Free access until 31 December 2026 · Physics and Maths questions and notes, updated continuously · Build a test · Get an email before anything changes

ExaminerPrep › Question bank › IAL Maths › M1.3 Dynamics: Newton's laws, connected particles and friction

Edexcel International A Level Maths · M1 · Mechanics 1

Dynamics: Newton's laws, connected particles and friction

28 free, high-quality past-paper-style questions on Dynamics: Newton's laws, connected particles and friction 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 · 252 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

1IALM-M1-3-B-01 M1 · ASEasy6 marks

A puck P of mass 1.5 kg slides on a smooth horizontal surface. The only horizontal forces acting on P are (i + 6j) N and (−6i − 4j) N, where i and j are perpendicular horizontal unit vectors. P is modelled as a particle.

(a)

Find the resultant force on P, in the form ai + bj.

(2)
(b)

Find the magnitude of the acceleration of P.

(2)
(c)

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

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
adding components: (1 − 6)i + (6 − 4)jM1
(−5i + 2j) NA1
Part (b)
|R| = √((−5)2 + 22) = √29, and a = √29 ÷ 1.5M1
= 3.59 m s−2A1
Part (c)
tan−1(2/5) = 21.8°M1
the acceleration points into the second quadrant, so the angle with i is 180° − 21.8° = 158.2° (accept 158°)A1

Exam tip: Sketch the resultant before finding its direction. A negative i-component and positive j-component put the vector in the second quadrant, so the angle with i is more than 90°.

2IALM-M1-3-B-02 M1 · ASEasy5 marks

A stone of mass 0.8 kg is released from rest at the surface of a pond and sinks vertically. The water exerts a constant resistance of 2.4 N on the stone (this includes every force from the water). The stone is modelled as a particle. Take g = 9.8 m s−2.

(a)

Find the acceleration of the stone.

(2)
(b)

The pond is 1.7 m deep. Find the time the stone takes to reach the bottom.

(2)
(c)

Find the speed of the stone as it reaches the bottom.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
0.8g − 2.4 = 0.8aM1
a = 6.8 m s−2A1
Part (b)
1.7 = ½ × 6.8 × t2M1
t = √0.5 = 0.707 sA1
Part (c)
v = 6.8 × 0.7071 (or v2 = 2 × 6.8 × 1.7) = 4.81 m s−1 (ft from (a))B1

Common error: Using a = g for anything that falls. The resistance reduces the resultant force, so the acceleration is less than 9.8 m s−2.

3IALM-M1-3-B-03 M1 · ASEasy6 marks

A goods lift of mass 500 kg carries a box of mass 30 kg on its floor. The lift is supported by a vertical cable. At one stage of the motion the tension in the cable is 4800 N. Take g = 9.8 m s−2.

30 kglift500 kg4800 Ncable
Figure: the lift carrying the box, supported by the cable; not to scale.
(a)

Find the magnitude and direction of the acceleration of the lift.

(3)
(b)

Find the magnitude of the force exerted by the floor of the lift on the box.

(2)
(c)

State, with a reason, whether it is possible to tell from this information if the lift is moving up or down.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
lift and box together, upwards positive: 4800 − 530g = 530aM1
a = −0.743A1
so 0.743 m s−2 downwardsA1
Part (b)
box alone: R − 30g = 30 × (−0.7434)M1
R = 272 NA1
Part (c)
no: a downward acceleration fits a lift moving down and speeding up or a lift moving up and slowing downB1

Exam tip: Treat the lift and its contents as one particle for the cable tension, then the contents alone for the force from the floor.

4IALM-M1-3-B-04 M1 · ASEasy6 marks

A small box slides in a straight line across a rough horizontal warehouse floor. It passes a point A with speed 6 m s−1 and comes to rest 9 m beyond A. No force other than friction acts on the box horizontally. The box is modelled as a particle. Take g = 9.8 m s−2.

(a)

Find the deceleration of the box.

(2)
(b)

Find the coefficient of friction between the box and the floor.

(3)
(c)

A second box of twice the mass slides on the same floor with the same coefficient of friction. It passes A with speed 6 m s−1. State how far beyond A it comes to rest, giving a reason.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
0 = 62 − 2 × a × 9M1
a = 2 m s−2A1
Part (b)
resolving vertically: R = mgB1
μmg = m × 2M1
μ = 2 ÷ 9.8 = 0.204A1
Part (c)
9 m, because the mass cancels: the deceleration μg does not depend on the massB1

Exam tip: When the mass is not given, keep it as m. It usually cancels, as it does here.

5IALM-M1-3-FB-01 M1 · ASEasy6 marks

A particle P of mass 0.5 kg lies on a smooth horizontal surface. Figure 1 shows the three horizontal forces F1, F2 and F3 acting on P, drawn on a grid in which one square represents 1 N. The unit vectors i and j are horizontal and perpendicular, as shown.

ijOF₁F₂F₃one grid square = 1 N
(a)

Write down each force in the form (xi + yj) N and hence find the resultant force on P.

(2)
(b)

Find the magnitude of the acceleration of P.

(2)
(c)

A fourth horizontal force F4 is now also applied to P. P then accelerates at 2i m s−2. Find F4.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
F1 = 3i + 2j, F2 = −i + 3j, F3 = −4i − 2j (all three)B1
resultant = (−2i + 3j) NB1
Part (b)
|R| = √((−2)2 + 32) = √13 and a = √13 ÷ 0.5M1
= 7.21 m s−2A1
Part (c)
(−2i + 3j) + F4 = 0.5 × 2iM1
F4 = (3i − 3j) NA1

Exam tip: Read each force from the tail at O to the tip of its arrow, counting squares across (i) then up or down (j). Write the components down before you add them.

