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Edexcel International A Level Maths · M1 · Mechanics 1

Momentum and impulse

28 free, high-quality past-paper-style questions on Momentum and impulse 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 · 235 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

1IALM-M1-4-B-01 M1 · ASEasy5 marks

A football of mass 0.45 kg is kicked horizontally at a vertical wall. It hits the wall at right angles with speed 10 m s−1 and rebounds along the same line with speed 6 m s−1. The football is modelled as a particle.

before10 m s⁻¹after6 m s⁻¹0.45 kg
Figure: the football just before and just after hitting the wall; not to scale.
(a)

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

(2)
(b)

State the direction of this impulse.

(1)
(c)

The football is in contact with the wall for 0.08 s. Modelling the force of the wall on the football as constant, find its magnitude.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
taking away from the wall as positive: I = 0.45 × 6 − 0.45 × (−10)M1
I = 7.2 N sA1
Part (b)
perpendicular to the wall, away from the wall (in the direction of the rebound)B1
Part (c)
7.2 = F × 0.08 ft their impulseM1
F = 90 NA1

Common error: Subtracting the speeds (10 − 6). The ball reverses, so its velocities have opposite signs and the speeds add in v − u.

2IALM-M1-4-B-02 M1 · ASEasy6 marks

Two trolleys run on a straight horizontal track, which is smooth. Trolley A, of mass 4 kg, moves at 3 m s−1 and catches up with trolley B, of mass 2 kg, which is moving at 1.5 m s−1 in the same direction. The trolleys collide and couple together. Both trolleys are modelled as particles.

A 4 kgB 2 kg3 m s⁻¹1.5 m s⁻¹
Figure: the trolleys just before they collide; not to scale.
(a)

Find the speed of the trolleys immediately after they couple.

(2)
(b)

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

(2)
(c)

Explain what the word ‘smooth’ tells you about the track, and how this was used in part (a).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
4 × 3 + 2 × 1.5 = 6vM1
v = 2.5 m s−1A1
Part (b)
I = 2 × 2.5 − 2 × 1.5 (or 4 × 3 − 4 × 2.5)M1
= 2 N sA1
Part (c)
there is no friction between the trolleys and the trackB1
so no horizontal external force/impulse acts on the trolleys and their total momentum is conservedB1

Exam tip: When particles couple or coalesce, write the total mass times one common velocity on the right-hand side: (m1 + m2)v.

3IALM-M1-4-B-03 M1 · ASEasy5 marks

A boat of mass 225 kg floats at rest on still water, with a man of mass 75 kg standing at rest in it. The man jumps horizontally from the boat, leaving it with speed 3 m s−1. The resistance of the water is negligible, and the man and the boat are modelled as particles moving in the same straight line.

(a)

Find the speed of the boat immediately after the man jumps, and state its direction of motion.

(3)
(b)

Find the magnitude of the impulse exerted by the man on the boat.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
total momentum before is zero: 0 = 75 × 3 + 225vM1
v = −1, so speed 1 m s−1A1
in the direction opposite to the man's motionA1
Part (b)
I = 225 × 1 − 0 (or 75 × 3 − 0)M1
= 225 N sA1

Exam tip: If everything starts at rest, the total momentum afterwards is zero, so the two parts must move in opposite directions.

4IALM-M1-4-B-04 M1 · ASEasy6 marks

A particle P of mass 0.4 kg is moving on a smooth horizontal surface with velocity (5i + 2j) m s−1, where i and j are perpendicular horizontal unit vectors. P receives an impulse of (−2.8i + 1.2j) N s.

(a)

Find the velocity of P immediately after the impulse.

(3)
(b)

Find the speed of P immediately after the impulse.

(1)
(c)

Find the angle between the direction of motion of P immediately after the impulse and the vector i.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
−2.8i + 1.2j = 0.4v − 0.4(5i + 2j)M1
0.4v = −0.8i + 2jA1
v = (−2i + 5j) m s−1A1
Part (b)
√((−2)2 + 52) = √29 = 5.39 m s−1 (3 s.f.) ft their vB1
Part (c)
tan−1(5 ÷ 2) = 68.2°M1
angle with i = 180° − 68.2° = 112° (3 s.f.)A1

Common error: Giving 68.2° as the answer. The velocity −2i + 5j points up and to the left, so the angle with i is obtuse: sketch the vector first.

5IALM-M1-4-FB-01 M1 · ASEasy6 marks

A puck of mass 0.8 kg is at rest on smooth horizontal ice. From time t = 0 it is acted on by a horizontal force in a fixed line. The graph shows the component F newtons of this force, in a fixed positive direction along the line, against time t seconds. The force is constant within each interval shown, and is zero for t > 3. The puck is modelled as a particle.

0 0.5 1 1.5 2 2.5 3 3.5 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 F / N t / s
(a)

Find the total impulse of the force on the puck from t = 0 to t = 2.

(2)
(b)

Find the speed of the puck at t = 2.

