question 8

A ball moving in a horizontal plane strikes a smooth vertical wall, then a second smooth vertical wall at right angles to the first, with the same coefficient e at each. Show that it ends up moving in exactly the opposite direction to its original one.

Accepted Answer

Take the two walls along the axes, so the velocity is (u, v). The first wall reverses and scales the component perpendicular to it, giving (−eu, v), and leaves the other component alone.
The second wall does the same to the other component, giving (−eu, −ev) = −e(u, v).
That is a negative scalar multiple of the original velocity, so the direction is exactly reversed, and the speed is e times the original. M1 A1 for the effect of the first wall, M1 A1 for the second and the conclusion. It is why a ball played into the corner of a court comes straight back.

Question 9

A ball moving in a horizontal plane with velocity (6i + 4j) m/s strikes a smooth vertical wall lying along the direction of i, with coefficient of restitution 0.5. It then strikes a second smooth vertical wall lying along the direction of j, with coefficient of restitution 0.75. Find the velocity of the ball after the second impact, and determine whether it is then moving parallel to its original line of motion.

Accepted Answer

The first wall acts on the j component only: (6i + 4j) becomes (6i − 2j). M1 for reversing and scaling the perpendicular component, A1 for (6i − 2j). The second wall acts on the i component only: (6i − 2j) becomes (−4.5i − 2j) m/s. M1 for the second wall, A1 for the vector.
The original line of motion has gradient 4/6 = 2/3; the final one has gradient (−2)/(−4.5) = 4/9. These differ, so the ball is not parallel to its original line; the two lines are about 9.7° apart. Each component has been multiplied by one coefficient, so the gradient is scaled by 0.5/0.75 = 2/3, and only equal coefficients would leave the line unchanged in direction.

Question 10

Two smooth spheres A and B, of equal radius and masses 2 kg and 1 kg, move on a smooth horizontal table and collide. At the instant of impact the line of centres is parallel to i, and the velocities are (3i + 2j) m/s for A and (−i + j) m/s for B. The coefficient of restitution is 0.5. Find the velocity of each sphere after the impact and the kinetic energy lost.

Accepted Answer

The spheres are smooth, so the impulse between them is along the line of centres, the i direction, and the j components are unchanged: A keeps 2 and B keeps 1. B1 for the j components carried through. Along i, momentum: 2(3) + 1(−1) = 2a + b, so 2a + b = 5. M1. Restitution along i: b − a = 0.5(3 − (−1)) = 2. M1. Solving, a = 1 and b = 3, so A has velocity (i + 2j) m/s and B has (3i + j) m/s; A1 for both.
Kinetic energy before: ½(2)(9 + 4) + ½(1)(1 + 1) = 13 + 1 = 14 J. After: ½(2)(1 + 4) + ½(1)(9 + 1) = 5 + 5 = 10 J. The loss is 4 J; A1. Only the i components changed, so the loss could equally be found from them alone: ½(2)(9 − 1) + ½(1)(1 − 9) = 8 − 4 = 4 J. The restitution equation uses the relative velocity along the line of centres, 3 − (−1) = 4, not A's velocity alone, since B was moving towards A.

Question 11

A ball falling vertically at 7 m/s strikes a smooth plane inclined at 30° to the horizontal. The coefficient of restitution is 0.5. Find the speed of the ball immediately after the impact and the angle its path then makes with the plane.

Accepted Answer

Resolve the velocity along and perpendicular to the plane, not horizontally and vertically. The vertical makes 60° with the plane, so the component along the plane is 7cos60° = 7sin30° = 3.5 m/s and the component into the plane is 7cos30° = 6.06 m/s. M1 for resolving relative to the plane, A1 for the two components. The plane is smooth, so 3.5 m/s survives and 6.06 becomes 0.5 × 6.06 = 3.03 m/s away from the plane. Speed = √(3.5² + 3.03²) = 4.63 m/s; A1. The angle with the plane is arctan(3.03/3.5) = 40.9°; A1. The ball arrived at 60° to the plane and leaves at 40.9°, closer to it, as it must for e < 1.

Question 12

Two smooth spheres A and B of equal mass and equal radius lie on a smooth horizontal table. B is at rest, and A is moving at 6 m/s in a direction making an angle of 30° with the line of centres at the moment of impact. The coefficient of restitution between them is 0.5. Find the speed and direction of each sphere after the impact, and the fraction of the kinetic energy lost.

Accepted Answer

Resolve A's velocity along and perpendicular to the line of centres. That line, not the direction of motion, is where the impulse acts.
Along: 6cos30° = 3√3 ≈ 5.196 m/s. Perpendicular: 6sin30° = 3 m/s.
The spheres are smooth, so the perpendicular component of A is unchanged and B gains none. Everything else happens along the line of centres.
Taking unit mass, conservation of momentum along that line gives v_A + v_B = 3√3, and the law of restitution gives v_B − v_A = 0.5(3√3) = 1.5√3.
Adding and halving: v_B = 2.25√3 ≈ 3.90 m/s along the line of centres. Subtracting: v_A = 0.75√3 ≈ 1.30 m/s along it.
B moves along the line of centres at 3.90 m/s, since it had no perpendicular component to keep.
A has components 1.30 along and 3 perpendicular, so its speed is √(1.6875 + 9) = √10.6875 = 3.27 m/s, at arctan(3/1.30) = 66.6° to the line of centres. A is deflected much further from that line than it arrived.
Kinetic energy, per unit mass and dropping the common ½: before, 6² = 36. After, 10.6875 + 15.1875 = 25.875, taking B's contribution as (2.25√3)² = 15.1875.
The fraction lost is (36 − 25.875)/36 = 10.125/36 = 9/32, that is 0.28125 or 28.1%. M1 A1 for resolving along and perpendicular to the line of centres, M1 for conservation of momentum, M1 for the restitution equation, A1 A1 for the two speeds along that line, A1 for A's speed and direction, A1 for the fraction lost.

