OCR A Physics · Module 5: Newtonian world and astrophysics
Circular motion: practice questions with answers
6 exam-style questions on circular motion. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.
- Unit prefixes2 marks
A 1100 kg car travels at 15 m s⁻¹ round a circular bend of radius 45 m. Calculate the centripetal force needed, in kN.
F = kN
Show the answer and working
Answer F = 5.5 kN
F = mv²/r = 1100 × 15² ÷ 45 = 5500 N = 5.5 kN.
- Got 0.367 kN? You did not square the speed. F = mv² ÷ r = 1100 × 15² ÷ 45 = 5.5 kN.
- Got 11100 kN? You multiplied by the radius instead of dividing. F = mv² ÷ r = 5.5 kN.
- Radians vs degrees2 marks
A satellite orbits the Earth once every 92 minutes. Calculate its angular speed.
ω = rad s⁻¹
Show the answer and working
Answer ω = 0.00114 rad s⁻¹
T = 92 × 60 = 5520 s. ω = 2π ÷ T = 1.14 × 10⁻³ rad s⁻¹.
- Got 0.0683 rad s⁻¹? You used the period in minutes. T = 92 × 60 = 5520 s, so ω = 2π ÷ T = 1.14 × 10⁻³ rad s⁻¹.
- Got 0.0652 rad s⁻¹? You used 360 instead of 2π. Angular speed in rad s⁻¹ uses 2π radians per revolution: ω = 1.14 × 10⁻³ rad s⁻¹.
- Radians vs degrees3 marks
A 0.40 kg mass on a string moves in a horizontal circle of radius 0.80 m at 2.5 revolutions per second. Calculate the tension in the string.
T = N
Show the answer and working
Answer T = 79 N
ω = 2π × 2.5 = 15.7 rad s⁻¹. Tension = mω²r = 0.40 × 15.7² × 0.80 = 79 N.
- Got 2 N? You used the frequency as ω. Convert: ω = 2πf = 15.7 rad s⁻¹, so F = mω²r = 79 N.
- Got 5.03 N? You did not square ω. F = mω²r = 0.40 × 15.7² × 0.80 = 79 N.
- Rearranging equations3 marks
A car crosses a hump-backed bridge that is an arc of a circle of radius 40 m. Calculate the speed at which the car just loses contact with the road at the top.
v = m s⁻¹
Show the answer and working
Answer v = 19.8 m s⁻¹
Contact force is zero, so weight provides the centripetal force: mg = mv²/r, v = √(9.81 × 40) = 19.8 m s⁻¹.
- Got 392 m s⁻¹? You did not take the square root. At the top mg = mv²/r, so v = √(gr) = √392 = 19.8 m s⁻¹.
- Got 14 m s⁻¹? You included a factor of ½. Setting weight equal to centripetal force gives mg = mv²/r, so v = √(gr) = 19.8 m s⁻¹.
- Using the right equation1 mark
What is the direction of the acceleration of an object moving at constant speed in a circle?
- A Along the tangent, in the direction of motion.
- B Towards the centre of the circle.
- C Away from the centre of the circle.
Show the answer and working
Answer B: Towards the centre of the circle.
The velocity changes direction towards the centre, so a = v²/r points inwards.
- Why not A? The velocity is along the tangent; the acceleration is perpendicular to it as the velocity changes direction.
- Why not C? The resultant force and acceleration are directed inwards, not outwards.
- Using the right equation1 mark
A satellite moves in a circular orbit at constant speed. Which statement is correct?
- A No force acts on it.
- B Its kinetic energy increases steadily.
- C Gravity does no work on it.
Show the answer and working
Answer C: Gravity does no work on it.
The gravitational force is perpendicular to its velocity, so the work done is zero and speed is constant.
- Why not A? Gravity provides the centripetal force, so a force does act.
- Why not B? Constant speed means constant kinetic energy.