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Edexcel International A Level Maths · M2 · Mechanics 2

Collisions and restitution

28 free, high-quality past-paper-style questions on Collisions and restitution for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 248 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

1IALM-M2-5-B-01 M2 · A2Easy6 marks

A puck of mass 0.5 kg slides on smooth horizontal ice with velocity (3i + 4j) m s−1. It is struck by a stick and receives an impulse of (2i − 6j) N s. The puck is modelled as a particle.

(a)

Find the velocity of the puck immediately after the impulse.

(3)
(b)

Find the speed of the puck immediately after the impulse.

(1)
(c)

Find the increase in the kinetic energy of the puck caused by the impulse.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
use of I = mv − mu: 2i − 6j = 0.5v − 0.5(3i + 4j)M1
0.5v = 3.5i − 4jA1
v = (7i − 8j) m s−1A1
Part (b)
√(72 + 82) = √113 = 10.6 m s−1 ftA1
Part (c)
½(0.5)(113) − ½(0.5)(32 + 42)M1
= 28.25 − 6.25 = 22 JA1

Common error: Adding the impulse straight to the velocity. The impulse changes the momentum, so add it to mu (or divide it by the mass first).

2IALM-M2-5-B-02 M2 · A2Easy5 marks

A model boat of mass 2 kg is drifting on a still pond with velocity (4i − j) m s−1. A sudden gust of wind acts on it, and immediately afterwards its velocity is (−2i + 5j) m s−1. The boat is modelled as a particle and the gust as an impulse; resistance from the water is ignored during the gust.

(a)

Find the impulse exerted on the boat by the gust.

(2)
(b)

Find the magnitude of this impulse.

(1)
(c)

Find the angle between the direction of the impulse and the vector i.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
I = 2[(−2i + 5j) − (4i − j)]M1
= (−12i + 12j) N sA1
Part (b)
√(122 + 122) = 12√2 = 17.0 N s ftA1
Part (c)
tan−1(12/12) = 45° to the negative i directionM1
angle with i = 135°A1

Exam tip: For an angle with i, sketch the vector first: here it points up and to the left, so the angle with the positive i direction is obtuse.

3IALM-M2-5-B-03 M2 · A2Easy6 marks

Two beads A and B, of masses 0.2 kg and 0.3 kg, are threaded on a smooth straight horizontal wire. A moves at 5 m s−1 and B moves at 1 m s−1 in the same direction, with A behind B, so that A catches B and they collide. Immediately after the collision, B moves at 3.4 m s−1 in its original direction.

A0.2 kgB0.3 kg5 m s⁻¹1 m s⁻¹
Figure: the beads on the wire just before A catches B; not to scale.
(a)

Find the velocity of A immediately after the collision.

(2)
(b)

Find the coefficient of restitution between A and B.

(2)
(c)

Find the kinetic energy lost in the collision.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
0.2 × 5 + 0.3 × 1 = 0.2v + 0.3 × 3.4M1
v = 1.4 m s−1 in the original directionA1
Part (b)
speed of separation 3.4 − 1.4 = 2, speed of approach 5 − 1 = 4M1
e = 2 ÷ 4 = 0.5 ftA1
Part (c)
before ½(0.2)(52) + ½(0.3)(12) = 2.65 J; after ½(0.2)(1.42) + ½(0.3)(3.42) = 1.93 JM1
loss = 0.72 JA1

Exam tip: When one velocity after the impact is given, momentum gives the other one; restitution is then only needed to find e.

4IALM-M2-5-B-04 M2 · A2Easy5 marks

A small smooth sphere of mass 0.3 kg slides on a smooth horizontal floor with speed 4 m s−1, moving at right angles towards a fixed smooth vertical wall. It rebounds from the wall along the same line with speed 2.8 m s−1.

(a)

Find the coefficient of restitution between the sphere and the wall.

(1)
(b)

Find the magnitude of the impulse exerted by the wall on the sphere.

(2)
(c)

Find the kinetic energy lost in the impact.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
e = 2.8 ÷ 4 = 0.7B1
Part (b)
I = 0.3(2.8 − (−4))M1
= 2.04 N sA1
Part (c)
½(0.3)(42) − ½(0.3)(2.82)M1
= 2.4 − 1.176 = 1.22 J (1.224)A1

Common error: Using 0.3 × (4 − 2.8) for the impulse. The sphere reverses direction, so the velocities have opposite signs and the speeds add.

5IALM-M2-5-FB-01 M2 · A2Easy7 marks

A small ball of mass 0.2 kg rolls on smooth horizontal ground, where i and j are perpendicular horizontal unit vectors. The ball is kicked. Figure 1 shows the velocity of the ball immediately before the kick, 10 m s−1 at an angle θ above the i direction, and immediately after the kick, 13 m s−1 at an angle α above the −i direction, where tan θ = ¾ and tan α = 12/5. The ball is modelled as a particle.

Figure 1ijbefore: 10 m s⁻¹after: 13 m s⁻¹θα
(a)

Write the velocity of the ball immediately before and immediately after the kick in the form (ai + bj) m s−1.

(2)
(b)

Find the impulse exerted on the ball by the kick.

(2)
(c)

Find the magnitude of this impulse.

