The questions are on the projectile motion test. Take it under real conditions before reading the mark scheme.
Quick answer: Projectile motion combines a constant horizontal velocity with a vertical motion of constant acceleration g = 9.81 m/s² downwards. This mark scheme gives the model answers and the marks for the 8 questions of the Physics 11-12 test on projectile motion, from horizontal launch to best angle, with the equivalent answers to accept.
Marks are given for correct method, 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: A Level 3.4
Question 1 Horizontal launch (2 marks)
Horizontal acceleration is zero. Vertical acceleration is 9.81 m/s² downwards.Marking: 1 mark for each correct value with direction.
Question 2 Rolling off a platform (3 marks)
- t = √(2h ÷ g) = √(3.6 ÷ 9.81) = 0.61 s.
- x = 4.0 × 0.606 = 2.4 m.
Marking: (a) 1 mark for the equation, 1 for the answer; (b) 1 mark.
Question 3 Components (2 marks)
Horizontal: 30 cos 42° = 22.3 m/s. Vertical: 30 sin 42° = 20.1 m/s.Marking: 1 mark for each component.
Question 4 Full trajectory (4 marks)
- uy = 24 sin 38° = 14.78 m/s; T = 2 × 14.78 ÷ 9.81 = 3.01 s.
- ux = 24 cos 38° = 18.91 m/s; R = 18.91 × 3.012 = 57.0 m.
- H = 14.78² ÷ 19.62 = 11.1 m.
Marking: (a) 1 mark for uy, 1 for T; (b) 1 mark; (c) 1 mark.
Question 5 Equal ranges (3 marks)
R = u² sin 2θ ÷ g. For 30°: sin 60° = 0.866. For 60°: sin 120° = 0.866. Both give R = 324 × 0.866 ÷ 9.81 = 28.6 m.Marking: 1 mark for using the range formula; 1 mark for showing both sines equal; 1 mark for the value.
Question 6 Air resistance (3 marks)
Any three: the range is shorter; the maximum height is lower; the descent is steeper than the climb; the landing speed is smaller than the launch speed; the best launch angle for range is below 45°.Marking: 1 mark for each valid effect, maximum 3.
Question 7 Velocity at the top (2 marks)
At the top the vertical velocity is zero, but the horizontal velocity is unchanged, so the speed equals the horizontal component, which is the least value of the speed.Marking: 1 mark for vertical component zero; 1 mark for the horizontal component constant and non-zero.
Question 8 Best angle (1 mark)
45°, because sin 2θ = 1.Marking: 1 mark.