6IALM-M1-3-FB-02 M1 · ASEasy7 marks

A sledge of mass 40 kg is pulled from rest across rough horizontal snow by a horizontal rope with constant tension. After 6 s the rope is released and the sledge slides on until it comes to rest. Figure 1 is the velocity–time graph of the motion. The sledge is modelled as a particle and the coefficient of friction between the sledge and the snow is constant. Take g = 9.8 m s−2.

0246810121416012345t / sv / m s⁻¹(6, 4.5)(15, 0)rope released
(a)

Use the graph to find the deceleration of the sledge after the rope is released, and hence find the frictional force on the sledge.

(2)
(b)

Find the coefficient of friction between the sledge and the snow.

(2)
(c)

Find the tension in the rope.

(2)
(d)

Find the total distance travelled by the sledge.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
deceleration = 4.5 ÷ (15 − 6) = 0.5 m s−2B1
friction = 40 × 0.5 = 20 NB1
Part (b)
R = 40g = 392 N and 20 = μ × 392M1
μ = 0.0510A1
Part (c)
acceleration = 4.5 ÷ 6 = 0.75 m s−2, and T − 20 = 40 × 0.75M1
T = 50 NA1
Part (d)
area under the graph = ½ × 15 × 4.5 = 33.75 m (accept 33.8 m)B1

Exam tip: Start with the stage where only one unknown force acts. After the rope is released, friction alone causes the deceleration, so that stage gives the friction directly.

7IALM-M1-3-FB-03 M1 · ASEasy6 marks

A crate of mass 30 kg rests on a rough horizontal floor. A horizontal force of magnitude P newtons is applied to the crate and P is slowly increased from zero. Figure 1 shows how the magnitude F newtons of the friction force on the crate varies with P. The crate is modelled as a particle and, once it slides, the friction on it keeps its limiting value. Take g = 9.8 m s−2.

03060901201500102030405060708090100P / NF / Ncrate at restcrate sliding
(a)

Explain why F = P for values of P less than 90.

(1)
(b)

Find the coefficient of friction between the crate and the floor.

(3)
(c)

Find the acceleration of the crate when P = 135.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
the crate stays at rest, so the horizontal forces balance and friction equals the applied forceB1
Part (b)
limiting friction = 90 N, read from the graphB1
R = 30g = 294 N and 90 = μ × 294M1
μ = 0.306A1
Part (c)
135 − 90 = 30aM1
a = 1.5 m s−2A1

Common error: Taking the friction to be μR while the crate is still at rest. Below the limit, friction is only as large as it needs to be to stop the crate moving, which is why the graph starts as F = P.

8IALM-M1-3-FB-04 M1 · ASEasy7 marks

A student of mass 60 kg stands on a set of bathroom scales on the floor of a lift. The scales show the magnitude of the normal reaction on the student. The lift starts from rest at time t = 0 and moves vertically upwards. Figure 1 shows the reading on the scales for 0 ≤ t ≤ 12, where t is in seconds. The student is modelled as a particle. Take g = 9.8 m s−2.

0123456789101112500520540560580600620640660680700t / sreading / N660 N588 N540 N588 N
(a)

Explain how the graph shows that the lift is moving with constant velocity for 2 < t < 8.

(1)
(b)

Find the acceleration of the lift for 0 < t < 2.

(2)
(c)

Find the deceleration of the lift for 8 < t < 11 and show that the lift is at rest at t = 11.

(2)
(d)

The lift has mass 740 kg and the scales have negligible mass. Find the tension in the lift cable for 0 < t < 2.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
the reading is 588 N = 60g, so the resultant force on the student is zero and the acceleration is zeroB1
Part (b)
660 − 60g = 60aM1
a = 1.2 m s−2 upwardsA1
Part (c)
60g − 540 = 60d, so d = 0.8 m s−2M1
speed for 2 < t < 8 is 1.2 × 2 = 2.4 m s−1, and 2.4 − 0.8 × 3 = 0A1
Part (d)
lift and student together: T − 800g = 800 × 1.2M1
T = 8800 NA1

Exam tip: A scale reading above the weight means an upward acceleration; below the weight means a downward acceleration (here, the lift slowing while still moving up).

9IALM-M1-3-B-05 M1 · ASMedium8 marks

A trunk of mass 30 kg is dragged across a rough horizontal floor by a rope. The rope is inclined at 20° above the horizontal and the tension in it is T newtons. The coefficient of friction between the trunk and the floor is 0.35. The trunk is modelled as a particle and the rope as light. Take g = 9.8 m s−2.

30 kgT N20°30g NRF
Figure: forces on the trunk, with the rope at 20° above the horizontal; not to scale.
(a)

Find, in terms of T, the normal reaction of the floor on the trunk.

(2)
(b)

The trunk moves at constant speed. Find T.

(3)
(c)

The tension is increased to 130 N, with the rope at the same angle. Find the acceleration of the trunk.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
resolving vertically: R + T sin 20° = 30gM1
R = 294 − T sin 20° (= 294 − 0.342T)A1
Part (b)
friction = 0.35(294 − T sin 20°)B1
resolving horizontally with zero acceleration: T cos 20° = 0.35(294 − T sin 20°)M1
T = 97.1A1
Part (c)
R = 294 − 130 sin 20° = 249.5 NM1
130 cos 20° − 0.35 × 249.5 = 30aM1
a = 1.16 m s−2A1

Common error: Keeping R = 30g when the rope is at an angle. The upward component of the tension reduces the normal reaction, and the friction with it.