(1)
(d)

Find the kinetic energy gained by the ball as a result of the kick.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
before: 10(0.8i + 0.6j) = (8i + 6j) m s−1B1
after: 13(−5/13i + 12/13j) = (−5i + 12j) m s−1B1
Part (b)
I = 0.2[(−5i + 12j) − (8i + 6j)]M1
= (−2.6i + 1.2j) N sA1
Part (c)
√(2.62 + 1.22) = √8.2 = 2.86 N s ftA1
Part (d)
½(0.2)(132) − ½(0.2)(102)M1
= 16.9 − 10 = 6.9 JA1

Common error: Giving the after-velocity a positive i component. The angle α is measured from the −i direction, so the ball now moves to the left: −5i, not +5i.

6IALM-M2-5-FB-02 M2 · A2Easy7 marks

Two gliders A and B move along a smooth straight horizontal air track. Figure 1 shows their masses and velocities immediately before and immediately after they collide. The gliders are modelled as particles.

Figure 1BeforeA0.6 kgB0.4 kg4 m s⁻¹1.5 m s⁻¹AfterA0.6 kgB0.4 kg0.7 m s⁻¹v m s⁻¹
(a)

Find the value of v.

(2)
(b)

Find the coefficient of restitution between A and B.

(2)
(c)

Find the kinetic energy lost in the collision.

(2)
(d)

Find the magnitude of the impulse exerted on B in the collision.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
0.6(4) + 0.4(−1.5) = 0.6(0.7) + 0.4vM1
1.8 = 0.42 + 0.4v ⇒ v = 3.45A1
Part (b)
speed of separation 3.45 − 0.7 = 2.75; speed of approach 4 + 1.5 = 5.5M1
e = 2.75 ÷ 5.5 = 0.5 ftA1
Part (c)
before ½(0.6)(42) + ½(0.4)(1.52) = 5.25 J; after ½(0.6)(0.72) + ½(0.4)(3.452) = 2.5275 JM1
loss = 2.72 J (2.7225)A1
Part (d)
0.4(3.45 − (−1.5)) = 1.98 N s ftA1

Common error: Using 4 − 1.5 as the speed of approach. The gliders move towards each other, so their speeds add: the approach speed is 4 + 1.5.

7IALM-M2-5-FB-03 M2 · A2Easy7 marks

A ball of mass 0.25 kg is released from rest above a horizontal floor and bounces. Figure 1 shows the velocity–time graph of the ball for the first 1.2 s, taking upwards as positive; labelled points give exact values. The ball is modelled as a particle moving vertically, with air resistance ignored. Take g = 9.8 m s−2.

Figure 10.00.10.20.30.40.50.60.70.80.91.01.11.2−5−4−3−2−10123v / m s⁻¹t / s(0.5, −4.9)(0.5, 2.94)(1.1, −2.94)(1.1, 1.764)
(a)

Find the coefficient of restitution between the ball and the floor.

(2)
(b)

Find the height above the floor from which the ball was released.

(2)
(c)

Find the magnitude of the impulse exerted on the ball by the floor at the first bounce.

(2)
(d)

Find the greatest height reached by the ball between the first and second bounces.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
speed just before the first bounce 4.9, just after 2.94M1
e = 2.94 ÷ 4.9 = 0.6A1
Part (b)
area of the first triangle: ½ × 0.5 × 4.9 (or 4.92 ÷ (2 × 9.8))M1
= 1.23 m (1.225)A1
Part (c)
0.25(2.94 − (−4.9))M1
= 1.96 N sA1
Part (d)
area ½ × (0.8 − 0.5) × 2.94 = 0.441 mB1

Exam tip: On a velocity–time graph a bounce shows as a vertical jump. The area between the line and the time axis gives the distance fallen or risen, and the ratio of the two speeds at the jump is e.

8IALM-M2-5-FB-04 M2 · A2Easy7 marks

Two small lumps of putty A and B move towards each other along a straight line on a smooth horizontal floor, as shown in Figure 1. They collide and stick together to form a single lump C. Later, C strikes a fixed smooth vertical wall at right angles and rebounds along the same line with speed 1.5 m s−1. The lumps are modelled as particles.

Figure 1AB3 kg1 kg4 m s⁻¹2 m s⁻¹wall
(a)

Find the speed of C immediately after A and B collide.

(2)
(b)

Find the kinetic energy lost when A and B collide.

(2)
(c)

Find the coefficient of restitution between C and the wall.

(1)
(d)

Find the magnitude of the impulse exerted by the wall on C.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
3(4) + 1(−2) = 4vM1
v = 2.5 m s−1 towards the wallA1
Part (b)
½(3)(42) + ½ × 1 × 22 − ½(4)(2.52) = 26 − 12.5M1
= 13.5 JA1
Part (c)
e = 1.5 ÷ 2.5 = 0.6 ftA1
Part (d)
4(1.5 − (−2.5))M1
= 16 N s ftA1

Exam tip: When particles stick together (e = 0) there is a single unknown velocity, so momentum alone finds it. Restitution is only needed for the impact with the wall.

9IALM-M2-5-B-05 M2 · A2Medium8 marks

Two model railway trucks A and B, of masses 4 kg and 1 kg, run towards each other along a straight horizontal track. Just before they collide, A has speed 3 m s−1 and B has speed 5 m s−1. The coefficient of restitution between the trucks is 0.4. The trucks are modelled as particles and the track as smooth.

A 4 kgB 1 kg3 m s⁻¹5 m s⁻¹
Figure: the trucks just before they collide; e = 0.4; not to scale.
(a)

Find the velocities of A and B immediately after the collision, stating clearly the direction of motion of each truck.