(2)
(c)

Find the velocity of the puck at t = 3, stating its direction of motion.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
I = 6 × 0.5 + 2 × 1.5 areas read from the graphM1
I = 6 N sA1
Part (b)
6 = 0.8v − 0.8 × 0 ft their impulseM1
v = 7.5 m s−1A1
Part (c)
(−4) × 1 = 0.8w − 0.8 × 7.5M1
w = 2.5, so 2.5 m s−1 in the positive direction (still moving the original way)A1

Exam tip: For a force that is constant over an interval, impulse = force × time, which is the area of that strip of the F–t graph. Strips below the axis give negative impulse.

6IALM-M1-4-FB-02 M1 · ASEasy6 marks

Two particles A and B move along the same straight line on a smooth horizontal surface, and collide directly at t = 2. The graph shows the velocity v m s−1 of each particle, in a fixed positive direction along the line, against time t seconds. The mass of A is 1.2 kg.

0 0.5 1 1.5 2 2.5 3 3.5 4 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 v / m s⁻¹ t / s A A B B A B collision
(a)

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

(2)
(b)

Find the mass of B.

(3)
(c)

Describe the direction of motion of each particle immediately after the collision, compared with before it.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
I = 1.2 × (−0.5) − 1.2 × 4 velocities read from the graphM1
magnitude 5.4 N sA1
Part (b)
1.2 × 4 + m(−2) = 1.2 × (−0.5) + m × 1, or 5.4 = m(1 − (−2))M1
5.4 = 3mA1
m = 1.8 kgA1
Part (c)
both particles reverse direction: A now moves in the negative direction (at 0.5 m s−1) and B in the positive direction (at 1 m s−1)B1

Common error: Ignoring the signs on the graph. A line below the t-axis means motion in the negative direction, so B's velocity before the collision is −2, not 2.

7IALM-M1-4-FB-03 M1 · ASEasy6 marks

A ball of mass 0.25 kg moves on a smooth horizontal surface. It is struck, and its velocity changes from u to v. The figure shows u and v drawn to scale on a grid of squares of side 1 m s−1, together with the perpendicular horizontal unit vectors i and j. The ball is modelled as a particle.

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 u v i j 1 square = 1 m s⁻¹
(a)

Write down u and v in terms of i and j.

(2)
(b)

Find the impulse exerted on the ball, in terms of i and j.

(2)
(c)

Find the magnitude of this impulse.

(1)
(d)

The ball is in contact with the striker for 0.02 s. Modelling the force on the ball as constant, find its magnitude.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
u = (6i + 2j) m s−1B1
v = (2i + 5j) m s−1B1
Part (b)
I = 0.25(2i + 5j) − 0.25(6i + 2j)M1
I = (−i + 0.75j) N sA1
Part (c)
|I| = √(12 + 0.752) = 1.25 N s ft their IA1
Part (d)
1.25 ÷ 0.02 = 62.5 N ft their magnitudeA1

Exam tip: Impulse is mv − mu, worked component by component. The impulse is parallel to v − u, not to v, so do not expect it to point along either velocity arrow.

8IALM-M1-4-FB-04 M1 · ASEasy6 marks

Two toy trucks P and Q are held at rest on a smooth straight horizontal track, with a compressed light spring between them. A buffer is fixed across the track at each end. The figure shows the masses of the trucks and the distance of each truck from the buffer it faces. The trucks are released, and the spring pushes them apart instantaneously. Immediately afterwards Q moves with speed 1.2 m s−1. The trucks are modelled as particles.

P Q 1.5 kg 2.5 kg 3.6 m 2.4 m buffer buffer smooth horizontal track
(a)

Find the speed of P immediately after the trucks are released.

(2)
(b)

Find the magnitude of the impulse exerted by the spring on P.

(1)
(c)

Determine which truck reaches its buffer first, and find how much later the other truck reaches its buffer.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
0 = 2.5 × 1.2 + 1.5v masses from the figureM1
v = −2, so speed 2 m s−1A1
Part (b)
I = 1.5 × 2 − 0 = 3 N s ft their speedA1
Part (c)
P: 3.6 ÷ 2 = 1.8 s distances from the figureM1
Q: 2.4 ÷ 1.2 = 2 sA1
P arrives first, and Q arrives 0.2 s laterA1

Exam tip: On a smooth track nothing changes the trucks' speeds once the spring has acted, so each journey time is simply distance ÷ speed.

9IALM-M1-4-B-05 M1 · ASMedium7 marks

Two particles A and B, of masses 2m and 3m, move towards each other along the same straight line on a smooth horizontal plane. Just before they collide, A has speed 4u and B has speed 2u. In the collision the direction of motion of A is reversed, and immediately afterwards A has speed u.

beforeAB4u2u2m3mafterABu?
Figure: speeds of A and B just before and just after the collision; not to scale.
(a)

Find the velocity of B immediately after the collision, in terms of u, stating its direction.

(4)
(b)

Find, in terms of m and u, the magnitude of the impulse exerted on B in the collision.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
positive in A's original direction: 2m(4u) + 3m(−2u) = 2m(−u) + 3mvM1
2mu = −2mu + 3mv, with all signs correctA1
v = 4u/3A1
in the direction in which A was originally moving, so B's direction is also reversedA1
Part (b)
I = 3m(v − (−2u))M1
= 3m(4u/3 + 2u) ft their vA1
= 10muA1

Exam tip: With letters for masses and speeds, m cancels from the momentum equation. Keep it in the impulse, though: impulse depends on mass.

