Topic 3 Practice Paper: Momentum — mark scheme (print)

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QQ/IGCSE-PHYS/T3

IGCSE · Cambridge IGCSE Physics (Extended (Core + Supplement))

Mark scheme

Momentum

Topic 3 Practice Paper · Maximum mark: 30

Notes for markers

Award one mark for each point marked [1], up to the marks shown for the part.
Accept answers worded differently that show the same understanding.
Units are required where the answer line shows one. ecf: allow an error carried forward.

Section A: Multiple-choice questions[10 marks]

Question	Answer and marking points	Marks
	Total for Question 1	1
	Total for Question 2	1
	Total for Question 3	1
	Total for Question 4	1
	Total for Question 5	1
	Total for Question 6	1
	Total for Question 7	1
	Total for Question 8	1
	Total for Question 9	1
	Total for Question 10	1

Section B: Structured questions[20 marks]

Question	Answer and marking points	Marks
11(a)	p = 18 × 1.5 = 27 kg m/s. substitution 18 × 1.5 [1] 27 kg m/s [1]	2
11(b)	No external horizontal force acts on the trolleys, so momentum is conserved. no external force → momentum conserved [1]	1
11(c)	Total mass 30 kg: v = 27 ÷ 30 = 0.90 m/s in the original direction. total mass 30 kg [1] 0.90 m/s [1]	2
11(d)	It decreases — some is transferred to heat and sound as the trolleys couple. kinetic energy decreases [1]	1
	Total for Question 11	6
12(a)	Impulse = FΔt = 400 × 0.0050 = 2.0 N s. use of FΔt [1] 2.0 N s [1]	2
12(b)	2.0 kg m/s, in the direction of the hit. [1]	1
12(c)	v = p/m = 2.0 ÷ 0.058 = 34 m/s (2 s.f.). division by 0.058 [1] 34 m/s [1]	2
12(d)	It halves (to about 17 m/s): the impulse equals the momentum gained, and half the momentum with the same mass means half the speed. speed halves with reason [1]	1
	Total for Question 12	6
13(a)	Δp = 0 − (750 × 20) = −15 000 kg m/s; magnitude 15 000 kg m/s. 750 × 20 = 15 000 seen [1] change of 15 000 kg m/s stated [1]	2
13(b)	F = Δp/Δt = 15 000 ÷ 0.25 = 60 000 N, opposite to the trolley’s motion. use of F = Δp/Δt [1] 60 000 N [1]	2
13(c)	30 000 N — double the time gives half the force for the same momentum change. [1]	1
	Total for Question 13	5
14(a)	It was zero before the throw (both at rest) and no external horizontal force acts, so conservation of momentum keeps the total at zero. zero before and momentum conserved [1]	1
14(b)	55v + (4.0 × 3.0) = 0, so v = −12 ÷ 55 = −0.22 m/s: 0.22 m/s in the opposite direction to the bag. equation 55v + 12 = 0 [1] 0.22 m/s opposite to the bag [1]	2
	Total for Question 14	3

— End of mark scheme —