10IALM-M1-3-B-06 M1 · ASMedium8 marks

A particle of mass 2.5 kg is on a smooth plane inclined at 25° to the horizontal. It is pushed up the plane by a horizontal force of magnitude 30 N, which acts in the vertical plane containing a line of greatest slope. The particle starts from rest. Take g = 9.8 m s−2.

25°2.5 kg30 NR2.5g N
Figure: the particle on the smooth plane, pushed by the horizontal force of 30 N; not to scale.
(a)

Find the normal reaction of the plane on the particle.

(3)
(b)

Find the acceleration of the particle.

(3)
(c)

Find the distance the particle moves up the plane in the first 2 s.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
resolving perpendicular to the plane: R = 2.5g cos 25° + 30 sin 25°M1
both components correctA1
R = 34.9 NA1
Part (b)
resolving parallel to the plane: 30 cos 25° − 2.5g sin 25° = 2.5aM1
both components correctA1
a = 6.73 m s−2A1
Part (c)
s = ½ × 6.734 × 22M1
= 13.5 mA1

Exam tip: A horizontal force on a slope has a component along the plane (cos of the slope angle) and a component into the plane (sin of the slope angle). Draw the angle between the force and the plane before resolving.

11IALM-M1-3-B-07 M1 · ASMedium9 marks

A car of mass 1350 kg tows a trailer of mass 450 kg up a straight road inclined at an angle α to the horizontal, where sin α = 1⁄15. The tow bar is light, rigid and parallel to the road. The resistances to motion are constant: 240 N on the car and 160 N on the trailer. The engine of the car produces a constant driving force of 3600 N. Take g = 9.8 m s−2.

α450 kg1350 kg3600 N240 N160 N
Figure: the car towing the trailer up the slope, with the driving force and resistances; sin α = 1/15, angle exaggerated; not to scale.
(a)

Find the acceleration of the car and trailer.

(3)
(b)

Find the tension in the tow bar.

(3)
(c)

When the car and trailer are moving at 12 m s−1, the tow bar breaks. Find the distance the trailer then travels up the road before it comes to rest.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
whole system, up the road: 3600 − 240 − 160 − 1800g sin α = 1800aM1
weight component 1800 × 9.8 ÷ 15 = 1176 NA1
a = 2024 ÷ 1800 = 1.12 m s−2A1
Part (b)
trailer: T − 160 − 450g sin α = 450aM1
T − 160 − 294 = 506A1
T = 960 NA1
Part (c)
trailer alone: deceleration = (160 + 294) ÷ 450 = 1.009 m s−2M1
0 = 122 − 2 × 1.009 × sM1
s = 71.4 mA1

Exam tip: When the tow bar breaks, the tension disappears and nothing else changes: the trailer keeps its speed, its weight component and its resistance.

12IALM-M1-3-B-08 M1 · ASMedium9 marks

Two particles, of masses 0.6 kg and 0.4 kg, are attached to the ends of a light inextensible string that passes over a small smooth peg fixed above a horizontal floor. The particles hang with the hanging parts of the string vertical, the 0.6 kg particle being 0.9 m above the floor. The system is released from rest with the string taut. Take g = 9.8 m s−2.

0.6 kg0.4 kg0.9 m
Figure: the particles hanging over the smooth peg, released from rest; not to scale.
(a)

Find the acceleration of the particles and the tension in the string.

(4)
(b)

Find the magnitude of the force exerted by the string on the peg.

(2)
(c)

The 0.6 kg particle hits the floor and does not rebound. The 0.4 kg particle does not reach the peg. Find the greatest height of the 0.4 kg particle above its starting position.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
0.6 kg particle: 0.6g − T = 0.6aM1
0.4 kg particle: T − 0.4g = 0.4aM1
a = 1.96 m s−2A1
T = 4.70 NA1
Part (b)
both parts of the string are vertical, so the force is 2TM1
= 9.41 NA1
Part (c)
speed when the string goes slack: v2 = 2 × 1.96 × 0.9 = 3.528M1
it then moves freely under gravity: further rise = 3.528 ÷ (2 × 9.8) = 0.18 mM1
greatest height = 0.9 + 0.18 = 1.08 mA1

Common error: Stopping at 0.9 m. When the heavier particle lands, the lighter one is still moving upwards and carries on rising, with deceleration g, until it stops.

13IALM-M1-3-B-09 M1 · ASMedium8 marks

In a classroom demonstration, a block A of mass 5 kg lies on a rough horizontal bench. A light inextensible string attached to A passes over a small smooth pulley fixed at the edge of the bench and carries a block B of mass 3 kg, which hangs freely. The string between A and the pulley is horizontal and perpendicular to the edge of the bench. When the blocks are released from rest, a motion sensor shows that they accelerate at 2.1 m s−2. The blocks are modelled as particles. Take g = 9.8 m s−2.

A 5 kgB3 kg2.1 m s⁻²2.1 m s⁻²
Figure: block A on the rough bench connected over the pulley to block B; not to scale.
(a)

Find the tension in the string.

(2)
(b)

Find the coefficient of friction between A and the bench.