10IALM-M1-4-B-06 M1 · ASMedium7 marks

A particle A of mass 0.5 kg lies at rest on a smooth horizontal table. It is attached to one end of a light inextensible string, which passes over a small smooth pulley fixed at the edge of the table. The other end of the string is attached to a particle B of mass 0.3 kg. B is held at rest vertically below the pulley with the string slack, and is then released. B falls freely through 0.4 m, at which instant the string becomes taut. A does not reach the pulley in the motion considered.

A0.5 kgB0.3 kg0.4 m
Figure: B falls freely through 0.4 m before the string becomes taut; not to scale.
(a)

Find the speed of B just before the string becomes taut.

(2)
(b)

Find the common speed of A and B immediately after the string becomes taut.

(4)
(c)

Find the magnitude of the impulse in the string when it becomes taut.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
v2 = 0 + 2 × 9.8 × 0.4M1
v = 2.8 m s−1A1
Part (b)
impulse J in the string; for B (downwards): −J = 0.3V − 0.3 × 2.8M1
for A (towards the pulley): J = 0.5VM1
eliminating J: 0.84 = 0.8VM1
V = 1.05 m s−1A1
Part (c)
J = 0.5 × 1.05 = 0.525 N s ft their VA1

Exam tip: Over a pulley the particles move in different directions, so do not add their momenta. Write impulse = change in momentum for each particle, with the same impulse J in both equations.

11IALM-M1-4-B-07 M1 · ASMedium8 marks

A hammer head of mass 0.6 kg, moving horizontally at 9 m s−1, strikes a nail of mass 0.015 kg that is at rest and pointing horizontally into a vertical wooden board. Immediately after the impact the hammer head and nail move together, and they come to rest after the nail has moved 0.025 m into the board. The board exerts a constant resistance on the nail, and all other horizontal forces are negligible. The hammer head and nail are modelled as particles.

(a)

Find the common speed of the hammer head and nail immediately after the impact.

(2)
(b)

Find the magnitude of the resistance exerted by the board.

(4)
(c)

Find the time taken for the hammer head and nail to come to rest.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
0.6 × 9 = 0.615VM1
V = 8.78 m s−1 (3 s.f.)A1
Part (b)
0 = V2 − 2a × 0.025M1
a = 1540 m s−2 (1541.9…)A1
R = 0.615aM1
R = 948 N (3 s.f.)A1
Part (c)
0 = V − at, or 0.025 = ½VtM1
t = 0.00569 s (3 s.f.)A1

Exam tip: Carry the unrounded common speed (8.7804…) into the later parts. Rounding early, especially to 2 s.f., can change a final answer at 3 s.f.

12IALM-M1-4-B-08 M1 · ASMedium8 marks

A small box of mass 4 kg is at rest at the bottom of a rough plane inclined at an angle α to the horizontal, where tan α = ¾. The coefficient of friction between the box and the plane is 0.5. The box is given an impulse of magnitude 24 N s directed up a line of greatest slope of the plane. The box is modelled as a particle.

α4 kg24 N s
Figure: the box at the bottom of the rough plane receives an impulse up the slope; tan α = 3/4; not to scale.
(a)

Find the speed of the box immediately after the impulse.

(2)
(b)

Find the distance the box moves up the plane before coming to instantaneous rest.

(4)
(c)

Determine whether the box remains at rest after it stops, or slides back down the plane.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
24 = 4u − 0M1
u = 6 m s−1A1
Part (b)
R = 4g cos α = 31.36 N, so F = 0.5 × 31.36 = 15.68 NM1
4g sin α + F = 4a: 23.52 + 15.68 = 4aM1
a = 9.8 m s−2A1
0 = 62 − 2 × 9.8 × s → s = 1.84 m (3 s.f.)A1
Part (c)
component of weight down the plane 4g sin α = 23.52 N; maximum friction μR = 15.68 NM1
23.52 > 15.68, so friction cannot hold the box: it slides back downA1

Common error: Forgetting that friction acts down the plane while the box moves up. Both friction and the weight component oppose the motion on the way up.

13IALM-M1-4-B-09 M1 · ASMedium7 marks

A firework of mass 0.9 kg is moving vertically upwards with speed 14 m s−1 when it explodes into two parts, P of mass 0.3 kg and Q of mass 0.6 kg. Immediately after the explosion P moves vertically upwards with speed 35 m s−1. The explosion is instantaneous, air resistance is negligible, and P and Q are modelled as particles.

(a)

Find the velocity of Q immediately after the explosion.

(3)
(b)

Find the magnitude of the impulse exerted on P by the explosion.

(2)
(c)

Find the further height risen by Q after the explosion.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
upwards positive: 0.9 × 14 = 0.3 × 35 + 0.6vM1
12.6 = 10.5 + 0.6vA1
v = 3.5 m s−1 upwardsA1
Part (b)
I = 0.3 × 35 − 0.3 × 14M1
= 6.3 N sA1
Part (c)
0 = 3.52 − 2 × 9.8 × h ft their v if upwardsM1
h = 0.625 mA1

Exam tip: An explosion of a moving body still conserves momentum: the total after equals the momentum before, not zero.

