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Edexcel IAL Mechanics 1 mock exam Free full practice test

Written by STEM-Tutor's tutors · Updated October 2026 · 30 questions and 4 structured exam questions · about 100 minutes

A full-length practice test built from our exam-style questions on every chapter of Edexcel IAL Mechanics 1. Each answer is marked as you go, and every question has a worked solution to open afterwards.

On this page
  1. Before you start
  2. Section I
  3. Section II
  4. After the mock

Before you start

Section I: 30 exam-style questions

Q1Exam-styleVertical motion
A ball is projected vertically upwards at 14.7 m s−1. Modelling the ball as a particle moving freely under gravity, find the greatest height it reaches above the point of projection.
m

Hint

At the greatest height v = 0. Take upwards as positive, so a = −9.8.

Worked solution
  1. Upwards positive: u = 14.7, v = 0, a = −9.8.
  2. v2 = u2 + 2as: 0 = 14.72 − 2 × 9.8 × s.
  3. s = 216.09/19.6 = 11.0 m (3 s.f.).
Q2Exam-styleVertical motion
For the same ball, find the time from projection until it returns to the point of projection.
s

Hint

When it returns, the displacement s = 0.

Worked solution
  1. s = ut + ½at2: 0 = 14.7t − 4.9t2 = t(14.7 − 4.9t).
  2. t = 0 is the start, so t = 14.7/4.9 = 3 s.
Q3Exam-style
A cyclist moves with constant deceleration along a straight road. She passes A at 20 m s−1 and B at 8 m s−1, where AB = 84 m. Find the deceleration.
m s−2

Hint

You know u, v and s. Leave out t.

Worked solution
  1. v2 = u2 + 2as: 82 = 202 + 2 × a × 84.
  2. 64 = 400 + 168a, so a = −2.
  3. The deceleration is 2 m s−2.
Q4Exam-style
A stone is thrown vertically downwards at 4 m s−1 from the top of a cliff 50 m above the sea. Find its speed as it hits the sea.
m s−1

Hint

Take downwards as positive, so a = +9.8 and s = +50.

Worked solution
  1. Downwards positive: u = 4, a = 9.8, s = 50.
  2. v2 = 42 + 2 × 9.8 × 50 = 996.
  3. v = 31.6 m s−1 (3 s.f.).
Q5Challenge
Car P starts from rest at a set of traffic lights and accelerates at 2 m s−2. At the same moment car Q passes the lights at a constant 12 m s−1. How long is it until P catches Q?
s

Hint

When P catches Q they have both travelled the same distance from the lights.

Worked solution
  1. Distance for P: s = ½ × 2 × t2 = t2. Distance for Q: s = 12t.
  2. t2 = 12t, so t(t − 12) = 0.
  3. t = 0 is the start, so P catches Q after 12 s, 144 m from the lights.
Q6Exam-styleLifts
A woman of mass 60 kg stands in a lift that is accelerating upwards at 1.2 m s−2. Find the normal reaction from the floor of the lift on her.
N

Hint

Draw her forces: R up, 60g down. The resultant is upwards.

Worked solution
  1. Resolve upwards, F = ma: R − 60g = 60 × 1.2.
  2. R = 60 × 9.8 + 72 = 660 N.
Q7Exam-style
A box of mass 2 kg is pulled along a rough horizontal floor by a horizontal force of 15 N. The coefficient of friction is 0.4. Find the acceleration of the box.
m s−2

Hint

First find R by resolving vertically, then friction = μR because the box is moving.

Worked solution
  1. Vertically: R = 2g = 19.6 N, so friction F = 0.4 × 19.6 = 7.84 N.
  2. Horizontally: 15 − 7.84 = 2a.
  3. a = 3.58 m s−2.
Q8Exam-styleInclined plane
A particle slides down a rough plane inclined at 30° to the horizontal. The coefficient of friction is 0.2. Find its acceleration.
m s−2

Hint

Resolve perpendicular to the plane for R, then along the plane. Does the mass matter?

Worked solution
  1. Perpendicular: R = mg cos 30°. Friction acts up the plane: F = 0.2mg cos 30°.
  2. Along the plane: mg sin 30° − 0.2mg cos 30° = ma, and m cancels.
  3. a = 9.8(0.5 − 0.2 × 0.866) = 3.20 m s−2.
Q9Exam-styleConnected particles
Particle A of mass 3 kg lies on a smooth horizontal table. It is joined by a light inextensible string, passing over a smooth pulley at the edge of the table, to particle B of mass 2 kg hanging freely. The system is released from rest. Find the acceleration.
m s−2

Hint

Write F = ma for each particle separately, then add the equations to remove T.

