Topic 2 Practice Paper: Forces — mark scheme (print)

Back to Topic 2

Question paper PDF

QuizQuipPractice Series

QQ/IGCSE-PHYS/T2

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

Mark scheme

Forces

Topic 2 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
1	CConstant velocity means no acceleration, so the driving force and drag cancel and the resultant is zero.	1
2	DRearranging F = ma gives F = 6.0 × 5.0 = 30 N, in the direction of the acceleration.	1
3	DConvert 16 mm to 0.016 m, then k = 4.0 ÷ 0.016 = 250 N/m.	1
4	CPast the limit of proportionality the load–extension graph curves: the simple proportionality has ended.	1
5	BDrag is the friction-like force a fluid (gas or liquid) exerts on an object moving through it — air resistance is drag.	1
6	BMoment = force × perpendicular distance = 8.0 × 0.50 = 4.0 N m.	1
7	A25 × 1.6 = 40 × d, so d = 40 ÷ 40 = 1.0 m.	1
8	CEquilibrium needs both a zero resultant force and a zero resultant moment; the body can still be moving at constant velocity.	1
9	BA freely suspended object rests with its centre of gravity directly below the point of support, so there is no turning moment.	1
10	AA uniform rule has its mass spread evenly, so its weight acts at its centre — the 50 cm mark — which is exactly where the rule balances.	1

Section B: Structured questions[20 marks]

Question	Answer and marking points	Marks
11(a)	Extensions: 6.0 mm at 2.0 N, 12 mm at 4.0 N, 18 mm at 6.0 N (length − 84 mm). subtraction of 84 mm used [1] all three extensions correct [1]	2
11(b)	Extension is proportional to load — doubling the load from 2.0 N to 4.0 N doubles the extension from 6.0 mm to 12 mm. extension proportional to load [1]	1
11(c)	k = F/x = 2.0 ÷ 0.0060 = 330 N/m (2 s.f.). use of k = F/x with extension in metres [1] 330 N/m [1]	2
11(d)	limit of proportionality [1]	1
	Total for Question 11	6
12(a)	50 − 14 = 36 N forwards. [1]	1
12(b)	a = F/m = 36 ÷ 18 = 2.0 m/s². use of a = F/m [1] 2.0 m/s² [1]	2
12(c)	The only horizontal force is now 14 N backwards, so the sledge decelerates at 14 ÷ 18 ≈ 0.78 m/s² and eventually stops. a resultant force now acts backwards [1] sledge slows / decelerates and stops [1]	2
12(d)	It produces heating (warms the surfaces / melts a thin layer of snow). heating [1]	1
	Total for Question 12	6
13(a)	The 5.0 N weight is 20 cm from the pivot; the unknown weight is 25 cm from the pivot. 20 cm and 25 cm [1]	1
13(b)	5.0 × 20 = W × 25, so W = 100 ÷ 25 = 4.0 N. moments equated 5.0 × 20 = W × 25 [1] 4.0 N [1]	2
13(c)	The total anticlockwise moment grows (5.0 N acts 10 cm from the pivot), so the anticlockwise moment now exceeds the clockwise moment and the rule rotates anticlockwise. extra weight adds to the anticlockwise moment [1] rule tips anticlockwise / towards the 30–40 cm side [1]	2
	Total for Question 13	5
14(a)	The centre of gravity rises (moves up the stem). [1]	1
14(b)	With a higher centre of gravity, a smaller tilt is enough for the vertical line through the centre of gravity to fall outside the base; once it does, the weight creates a toppling moment and the lamp falls. line of weight leaves the base after a smaller tilt [1] weight then produces a toppling moment [1]	2
	Total for Question 14	3

— End of mark scheme —