14IALM-M1-4-B-10 M1 · ASMedium7 marks

Three particles A, B and C, of masses 2 kg, 1 kg and 3 kg, lie in a straight line, in that order, on a smooth horizontal surface. B and C are at rest. A is projected towards B with speed 6 m s−1. A collides directly with B, and immediately afterwards A moves in its original direction with speed 2.5 m s−1. B then collides with C, and B and C coalesce.

A2 kgB1 kgC3 kg6 m s⁻¹at restat rest
Figure: the three particles before A is projected towards B; not to scale.
(a)

Find the speed of B immediately after its collision with A.

(3)
(b)

Find the speed of the combined particle formed by B and C.

(2)
(c)

Explain why there is a further collision.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
2 × 6 = 2 × 2.5 + 1 × vBM1
12 = 5 + vBA1
vB = 7 m s−1A1
Part (b)
1 × 7 = 4V ft their vBM1
V = 1.75 m s−1A1
Part (c)
A moves at 2.5 m s−1 and the combined particle at 1.75 m s−1, in the same directionM1
A is faster and behind, and the surface is smooth so neither slows: A catches the combined particleA1

Exam tip: In a chain of collisions, finish by comparing velocities: a particle behind that is moving faster in the same direction will always catch up on a smooth surface.

15IALM-M1-4-FB-05 M1 · ASMedium8 marks

Two particles P and Q move along a straight line on a smooth horizontal surface. The line passes through a fixed point O. The graph shows the displacement s metres of each particle from O against time t seconds. The particles collide at t = 4 and coalesce, and the graph also shows the displacement of the combined particle up to t = 10. The mass of P is 3 kg.

0 1 2 3 4 5 6 7 8 9 10 0 2 4 6 8 10 12 14 16 s / m t / s P Q combined
(a)

Find the velocity of P and the velocity of Q before the collision.

(2)
(b)

Find the velocity of the combined particle.

(1)
(c)

Find the mass of Q.

(3)
(d)

Assuming the combined particle continues to move in the same way, find the value of t when it is 16 m from O.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
P: 10 ÷ 4 = 2.5 m s−1B1
Q: (10 − 14) ÷ 4 = −1 m s−1, i.e. 1 m s−1 towards OB1
Part (b)
(13 − 10) ÷ (10 − 4) = 0.5 m s−1B1
Part (c)
3 × 2.5 + m(−1) = (3 + m) × 0.5 ft their velocitiesM1
7.5 − m = 1.5 + 0.5mA1
m = 4 kgA1
Part (d)
10 + 0.5(t − 4) = 16 ft their velocityM1
t = 16A1

Exam tip: On a displacement–time graph the velocity is the gradient. A line sloping downwards means the particle is moving in the negative direction, so its velocity goes into the momentum equation with a minus sign.

16IALM-M1-4-FB-06 M1 · ASMedium7 marks

A bullet is fired horizontally along a straight line on which two blocks A and B lie at rest on a smooth horizontal surface. The figure shows three stages of the motion, with the masses and the known speeds. In stage 2 the bullet has passed through A, and in stage 3 it has become embedded in B. The bullet passes through A in a negligible time and loses no mass. The bullet and blocks are modelled as particles moving in the same straight line.

Stage 1 Stage 2 Stage 3 A B 300 m s⁻¹ 1.5 kg 2 kg bullet 0.025 kg at rest at rest A vA 180 m s⁻¹ B at rest A vA B V bullet embedded in B
(a)

Find the speed vA of A after the bullet has passed through it.

(2)
(b)

Find the magnitude of the impulse exerted by A on the bullet.

(2)
(c)

Find the speed V of B immediately after the bullet becomes embedded in it.

(2)
(d)

Explain whether A will later collide with B.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
0.025 × 300 = 0.025 × 180 + 1.5vAM1
vA = 2 m s−1A1
Part (b)
I = 0.025 × 300 − 0.025 × 180M1
= 3 N sA1
Part (c)
0.025 × 180 = (2 + 0.025)VM1
V = 2.22 m s−1 (3 s.f.)A1
Part (d)
B (with the bullet) is ahead of A and moves in the same direction at 2.22 m s−1 > 2 m s−1; the surface is smooth, so A never catches B: no collision ft their speedsB1

Common error: Using the mass of B alone in part (c). Once the bullet is embedded it moves with B, so the mass on the right-hand side is 2 + 0.025 = 2.025 kg.

17IALM-M1-4-FB-07 M1 · ASMedium9 marks

A puck of mass 0.2 kg is at rest on a rough horizontal ice rink. At time t = 0 it is struck by a stick and slides in a straight line towards a vertical board. It hits the board at right angles and rebounds along the same line. The graph shows the velocity v m s−1 of the puck, with the direction away from the stick as positive, against time t seconds, until the puck comes to rest. The contacts with the stick and with the board are modelled as instantaneous, and the puck is modelled as a particle.

0 0.5 1 1.5 2 2.5 3 3.5 4 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10 v / m s⁻¹ t / s hits board
(a)

Find the magnitude of the impulse exerted on the puck by the stick.

(2)
(b)

Find the coefficient of friction between the puck and the ice.

(3)
(c)

Find the magnitude and direction of the impulse exerted on the puck by the board.