(5)
(b)

Find the magnitude of the impulse exerted on B in the collision.

(1)
(c)

Find the kinetic energy lost in the collision.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
(A's direction positive) 4(3) + 1(−5) = 4vA + vBM1
4vA + vB = 7A1
vB − vA = 0.4(3 + 5) = 3.2M1
solving: vA = 0.76 m s−1, in its original directionA1
vB = 3.96 m s−1, direction of motion reversedA1
Part (b)
1 × (3.96 + 5) = 8.96 N s ftA1
Part (c)
before ½ × 4 × 32 + ½ × 1 × 52 = 30.5 J; after ½ × 4 × 0.762 + ½ × 1 × 3.962 = 8.996 JM1
loss = 21.5 J (21.504)A1

Exam tip: Draw both unknown velocities in the positive direction. A positive answer keeps that direction; a negative one means the truck moves the other way.

10IALM-M2-5-B-06 M2 · A2Medium9 marks

Two smooth spheres A and B, of equal radius and masses 3m and m, move towards each other along the same straight line on a smooth horizontal surface. Just before they collide, A has speed 2u and B has speed u. The coefficient of restitution between them is e.

(a)

Show that the speed of B immediately after the collision is ¼u(5 + 9e), and find the velocity of A in terms of u and e.

(4)
(b)

Explain why the direction of motion of A is not changed by the collision.

(1)
(c)

Given that the speed of B immediately after the collision is twice its speed immediately before, find the value of e.

(2)
(d)

Using your value of e, find the kinetic energy lost in the collision, in terms of m and u.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
(A's direction positive) 3m(2u) − mu = 3mvA + mvBM1
vB − vA = e(2u + u) = 3euM1
5u = 4vA + 3eu ⇒ vA = ¼u(5 − 3e) in A's original directionA1
vB = ¼u(5 − 3e) + 3eu = ¼u(5 + 9e)A1
Part (b)
e ≤ 1, so 5 − 3e ≥ 2 > 0 and vA is positiveB1
Part (c)
¼u(5 + 9e) = 2uM1
e = ⅓A1
Part (d)
vA = u, vB = 2u; before ½(3m)(4u2) + ½mu2 = 6.5mu2, after 1.5mu2 + 2mu2 = 3.5mu2M1
loss = 3mu2A1

Exam tip: In a “show that” with letters, keep e as a symbol until the end and use 0 ≤ e ≤ 1 to justify any direction.

11IALM-M2-5-B-07 M2 · A2Medium7 marks

Two pucks A and B slide on a smooth horizontal air table. A has mass 2 kg and velocity (3i + 2j) m s−1; B has mass 3 kg and velocity (−i + 4j) m s−1. The pucks collide and stick together. The pucks are modelled as particles.

(a)

Find the velocity of the combined pucks immediately after the collision.

(3)
(b)

Find the speed of the combined pucks immediately after the collision.

(1)
(c)

Find the magnitude of the impulse exerted on A in the collision.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
2(3i + 2j) + 3(−i + 4j) = 5vM1
5v = 3i + 16jA1
v = (0.6i + 3.2j) m s−1A1
Part (b)
√(0.62 + 3.22) = √10.6 = 3.26 m s−1 ftA1
Part (c)
I = 2[(0.6i + 3.2j) − (3i + 2j)]M1
= −4.8i + 2.4jA1
|I| = √28.8 = 5.37 N sA1

Exam tip: Momentum is conserved as a vector, so the same equation handles the i and j components at once. The impulse on B is exactly the negative of the impulse on A.

12IALM-M2-5-B-08 M2 · A2Medium8 marks

Two smooth spheres P and Q, of equal radius and masses 0.5 kg and 1.5 kg, lie on a smooth horizontal floor. Q is at rest between P and a fixed smooth vertical wall, and the line of centres of P and Q is perpendicular to the wall. P is projected towards Q with speed 6 m s−1. The coefficient of restitution between P and Q is 0.6, and between Q and the wall it is 0.4.

PQ0.5 kg1.5 kg6 m s⁻¹at rest
Figure: P projected towards Q, which is at rest between P and the wall; not to scale.
(a)

Find the velocities of P and Q immediately after they collide.

(4)
(b)

Find the speed of Q immediately after it hits the wall.

(1)
(c)

Show that there are no further collisions between P and Q.

(2)
(d)

The wall is replaced by a different fixed smooth wall with coefficient of restitution f. Find the range of values of f for which P and Q would collide again.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
(towards the wall positive) 0.5(6) = 0.5vP + 1.5vQM1
vQ − vP = 0.6 × 6 = 3.6M1
vP = −1.2: P moves away from the wall at 1.2 m s−1A1
vQ = 2.4 m s−1 towards the wallA1
Part (b)
0.4 × 2.4 = 0.96 m s−1 ftA1
Part (c)
after rebounding, P and Q both move away from the wallM1
Q's speed 0.96 < P's speed 1.2, so Q cannot catch P: no further collisionA1
Part (d)
2.4f > 1.2 ⇒ 0.5 < f ≤ 1B1

Exam tip: “No further collisions” is a comparison of velocities: once both spheres move the same way, the one behind must be slower.

13IALM-M2-5-B-09 M2 · A2Medium7 marks

A tennis ball of mass 0.2 kg is released from rest at a height of 3.6 m above a horizontal concrete path. It bounces and first rises to a height of 1.6 m. The ball is modelled as a particle moving vertically, and air resistance is ignored.