(3)
(c)

Find the magnitude and direction of the force exerted by the string on the pulley.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
B: 3g − T = 3 × 2.1M1
T = 23.1 NA1
Part (b)
resolving vertically for A: R = 5g = 49 NB1
A: 23.1 − μ × 49 = 5 × 2.1M1
μ = 12.6 ÷ 49 = 0.257A1
Part (c)
two tensions of 23.1 N at right angles: √(23.12 + 23.12)M1
= 32.7 NA1
at 45° below the horizontal, towards the benchB1

Exam tip: The two parts of the string pull the pulley along each string, away from the pulley. Add them as vectors: at right angles the resultant is T√2, along the bisector.

14IALM-M1-3-B-10 M1 · ASMedium10 marks

A particle of mass 4 kg moves on a smooth horizontal plane under the action of three forces: F1 = (pi + 3j) N, F2 = (−2i + qj) N and F3 = (5i − 7j) N, where p and q are constants. The particle starts from rest at time t = 0 and its acceleration is (1.5i − 0.5j) m s−2.

(a)

Find the values of p and q.

(3)
(b)

Find the speed of the particle when t = 3 s.

(2)
(c)

At t = 3 s the force F3 is removed. Find the new acceleration of the particle.

(2)
(d)

Find the time, after F3 is removed, at which the particle is moving parallel to the vector (i + j).

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
resultant = 4(1.5i − 0.5j) = (6i − 2j) NM1
p − 2 + 5 = 6 ⇒ p = 3A1
3 + q − 7 = −2 ⇒ q = 2A1
Part (b)
v = 3(1.5i − 0.5j) = (4.5i − 1.5j) m s−1M1
speed = √(4.52 + 1.52) = 4.74 m s−1A1
Part (c)
F1 + F2 = (i + 5j) NM1
a = (0.25i + 1.25j) m s−2A1
Part (d)
T s after removal: v = (4.5 + 0.25T)i + (−1.5 + 1.25T)jM1
parallel to i + j: 4.5 + 0.25T = −1.5 + 1.25TM1
T = 6 sA1

Exam tip: “Parallel to i + j” means the two components of the velocity are equal. Check they are also positive, so the particle really moves in the direction of i + j.

15IALM-M1-3-FB-05 M1 · ASMedium10 marks

Figure 1 shows particles A, B and C, of masses 2.2 kg, 1.5 kg and 1 kg. A and B are attached to the ends of a light inextensible string that passes over a smooth pulley fixed above them. C hangs below B, attached to it by a second light inextensible string. All hanging parts of the strings are vertical. The system is released from rest with both strings taut. No particle reaches the pulley or the ground during the motion described. Take g = 9.8 m s−2.

A2.2 kgB1.5 kgC1 kgsmooth pulley · light inextensible strings
(a)

Find the acceleration of the particles.

(3)
(b)

Find the tension in the string joining B and C.

(2)
(c)

Find the magnitude of the force exerted on the pulley by the string passing over it.

(2)
(d)

After 2 s the string joining B and C is cut. Find the further distance that B moves downwards before it comes to instantaneous rest.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
B and C together: 2.5g − T = 2.5aM1
A: T − 2.2g = 2.2aM1
a = 0.3g ÷ 4.7 = 0.626 m s−2A1
Part (b)
C: g − S = 1 × 0.6255M1
S = 9.17 NA1
Part (c)
T = 2.2(9.8 + 0.6255) = 22.94 N, and both parts of the string are vertical, so the force is 2TM1
= 45.9 NA1
Part (d)
speed when the string is cut = 0.6255 × 2 = 1.251 m s−1B1
A and B only: deceleration = (2.2 − 1.5)g ÷ 3.7 = 1.854 m s−2M1
0 = 1.2512 − 2 × 1.854 × s, s = 0.422 mA1

Exam tip: When the lower string is cut, the speed does not change but the side that was heavier becomes lighter. B keeps moving down while it slows, so use the new deceleration with the speed at the cut.

16IALM-M1-3-FB-06 M1 · ASMedium9 marks

A tractor pulls two trailers P and Q along a straight horizontal road. The tractor is joined to P, and P is joined to Q, by light rigid horizontal couplings. Figure 1 shows the masses of the three vehicles and the constant resistances to motion acting on them. The tractor's engine produces a constant driving force D newtons and the vehicles accelerate at 0.5 m s−2.

trailer Q800 kg200 Ntrailer P1200 kg300 Ntractor2400 kg600 NDacceleration 0.5 m s⁻²resistances to motion (arrows below each vehicle)
(a)

Find D.

(2)
(b)

Find the tension in the coupling between P and Q and the tension in the coupling between the tractor and P.

(4)
(c)

Later the driving force is removed and the tractor's brakes are applied. The vehicles decelerate at 1.5 m s−2 and the resistances are unchanged. Find the magnitude of the force in the coupling between P and Q, and state whether it is a tension or a thrust.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
whole system: D − 600 − 300 − 200 = 4400 × 0.5M1
D = 3300A1
Part (b)
Q: T1 − 200 = 800 × 0.5M1
T1 = 600 NA1
P: T2 − T1 − 300 = 1200 × 0.5M1
T2 = 1500 NA1
Part (c)
Q, with X the force on Q from the coupling in the direction of motion: X − 200 = 800 × (−1.5)M1
X = −1000A1
so the coupling pushes Q backwards: a thrust of 1000 NA1

Exam tip: For the force in a coupling, isolate the vehicle behind it: Q alone has only one unknown force, so start there.