(2)
(d)

Find the distance of the puck from its starting point when it comes to rest.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
I = 0.2 × 9 − 0 9 read from the graphM1
= 1.8 N sA1
Part (b)
deceleration = (9 − 6) ÷ 1.5 = 2 m s−2M1
μ × 0.2g = 0.2 × 2M1
μ = 0.204 (3 s.f.)A1
Part (c)
J = 0.2 × (−4) − 0.2 × 6M1
magnitude 2 N s, directed away from the board (negative direction)A1
Part (d)
towards the board: ½(9 + 6) × 1.5 = 11.25 m; back: ½ × 2 × 4 = 4 m areasM1
distance = 7.25 mA1

Exam tip: After the rebound the gradient is positive. Friction has the same magnitude as before but now acts in the positive direction, because the puck is moving the other way.

18IALM-M1-4-FB-08 M1 · ASMedium8 marks

A ball of mass 0.4 kg is released from rest and falls vertically onto horizontal ground. It bounces and rises vertically. The figure shows the height from which the ball is released and the greatest height it reaches after the bounce. Air resistance is negligible, and the ball is modelled as a particle.

3.6 m 2.5 m released from rest greatest height after the bounce ground
(a)

Find the speed of the ball just before it hits the ground.

(2)
(b)

Find the speed of the ball just after it leaves the ground.

(2)
(c)

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

(2)
(d)

The ball is in contact with the ground for 0.05 s. Modelling the force exerted by the ground as constant, find its magnitude.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
v2 = 0 + 2 × 9.8 × 3.6M1
v = 8.4 m s−1A1
Part (b)
0 = w2 − 2 × 9.8 × 2.5M1
w = 7 m s−1A1
Part (c)
upwards positive: I = 0.4 × 7 − 0.4 × (−8.4)M1
I = 6.16 N sA1
Part (d)
(R − 0.4g) × 0.05 = 6.16 includes the weightM1
R = 123.2 + 3.92 = 127 N (3 s.f.)A1

Common error: Taking 6.16 ÷ 0.05 = 123 N as the force from the ground. Over 0.05 s the weight also acts, so the resultant force is R − 0.4g, and the ground must supply the extra 3.92 N.

19IALM-M1-4-FB-09 M1 · ASMedium8 marks

A particle of mass 0.5 kg moves on a smooth horizontal plane. The unit vectors i and j are in the directions of the x- and y-axes shown in the figure, and distances are in metres. At time t = 0 the particle is at O. It moves with constant velocity to P, which it reaches at t = 2. At P it receives an impulse, and it then moves with a new constant velocity to Q, which it reaches at t = 4. The figure shows O, P and Q on a grid of 1 m squares, with t in seconds.

0 2 4 6 8 0 2 4 6 8 10 12 14 x / m y / m O (t = 0) P (t = 2) Q (t = 4) i j t in seconds
(a)

Find the velocity of the particle before it reaches P.

(2)
(b)

Find the velocity of the particle after it leaves P.

(2)
(c)

Find the impulse received by the particle at P, in terms of i and j.

(2)
(d)

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

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
(6i + 4j) ÷ 2 coordinates from the figureM1
= (3i + 2j) m s−1A1
Part (b)
((4 − 6)i + (14 − 4)j) ÷ (4 − 2)M1
= (−i + 5j) m s−1A1
Part (c)
I = 0.5(−i + 5j) − 0.5(3i + 2j) ft their velocitiesM1
I = (−2i + 1.5j) N sA1
Part (d)
|I| = √(22 + 1.52) = 2.5 N s ftB1
angle = 180° − tan−1(1.5 ÷ 2) = 143° (3 s.f.) ftB1

Exam tip: Read the coordinates of the labelled points, then velocity = change in position ÷ time taken. Use the time for each stage separately: the second stage lasts 4 − 2 = 2 s, not 4 s.

20IALM-M1-4-FB-10 M1 · ASMedium8 marks

A truck moves along a straight horizontal railway track. A sack is released from rest from a hopper and falls vertically through the height shown in the figure onto the floor of the truck. The sack does not bounce, and immediately moves with the truck. The figure shows the masses and the speed of the truck just before the sack lands. The track is smooth, air resistance is negligible, the impact is instantaneous, and the truck and sack are modelled as particles. The unit vectors i and j are horizontal in the direction of motion of the truck and vertically upwards respectively.

truck 400 kg 3 m s⁻¹ hopper sack 100 kg 2.5 m i j
(a)

Find the speed of the sack just before it lands on the truck.

(2)
(b)

Find the speed of the truck immediately after the sack lands.

(2)
(c)

Large vertical forces act during the impact. Explain why horizontal momentum is still conserved.

(1)
(d)

Find the impulse exerted on the sack by the truck, in terms of i and j, and find its magnitude.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
v2 = 0 + 2 × 9.8 × 2.5M1
v = 7 m s−1A1
Part (b)
horizontally: 400 × 3 + 100 × 0 = 500VM1
V = 2.4 m s−1A1
Part (c)
the track is smooth, so the forces from the track (and the weights) are all vertical: there is no horizontal external impulseB1
Part (d)
I = 100(2.4i) − 100(−7j) ft their speedsM1
I = (240i + 700j) N sA1
|I| = √(2402 + 7002) = 740 N sA1

Exam tip: Only the horizontal component of momentum is conserved here. The sack's vertical velocity is destroyed by the truck, which is why the impulse on the sack has a large upward component.