Worked solution
  1. For A (horizontal): T = 3a. For B (downwards): 2g − T = 2a.
  2. Adding: 2g = 5a.
  3. a = 19.6/5 = 3.92 m s−2.
Q10ChallengePulleys
Particles of mass 5 kg and 3 kg hang on either side of a smooth fixed pulley, joined by a light inextensible string. The system is released from rest. Find the tension in the string.
N

Hint

Find the acceleration first, from the two equations of motion, then substitute back.

Worked solution
  1. 5 kg (down): 5g − T = 5a. 3 kg (up): T − 3g = 3a.
  2. Adding: 2g = 8a, so a = 2.45 m s−2.
  3. T = 3(9.8 + 2.45) = 36.8 N (3 s.f.).
Q11Exam-styleLimiting friction
A particle rests on a rough plane inclined at 25° to the horizontal and is on the point of slipping down it. Find the coefficient of friction.

Hint

Resolve along and perpendicular to the plane, with friction acting up the plane.

Worked solution
  1. Perpendicular: R = mg cos 25°. Along the plane: F = mg sin 25°.
  2. Limiting, so F = μR: mg sin 25° = μmg cos 25°.
  3. μ = tan 25° = 0.466 (3 s.f.).
Q12Exam-style
A particle of mass 5 kg rests on a smooth plane inclined at 20° to the horizontal. It is held in equilibrium by a horizontal force P. Find P.
N

Hint

Resolve along the plane so that R does not appear.

Worked solution
  1. Along the plane: P cos 20° = 5g sin 20°.
  2. P = 49 tan 20°.
  3. P = 17.8 N (3 s.f.).
Q13Exam-style
A particle of mass 4 kg hangs in equilibrium from two light strings. One string makes 30° with the horizontal and the other makes 60° with the horizontal. Find the tension in the string at 60°.
N

Hint

Resolve horizontally and vertically, then solve the two equations simultaneously.

Worked solution
  1. Horizontally: T1 cos 30° = T2 cos 60°, so T2 = √3 T1.
  2. Vertically: T1 sin 30° + T2 sin 60° = 4g, so 0.5T1 + 1.5T1 = 39.2 and T1 = 19.6 N.
  3. T2 = 19.6√3 = 33.9 N (3 s.f.).
Q14Exam-style
Forces (3i − 2j) N, (ai + 5j) N and (−4i + bj) N act on a particle in equilibrium. Find a and b.

Hint

In equilibrium the resultant is zero, so each component adds to zero.

Worked solution
  1. i: 3 + a − 4 = 0, so a = 1.
  2. j: −2 + 5 + b = 0, so b = −3.
  3. The answer is A.
Q15Challenge
A box of mass 2 kg rests on rough horizontal ground, with coefficient of friction 0.5. A force P acting at 20° above the horizontal is just enough to make the box start to move. Find P.
N

Hint

P reduces the normal reaction. Write R in terms of P first.

Worked solution
  1. Vertically: R + P sin 20° = 2g, so R = 19.6 − P sin 20°.
  2. Horizontally, limiting: P cos 20° = 0.5(19.6 − P sin 20°).
  3. P(cos 20° + 0.5 sin 20°) = 9.8, so P = 9.8/1.1107 = 8.82 N (3 s.f.).
Q16Exam-style
Particle P, of mass 2 kg, moving at 6 m s−1, hits particle Q, of mass 3 kg, which is at rest. After the collision P moves on in the same direction at 1.5 m s−1. Find the speed of Q.
m s−1

Hint

Momentum before = momentum after, with the direction of P's motion positive.

Worked solution
  1. Before: 2 × 6 + 0 = 12. After: 2 × 1.5 + 3v = 3 + 3v.
  2. 12 = 3 + 3v.
  3. v = 3 m s−1.
Q17Exam-style
In the same collision, find the magnitude of the impulse exerted by P on Q.
N s

Hint

Find the change in momentum of Q.

Worked solution
  1. Impulse on Q = 3 × 3 − 3 × 0.
  2. = 9 N s, equal and opposite to the impulse on P.
Q18Exam-style
A ball of mass 0.4 kg hits a vertical wall at 10 m s−1 and rebounds along the same line at 6 m s−1. Find the magnitude of the impulse on the ball.
N s

Hint

The velocity changes direction, so one of them is negative.

