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Topic 4 Practice Paper: Energy, work and power — mark scheme

QuizQuipPractice Series
QQ/IGCSE-PHYS/T4

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

Mark scheme

Energy, work and power

Topic 4 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]

QuestionAnswer and marking pointsMarks
Total for Question 11
Total for Question 21
Total for Question 31
Total for Question 41
Total for Question 51
Total for Question 61
Total for Question 71
Total for Question 81
Total for Question 91
Total for Question 101

Section B: Structured questions[20 marks]

QuestionAnswer and marking pointsMarks
11(a)

Ek = ½mv² = ½ × 6.0 × 4.0² = 48 J.

  • use of ½mv² [1]
  • 48 J [1]
2
11(b)

New Ek = ½ × 6.0 × 3.0² = 27 J; transferred = 48 − 27 = 21 J.

  • 27 J found [1]
  • difference 21 J [1]
2
11(c)

The internal (thermal) store of the lane, air and ball — friction and drag warm them slightly. [1]

1
11(d)

The principle of conservation of energy. [1]

1
Total for Question 116
12(a)

W = Fd = 220 × 15 = 3300 J.

  • use of Fd [1]
  • 3300 J [1]
2
12(b)

Efficiency = (3300 ÷ 6000) × 100% = 55%.

  • ratio 3300/6000 [1]
  • 55% [1]
2
12(c)

P = W/t = 3300 ÷ 30 = 110 W.

  • division by 30 [1]
  • 110 W [1]
2
Total for Question 126
13(a)

Solar cells: no output at night and reduced output in cloudy weather, whereas diesel is always available and tides are predictable.

  • solar cells identified [1]
  • night/weather dependence explained [1]
2
13(b)

The tidal barrage: tides are driven mainly by the Moon’s gravity. (Diesel is ancient solar energy; solar cells use sunlight directly.)

  • tidal identified [1]
  • Moon’s gravity reason [1]
2
13(c)

It produces no CO₂ or polluting gases at generation (it is renewable and does not burn fuel).

  • no combustion emissions / renewable [1]
1
Total for Question 135
14(a)

ΔEp = mgΔh = 500 × 9.8 × 300 = 1 470 000 J. [1]

1
14(b)

Input = useful ÷ 0.40 = 1 470 000 ÷ 0.40 = 3 675 000 J.

  • division by 0.40 [1]
1
14(c)

Time = 300 s; P = 3 675 000 ÷ 300 = 12 250 W ≈ 12 kW. [1]

1
Total for Question 143

— End of mark scheme —