21IALM-M1-4-B-11 M1 · ASHard12 marks

Two particles A and B, of masses 0.4 kg and 0.6 kg, lie on a rough horizontal floor. B is at rest. A is projected along the floor directly towards B with speed 5 m s−1 from a point 3 m from B. The coefficient of friction between each particle and the floor is μ. A collides directly with B. Immediately after the collision A is at rest. B then slides 1.5 m before coming to rest.

AB0.4 kg0.6 kg5 m s⁻¹3 m
Figure: A projected towards B along the rough floor; not to scale.
(a)

Show that the speed v of A just before the collision satisfies v2 = 25 − 6μg.

(2)
(b)

Show that the speed of B immediately after the collision is ⅔v.

(2)
(c)

Find the value of μ.

(4)
(d)

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

(2)
(e)

Find the total time from the moment A is projected until B comes to rest.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
F = μ × 0.4g = 0.4a, so the deceleration is μgM1
v2 = 52 − 2μg × 3 = 25 − 6μg (given answer)A1
Part (b)
0.4v = 0.4 × 0 + 0.6wM1
w = ⅔v (given answer)A1
Part (c)
B also decelerates at μg: (⅔v)2 = 2μg × 1.5M1
v2 = 27μg/4A1
25 − 6μg = 6.75μgM1
μ = 25 ÷ (12.75 × 9.8) = 0.200 (3 s.f.)A1
Part (d)
v = 3.638…, w = ⅔v = 2.425… m s−1M1
I = 0.6 × 2.425… = 1.46 N s (3 s.f.)A1
Part (e)
A: (5 − 3.638…) ÷ μg = 0.695 s; B: 2.425… ÷ μg = 1.24 sM1
total = 1.93 s (3 s.f.)A1

Exam tip: When the coefficient of friction is unknown, keep μg as a single symbol through every stage and solve for it at the end.

22IALM-M1-4-B-12 M1 · ASHard13 marks

A pile driver of mass 400 kg is released from rest at a height of 1.8 m above the top of a pile of mass 600 kg, which is at rest with its lower end in the ground. The driver falls freely, strikes the pile and does not rebound, so that the driver and pile move downwards together. The ground exerts a constant resistance of 45 000 N on the pile while it moves. The driver and pile are modelled as particles.

pile 600 kg400 kgdriver1.8 m
Figure: the pile driver released from rest 1.8 m above the top of the pile; not to scale.
(a)

Find the speed of the driver just before it strikes the pile.

(2)
(b)

Find the speed of the driver and pile immediately after the impact, and explain why the weights and the resistance of the ground can be ignored in this calculation.

(3)
(c)

Find the deceleration of the driver and pile while they move together.

(3)
(d)

Find the distance the pile moves into the ground after one blow.

(2)
(e)

The driver is to be dropped from a greater height so that the pile moves 0.12 m in one blow. All other conditions are unchanged. Find this height.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
U2 = 2 × 9.8 × 1.8 = 35.28M1
U = 5.94 m s−1 (3 s.f.)A1
Part (b)
400U = 1000VM1
V = 2.38 m s−1 (3 s.f.)A1
the impact lasts a very short time, so the impulses of these forces are negligible compared with the impulse between driver and pileB1
Part (c)
upwards positive: 45 000 − 1000g = 1000a three termsM1
45 000 − 9800 correctA1
a = 35.2 m s−2A1
Part (d)
0 = V2 − 2 × 35.2 × dM1
d = 0.0802 m (3 s.f.)A1
Part (e)
V2 = 2 × 35.2 × 0.12 = 8.448M1
U = 2.5V, so U2 = 6.25 × 8.448 = 52.8M1
h = 52.8 ÷ (2 × 9.8) = 2.69 m (3 s.f.)A1

Common error: Leaving out the weight of the moving driver and pile. The resistance must both stop them and support them, so the equation of motion has three terms.

23IALM-M1-4-B-13 M1 · ASHard11 marks

Two particles A and B, of masses 3 kg and 2 kg, lie at rest on a rough horizontal floor, 1.2 m apart. They are connected by a light inextensible string of length 2 m, which is slack. The coefficient of friction between each particle and the floor is 0.25. A is given an impulse of magnitude 18 N s in the direction BA, directly away from B. When the string becomes taut, the particles move off together with the same speed.

BA2 kg3 kg18 N s1.2 m
Figure: A and B joined by the slack string (length 2 m); A receives an impulse directly away from B; not to scale.
(a)

Find the speed of A immediately after the impulse.

(2)
(b)

Find the speed of A just before the string becomes taut.

(3)
(c)

Find the common speed of A and B immediately after the string becomes taut.

(2)
(d)

Find the magnitude of the impulse in the string when it becomes taut.

