Mark-by-Mark · Cambridge IGCSE Physics 0625

0625/42 May/Jun 2025: where students lost marks

The 4 questions the June 2025 examiners' report flagged on this paper. Full question, mark scheme and video walkthrough, free. No sign-in needed.

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Chapters

Question 2(c)(i)0:32 Question 3(b)2:20 Question 5(b)(iii)4:09 Question 8(b)(i)(ii)5:34

Question 2(c)(i)

3 marks · Forces

The question

Where students lost marks

Candidates had to equate the sum of the clockwise moments with the anticlockwise moment, and the clockwise side included both the 0.12 N force and the weight of the ruler itself. Those with a correct equation usually went on to divide the weight by g. Stronger answers worked line by line and labelled every distance on the diagram. Weaker candidates worked backwards from the given mass, which earned no credit for multiplying the mass by g.

The trap: The ruler is uniform, so its weight acts at 50 cm, 8 cm right of the pivot: that is a clockwise moment too. And a show-that goes forwards: 0.081 × 9.8 earns nothing.

Source: Paper 0625/42 (Theory, Extended), Question 2(c)(i), June 2025 examiners' report (paraphrased)

Mark scheme, mark by mark

B1sum of clockwise moments = sum of anticlockwise moments

B1(mg × 8) + (0.12 × 38) = 0.34 × 32

B1mg = 6.32 ÷ 8 = 0.79 N, mass = 0.79 ÷ 9.8 = 0.081 kg

Teacher's note: Label the distances from the pivot first: 32 cm, 38 cm, and 8 cm for the ruler's own weight. Then the equation writes itself.

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Question 3(b)

4 marks · Momentum

The question

Where students lost marks

Many answered well, stating conservation of momentum and working logically. Some added the two initial momenta without taking account of the opposite directions; both velocity and momentum are vectors. Some did not combine the masses after the collision, or gave two separate velocities. The weakest simply added or subtracted velocities, and there were significant-figure and rounding errors.

The trap: Give each momentum a sign before adding: train B moves left, so its momentum is negative. The trains stick, so after the collision it is one mass, 0.66 kg, with one velocity.

Source: Paper 0625/42 (Theory, Extended), Question 3(b), June 2025 examiners' report (paraphrased)

Mark scheme, mark by mark

C1momentum before = (0.45 × 0.34) − (0.21 × 0.12) = 0.1278

C1momentum after = 0.66 × v

C1momentum before = momentum after

A1v = 0.19 m/s to the right

Teacher's note: Sense check: train A has more momentum (0.153 against 0.025), so the pair must carry on to the right.

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Question 5(b)(iii)

3 marks · Electromagnetic spectrum

The question

Where students lost marks

The equation v = fλ was well known by most candidates, and many rearranged it correctly. Fewer were able to convert GHz to Hz: the prefix 10⁹ was not well known.

The trap: Giga means × 10⁹. Use 2.48 on its own and you get a wavelength of about 120 000 000 m, for a phone signal.

Source: Paper 0625/42 (Theory, Extended), Question 5(b)(iii), June 2025 examiners' report (paraphrased)

Mark scheme, mark by mark

C1v = fλ, so λ = v ÷ f

C1λ = (3.0 × 10⁸) ÷ (2.48 × 10⁹)

A1λ = 0.12 m

Teacher's note: Learn the prefixes as powers of ten: kilo 10³, mega 10⁶, giga 10⁹. A Bluetooth wavelength of 12 cm is about the size of a phone, which is why the answer is believable.

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Question 8(b)(i)(ii)

3 marks · Electric circuits

The question

Where students lost marks

Two key points identified the current in resistor B: the p.d. is the same across each branch, and the resistance of the second branch is the sum of B and C. The strongest candidates used this to explain 1.2 A; some left out the unit. Common misconceptions were that the current was split between the resistors, or was the same in every branch. For the total current, only stronger candidates realised I is the sum of the currents in all the branches; a common error was multiplying the answer to (i) by 4.

The trap: Parallel branches share the voltage, not the current. B and C in series double the resistance, so half the current: 1.2 A. Then I is the branches added: 2.4 + 1.2 + 2.4.

Source: Paper 0625/42 (Theory, Extended), Question 8(b)(i)(ii), June 2025 examiners' report (paraphrased)

Mark scheme, mark by mark

B1current in B = 1.2 A

B1two identical resistors in series: resistance doubled (same p.d. across each branch)

B1I = 2.4 + 1.2 + 2.4 = 6.0 A

Teacher's note: D is identical to A and alone in its branch, so it also carries 2.4 A. That is the step the ×4 answers missed.

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