Worked solution
  1. Take the rebound direction as positive: u = −10, v = 6.
  2. I = 0.4 × 6 − 0.4 × (−10) = 2.4 + 4.
  3. I = 6.4 N s.
Q19Exam-style
Two particles collide on a smooth surface. Which statement is always true?

Hint

Think about Newton's third law during the collision.

Worked solution
  1. The impulses on the two particles are equal and opposite, so the total momentum does not change.
  2. Kinetic energy is usually lost. The answer is B.
Q20Challenge
Particle A, of mass 4 kg, moving at 5 m s−1, and particle B, of mass 2 kg, moving at 3 m s−1, move towards each other along a straight line. They collide and join together. Find their common speed.
m s−1

Hint

B moves in the opposite direction, so its momentum is negative.

Worked solution
  1. Take A's direction as positive. Before: 4 × 5 + 2 × (−3) = 14.
  2. After: 6v = 14.
  3. v = 7/3 = 2.33 m s−1 (3 s.f.), in A's original direction.
Q21Exam-style
A uniform beam AB, of length 6 m and mass 20 kg, rests horizontally on supports at A and B. A man of mass 30 kg stands on the beam 2 m from A. Find the reaction at A.
N

Hint

Take moments about B so that the reaction at B does not appear.

Worked solution
  1. Moments about B: RA × 6 = 20g × 3 + 30g × 4.
  2. RA = 180g/6 = 30g.
  3. RA = 294 N.
Q22Exam-style
For the same beam, find the reaction at B.
N

Hint

Resolve vertically.

Worked solution
  1. RA + RB = 50g, so RB = 50g − 30g = 20g.
  2. RB = 196 N.
Q23Exam-style
A plank rests on two supports, C and D. A person walks along it past D until the plank is on the point of tilting about D. What is the reaction at C at that moment?

Hint

On the point of tilting, the plank is about to lift off one support.

Worked solution
  1. About to tilt about D means the plank is about to leave C.
  2. So the reaction at C is zero: A.
Q24Exam-style
A uniform plank AB, of length 5 m and mass 40 kg, rests on supports at C and D, where AC = 1 m and DB = 1.5 m. A boy of mass 60 kg walks from D towards B. How far beyond D can he go before the plank tilts?
m

Hint

The plank's weight acts 2.5 m from A, which is 1 m from D. Set the reaction at C to zero.

Worked solution
  1. D is 3.5 m from A, and the centre of the plank is 2.5 m from A, so 1 m on the other side of D.
  2. On the point of tilting, RC = 0. Moments about D: 40g × 1 = 60g × x.
  3. x = 40/60 = 0.667 m (3 s.f.), still on the plank since DB = 1.5 m.
Q25ChallengeNon-uniform rod
A non-uniform rod AB, of length 4 m and mass 12 kg, rests horizontally on supports at A and B. The reaction at A is twice the reaction at B. Find the distance of the rod's centre of mass from A.
m

Hint

Resolve vertically first to find each reaction, then take moments about A.

Worked solution
  1. RA + RB = 12g with RA = 2RB, so RB = 4g.
  2. Moments about A: 12g × x = 4g × 4.
  3. x = 16/12 = 1.33 m (3 s.f.).
Q26Exam-style
Forces (4i + 2j) N and (2i + 6j) N act on a particle of mass 2 kg. Find the magnitude of its acceleration.
m s−2

Hint

Find the resultant force, then use F = ma.

Worked solution
  1. Resultant = (6i + 8j) N, with magnitude √(36 + 64) = 10 N.
  2. a = 10/2 = 5 m s−2.
Q27Exam-style
A boat moves with velocity (−3i + 3j) km h−1, where i points east and j points north. On what bearing is it moving?

Hint

West 3 and north 3: sketch it, then measure clockwise from north.

Worked solution
  1. The boat moves north-west, at 45° between west and north.
  2. Clockwise from north, that is 360° − 45° = 315°: D.
Q28Exam-style
A particle starts at the point with position vector (2i − 3j) m and moves with constant velocity (3i + 4j) m s−1. Find its distance from the origin after 2 s.
m

Hint

Use r = r0 + vt, then find the magnitude.

Worked solution
  1. r = (2i − 3j) + 2(3i + 4j) = (8i + 5j) m.
  2. Distance = √(64 + 25) = √89 = 9.43 m (3 s.f.).
Q29Exam-style
A particle P starts at (−4i + 6j) m and moves with constant velocity (3i − j) m s−1, where i is east and j is north. Find the time at which P is due east of the origin.
s

Hint

Due east of the origin means the j component is zero and the i component is positive.

