The questions are on the friction and inclined planes test. Take it under real conditions before reading the mark scheme.
Quick answer: Newton's second law states that the resultant force equals mass times acceleration, ΣF = ma. This mark scheme gives the model answers and the marks for the 8 questions of the Physics 11-12 test on friction and inclined planes, from Newton’s second law to normal reaction on a slope, with the equivalent answers to accept.
Marks are given for the correct equation, substitution and answer with units. Follow-through marks apply when an earlier error is carried into a later part correctly. The marking notes show where each mark is earned.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB A.2
Question 1 Newton’s second law (1 mark)
ΣF = ma, where ΣF is the resultant force, m the mass and a the acceleration.Marking: 1 mark for the equation with symbols defined.
Question 2 Friction and direction (2 marks)
Friction acts opposite to the sliding direction. Static friction adjusts to equal the applied push, up to a limit, so while the box is not moving it is only as large as needed.Marking: 1 mark for direction; 1 mark for static friction matching the push.
Question 3 Pushed crate (3 marks)
- f = 0.25 × 30 × 9.81 = 73.6 N.
- Resultant = 150 − 73.6 = 76.4 N; a = 76.4 ÷ 30 = 2.5 m/s².
Marking: (a) 1 mark; (b) 1 mark for the resultant, 1 mark for the acceleration.
Question 4 Block on a slope (4 marks)
- mg sin 35° = 58.86 × 0.5736 = 33.8 N.
- N = mg cos 35° = 58.86 × 0.8192 = 48.2 N.
- f = 0.30 × 48.2 = 14.5 N.
- a = (33.8 − 14.5) ÷ 6.0 = 3.2 m/s².
Marking: 1 mark for each part.
Question 5 Joined trolleys (3 marks)
- a = 15 ÷ (4.0 + 1.0) = 3.0 m/s².
- The string accelerates only the 1.0 kg trolley: T = 1.0 × 3.0 = 3.0 N.
- The string is taut and does not stretch, so the trolleys move together.
Marking: 1 mark for each part.
Question 6 Lift passenger (3 marks)
- mg − N = ma, so N = 60 × (9.81 − 2.0) = 469 N.
- N = mg = 60 × 9.81 = 589 N.
Marking: (a) 1 mark for the equation, 1 for the answer; (b) 1 mark.
Question 7 Static and kinetic (2 marks)
With locked wheels the tyres slide, so the smaller kinetic friction acts. With gradual braking the tyres roll without slipping, so the larger static friction up to μs N is available.Marking: 1 mark for kinetic friction on sliding tyres; 1 mark for the larger static friction.
Question 8 Normal reaction on a slope (2 marks)
The weight acts vertically, but the surface only has to balance the component perpendicular to the slope, mg cos θ. The component along the slope, mg sin θ, is not balanced by the normal reaction, so N = mg cos θ is less than mg.Marking: 1 mark for the perpendicular component; 1 mark for N = mg cos θ.