17IALM-M1-3-FB-07 M1 · ASMedium9 marks

A particle is projected up a line of greatest slope of a rough plane inclined at an angle α to the horizontal. The coefficient of friction between the particle and the plane is μ. Figure 1 is the velocity–time graph for the first 1.5 s of the motion, taking the direction up the plane as positive. Take g = 9.8 m s−2.

00.250.50.7511.251.5−4−3−2−11234567t / sv / m s⁻¹ (up the plane +)(0, 6)(0.75, 0)(1.5, −3)
(a)

Use the graph to find the magnitude of the acceleration of the particle while it moves up the plane and while it moves down the plane.

(2)
(b)

By resolving parallel to the plane for each stage, find α.

(3)
(c)

Find μ.

(2)
(d)

Find the speed of the particle when it returns to its point of projection.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
up: 6 ÷ 0.75 = 8 m s−2B1
down: 3 ÷ 0.75 = 4 m s−2B1
Part (b)
moving up, friction acts down the plane: g sin α + μg cos α = 8M1
moving down, friction acts up the plane: g sin α − μg cos α = 4M1
adding: sin α = 12 ÷ 19.6, α = 37.8°A1
Part (c)
subtracting: 2μg cos α = 4M1
μ = 2 ÷ (9.8 × 0.7907) = 0.258A1
Part (d)
distance up the plane = ½ × 0.75 × 6 = 2.25 mM1
v2 = 2 × 4 × 2.25 = 18, v = 4.24 m s−1A1

Common error: Using the same friction direction for both stages. Friction always opposes the motion, so it changes direction when the particle turns round, and the two gradients differ.

18IALM-M1-3-FB-08 M1 · ASMedium8 marks

A barge of mass 12 000 kg is pulled from rest along a straight canal by two horizontal ropes attached at its bow. Figure 1, a view from above, shows the forces on the barge: the ropes make angles of 30° and 40° with the centre line of the barge, on opposite sides, and a resistance of 2500 N acts along the centre line. The tension in the first rope is 4000 N and the tension in the second is T2 newtons. The barge moves along its centre line. The barge is modelled as a particle.

bargecentre line4000 NT₂30°40°2500 Ndirection of motionview from above · all forces horizontal
(a)

Find T2.

(3)
(b)

Find the acceleration of the barge.

(3)
(c)

Find the time taken for the barge to reach a speed of 1.5 m s−1.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
no acceleration perpendicular to the centre line, so resolve perpendicular to itM1
4000 sin 30° = T2 sin 40°A1
T2 = 3110 (3111)A1
Part (b)
along the centre line: 4000 cos 30° + T2 cos 40° − 2500 = 12 000aM1
3464 + 2384 − 2500 = 12 000aA1
a = 0.279 m s−2A1
Part (c)
1.5 = 0 + 0.2790tM1
t = 5.38 sA1

Exam tip: If a body moves in a straight line, the resultant force has no component perpendicular to that line. Resolving perpendicular to the motion first gives an equation without the acceleration.

19IALM-M1-3-FB-09 M1 · ASMedium9 marks

Two boxes A and B, of masses 4 kg and 6 kg, are at rest in contact on a rough horizontal floor. The coefficient of friction between each box and the floor is 0.1. A horizontal force of magnitude P newtons is applied to A, pushing it against B as shown in Figure 1, which also shows how P varies with the time t seconds. The boxes are modelled as particles moving along the same straight line. Take g = 9.8 m s−2.

A4 kgB6 kgP01234567891011121305101520253035t / sP / N30 N12 N
(a)

Find the acceleration of the boxes for 0 < t < 4.

(2)
(b)

Find the magnitude of the force exerted by A on B for 0 < t < 4.

(2)
(c)

Find the speed of the boxes at t = 10.

(3)
(d)

For t > 10, explain why the force between the boxes is zero, and find the time taken for the boxes to come to rest after t = 10.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
both boxes: 30 − 0.1 × 10g = 10aM1
a = 2.02 m s−2A1
Part (b)
B: C − 0.1 × 6g = 6 × 2.02M1
C = 18.0 NA1
Part (c)
at t = 4: v = 2.02 × 4 = 8.08 m s−1B1
for 4 < t < 10: 12 − 9.8 = 10a, a = 0.22 m s−2M1
v = 8.08 + 0.22 × 6 = 9.40 m s−1A1
Part (d)
each box on its own would decelerate at μg = 0.98 m s−2, the same for both, so neither needs a push from the otherB1
t = 9.40 ÷ 0.98 = 9.59 sA1

Exam tip: For the force between two boxes in contact, apply Newton's second law to the box that is pushed (B here): only the contact force and its own friction act on it horizontally.

20IALM-M1-3-FB-10 M1 · ASMedium9 marks

Figure 1 shows a block of mass 4 kg on a rough plane inclined at 35° to the horizontal. The coefficient of friction between the block and the plane is 0.2. A horizontal force of magnitude 15 N acts on the block, towards the plane, in the vertical plane containing a line of greatest slope. The block is released from rest and slides down the plane. The block is modelled as a particle. Take g = 9.8 m s−2.

4 kg15 Nmotion35°rough plane, μ = 0.2
(a)

Find the normal reaction of the plane on the block.

(3)
(b)

Find the acceleration of the block.