Worked solution
  1. r = (−4 + 3t)i + (6 − t)j.
  2. Due east: 6 − t = 0, so t = 6. Check: the i component is −4 + 18 = 14 > 0.
  3. t = 6 s.
Q30Challenge
A particle has initial velocity (2i − 5j) m s−1 and constant acceleration (i + 2j) m s−2. Find its speed after 4 s.
m s−1

Hint

Use v = u + at with vectors, then find the magnitude.

Worked solution
  1. v = (2i − 5j) + 4(i + 2j) = (6i + 3j) m s−1.
  2. Speed = √(36 + 9) = √45 = 6.71 m s−1 (3 s.f.).

Section II: 4 structured exam questions

Structured exam question 1: Constant acceleration

Structured exam question style · 7 points · about 10 minutes

A train starts from rest at station A. It accelerates uniformly for 20 s to a speed of V m s−1, travels at this speed for 60 s, then decelerates uniformly for 10 s, coming to rest at station B. The distance from A to B is 1.8 km.

  1. (a) Sketch a speed-time graph for the journey.
  2. (b) Find the value of V.
  3. (c) Find the deceleration of the train.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

A trapezium: a straight line from (0, 0) up to (20, V), horizontal to (80, V), then straight down to (90, 0).

Part (b)

Area = ½(90 + 60)V = 75V = 1800, so V = 24.

Part (c)

Deceleration = 24/10 = 2.4 m s−2.

Your mark: 0 of 7

Structured exam question 2: Statics of a particle

Structured exam question style · 7 points · about 12 minutes

A particle of mass 2 kg rests on a rough plane inclined at 30° to the horizontal. The coefficient of friction between the particle and the plane is 0.25. The particle is held in equilibrium by a force of magnitude P newtons acting up the plane, along a line of greatest slope.

  1. (a) Find the normal reaction on the particle.
  2. (b) Find the least value of P for which the particle stays at rest.
  3. (c) Find the greatest value of P for which the particle stays at rest.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

R = 2g cos 30° = 16.97… = 17.0 N.

Part (b)

About to slip down, so friction acts up the plane: P + 0.25R = 2g sin 30°, P = 9.8 − 4.24 = 5.56 N.

Part (c)

About to move up, so friction acts down the plane: P = 9.8 + 4.24 = 14.0 N.

Your mark: 0 of 7

Structured exam question 3: Momentum and impulse

Structured exam question style · 6 points · about 10 minutes

Two particles A and B move towards each other along the same straight line on a smooth horizontal surface. A has mass 2 kg and speed 5 m s−1; B has mass m kg and speed 3 m s−1. In the collision the direction of each particle is reversed. Immediately after, A has speed 1 m s−1 and B has speed 2 m s−1.

  1. (a) Find the value of m.
  2. (b) Find the magnitude of the impulse exerted on A in the collision.
  3. (c) State what is meant by modelling the surface as smooth.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Taking A's original direction positive: 2(5) + m(−3) = 2(−1) + m(2). So 12 = 5m and m = 2.4.

Part (b)

I = 2(−1) − 2(5) = −12, so the magnitude is 12 N s.

Part (c)

There is no friction between the particles and the surface.

Your mark: 0 of 6

Structured exam question 4: Vectors in mechanics

Structured exam question style · 7 points · about 10 minutes

A particle P of mass 0.5 kg is acted on by two forces, F1 = (2i − 3j) N and F2 = (4i + qj) N, where q is a constant. The resultant of the two forces is parallel to the vector (i + 2j).

  1. (a) Find the value of q.
  2. (b) Find the magnitude of the acceleration of P.
  3. (c) Find the angle between the resultant force and the vector i.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Resultant = 6i + (q − 3)j. Parallel to i + 2j, so (q − 3)/6 = 2 and q = 15.

Part (b)

Resultant = (6i + 12j) N, magnitude √180 = 13.4 N. a = 13.4/0.5 = 26.8 m s−2.

Part (c)

tan θ = 12/6 = 2, so θ = 63.4°.

Your mark: 0 of 7

After the mock

Note every question you missed and its chapter, then go back to that chapter's practice page: all Edexcel IAL Mechanics 1 chapters. Then try the real thing with official Edexcel IAL Mechanics 1 past papers.