Question 13

A ball strikes a smooth plane surface with speed u, its direction of motion making an angle θ with the surface, and the coefficient of restitution is e. Show that the fraction of the kinetic energy lost in the impact is (1 − e²) sin²θ. Given that e = 0.5, find the angle θ for which exactly half the kinetic energy is lost.

Accepted Answer

Resolve along and perpendicular to the surface: u cos θ along it, unchanged by the impact, and u sin θ into it, which becomes eu sin θ outwards. M1 for the two components with e applied to the perpendicular one only.
The squared speed after is u² cos²θ + e²u² sin²θ, so the kinetic energy after is ½mu²(cos²θ + e² sin²θ). A1. The fraction lost is 1 − (cos²θ + e² sin²θ) = sin²θ − e² sin²θ, using cos²θ = 1 − sin²θ. M1 for the fraction as one minus the ratio of squared speeds. That is (1 − e²) sin²θ, as required. A1.
With e = 0.5, half lost means 0.75 sin²θ = 0.5, so sin²θ = 2/3 and θ = arcsin √(2/3) = 54.7°. M1 for setting the fraction to ½, A1 for 54.7°.
The formula reads as a sentence: the parallel component loses nothing, and the perpendicular component, which holds the fraction sin²θ of the energy, keeps e² of its share. A glancing blow with θ small loses almost nothing whatever e is, and a head-on strike with θ = 90° loses the full 1 − e².

Question 14

A smooth sphere A moving with speed u strikes a smooth sphere B of the same mass and radius which is at rest. At the instant of impact the direction of motion of A makes an angle θ with the line of centres, where 0 < θ < 90°, and the coefficient of restitution is e. Prove that after the impact the two spheres move at right angles to each other if and only if e = 1.

Accepted Answer

Resolve A's velocity into u cos θ along the line of centres and u sin θ perpendicular to it. The spheres are smooth, so A keeps u sin θ and B gains no perpendicular component: B moves along the line of centres. M1 for resolving and carrying the perpendicular components through.
Along the line of centres, with equal masses, momentum gives a + b = u cos θ and restitution gives b − a = eu cos θ. M1 for the two equations. Hence a = ½(1 − e)u cos θ and b = ½(1 + e)u cos θ. A1.
B's path is the line of centres, so the angle between the two paths is the angle A's velocity (a, u sin θ) makes with that line. Since u sin θ ≠ 0, this angle is 90° exactly when a = 0, that is when (1 − e)u cos θ = 0. M1 for reducing the right-angle condition to a = 0. As u cos θ ≠ 0 for 0 < θ < 90°, a = 0 if and only if e = 1. A1.
For e < 1, a is positive, so A still has a forward component along the line of centres and the angle between the paths is acute. This is the equal-mass result that a perfectly elastic glancing blow sends the two spheres off at right angles, and it fails for unequal masses because a then contains the mass ratio as well as e.

Question 15

A ball moves between two smooth parallel vertical walls that are 2 m apart. Immediately before its first impact its velocity has component 3 m/s parallel to the walls and 4 m/s perpendicular to them, and the coefficient of restitution at each wall is 0.5. Show that the distances the ball travels parallel to the walls between successive impacts form a geometric sequence, and find the distance it travels parallel to the walls between the first impact and the fourth. Find also the angle its path makes with the wall at the fourth impact.

Accepted Answer

The walls are smooth, so the 3 m/s parallel to them is never changed. Each impact multiplies the perpendicular component by 0.5, so after k impacts it is 4(0.5)^k m/s. M1 for the perpendicular component after k impacts, A1 for 4(0.5)^k with the parallel component constant.
Between impact k and impact k + 1 the ball crosses the 2 m gap at 4(0.5)^k m/s, taking 2/(4(0.5)^k) = 0.5(2)^k s, and in that time travels 3 × 0.5(2)^k = 1.5(2)^k m parallel to the walls. M1 for distance as parallel speed times crossing time. The distances are 3, 6, 12, … m: a geometric sequence with common ratio 2, because each impact halves the perpendicular speed and so doubles the crossing time. A1.
From the first impact to the fourth there are three crossings: 3 + 6 + 12 = 21 m. A1.
Arriving at the fourth impact the ball has made three impacts, so its perpendicular component is 4(0.5)³ = 0.5 m/s and the angle with the wall is arctan(0.5/3) = 9.46°, against arctan(4/3) = 53.1° at the first. A1.
The sequence of distances diverges even though the perpendicular speed tends to zero: the ball keeps its 3 m/s along the walls and the crossings take longer and longer, so it never settles against either wall. Counting the fourth impact as after four crossings, or using the component before the first impact for the fourth, gives 45 m or 53.1° in place of the answers.