(3)
(c)

The magnitude of the horizontal force is changed to H newtons, still acting in the same direction. The block now slides down the plane at constant speed. Find H.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
resolving perpendicular to the plane: R = 4g cos 35° + 15 sin 35°M1
both components correctA1
R = 40.7 NA1
Part (b)
resolving down the plane: 4g sin 35° − 15 cos 35° − 0.2 × 40.71 = 4aM1
all terms correctA1
a = 0.514 m s−2A1
Part (c)
4g sin 35° − H cos 35° − 0.2(4g cos 35° + H sin 35°) = 0M1
correct equation, with R depending on HA1
H = 17.2A1

Common error: Using R = 4g cos 35° only. The horizontal force has a component into the plane, 15 sin 35°, which increases the normal reaction and so the friction.

21IALM-M1-3-B-11 M1 · ASHard14 marks

A particle A of mass 3 kg is held at rest on a rough plane inclined at angle α to the horizontal, where tan α = ¾. The coefficient of friction between A and the plane is 1⁄6. A is attached to one end of a light inextensible string, which lies along a line of greatest slope of the plane and passes over a small smooth pulley fixed at the top of the plane. The other end of the string is attached to a particle B of mass 4 kg, which hangs freely 2 m above horizontal ground. A is released and the system moves with the string taut. A does not reach the pulley. Take g = 9.8 m s−2.

αA3 kgB4 kg2 m
Figure: particle A on the rough plane, connected over the pulley to B; tan α = 3/4; not to scale.
(a)

Find the normal reaction on A and the frictional force on A while it moves.

(2)
(b)

Find the acceleration of B and the tension in the string.

(4)
(c)

Find the magnitude of the force exerted by the string on the pulley.

(3)
(d)

Find the speed of B as it hits the ground.

(2)
(e)

B does not rebound. Find the further distance A moves up the plane before it first comes to rest.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
R = 3g cos α = 3 × 9.8 × 0.8 = 23.52 NM1
F = 1⁄6 × 23.52 = 3.92 N, with R = 23.5 NA1
Part (b)
B: 4g − T = 4aM1
A: T − 3g sin α − 3.92 = 3aM1
a = (39.2 − 17.64 − 3.92) ÷ 7 = 2.52 m s−2A1
T = 4(9.8 − 2.52) = 29.1 N (29.12)A1
Part (c)
the angle between the two parts of the string is 90° − α = 53.1°B1
magnitude = 2T cos(½(90° − α)) = 2 × 29.12 × cos 26.57°M1
= 52.1 NA1
Part (d)
v2 = 2 × 2.52 × 2 = 10.08M1
v = 3.17 m s−1A1
Part (e)
string slack, so for A: deceleration = g sin α + 1⁄6g cos α = 7.187 m s−2M1
0 = 10.08 − 2 × 7.187 × sM1
s = 0.701 mA1

Exam tip: After B lands, the tension is zero. A is then slowed by both its weight component and friction, which acts down the plane because A is still moving up it.

22IALM-M1-3-B-12 M1 · ASHard12 marks

A hospital lift of mass 800 kg carries one passenger of mass 65 kg. The lift starts from rest and moves vertically upwards. It accelerates uniformly for 3 s until its speed is 2.4 m s−1, travels at 2.4 m s−1 for 8 s, then decelerates uniformly, coming to rest after a further 2 s. The lift is raised by a vertical cable. Take g = 9.8 m s−2.

311132.4t (s)v (m s⁻¹)
Figure: velocity–time graph of the lift, drawn from the information given; not to scale.
(a)

Find the tension in the cable during the first 3 s.

(3)
(b)

Find the force exerted on the passenger by the floor of the lift during the last 2 s.

(3)
(c)

Find the total distance travelled by the lift.

(2)
(d)

The cable can safely carry a tension of at most 9500 N. With the same passenger, find the greatest possible upward acceleration of the lift.

(2)
(e)

Find the least time in which the lift could reach 2.4 m s−1 from rest without exceeding this tension.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
acceleration = 2.4 ÷ 3 = 0.8 m s−2B1
lift and passenger: T − 865g = 865 × 0.8M1
T = 865 × 10.6 = 9169 = 9170 N (3 s.f.)A1
Part (b)
deceleration = 2.4 ÷ 2 = 1.2 m s−2B1
passenger: R − 65g = 65 × (−1.2)M1
R = 559 NA1
Part (c)
area under the velocity–time graph: ½ × 3 × 2.4 + 8 × 2.4 + ½ × 2 × 2.4M1
= 25.2 mA1
Part (d)
9500 − 865g = 865aM1
a = 1.18 m s−2A1
Part (e)
t = 2.4 ÷ 1.1827M1
= 2.03 sA1

Common error: Using the passenger's mass alone for the cable tension. The cable supports and accelerates the lift and everything in it, 865 kg in total.

23IALM-M1-3-B-13 M1 · ASHard12 marks

A box of mass 12 kg is pulled up a line of greatest slope of a rough ramp inclined at 15° to the horizontal. The pulling rope is light and makes an angle of 25° with the ramp, in the vertical plane containing the line of greatest slope, above the ramp. The tension in the rope is 90 N and the coefficient of friction between the box and the ramp is 0.3. The box starts from rest. After 4 s the rope is let go. The box is modelled as a particle. Take g = 9.8 m s−2.

15°12 kg90 N25°
Figure: the box on the rough ramp, pulled by the rope at 25° to the ramp; not to scale.
(a)

Find the normal reaction of the ramp on the box while the rope is pulling.

(3)
(b)

Find the acceleration of the box while the rope is pulling.

