| Syllabus | Alternate Universe Cambridge 0625 |
| Subject | Physics (0625) Core |
| Paper | 0625/31/M/J/25 · Paper 3, Variant 1 |
| Also listed as | 0625_s25_qp_31 · 0625_s25_ms_31 |
| Session | May/June 2025 |
| Questions | 11, multiple choice, with answers and explanations |
| Cost | Free |
0625/31/M/J/25 answer key
Preview the first questions
Question 1
A car decelerates uniformly from $20\ \text{m/s}$ to $0\ \text{m/s}$ over a time interval of $15\ \text{s}$. What is the distance the car travels during this $15\ \text{s}$ interval?
- $150\ \text{m}$
- $300\ \text{m}$
- $75\ \text{m}$
- $35\ \text{m}$
Show the answer and explanation
Correct answer: A
Step 1: On a speed-time graph, the distance travelled is the area between the line and the time axis. Step 2: For uniform deceleration from $20\ \text{m/s}$ to $0\ \text{m/s}$ over $15\ \text{s}$, the area is a triangle. Step 3: Area $= \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 15 \times 20 = 150\ \text{m}$.
Question 2
A pendulum is timed over $15$ complete oscillations and takes a total time of $11.7\ \text{s}$. What is the average time for one complete oscillation?
- $0.78\ \text{s}$
- $1.28\ \text{s}$
- $175.5\ \text{s}$
- $11.7\ \text{s}$
Show the answer and explanation
Correct answer: A
Step 1: Average period $T = \frac{\text{total time}}{\text{number of oscillations}}$. Step 2: $T = 11.7 / 15 = 0.78\ \text{s}$.
Question 3
A horizontal beam balances on a central pivot. Child A sits at a distance of $1.6\ \text{m}$ from the pivot. Child B has a weight of $360\ \text{N}$ and sits at a distance of $1.2\ \text{m}$ from the pivot on the other side. The see-saw is balanced. What is the weight $W$ of child A?
- $270\ \text{N}$
- $480\ \text{N}$
- $360\ \text{N}$
- $240\ \text{N}$
Show the answer and explanation
Correct answer: A
Step 1: Principle of moments: at balance, total clockwise moment = total anticlockwise moment about the pivot. Step 2: $W \times 1.6 = 360 \times 1.2$. Step 3: $W = \frac{360 \times 1.2}{1.6} = \frac{432}{1.6} = 270\ \text{N}$.
Worked answers: questions 4 to 11
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Question 4 · Answer A
Step 1: Work done $W = F \times d$, where $F$ is the force and $d$ is the distance in the direction of the force. Step 2: Here $F$ is the weight ($15\ \text{N}$) and $d$ is the vertical height ($0.80\ \text{m}$). Step 3: $W = 15 \times 0.80 = 12\ \text{J}$.
Question 5 · Answer A
Step 1: Dull dark surfaces are good absorbers of infrared radiation. Step 2: Shiny light surfaces reflect most infrared radiation away. Step 3: With the same Bunsen burner energy reaching each container, the dull black surface absorbs more energy and heats the air inside it more than the shiny white one does.
Question 6 · Answer A
Step 1: Light speeds up when going from a denser medium (glass) to a less dense medium (air). Step 2: The rule is: light bends away from the normal when entering a less dense medium. Step 3: Total internal reflection happens only above the critical angle; for ordinary angles the ray refracts and emerges into the air.
Question 7 · Answer A
Step 1: Wave equation $v = f\lambda$, so $f = v/\lambda$. Step 2: $f = \frac{3.0 \times 10^8}{5.8 \times 10^{-7}} = 5.17 \times 10^{14}\ \text{Hz}$. Step 3: To $2$ significant figures: $f \approx 5.2 \times 10^{14}\ \text{Hz}$.
Question 8 · Answer A
Step 1: For resistors in series, the combined resistance is the sum: $R = R_1 + R_2$. Step 2: $R = 16 + 8.0 = 24\ \Omega$.
Question 9 · Answer A
Step 1: Turns ratio: $\frac{V_p}{V_s} = \frac{N_p}{N_s}$. Step 2: Rearrange: $N_s = N_p \times \frac{V_s}{V_p}$. Step 3: $N_s = 3600 \times \frac{12}{240} = 3600 \times 0.05 = 180$ turns.
Question 10 · Answer A
Step 1: Number of half-lives in $42$ years $= 42 / 14 = 3$. Step 2: After each half-life the mass halves: $72 \to 36 \to 18 \to 9\ \text{mg}$.
Question 11 · Answer A
Step 1: Although the word year suggests a time, the unit light-year is actually a distance. Step 2: It is defined as the distance light travels in one year through the vacuum of space, about $9.5 \times 10^{15}\ \text{m}$.