(a)

Find the speed of the ball just before and just after its first bounce.

(2)
(b)

Find the coefficient of restitution between the ball and the path.

(1)
(c)

Find the magnitude of the impulse exerted by the path on the ball.

(2)
(d)

Find the height to which the ball rises after its second bounce.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
v2 = 2(9.8)(3.6) ⇒ 8.4 m s−1 beforeB1
0 = w2 − 2(9.8)(1.6) ⇒ 5.6 m s−1 afterB1
Part (b)
e = 5.6 ÷ 8.4 = ⅔ ftA1
Part (c)
0.2(5.6 − (−8.4))M1
= 2.8 N sA1
Part (d)
speed after second bounce = ⅔ × 5.6 = 3.733… m s−1 (56/15)M1
height = (56/15)2 ÷ 19.6 = 1.6 × (⅔)2 = 0.711 mA1

Exam tip: Each bounce multiplies the speed by e, so each rebound height is e2 times the previous one: here 1.6 × (⅔)2.

14IALM-M2-5-B-10 M2 · A2Medium7 marks

Two small blocks A and B slide on smooth horizontal ice along the same straight line. A has mass 2 kg and moves at 5 m s−1 towards B, which is at rest. After the collision A moves back along the line with speed 1 m s−1 and B moves forward with speed 2 m s−1. The blocks are modelled as particles.

(a)

Find the mass of B.

(2)
(b)

Find the coefficient of restitution between A and B.

(2)
(c)

Find the kinetic energy lost in the collision.

(2)
(d)

Find the magnitude of the impulse exerted on B.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
2(5) = 2(−1) + 2MM1
M = 6 kgA1
Part (b)
separation 2 − (−1) = 3, approach 5M1
e = 0.6A1
Part (c)
before ½(2)(25) = 25 J; after ½(2)(12) + ½(6)(22) = 13 JM1
loss = 12 J ftA1
Part (d)
6 × 2 = 12 N s ftA1

Common error: Writing A's velocity after the collision as +1. A has reversed, so it is −1 in the momentum equation, and the separation speed is 2 + 1.

15IALM-M2-5-FB-05 M2 · A2Medium9 marks

Two particles A and B move along a straight line on a smooth horizontal surface and collide. Figure 1 shows the displacement–time graphs of A and B, where x metres is the displacement from a fixed point O on the line. The mass of A is 4 kg.

Figure 101234502468101214161820x / mt / sABAB
(a)

Use the graph to find the velocities of A and B immediately before and immediately after the collision.

(3)
(b)

Find the mass of B.

(2)
(c)

Find the coefficient of restitution between A and B.

(2)
(d)

Find the kinetic energy lost in the collision.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
before: uA = 12 ÷ 3 = 4 m s−1, uB = (12 − 18) ÷ 3 = −2 m s−1B1
after: vA = (10 − 12) ÷ 2 = −1 m s−1B1
vB = (16 − 12) ÷ 2 = 2 m s−1B1
Part (b)
4(4) + M(−2) = 4(−1) + M(2)M1
M = 5 kgA1
Part (c)
separation 2 − (−1) = 3; approach 4 − (−2) = 6M1
e = 0.5A1
Part (d)
before ½(4)(42) + ½(5)(22) = 42 J; after ½(4)(12) + ½(5)(22) = 12 JM1
loss = 30 J ftA1

Exam tip: The gradient of a displacement–time graph is the velocity, sign included. A line sloping down means motion in the negative direction.

16IALM-M2-5-FB-06 M2 · A2Medium9 marks

A ball of mass 0.5 kg moves on smooth horizontal ground. Figure 1 is a plan view, drawn on a 1 m grid, with i and j the unit vectors in the directions of the x- and y-axes. The ball passes through O at time t = 0 and moves with constant velocity to A, which it reaches when t = 2 s. At A it receives an impulse, and it then moves with constant velocity to B, which it reaches when t = 3.5 s. The ball is modelled as a particle.

Figure 102468100246810121416y / mx / mO(t = 0)A(t = 2 s)B(t = 3.5 s)
(a)

Find the velocity of the ball as it moves from O to A.

(2)
(b)

Find the velocity of the ball as it moves from A to B.

(2)
(c)

Find the impulse received by the ball at A, and its magnitude.

(3)
(d)

Find the change in the kinetic energy of the ball caused by the impulse.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
displacement 8i + 6j in 2 sM1
u = (4i + 3j) m s−1A1
Part (b)
displacement (5 − 8)i + (15 − 6)j = −3i + 9j in 1.5 sM1
v = (−2i + 6j) m s−1A1
Part (c)
I = 0.5[(−2i + 6j) − (4i + 3j)]M1
= (−3i + 1.5j) N sA1
|I| = √11.25 = 3.35 N sA1
Part (d)
½(0.5)(22 + 62) − ½(0.5)(42 + 32) = 10 − 6.25M1
increase of 3.75 JA1

Common error: Dividing the displacement from A to B by 3.5 s. The ball takes only 3.5 − 2 = 1.5 s to travel from A to B.

17IALM-M2-5-FB-07 M2 · A2Medium7 marks

A small smooth sphere of mass 0.4 kg moves on a smooth horizontal floor between two fixed smooth parallel vertical walls W1 and W2. The sphere moves along a line perpendicular to both walls. At time t = 0 it is at the point A shown in Figure 1, moving towards W2. The coefficient of restitution between the sphere and each wall is 0.5. The sphere is modelled as a particle.