(2)
(e)

Both particles then decelerate at the same rate. Find the distance moved by B after the string becomes taut.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
18 = 3u − 0M1
u = 6 m s−1A1
Part (b)
friction 0.25 × 3g = 3a, so the deceleration is 2.45 m s−2M1
A slides 2 − 1.2 = 0.8 m: v2 = 36 − 2 × 2.45 × 0.8 = 32.08M1
v = 5.66 m s−1 (3 s.f.)A1
Part (c)
3 × 5.663… = (3 + 2)VM1
V = 3.40 m s−1 (3 s.f.)A1
Part (d)
J = 2 × 3.398… − 0 (or 3 × (5.663… − 3.398…))M1
J = 6.80 N s (3 s.f.)A1
Part (e)
0 = 3.398…2 − 2 × 2.45 × sM1
s = 2.36 m (3 s.f.)A1

Exam tip: While B is at rest and the string is slack, nothing pulls B, so friction on B is zero. Friction only starts to act on B once B moves.

24IALM-M1-4-B-14 M1 · ASHard13 marks

A rifle of mass 3 kg fires a bullet of mass 0.02 kg horizontally with speed 400 m s−1. The rifle is free to move horizontally and is initially at rest. The bullet travels horizontally, with no change in speed, and embeds itself in a block of wood of mass 1.98 kg, which is at rest on a rough horizontal table. The coefficient of friction between the block and the table is 0.4. The bullet and block are modelled as particles.

(a)

Find the speed with which the rifle recoils.

(2)
(b)

Find the speed of the block and bullet immediately after the bullet embeds itself.

(2)
(c)

Find the magnitude of the impulse exerted by the bullet on the block.

(2)
(d)

Find the distance the block slides before coming to rest.

(4)
(e)

When the block is at rest, an identical second bullet with the same speed embeds itself in the block in the same way. Find the distance the block slides this time.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
0 = 0.02 × 400 + 3wM1
recoil speed = 2.67 m s−1 (3 s.f.)A1
Part (b)
0.02 × 400 = 2VM1
V = 4 m s−1A1
Part (c)
I = 1.98 × 4 − 0 ft their VM1
= 7.92 N sA1
Part (d)
R = 2g = 19.6 NM1
F = 0.4 × 19.6 = 7.84 N and 7.84 = 2a, a = 3.92 m s−2M1
0 = 42 − 2 × 3.92 × sM1
s = 2.04 m (3 s.f.)A1
Part (e)
0.02 × 400 = 2.02V′, V′ = 3.960… m s−1M1
deceleration still 0.4g = 3.92 m s−2 (mass cancels)B1
s′ = 3.960…2 ÷ 7.84 = 2.00 m (3 s.f.)A1

Exam tip: On a rough horizontal surface the deceleration from friction alone is μg, whatever the mass. Spotting this saves time when the mass changes between stages.

25IALM-M1-4-FB-11 M1 · ASHard11 marks

Two small blocks A and B are held at rest together at a point O on a rough horizontal floor, with a compressed light spring between them. The spring is released and pushes the blocks apart instantaneously along a straight line. Each block then slides until it comes to rest. The figure, drawn to scale, shows where A and B come to rest. The coefficient of friction between each block and the floor is 0.3, and the total mass of A and B is 5 kg. The blocks are modelled as particles.

A B O A B 2.0 m 4.5 m A stops B stops start rough horizontal floor (scale drawing)
(a)

Show that the speeds uA and uB of A and B immediately after the spring is released satisfy uA : uB = 2 : 3.

(3)
(b)

Find the mass of A and the mass of B.

(3)
(c)

Find the magnitude of the impulse exerted by the spring on B.

(3)
(d)

Determine which block comes to rest first, and find how much sooner it stops.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
each block decelerates at 0.3g = 2.94 m s−2, the same for bothM1
uA2 = 2 × 2.94 × 2.0 and uB2 = 2 × 2.94 × 4.5 distances from the figureM1
uA2 : uB2 = 2 : 4.5 = 4 : 9, so uA : uB = 2 : 3 (given answer)A1
Part (b)
0 = mAuA − mBuBM1
mA : mB = uB : uA = 3 : 2A1
mA = 3 kg, mB = 2 kgA1
Part (c)
uB2 = 2 × 2.94 × 4.5 = 26.46M1
uB = 5.143… m s−1A1
I = 2 × 5.143… = 10.3 N s (3 s.f.) ft their massA1
Part (d)
tA = 3.429… ÷ 2.94 = 1.166… s and tB = 5.143… ÷ 2.94 = 1.749… sM1
A stops first, 0.583 s sooner (3 s.f.)A1

Exam tip: With the same coefficient of friction the deceleration μg is the same for both blocks, so u2 is proportional to the stopping distance. Working in ratios means you need μ only at the end.

26IALM-M1-4-FB-12 M1 · ASHard11 marks

The figure shows a particle P of mass 0.5 kg held at rest at a point A on a rough plane inclined at an angle α to the horizontal, where tan α = ¾. The plane meets a rough horizontal floor at the point F, and AF is a line of greatest slope. A particle Q of mass 0.9 kg is at rest on the floor at the point shown, with F and Q in the same vertical plane as A. The coefficient of friction between P and the plane is 0.25, and between each particle and the floor is 0.2. P is released, slides down the plane and passes onto the floor at F with no change in speed. It then slides along the floor, collides directly with Q, and the two particles coalesce.

α P Q A F 2.25 m 2.5 m rough floor rough plane
(a)

Find the acceleration of P down the plane.

(3)
(b)

Find the speed of P when it reaches F.

(2)
(c)

Find the speed of P just before it hits Q.

