May/June 2025 Paper 32 Worked Answers (IGCSE Physics 0625 Core)

11 questions · 11 marks · 75 minutes

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1. Question 1

   1 marksDistance from uniform acceleration
   Step 1: The distance is the area under the speed-time line. For uniform acceleration from rest, this is a right-angled triangle. Step 2: Area $= \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 4.0 \times 22 = 44\ \text{m}$.
   Method:

   Apply the triangle area formula to the speed-time graph.
   Examiner tips

   • Constant acceleration from rest: distance = $\tfrac{1}{2} v t$.

2. Question 2

   1 marksMass from weight
   Step 1: Weight $W = mg$, so mass $m = W/g$. Step 2: $m = 25\,000 / 9.8 \approx 2551\ \text{kg}$, which is $2600\ \text{kg}$ to 2 significant figures.
   Method:

   Divide weight by gravitational field strength.
   Examiner tips

   • Keep track of where the value of $g$ goes when rearranging.

3. Question 3

   1 marksPressure of a block on the ground
   Step 1: Pressure $p = F / A$. Step 2: Area $A = 0.54 \times 0.18 = 0.0972\ \text{m}^2$. Step 3: $p = 890 / 0.0972 \approx 9200\ \text{N/m}^2$.
   Method:

   Find the contact area, then divide weight by area.
   Examiner tips

   • Multiply two side lengths to get the contact area before dividing.

4. Question 4

   1 marksAbsolute zero
   Step 1: The lowest possible temperature is the point at which particles have no more kinetic energy to lose. Step 2: This is called absolute zero and equals $-273\ ^\circ\text{C}$ (or $0\ \text{K}$).
   Method:

   Recall the name and Celsius value of absolute zero.
   Examiner tips

   • $0\ \text{K}$ and $-273\ ^\circ\text{C}$ are the same temperature.

5. Question 5

   1 marksFrequency from speed and wavelength
   Step 1: Wave equation: $v = f\lambda$, so $f = v/\lambda$. Step 2: $f = 18 / 1.2 = 15\ \text{Hz}$.
   Method:

   Divide speed by wavelength.
   Examiner tips

   • Use consistent units; here speed and wavelength are both in cm.

6. Question 6

   1 marksThermal radiation by surface colour
   Step 1: Dull dark surfaces are good emitters of infrared radiation. Shiny light surfaces are poor emitters. Step 2: The dull black surface radiates more energy per second than the shiny white surface, so the thermometer in front of the dull black surface (X) receives more energy and reaches a higher temperature.
   Method:

   Link surface property to emission of infrared radiation.
   Examiner tips

   • Dull dark = best emitter; shiny light = poor emitter.

7. Question 7

   1 marksDistance to the Moon by light pulse
   Step 1: The light travels there and back, so it covers twice the Earth-Moon distance. Step 2: Total path = $v \times t = 3.0 \times 10^8 \times 2.5 = 7.5 \times 10^8\ \text{m}$. Step 3: One-way distance $= 7.5 \times 10^8 / 2 = 3.75 \times 10^8 \approx 3.8 \times 10^8\ \text{m}$.
   Method:

   Compute total distance, then halve for one-way.
   Examiner tips

   • Always halve the total time (or distance) for there-and-back signals.

8. Question 8

   1 marksCurrent from power and voltage
   Step 1: Power: $P = VI$, so $I = P/V$. Step 2: $I = 2000 / 230 \approx 8.7\ \text{A}$.
   Method:

   Rearrange $P = VI$ for the current.
   Examiner tips

   • Always start from $P = VI$ for mains appliances.

9. Question 9

   1 marksPotential difference across a series resistor
   Step 1: Ohm's law gives the p.d. across a resistor: $V = I R$. Step 2: $V = 0.20 \times 20 = 4.0\ \text{V}$.
   Method:

   Apply $V = IR$ to the resistor in question.
   Examiner tips

   • Series circuit: same current through all components.

10. Question 10

    1 marksHalf-life from mass decay
    Step 1: Ratio of masses: $60 / 7.5 = 8 = 2^3$, so $3$ half-lives have passed. Step 2: Half-life $= 11.5 / 3 \approx 3.83 \approx 3.8\ \text{days}$.
    Method:

    Count halvings, then divide total time by that count.
    Examiner tips

    • Each halving is one half-life; count them by ratio.

11. Question 11

    1 marksMain elements in the Sun
    Step 1: The Sun's mass is roughly $\,74\%$ hydrogen and $\,24\%$ helium, with only traces of heavier elements. Step 2: Energy comes from hydrogen fusion: hydrogen nuclei fuse to form helium nuclei.
    Method:

    Recall the composition of a main-sequence star.
    Examiner tips

    • Stars start with hydrogen and fuse it into helium.