Figure 1W1W2A6 m s⁻¹1.2 m1.2 m
(a)

Find the speed of the sphere immediately after its first impact and immediately after its second impact.

(2)
(b)

Find the value of t when the sphere first returns to A.

(3)
(c)

Find the total kinetic energy lost by the sphere between t = 0 and the instant it first returns to A.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
after hitting W2: 0.5 × 6 = 3 m s−1B1
after hitting W1: 0.5 × 3 = 1.5 m s−1B1
Part (b)
A to W2: 1.2 ÷ 6 = 0.2 s; W2 to W1: 2.4 ÷ 3 = 0.8 sM1
W1 back to A: 1.2 ÷ 1.5 = 0.8 sM1
t = 0.2 + 0.8 + 0.8 = 1.8 sA1
Part (c)
½(0.4)(62) − ½(0.4)(1.52) = 7.2 − 0.45M1
= 6.75 JA1

Exam tip: Between impacts the speed is constant on a smooth floor, so each stage of the journey is just distance ÷ speed, with the speed multiplied by e at every wall.

18IALM-M2-5-FB-08 M2 · A2Medium9 marks

Trolley A, of mass 1.5 kg, moves with speed 4 m s−1 along a smooth straight horizontal track towards trolley B, of mass 2.5 kg, which is at rest. The trolleys collide. Figure 1 shows how the force F exerted on B by A varies with time during the collision, which lasts 40 ms. The trolleys are modelled as particles.

Figure 1051015202530354045050100150200250300350F / Nt / ms
(a)

Use the graph to find the magnitude of the impulse exerted on B.

(2)
(b)

Find the speeds of A and B immediately after the collision.

(3)
(c)

Find the coefficient of restitution between the trolleys.

(2)
(d)

Find the kinetic energy lost in the collision.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
area under the graph: ½ × 0.040 × 300M1
= 6 N sA1
Part (b)
B: 2.5vB = 6 ⇒ vB = 2.4 m s−1M1 A1
A receives −6 N s: 1.5vA = 1.5 × 4 − 6 ⇒ A is brought to restA1
Part (c)
e = (2.4 − 0) ÷ (4 − 0)M1
= 0.6 ftA1
Part (d)
½(1.5)(42) − ½(2.5)(2.42) = 12 − 7.2M1
= 4.8 J ftA1

Common error: Using the time in milliseconds when finding the area. Convert 40 ms to 0.040 s first, or the impulse comes out 1000 times too large.

19IALM-M2-5-FB-09 M2 · A2Medium9 marks

Figure 1 shows a smooth track in a vertical plane: two curved ramps joined by a straight horizontal section of floor. A small ball A, of mass 0.2 kg, is released from rest at the top of the left-hand ramp. A small ball B, of mass 0.3 kg, is at rest on the floor at the foot of the right-hand ramp. A moves down the ramp and along the floor and collides directly with B. The coefficient of restitution between A and B is 0.6. The balls are modelled as particles. Take g = 9.8 m s−2.

Figure 1A (0.2 kg)B (0.3 kg)0.9 msmooth horizontal floor
(a)

Find the speed of A just before it collides with B.

(2)
(b)

Find the velocities of A and B immediately after the collision.

(4)
(c)

Find the greatest height above the floor reached by B on the right-hand ramp.

(2)
(d)

Explain why A and B must collide again.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
½v2 = 9.8 × 0.9M1
v = 4.2 m s−1A1
Part (b)
0.2(4.2) = 0.2vA + 0.3vBM1
vB − vA = 0.6 × 4.2 = 2.52M1
vA = 0.168 m s−1, still towards B's rampA1
vB = 2.69 m s−1 (2.688)A1
Part (c)
½(2.688)2 = 9.8hM1
h = 0.369 m ftA1
Part (d)
the track is smooth, so B returns to the floor with speed 2.688 m s−1 moving towards A, while A is still moving towards B: they meet againB1

Exam tip: Energy conservation on the smooth ramps and momentum with restitution at the impact are separate stages. Find the speed at the bottom first, then treat the collision on its own.

20IALM-M2-5-FB-10 M2 · A2Medium10 marks

A particle A, of mass 2 kg, moving with speed 3 m s−1 on a smooth horizontal surface, collides directly with a particle B, of mass m kg, which is at rest. The coefficient of restitution between A and B is e. Figure 1 shows how the kinetic energy L joules lost in the collision depends on e.

Figure 10.00.10.20.30.40.50.60.70.80.91.001234567L / Je(0, 6)(0.5, 4.5)
(a)

When e = 0 the particles move together after the collision. Use the value of L at e = 0 to show that m = 4.

(4)
(b)

For the case shown by the labelled point where e = 0.5, find the velocities of A and B immediately after the collision.

(3)
(c)

Show that, for this pair of particles, L = 6(1 − e2).

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
common velocity V: 2 × 3 = (2 + m)VM1
½ × 2 × 32 − ½(2 + m)V2 = 6, i.e. 9 − 18/(2 + m) = 6M1
18/(2 + m) = 3A1
m = 4A1
Part (b)
2(3) = 2vA + 4vBM1
vB − vA = 0.5 × 3 = 1.5M1
vA = 0 (A is brought to rest), vB = 1.5 m s−1A1
Part (c)
6 = 2vA + 4vB and vB − vA = 3e give vA = 1 − 2e, vB = 1 + eM1
KE after = (1 − 2e)2 + 2(1 + e)2 = 3 + 6e2M1
L = 9 − (3 + 6e2) = 6(1 − e2)A1

Exam tip: The greatest loss of kinetic energy happens when e = 0 and there is no loss when e = 1. A graph of L against e always runs between these two cases.