(2)
(d)

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

(2)
(e)

Find the distance moved by the combined particle after the collision.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
R = 0.5g cos α = 3.92 N, friction = 0.25 × 3.92 = 0.98 NM1
0.5g sin α − 0.98 = 0.5a, i.e. 2.94 − 0.98 = 0.5aM1
a = 3.92 m s−2A1
Part (b)
v2 = 0 + 2 × 3.92 × 2.25 = 17.64 length from the figureM1
v = 4.2 m s−1A1
Part (c)
deceleration 0.2g = 1.96 m s−2; w2 = 17.64 − 2 × 1.96 × 2.5 = 7.84M1
w = 2.8 m s−1A1
Part (d)
0.5 × 2.8 = (0.5 + 0.9)V ft their wM1
V = 1 m s−1A1
Part (e)
0 = 12 − 2 × 1.96 × s deceleration still 0.2gM1
s = 0.255 m (3 s.f.)A1

Exam tip: Treat each stage separately and carry the speed across each boundary: the speed at F starts the floor stage, and the speed just before Q goes into the momentum equation.

27IALM-M1-4-FB-13 M1 · ASHard13 marks

Two particles A and B, of masses 2 kg and 1 kg, move towards each other along a straight horizontal track. The figure shows their positions and velocities at time t = 0, with distances x measured in metres from the fixed point O in the direction from A to B. The track is smooth for x < 10 and rough for x ≥ 10, and the coefficient of friction between each particle and the rough part of the track is 0.25. The particles collide directly, and immediately after the collision B moves in the positive x direction with speed 4 m s−1. A and B are modelled as particles.

0 2 4 6 8 10 12 14 16 x / m O rough: μ = 0.25 smooth A B 4 m s⁻¹ 2 m s⁻¹ 2 kg 1 kg t = 0
(a)

Find the time at which the particles collide, and the value of x at which they collide.

(3)
(b)

Find the velocity of A immediately after the collision.

(3)
(c)

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

(2)
(d)

Find the value of x at the point where B comes to rest.

(3)
(e)

Show that A does not collide with B again, and find the distance between A and B when both are at rest.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
4t + 2t = 9 positions and speeds from the figureM1
t = 1.5 sA1
x = 4 × 1.5 = 6A1
Part (b)
2 × 4 + 1 × (−2) = 2v + 1 × 4M1
6 = 2v + 4A1
v = 1 m s−1 in the positive directionA1
Part (c)
I = 2 × 1 − 2 × 4 ft their vM1
magnitude 6 N sA1
Part (d)
on the rough part: deceleration 0.25g = 2.45 m s−2M1
0 = 42 − 2 × 2.45 × s, s = 3.265… mM1
x = 10 + 3.265… = 13.3 (3 s.f.)A1
Part (e)
A enters the rough part at 1 m s−1: s = 12 ÷ 4.9 = 0.204… m, so A stops at x = 10.2, which is short of 13.3, so no further collisionM1
distance = 13.265… − 10.204… = 3.06 m (3 s.f.)A1

Common error: Using friction on the smooth part of the track. Both particles keep a constant speed until they reach x = 10; the deceleration 0.25g only starts there.

28IALM-M1-4-FB-14 M1 · ASHard13 marks

Two particles A and B, of masses 3 kg and 2 kg, are attached to the ends of a light inextensible string that passes over a small smooth pulley fixed above a horizontal floor, as shown in the figure. B rests on the floor, vertically below the pulley. A is held at rest, with the string slack, at the height above the floor shown. A is released and falls vertically. When A has fallen 0.9 m, the string becomes taut and B is jerked into motion. The parts of the string on each side of the pulley are vertical, and air resistance is negligible.

B 2 kg A 3 kg 2.0 m 3.0 m floor
(a)

Find the speed of A just before the string becomes taut.

(2)
(b)

Find the common speed of A and B immediately after the string becomes taut, and the magnitude of the impulse in the string.

(4)
(c)

Find the acceleration of the particles while the string is taut.

(3)
(d)

Find the speed of A as it hits the floor.

(2)
(e)

A does not rebound. Find the greatest height of B above the floor in the subsequent motion.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
v2 = 0 + 2 × 9.8 × 0.9 = 17.64M1
v = 4.2 m s−1A1
Part (b)
A (downwards): −J = 3V − 3 × 4.2M1
B (upwards): J = 2V − 0M1
V = 12.6 ÷ 5 = 2.52 m s−1A1
J = 5.04 N sA1
Part (c)
A: 3g − T = 3aM1
B: T − 2g = 2aM1
a = g ÷ 5 = 1.96 m s−2A1
Part (d)
A is now 2.0 − 0.9 = 1.1 m above the floor: v2 = 2.522 + 2 × 1.96 × 1.1 = 10.6624 height from the figureM1
v = 3.27 m s−1 (3 s.f.)A1
Part (e)
string goes slack and B moves freely under gravity: h = 10.6624 ÷ (2 × 9.8) = 0.544 mM1
greatest height = 1.1 + 0.544 = 1.64 m (3 s.f.), below the pulley at 3.0 mA1

Common error: Assuming B stops when A hits the floor. The string goes slack, and B keeps rising under gravity alone until its speed is zero.