(3)
(c)

Find the speed of the box and the distance it has moved up the ramp when the rope is let go.

(2)
(d)

Find the further distance the box moves up the ramp after the rope is let go.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
resolving perpendicular to the ramp: R + 90 sin 25° = 12g cos 15°M1
both components correctA1
R = 75.6 NA1
Part (b)
resolving parallel to the ramp: 90 cos 25° − 12g sin 15° − 0.3 × 75.56 = 12aM1
all terms correctA1
a = 2.37 m s−2A1
Part (c)
v = 2.372 × 4 = 9.49 m s−1 (ft)B1
s = ½ × 2.372 × 42 = 19.0 m (ft)B1
Part (d)
the rope no longer lifts the box: R = 12g cos 15° = 113.6 NB1
−(12g sin 15° + 0.3 × 113.6) = 12aM1
a = −5.38 m s−2A1
0 = 9.4882 − 2 × 5.376 × s ⇒ s = 8.37 mA1

Common error: Keeping the old normal reaction after the rope is let go. Without the rope's upward component, R increases, and so does the friction.

24IALM-M1-3-B-14 M1 · ASHard12 marks

A fixed wedge has two plane faces that meet at a horizontal ridge. Face X is smooth and inclined at 50° to the horizontal; face Y is rough and inclined at 30° to the horizontal. A small smooth pulley is fixed at the ridge. Particle P, of mass 5 kg, rests on face X and particle Q, of mass 2 kg, rests on face Y. They are joined by a light inextensible string that passes over the pulley, each part of the string lying along a line of greatest slope of its face. The coefficient of friction between Q and face Y is 0.25. The particles are released from rest with the string taut; P moves down face X and Q moves up face Y. Neither particle reaches the pulley or the bottom of its face during the motion described. Take g = 9.8 m s−2.

50°30°PQ5 kg2 kgXY
Figure: the wedge with smooth face X and rough face Y, and the particles joined over the pulley; not to scale.
(a)

Find the acceleration of the particles.

(4)
(b)

Find the tension in the string.

(2)
(c)

After 1.5 s the string breaks. Find the speed of Q at that instant and the further distance Q moves up face Y before it comes to rest.

(4)
(d)

Find the acceleration of P after the string breaks.

(1)
(e)

State where in your working you used the fact that the pulley is smooth.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
for Q: R = 2g cos 30°, so friction = 0.25 × 16.97 = 4.24 NB1
P: 5g sin 50° − T = 5aM1
Q: T − 2g sin 30° − 4.24 = 2aM1
a = (37.54 − 9.8 − 4.24) ÷ 7 = 3.36 m s−2A1
Part (b)
T = 2 × 3.356 + 9.8 + 4.24 (or from P's equation)M1
= 20.8 NA1
Part (c)
v = 3.356 × 1.5 = 5.03 m s−1 (ft)B1
deceleration of Q = g sin 30° + 0.25g cos 30°M1
= 7.02 m s−2A1
0 = 5.0342 − 2 × 7.022 × s ⇒ s = 1.80 mA1
Part (d)
g sin 50° = 7.51 m s−2 down face XB1
Part (e)
the tension was taken to be the same in both parts of the stringB1

Exam tip: On a two-faced wedge, resolve along each face separately: each particle has its own slope angle, its own normal reaction and (if rough) its own friction.

25IALM-M1-3-FB-11 M1 · ASHard12 marks

Figure 1 shows two particles A and B on a rough plane inclined at 30° to the horizontal. A has mass 2 kg and the coefficient of friction between A and the plane is 0.3. B has mass 3 kg and the coefficient of friction between B and the plane is 0.1. A and B lie on the same line of greatest slope, with A higher than B, and are joined by a light rod parallel to the slope. The particles are released from rest and slide down the plane. Take g = 9.8 m s−2.

B3 kg, μ = 0.1A2 kg, μ = 0.3light rod30°
(a)

Find the frictional force on each particle.

(2)
(b)

Find the acceleration of the particles.

(3)
(c)

Find the magnitude of the force in the rod and state whether the rod is in tension or in thrust.

(3)
(d)

Explain why the motion would be the same if the rod were replaced by a light inextensible string.

(1)
(e)

The rod is removed. The particles are placed on the same line of greatest slope with B now 1.5 m higher up the plane than A, and both are released from rest at the same instant. The plane is long enough for B to reach A. Find the time taken for B to reach A.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
A: 0.3 × 2g cos 30° = 5.09 NB1
B: 0.1 × 3g cos 30° = 2.55 NB1
Part (b)
whole system down the plane: 5g sin 30° − 5.092 − 2.546 = 5aM1
all terms correctA1
a = 3.37 m s−2A1
Part (c)
B, with T the force from the rod up the plane: 3g sin 30° − 2.546 − T = 3 × 3.372M1
T = 2.04 NA1
T > 0, so the rod pulls B back: tensionA1
Part (d)
the rod is in tension, and a string can supply a tension, so the string stays taut and the equations are unchangedB1
Part (e)
separately: aA = g sin 30° − 0.3g cos 30° = 2.354 and aB = g sin 30° − 0.1g cos 30° = 4.051 m s−2M1
½ × 4.051t2 = 1.5 + ½ × 2.354t2M1
t = 1.33 sA1

Exam tip: To decide between tension and thrust, compare how the particles would move on their own. The lower particle, B, would accelerate faster, so the rod has to hold it back: tension.