21IALM-M2-5-B-11 M2 · A2Hard12 marks

Three small smooth spheres A, B and C, of equal radius and masses m, 2m and 4m, lie at rest in a straight line in that order on a smooth horizontal table. A is projected directly towards B with speed u. The coefficient of restitution between each pair of spheres is e.

ABCm2m4mu
Figure: A projected towards B; B and C are at rest; not to scale.
(a)

Show that, immediately after A collides with B, the velocity of A is ⅓u(1 − 2e) and the speed of B is ⅓u(1 + e).

(4)
(b)

B then collides with C. Show that, immediately after this collision, the velocity of B is 1/9u(1 + e)(1 − 2e), and find the speed of C.

(4)
(c)

Given that e < ½, show that A and B collide again.

(3)
(d)

Describe the motion of A and B after the collisions above when e = ½, and state whether there are any further collisions.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
mu = mvA + 2mvBM1
vB − vA = euM1
u = 3vA + 2eu ⇒ vA = ⅓u(1 − 2e)A1
vB = ⅓u(1 − 2e) + eu = ⅓u(1 + e)A1
Part (b)
2mvB = 2mwB + 4mwC and wC − wB = evBM1
vB = 3wB + 2evB ⇒ wB = ⅓vB(1 − 2e)A1
= 1/9u(1 + e)(1 − 2e)A1
wC = ⅓vB(1 + e) = 1/9u(1 + e)2A1
Part (c)
with e < ½, A and B both move towards C (both velocities positive)B1
wB = ⅓(1 + e) × vA, and ⅓(1 + e) ≤ ⅔ < 1M1
so B is slower than A, which is behind it: A catches B and they collide againA1
Part (d)
A and B are both at rest (vA = wB = 0) while C moves away, so there are no further collisionsB1

Exam tip: Write the second collision using B's velocity from part (a) as its “before” velocity. Factorising the answers in terms of vA makes the comparison in part (c) almost immediate.

22IALM-M2-5-B-12 M2 · A2Hard13 marks

A straight, narrow, smooth horizontal channel ends at a fixed vertical barrier that is perpendicular to the channel. Marble Q, of mass 0.2 kg, rests in the channel 2.8 m from the barrier. Marble P, of mass 0.3 kg and the same radius as Q, rolls along the channel towards Q and the barrier at 5 m s−1. For impacts between the marbles e = 0.5; for an impact between Q and the barrier e = 0.8. Model the marbles as particles and ignore their rotation.

PQ0.3 kg0.2 kg5 m s⁻¹2.8 m
Figure: P rolls towards Q, which rests 2.8 m from the barrier; not to scale.
(a)

Find the velocities of P and Q immediately after they first collide.

(4)
(b)

Find the speed of Q immediately after it rebounds from the barrier.

(1)
(c)

Find the kinetic energy lost when Q hits the barrier.

(2)
(d)

Show that P and Q collide for a second time at a distance of 1.0 m from the barrier.

(4)
(e)

Find the time between the first and second collisions of P and Q.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
(towards the barrier positive) 0.3(5) = 0.3vP + 0.2vQM1
vQ − vP = 0.5 × 5M1
vP = 2 m s−1 towards the barrierA1
vQ = 4.5 m s−1 towards the barrierA1
Part (b)
0.8 × 4.5 = 3.6 m s−1 ftA1
Part (c)
½(0.2)(4.52 − 3.62)M1
= 0.729 J ftA1
Part (d)
time for Q to reach the barrier 2.8 ÷ 4.5 = 0.622 s (28/45 s)M1
P has then moved 2 × 28/45 = 1.244 m, so is 1.556 m (14/9 m) from the barrierA1
closing speed 2 + 3.6 = 5.6, time to meet 14/9 ÷ 5.6 = 5/18 sM1
distance from barrier = 3.6 × 5/18 = 1.0 mA1
Part (e)
28/45 + 5/18M1
= 0.9 sA1

Exam tip: Fix one origin (the barrier) and write each marble's distance from it as a function of time. The second collision is where the two expressions are equal.

23IALM-M2-5-B-13 M2 · A2Hard12 marks

Two small smooth spheres A and B, of equal radius and masses m and 2m, move towards each other along the same straight line on a smooth horizontal floor, with speeds 2u and u respectively. The coefficient of restitution between the spheres is e, where 0 < e ≤ 1. A fixed smooth vertical wall stands perpendicular to the line of motion, on the far side of B from A.

ABm2m2uu
Figure: A and B moving towards each other, with the wall beyond B; not to scale.
(a)

Show that, immediately after the collision, A moves with speed 2eu and B moves with speed eu, and that both spheres have reversed their directions of motion.

(4)
(b)

Show that the kinetic energy lost in the collision is 3mu2(1 − e2).

(3)
(c)

Given that 75% of the kinetic energy is lost in the collision, find e.

(2)
(d)

Using your value of e, find the magnitude of the impulse exerted on B in the collision, in terms of m and u.

