Topic 1 Practice Paper: Motion — mark scheme (print)

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QQ/IGCSE-PHYS/T1

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

Mark scheme

Motion

Topic 1 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)	The student’s reaction time is a fixed delay of about 0.2 s however many swings are timed. Spreading it over 20 oscillations makes its effect on the calculated period 20 times smaller. reaction time is a fixed error (about 0.2 s) [1] effect on the calculated period is 20 times smaller / percentage error reduced [1]	2
11(b)	T = 34.0 ÷ 20 = 1.7 s. T = 1.7 s [1]	1
11(c)	Average the results: (34.0 + 34.4) ÷ 2 = 34.2 s, so T = 34.2 ÷ 20 = 1.71 s. take the mean of the total times [1] period = 1.71 s [1]	2
	Total for Question 11	5
12(a)	a = (8.0 − 0) ÷ 4.0 = 2.0 m/s². substitution (8.0 − 0) ÷ 4.0 [1] 2.0 m/s² [1]	2
12(b)	a = (0 − 8.0) ÷ 2.0 = −4.0 m/s², so she decelerates at 4.0 m/s². substitution (0 − 8.0) ÷ 2.0 [1] 4.0 m/s² deceleration [1]	2
12(c)	Area under the speed–time graph: triangle ½ × 4.0 × 8.0 = 16 m; rectangle 20 × 8.0 = 160 m; triangle ½ × 2.0 × 8.0 = 8.0 m. Total = 16 + 160 + 8.0 = 184 m. area of first triangle = 16 m [1] area of rectangle = 160 m and final triangle = 8.0 m [1] total = 184 m [1]	3
	Total for Question 12	7
13(a)	190 − 100 = 90 cm³. [1]	1
13(b)	ρ = m/V = 234 ÷ 90 = 2.6 g/cm³. substitution 234 ÷ 90 [1] 2.6 g/cm³ [1]	2
13(c)	The density stays the same. Density is a property of the material, and both the mass and the volume of the remaining glass have fallen by the same proportion. density unchanged [1] mass and volume reduce in the same proportion / it is the same material [1]	2
	Total for Question 13	5
14(a)	Volume = 4.0 × 4.0 × 4.0 = 64 cm³; ρ = 38 ÷ 64 = 0.59 g/cm³. volume = 64 cm³ [1] ρ = 0.59 g/cm³ [1]	2
14(b)	It floats, because 0.59 g/cm³ is less than the density of water, 1.0 g/cm³. floats because less dense than water [1]	1
	Total for Question 14	3

— End of mark scheme —