26IALM-M1-3-FB-12 M1 · ASHard12 marks

A block A of mass 4 kg lies on a rough horizontal table. It is attached to one end of a light inextensible string that 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, which hangs freely at a height h metres above the floor, as shown in Figure 1. The string between A and the pulley is horizontal. The system is released from rest with the string taut. B hits the floor and does not rebound; A does not reach the pulley. Figure 1 also shows the velocity–time graph of A from release until A comes to rest. A and B are modelled as particles. Take g = 9.8 m s−2.

A4 kgBhsmooth pulley, rough table00.511.522.5300.40.81.21.622.42.8t / sv / m s⁻¹(1.5, 2.4)(2.5, 0)
(a)

Find h.

(2)
(b)

Find the coefficient of friction between A and the table.

(3)
(c)

Find the tension in the string while B is falling.

(2)
(d)

Find the mass of B.

(3)
(e)

Find the least possible initial distance between A and the pulley.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
B hits the floor at t = 1.5; h = area = ½ × 1.5 × 2.4M1
h = 1.8A1
Part (b)
after B lands, deceleration of A = 2.4 ÷ 1 = 2.4 m s−2B1
string slack, so friction alone: μ × 4g = 4 × 2.4M1
μ = 0.245A1
Part (c)
A: T − 0.2449 × 4g = 4 × 1.6, where friction = 9.6 NM1
T = 16.0 NA1
Part (d)
acceleration = 2.4 ÷ 1.5 = 1.6 m s−2B1
B: mg − 16 = 1.6mM1
m = 16 ÷ 8.2 = 1.95 kgA1
Part (e)
total distance moved by A = area of the whole triangle = ½ × 2.5 × 2.4M1
= 3 mA1

Exam tip: Each line of a velocity–time graph is one stage of the motion. Use the stage with the fewest forces first: once B has landed, only friction acts on A horizontally, which gives μ straight away.

27IALM-M1-3-FB-13 M1 · ASHard10 marks

A particle P of mass 2.5 kg moves on a smooth horizontal plane. The only horizontal force on P is a constant force F newtons. Figure 1 shows the velocity of P at time t = 0 and at time t = 4 s, drawn on a grid in which one square represents 1 m s−1. The unit vectors i and j are horizontal and perpendicular, as shown.

ijOvelocity at t = 0velocity at t = 4 sone grid square = 1 m s⁻¹
(a)

Write down the velocity of P at t = 0 and at t = 4 in the form (xi + yj) m s−1.

(1)
(b)

Find F.

(3)
(c)

Find the time at which P is moving parallel to j, and the speed of P at that time.

(3)
(d)

Find the least speed of P during the motion.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
(6i − 4j) and (−2i + 4j) (both)B1
Part (b)
a = ((−2i + 4j) − (6i − 4j)) ÷ 4M1
= (−2i + 2j) m s−2A1
F = 2.5a = (−5i + 5j)A1
Part (c)
v = (6 − 2t)i + (−4 + 2t)jM1
i-component zero: t = 3 sA1
v = 2j, speed 2 m s−1A1
Part (d)
|v|2 = (6 − 2t)2 + (2t − 4)2 = 8t2 − 40t + 52M1
= 8(t − 2.5)2 + 2, least when t = 2.5M1
least speed = √2 = 1.41 m s−1A1

Common error: Setting the j-component to zero for “parallel to j”. A velocity parallel to j has no i-component, so it is the i-component that must vanish.

28IALM-M1-3-FB-14 M1 · ASHard13 marks

A car tows a trailer down a straight road inclined at an angle α to the horizontal, where sin α = 1⁄20. The tow bar is light, rigid and parallel to the road. Figure 1 shows the masses of the car and the trailer and the constant resistances to motion acting on them. The car's engine provides no driving force, and a braking force of magnitude B newtons acts on the car only, up the road. The car and trailer are modelled as particles. Take g = 9.8 m s−2.

αtrailer600 kg100 Ncar1200 kg200 Ntow barbraking force Bmotionsin α = 1/20 · diagram not to scale
(a)

The car and trailer move down the road at constant speed. Find B, and find the force in the tow bar, stating whether it is a tension or a thrust.

(4)
(b)

The braking force is increased to 2400 N. Find the deceleration of the car and trailer and the magnitude of the force in the tow bar.

(4)
(c)

The braking force of 2400 N is first applied when the speed is 15 m s−1. Find the distance travelled before the car and trailer stop.

(2)
(d)

The tow bar can withstand a thrust of at most 1000 N. Find the greatest braking force that can be applied to the car.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
whole system: 1800g sin α − 200 − 100 − B = 0M1
B = 882 − 300 = 582A1
trailer, with X the force from the tow bar up the road: 600g sin α − 100 − X = 0M1
X = 294 − 100 = 194 N, pushing the trailer back: a thrustA1
Part (b)
whole system: 2400 + 300 − 882 = 1800dM1
d = 1.01 m s−2A1
trailer: X + 100 − 294 = 600 × 1.01M1
X = 800 N (thrust)A1
Part (c)
0 = 152 − 2 × 1.01 × sM1
s = 111 m (111.4)A1
Part (d)
trailer: 1000 + 100 − 294 = 600d, d = 1.343 m s−2M1
whole system: B + 300 − 882 = 1800 × 1.343M1
B = 3000A1

Exam tip: When only the car brakes, the trailer would run on into the car, so the tow bar pushes it back: a thrust. Take the trailer on its own to find it.