(1)
(e)

B now rebounds from the wall. The coefficient of restitution between B and the wall is f. Show that, whatever the value of f, there is no further collision between A and B.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
(A's original direction positive) m(2u) + 2m(−u) = mvA + 2mvB = 0M1
vB − vA = e(2u + u) = 3euM1
vA = −2eu: speed 2eu, direction reversedA1
vB = eu: speed eu, now moving towards the wall, direction reversedA1
Part (b)
before: ½m(2u)2 + ½(2m)u2 = 3mu2B1
after: ½m(2eu)2 + ½(2m)(eu)2 = 3e2mu2M1
loss = 3mu2(1 − e2)A1
Part (c)
1 − e2 = 0.75M1
e = ½A1
Part (d)
2m(½u + u) = 3mu ftA1
Part (e)
after the wall, B moves away from the wall, the same direction as A, with speed f × ½u ≤ ½uM1
A moves away at speed u > ½u ≥ B's speed, so B never catches AA1

Exam tip: Notice when the total momentum is zero: the two momenta after the impact must then be equal and opposite, which makes the algebra very short.

24IALM-M2-5-B-14 M2 · A2Hard12 marks

A ball of mass 0.4 kg is rolling on smooth horizontal ground with velocity (12i + 5j) m s−1, where i and j are perpendicular horizontal unit vectors. It receives an impulse of magnitude 5.2 N s in the direction of the vector (−5i + 12j). The ball is modelled as a particle.

(a)

Show that the impulse is (−2i + 4.8j) N s.

(2)
(b)

Find the velocity of the ball immediately after the impulse.

(2)
(c)

Show that the direction of motion of the ball is turned through 45°.

(3)
(d)

Find the kinetic energy gained by the ball.

(2)
(e)

A second impulse K then acts on the ball, after which the ball moves in the direction of i with its original speed. Find the magnitude of K.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
unit vector in the direction: (−5i + 12j) ÷ 13M1
5.2 × (−5i + 12j)/13 = 0.4(−5i + 12j) = −2i + 4.8jA1
Part (b)
0.4v = 0.4(12i + 5j) + (−2i + 4.8j) = 2.8i + 6.8jM1
v = (7i + 17j) m s−1A1
Part (c)
direction before: tan−1(5/12) = 22.6° to iM1
direction after: tan−1(17/7) = 67.6° to iA1
difference = 45.0° (exactly: tan(β − α) = (17/7 − 5/12) ÷ (1 + 85/84) = 1)A1
Part (d)
½(0.4)(72 + 172) − ½(0.4)(122 + 52)M1
= 67.6 − 33.8 = 33.8 JA1
Part (e)
original speed √(122 + 52) = 13, so the final velocity is 13iB1
K = 0.4[13i − (7i + 17j)] = 2.4i − 6.8jM1
|K| = √52 = 7.21 N sA1

Exam tip: An impulse given as a magnitude and a direction vector: divide the direction vector by its own length, then multiply by the magnitude.

25IALM-M2-5-FB-11 M2 · A2Hard14 marks

Three small smooth spheres A, B and C, of equal radius, lie in a straight line on a smooth horizontal table. Their masses, the initial velocity of A and the distance between B and C are shown in Figure 1. B and C are at rest. The coefficient of restitution between any two of the spheres is 0.5. The spheres are modelled as particles.

Figure 1ABC2 kg1 kg5 kg6 m s⁻¹at restat rest1.8 m
(a)

Find the speeds of A and B immediately after A collides with B.

(4)
(b)

B then collides with C. Find the velocities of B and C immediately after this collision.

(3)
(c)

Show that A and B collide for a second time at a point 0.3 m from the initial position of C.

(3)
(d)

Find the velocities of A and B immediately after their second collision.

(2)
(e)

Show that there is exactly one more collision, and state the total number of collisions between the spheres.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
2(6) = 2vA + vBM1
vB − vA = 0.5 × 6 = 3M1
vA = 3 m s−1A1
vB = 6 m s−1A1
Part (b)
1(6) = wB + 5wC and wC − wB = 0.5 × 6 = 3M1
wB = −1.5: B moves back towards A with speed 1.5 m s−1A1
wC = 1.5 m s−1 away from BA1
Part (c)
B takes 1.8 ÷ 6 = 0.3 s to reach C; in this time A moves 3 × 0.3 = 0.9 m, so A and B are then 0.9 m apartM1
closing speed 3 + 1.5 = 4.5, so they meet after 0.9 ÷ 4.5 = 0.2 sM1
B has moved back 1.5 × 0.2 = 0.3 m from C's initial positionA1
Part (d)
2(3) + 1(−1.5) = 2xA + xB and xB − xA = 0.5(3 + 1.5) = 2.25M1
xA = 0.75 m s−1, xB = 3 m s−1, both towards CA1
Part (e)
B (3 m s−1) is faster than C (1.5 m s−1), so B hits C again: yB + 5yC = 10.5, yC − yB = 0.75 ⇒ yB = 1.125, yC = 1.875M1
now 0.75 < 1.125 < 1.875 in the same direction, so no sphere catches the one in front: 4 collisions in totalA1

Exam tip: To decide whether more collisions happen, list the final velocities in order along the line. The motion is over only when each sphere is no faster than the one in front of it.

26IALM-M2-5-FB-12 M2 · A2Hard12 marks

Two particles A and B move along a straight line on a smooth horizontal floor towards a fixed smooth vertical wall which is perpendicular to the line. B is between A and the wall. A collides with B, and later B strikes the wall and rebounds. Figure 1 shows the velocity–time graphs of A and B for the first 3.2 s, with velocities towards the wall taken as positive. The mass of A is 2 kg.

