Newton’s laws of motion: A Level Mathematics questions with answers
Free Newton’s laws of motion questions for A Level Mathematics (9709), part of Mechanics. Each question has the answer and the marking points an examiner would look for.
A car of mass 1500 kg tows a trailer of mass 600 kg along a straight horizontal road using a light rigid tow-bar. The resistances to motion are 500 N on the car and 200 N on the trailer.
(a)
The driving force of the car is 2500 N. Find the acceleration and the tension in the tow-bar.
[4]
(b)
The car is now braking with a braking force of 2000 N and no driving force. Find the force in the tow-bar and state whether it is a tension or a thrust.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
0.857 m s−2, 714 N
whole system equation M1 a = 0.857 A1 trailer equation M1 T = 714 A1
4
(b)
571 N, thrust
deceleration 1.29 B1 trailer equation M1 571 N thrust A1
3
Question 2Structured · 10 marksA* challenge
Particle A of mass 4 kg lies on a rough horizontal table. It is attached to a light inextensible string that passes over a smooth pulley at the edge of the table. Particle B of mass 2 kg hangs from the other end, 1 m above the floor. The coefficient of friction between A and the table is 0.25. The system is released from rest with the string taut. A does not reach the pulley.
(a)
Find the acceleration of the system and the tension in the string.
[5]
(b)
Find the speed of B when it hits the floor.
[2]
(c)
B does not rebound. Find the total distance moved by A before it comes to rest.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
1.67 m s−2, 16.7 N
friction = 0.25 × 40 = 10 B1 Newton’s second law for A M1 Newton’s second law for B M1 a = 1.67 A1 T = 16.7 A1
5
(b)
1.83 m s−1
v2 = 2as M1 1.83 A1
2
(c)
1.67 m
deceleration of A = 2.5 m s−2B1 extra distance = 0.667 m M1 total 1.67 A1
3
Question 3Structured · 6 marks
A block of mass 7 kg is released from rest on a rough plane inclined at 25° to the horizontal. The coefficient of friction is 0.28. Find the acceleration of the block and its speed after it has moved 3 m. [Use g = 10 m s−2]
[6]
Show answers and mark scheme
Part
Answer and marking points
Marks
a = 1.69 m s−2, v = 3.18 m s−1
R = 7g cos 25 B1 F = μR M1 Newton II along plane M1 a = 1.69 A1 v2 = 2as M1 3.18 A1
6
Question 4Structured · 5 marks
A block A of mass 4 kg lies on a rough horizontal table (coefficient of friction 0.3). It is connected by a light inextensible string over a smooth pulley at the edge to a particle B of mass 7 kg hanging freely. The system is released from rest. Find the acceleration and the tension in the string.
[5]
Show answers and mark scheme
Part
Answer and marking points
Marks
a = 5.27 m s−2; T = 33.1 N
friction = μ × 4g B1 Newton II for A M1 Newton II for B M1 a = 5.27 A1 T = 33.1 A1
5
Question 5Structured · 7 marksA* challenge
A car of mass 1500 kg tows a trailer of mass 500 kg along a straight horizontal road using a light rigid tow-bar. The resistances to motion are 300 N on the car and 100 N on the trailer.
(a)
The driving force of the car is 3600 N. Find the acceleration and the tension in the tow-bar.
[4]
(b)
The car is now braking with a braking force of 2500 N and no driving force. Find the force in the tow-bar and state whether it is a tension or a thrust.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
1.6 m s−2, 900 N
whole system equation M1 a = 1.6 A1 trailer equation M1 T = 900 A1
4
(b)
625 N, thrust
deceleration 1.45 B1 trailer equation M1 625 N thrust A1
3
Question 6Structured · 10 marksA* challenge
Particle A of mass 4 kg lies on a rough horizontal table. It is attached to a light inextensible string that passes over a smooth pulley at the edge of the table. Particle B of mass 2 kg hangs from the other end, 0.5 m above the floor. The coefficient of friction between A and the table is 0.2. The system is released from rest with the string taut. A does not reach the pulley.
(a)
Find the acceleration of the system and the tension in the string.
[5]
(b)
Find the speed of B when it hits the floor.
[2]
(c)
B does not rebound. Find the total distance moved by A before it comes to rest.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
2 m s−2, 16 N
friction = 0.2 × 40 = 8 B1 Newton’s second law for A M1 Newton’s second law for B M1 a = 2 A1 T = 16 A1
5
(b)
1.41 m s−1
v2 = 2as M1 1.41 A1
2
(c)
1 m
deceleration of A = 2 m s−2B1 extra distance = 0.5 m M1 total 1 A1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in