May/June 2025 Paper 43 Worked Answers (IGCSE Physics 0625 Extended)

11 questions · 80 marks · 75 minutes

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1. Question 1

   1 marksDistance from a speed-time graph
   Split the area under the speed-time graph into a rectangle and a triangle. Rectangle (constant speed 0 to 0.5 s): area = 20 x 0.5 = 10 m. Triangle (deceleration from 0.5 s to 4.0 s, base 3.5 s, height 20 m/s): area = (1/2) x 3.5 x 20 = 35 m. Total distance = 10 + 35 = 45 m.
   Method:

   Add the rectangle (10 m) and the triangle (35 m) to get 45 m.
   Examiner tips

   • Always split a piecewise speed-time graph into simple shapes whose areas you can add.

2. Question 2

   1 marksConservation of momentum (ejection)
   By conservation of momentum: total momentum before = total momentum after. 840 x 7800 = 720 v + 120 x 7850. So 6 552 000 = 720 v + 942 000, giving 720 v = 5 610 000 and v = 7790 m/s.
   Method:

   Apply conservation of momentum; v = (840 x 7800 - 120 x 7850) / 720 = 7790 m/s.
   Examiner tips

   • Apply momentum conservation in one dimension when an object separates into two pieces.

3. Question 3

   1 marksKinetic energy of an arrow
   Kinetic energy = (1/2) m v squared, so v = square root of (2 E / m) = square root of (2 x 110 / 0.030) = square root of 7333, which is about 86 m/s.
   Method:

   Apply v = square root of (2 x 110 / 0.030) = 86 m/s.
   Examiner tips

   • Do not forget the factor of 2 and the square root when finding speed from kinetic energy.

4. Question 4

   1 marksSpecific heat capacity from mixing
   Energy gained by the water = m_w x c_w x change in temperature = 50 x 4.2 x (31 - 22) = 1890 J. Energy lost by the metal = m_m x c_m x change in temperature = 54 x c_m x (100 - 31) = 3726 x c_m. Setting them equal: c_m = 1890 / 3726 = about 0.51 J/(g degrees C).
   Method:

   Equate 50 x 4.2 x 9 to 54 x c_m x 69 and solve for c_m = 0.51 J/(g degrees C).
   Examiner tips

   • Use the correct temperature change for each substance: the water warms up, the metal cools down.

5. Question 5

   1 marksRefractive index from the speed of light
   Refractive index n = (speed of light in air) divided by (speed of light in the medium) = (3.0 x 10^8) / (2.25 x 10^8) = about 1.33.
   Method:

   Divide the speed in air by the speed in water: 3.0 / 2.25 = about 1.33.
   Examiner tips

   • For any transparent medium, the refractive index from air is greater than 1.

6. Question 6

   1 marksUltrasound depth-finding
   The pulse travels down to the seabed and back. Total distance = speed x time = 1500 x 0.30 = 450 m. The depth is half of this because the pulse covers the depth twice: 450 / 2 = 225 m.
   Method:

   Calculate 1500 x 0.30 = 450 m round-trip; halve to get 225 m depth.
   Examiner tips

   • For echo methods, halve the total round-trip distance to get the depth.

7. Question 7

   1 marksResistor in a parallel branch
   The p.d. across the 6.0 ohm resistor = 0.50 x 6.0 = 3.0 V. The p.d. across the parallel combination = 12 - 3.0 = 9.0 V. Current in the 20 ohm branch = 9.0 / 20 = 0.45 A. So current in R = 0.50 - 0.45 = 0.05 A. Therefore R = 9.0 / 0.05 = 180 ohms.
   Method:

   V_6 = 3.0 V, V_parallel = 9.0 V, I_20 = 0.45 A, I_R = 0.05 A, R = 180 ohms.
   Examiner tips

   • Work through the series section first, then split into parallel branches.

8. Question 8

   1 marksCharge from energy and voltage
   Potential difference is energy per unit charge: V = E / Q. Rearranging gives Q = E / V = (4.5 x 10^11) / (2.9 x 10^8) = about 1550, which is about 1600 C.
   Method:

   Apply Q = E / V = 4.5 x 10^11 / 2.9 x 10^8 = about 1600 C.
   Examiner tips

   • Convert any MJ values to joules before substituting.

9. Question 9

   1 marksStep-up transformer voltage
   For an ideal transformer the voltage ratio equals the turns ratio: Vs / Vp = Ns / Np. So Vs = Vp x Ns / Np = 12 x 20 = 240 V.
   Method:

   Apply Vs = Vp x Ns / Np = 12 x 20 = 240 V.
   Examiner tips

   • Step-up means more secondary turns and a higher secondary voltage.

10. Question 10

    1 marksDistance from light travel time
    Distance = speed times time = (3.0 x 10^8) x 490 = 1.47 x 10^11 m, which is about 1.5 x 10^11 m.
    Method:

    Multiply the speed of light by 490 s to get about 1.5 x 10^11 m.
    Examiner tips

    • Astronomical distances are huge; expect a large power of ten.

11. Question 11

    1 marksBeta decay of protactinium
    In beta-minus decay a neutron changes into a proton and an electron, and the electron leaves as the beta particle. The proton number rises by 1 (from 91 to 92, so the element becomes uranium) and the nucleon number stays at 234.
    Method:

    Apply beta-minus rules: proton number 91 to 92, nucleon number stays 234, giving uranium-234.
    Examiner tips

    • Beta-minus decay: proton number plus 1, nucleon number unchanged.