The questions are on the circular motion test. Take it under real conditions before reading the mark scheme.
Quick answer: Circular motion at constant speed involves a constant angular velocity ω = 2π/T and a centripetal acceleration a = v²/r = ω²r directed towards the centre. This mark scheme gives the model answers and the marks for the 8 questions of the Physics 11-12 test on circular motion, from centripetal acceleration to scaling, with the equivalent answers to accept.
The model answers show full working and the final result in bold. Marks are given for the correct method as well as the answer, so show each step. Check whether you converted rpm to rad/s and whether the net force points to the centre.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.6
Question 1 Centripetal acceleration (2 marks)
The acceleration is directed towards the centre of the circle. The direction of the velocity changes continuously, and velocity is a vector, so there is an acceleration.Marking: 1 mark for direction; 1 mark for the change of velocity direction.
Question 2 Rotating disc (3 marks)
- ω = 480 × 2π ÷ 60 = 50.3 rad/s.
- v = r ω = 0.060 × 50.27 = 3.0 m/s.
- a = ω² r = 50.27² × 0.060 = 152 m/s².
Marking: 1 mark for each part.
Question 3 Work done (2 marks)
The force is always at right angles to the velocity, so the displacement along the force is zero. With no work done, the kinetic energy and speed remain constant.Marking: 1 mark for perpendicular to velocity; 1 mark for no work so constant kinetic energy.
Question 4 Car on a flat bend (3 marks)
- μ m g = m v² ÷ r, so v = √(μ g r) = √(0.60 × 9.81 × 50) = 17 m/s.
- Equal, because the mass cancels.
Marking: (a) 1 mark for the equation, 1 for the result; (b) 1 mark.
Question 5 Ball in a vertical circle (3 marks)
- T − m g = m v² ÷ r, so T = 0.40 × 9.81 + 0.40 × 36 ÷ 0.90 = 3.92 + 16.0 = 19.9 N.
- At the limit T = 0, so v = √(g r) = √(9.81 × 0.90) = 3.0 m/s.
Marking: (a) 1 mark for the equation, 1 for the result; (b) 1 mark.
Question 6 Banked road (3 marks)
- The horizontal component of the normal reaction, N sin θ.
- tan θ = v² ÷ (r g), so v = √(150 × 9.81 × tan 8.0°) = √(1471.5 × 0.1405) = 14 m/s.
Marking: (a) 1 mark; (b) 1 mark for the formula, 1 for the result.
Question 7 Mud from a tyre (2 marks)
The adhesion that gave the mud a centripetal force is no longer enough, so the mud gets no resultant force. By Newton’s first law it moves in a straight line along the tangent at the speed it had at that instant.Marking: 1 mark for insufficient centripetal force; 1 mark for tangent motion by Newton’s first law.
Question 8 Scaling (2 marks)
- F = m v² ÷ r, so doubling v gives four times the force.
- F is proportional to 1/r, so halving r gives twice the force.
Marking: 1 mark each.