Figure 100.511.522.533.5−3−2−10123456v / m s⁻¹t / sA (2 kg)B
(a)

Find the mass of B.

(2)
(b)

Find the coefficient of restitution between A and B.

(2)
(c)

Find the coefficient of restitution between B and the wall, and the distance from the wall at which A and B first collide.

(2)
(d)

Show that A and B collide again when t = 3.25 s, and find the distance from the wall at which this happens.

(4)
(e)

Find the total kinetic energy lost by the two particles between t = 0 and the instant before their second collision.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
2(5) + m(2) = 2(3.5) + m(5)M1
m = 1 kgA1
Part (b)
separation 5 − 3.5 = 1.5; approach 5 − 2 = 3M1
e = 0.5A1
Part (c)
2.5 ÷ 5 = 0.5B1
B travels at 5 m s−1 from t = 2 to t = 3: 5 mB1
Part (d)
at t = 3, A has moved 3.5 × 1 = 3.5 m, so it is 5 − 3.5 = 1.5 m from the wallM1
closing speed 3.5 + 2.5 = 6, so they meet after 1.5 ÷ 6 = 0.25 sM1
t = 3 + 0.25 = 3.25 sA1
distance from the wall = 2.5 × 0.25 = 0.625 mA1
Part (e)
initial ½(2)(52) + ½ × 1 × 22 = 27 J; just before the second collision ½(2)(3.52) + ½ × 1 × 2.52 = 15.375 JM1
loss = 11.6 J (11.625)A1

Exam tip: On a velocity–time graph each collision is a vertical jump. Read the velocities on either side of a jump for momentum and restitution, and use areas for the distances travelled between impacts.

27IALM-M2-5-FB-13 M2 · A2Hard12 marks

A particle A, of mass 3 kg, moves with speed 4 m s−1 on a smooth horizontal floor and collides directly with a particle B, of mass M kg, which is at rest. The coefficient of restitution between A and B is e. Figure 1 shows how the velocities of A and B immediately after the collision depend on e, with A's original direction of motion taken as positive.

Figure 10.00.10.20.30.40.50.60.70.80.91.0−1.5−1−0.500.511.522.533.5v / m s⁻¹eAB
(a)

Use the graph to find the value of M.

(3)
(b)

Show that, immediately after the collision, the velocity of A is (1.5 − 2.5e) m s−1, and find the velocity of B in terms of e.

(4)
(c)

State the value of e for which A is brought to rest by the collision, and describe the motion of A after the collision when e is greater than this value.

(2)
(d)

After the collision, B strikes a fixed smooth vertical wall perpendicular to its path and rebounds. The coefficient of restitution between B and the wall is 0.2. Find the range of values of e for which A and B collide again.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
when e = 0 both particles move at 1.5 m s−1B1
3 × 4 = (3 + M) × 1.5M1
M = 5A1
Part (b)
12 = 3vA + 5vBM1
vB − vA = 4eM1
12 = 8vA + 20e ⇒ vA = 1.5 − 2.5eA1
vB = (1.5 + 1.5e) m s−1A1
Part (c)
e = 0.6B1
A's velocity is negative: A rebounds, moving back along the line away from BB1
Part (d)
B's velocity after the wall is −0.2(1.5 + 1.5e) = −0.3(1 + e)M1
they collide again if A's velocity exceeds B's: 1.5 − 2.5e > −0.3(1 + e)M1
1.8 > 2.2e, so 0 ≤ e < 9/11A1

Exam tip: For “do they collide again?”, compare velocities with signs, not speeds: the particle behind must have a greater velocity in the direction of the one in front, whichever way both are moving.

28IALM-M2-5-FB-14 M2 · A2Hard12 marks

Two particles P and Q, of masses 0.5 kg and 1 kg, move on a smooth horizontal plane, with i and j perpendicular horizontal unit vectors. P receives an impulse, which changes its velocity as shown in Figure 1. P then collides directly with Q, whose velocity is shown in the same figure; P and Q move along the same line just before the collision. The coefficient of restitution between P and Q is 0.5.

Figure 1−4−3−2−11234567−11234500i component / m s⁻¹j component / m s⁻¹P beforeP afterQ
(a)

Find the impulse received by P.

(2)
(b)

Find the magnitude of this impulse and the angle it makes with i.

(2)
(c)

Find the velocities of P and Q immediately after their collision.

(4)
(d)

Find the kinetic energy lost in the collision.

(2)
(e)

Find the angle between the direction of motion of P before the impulse and its direction of motion after the collision with Q.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
I = 0.5[6i − (2i + 4j)]M1
= (2i − 2j) N sA1
Part (b)
|I| = √8 = 2.83 N sA1
at 45° to i, below the i direction (towards −j)A1
Part (c)
0.5(6) + 1(−3) = 0.5vP + vQ (= 0)M1
vQ − vP = 0.5(6 + 3) = 4.5M1
P: −3i m s−1A1
Q: 1.5i m s−1A1
Part (d)
before ½(0.5)(62) + ½ × 1 × 32 = 13.5 J; after ½(0.5)(32) + ½ × 1 × 1.52 = 3.375 JM1
loss = 10.1 J (10.125)A1
Part (e)
cos φ = (2i + 4j)·(−i) ÷ (√20 × 1) = −2/√20 (or 180° − tan−12)M1
φ = 117° (116.6°)A1

Exam tip: Once P moves along the i direction, the collision with Q is a direct impact on a line: use signed components along i for both momentum and restitution.