A Level · Physics 9702 · Fields & motion

Motion in a circle: A Level Physics questions with answers

Free Motion in a circle questions for A Level Physics (9702), part of Fields & motion. Each question has the answer and the marking points an examiner would look for.

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Motion in a circle practice questions

Question 1Structured · 4 marksA* challenge

A car travels over the top of a hump-backed bridge of radius 30 m.

(a)

Explain why the car may lose contact with the road at high speed.

[2]
(b)

Calculate the maximum speed at which the car stays in contact.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)normal force falls to zero
weight − normal force provides centripetal force B1
at high speed required centripetal force exceeds weight, so normal force would be negative B1
2
(b)17.2 m s−1
mg = mv2/r C1
17.2 m s−1 A1
2
Question 2Structured · 8 marksA* challenge

A ball of mass 0.2 kg on a string of length 0.6 m is whirled in a vertical circle. At the lowest point its speed is 6.0 m s−1.

(a)

Calculate the speed at the highest point.

[2]
(b)

Calculate the tension in the string at the highest point and at the lowest point.

[4]
(c)

Calculate the minimum speed at the lowest point for the string to stay taut at the top.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)3.53 m s−1
energy conservation over height 2r C1
3.53 A1
2
(b)2.19 N, 14 N
T + mg = mv2/r at top C1
2.19 A1
T − mg = mv2/r at bottom C1
14 A1
4
(c)5.42 m s−1
vtop2 = gr C1
5.42 A1
2
Question 3Structured · 3 marks

A mass of 1.2 kg moves in a horizontal circle of radius 0.6 m with a period of 2.8 s. Calculate the centripetal force.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
3.63 N
v = 2πr/T or ω = 2π/T C1
F = mv2/r C1
3.63 A1
3
Question 4Structured · 8 marks

A ball of mass 0.20 kg on a light string is whirled in a vertical circle of radius 0.50 m. At the lowest point its speed is 7.0 m s−1.

(a)

Define the radian.

[1]
(b)

Calculate the tension in the string at the lowest point.

[3]
(c)

Use conservation of energy to calculate the speed at the highest point.

[2]
(d)

Use your answer to (c) to calculate the tension at the highest point.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)angle subtended at centre by arc equal in length to radius
the angle subtended at the centre of a circle by an arc equal in length to the radius B1
1
(b)21.6 N
T − mg = mv2/r C1
T = 0.20 × 7.02/0.50 + 0.20 × 9.81 C1
21.6 N A1
3
(c)5.42 m s−1
½mvtop2 = ½mv2 − mg(2r) C1
5.42 m s−1 A1
2
(d)9.79 N
T + mg = mvtop2/r C1
9.79 N A1
2
Question 5Structured · 5 marks

A ball of mass 0.5 kg on a string moves in a horizontal circle of radius 0.6 m on a smooth table, making 2 revolutions per second.

(a)

Calculate the angular speed of the ball.

[1]
(b)

Calculate the tension in the string.

[2]
(c)

Explain why the ball accelerates even though its speed is constant.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)12.6 rad s−1
12.6 rad s−1 B1
1
(b)47.4 N
F = mrω2 C1
47.4 N A1
2
(c)direction changes; centripetal force
velocity changes direction continuously B1
resultant force towards centre gives centripetal acceleration B1
2
Question 6Structured · 9 marks

A car of mass 1200 kg travels round a horizontal bend that is an arc of a circle of radius 80 m.

(a)

Explain why the car is accelerating even if its speed is constant.

[2]
(b)

Calculate the centripetal force on the car at a speed of 10 m s−1.

[2]
(c)

State which force provides the centripetal force.

[1]
(d)

The maximum friction force is 0.6 times the weight of the car. Calculate the maximum speed at which the car can travel round the bend.

[3]
(e)

Explain why the maximum speed does not depend on the mass of the car.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)direction of velocity changes; acceleration towards centre
the direction of the velocity is changing (velocity is a vector) B1
so there is an acceleration (towards the centre of the circle) B1
2
(b)1500 N
F = mv2/r C1
1500 N A1
2
(c)friction between tyres and road
friction (sideways) between the tyres and the road B1
1
(d)21.7 m s−1
μmg = mv2/r C1
v = √(0.6 × 9.81 × 80) C1
21.7 m s−1 A1
3
(e)mass cancels
both friction and centripetal force are proportional to mass, so mass cancels B1
1

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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