May/June 2025 Paper 31 Worked Answers (IGCSE Physics 0625 Core)

11 questions · 11 marks · 75 minutes

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1. Question 1

   1 marksDistance from a uniform deceleration
   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 $= \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 15 \times 20 = 150\ \text{m}$.
   Method:

   Identify the triangular area and apply the triangle formula.
   Examiner tips

   • Always compute the area, not the speed times time, when motion is non-uniform.

2. Question 2

   1 marksPeriod from many oscillations
   Step 1: Average period $T = \frac{\text{total time}}{\text{number of oscillations}}$. Step 2: $T = 11.7 / 15 = 0.78\ \text{s}$.
   Method:

   Divide the total time by the count of oscillations.
   Examiner tips

   • Time many oscillations to improve accuracy of the period.

3. Question 3

   1 marksBalanced see-saw and the principle of moments
   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: $W = \frac{360 \times 1.2}{1.6} = \frac{432}{1.6} = 270\ \text{N}$.
   Method:

   Apply the balance condition and solve for $W$.
   Examiner tips

   • Quote the principle of moments before rearranging.

4. Question 4

   1 marksWork done lifting a load
   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: $W = 15 \times 0.80 = 12\ \text{J}$.
   Method:

   Multiply weight by height.
   Examiner tips

   • Keep units consistent: N multiplied by m gives J.

5. Question 5

   1 marksAbsorption of infrared by surface colour
   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.
   Method:

   Link the surface property to absorption of infrared radiation.
   Examiner tips

   • Best emitters and best absorbers are the same: dull dark surfaces.

6. Question 6

   1 marksRefraction from glass into air
   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.
   Method:

   Apply the refraction rule for less dense to denser media.
   Examiner tips

   • Use the rule: enter denser medium = bend towards normal; enter less dense = bend away.

7. Question 7

   1 marksFrequency of yellow light
   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: $f \approx 5.2 \times 10^{14}\ \text{Hz}$.
   Method:

   Rearrange the wave equation and divide carefully.
   Examiner tips

   • Take care with the signs of the exponents on powers of ten.

8. Question 8

   1 marksCombined resistance in series
   Step 1: For resistors in series, the combined resistance is the sum: $R = R_1 + R_2$. Step 2: $R = 16 + 8.0 = 24\ \Omega$.
   Method:

   Add the two resistances.
   Examiner tips

   • Series resistors: simply add the resistances.

9. Question 9

   1 marksTransformer turns ratio
   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.
   Method:

   Apply the turns ratio.
   Examiner tips

   • Step-down transformer: secondary turns less than primary.

10. Question 10

    1 marksHalf-life of plutonium-241
    Step 1: Number of half-lives in $42$ years $= 42 / 14 = 3$. Step 2: After each half-life the mass halves: $72 \to 36 \to 18 \to 9\ \text{mg}$.
    Method:

    Count half-lives and apply repeated halving.
    Examiner tips

    • Each half-life halves the amount; do not subtract a fixed quantity each time.

11. Question 11

    1 marksMeaning of light-year
    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 $9.5 \times 10^{15}\ \text{m}$.
    Method:

    Recall the definition and pick the only option that gives a distance.
    Examiner tips

    • Watch out: light-year is a distance, despite its name.