0625/42 May/June 2025 Worked Answers (IGCSE Physics 0625 Extended)

10 questions · 80 marks · 75 minutes

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1. Question 1

   1 marksDistance and deceleration from a speed-time graph
   For motion at constant speed, distance = speed times time = 11 x 3.0 = 33 m. This is also the area under a horizontal speed-time graph of height 11 m/s over 3.0 s.
   Method:

   Multiply constant speed by time: 11 x 3.0 = 33 m.
   Examiner tips

   • Distance is the area under a speed-time graph.

2. Question 2

   1 marksDensity of wood
   Volume of the ruler = length x width x thickness = 1.0 x 2.6 x 10^-2 x 6.0 x 10^-3 = 1.56 x 10^-4 cubic metres. Density = mass divided by volume = 0.081 / (1.56 x 10^-4) = about 520 kg/m^3.
   Method:

   Volume = 1.0 x 0.026 x 0.006 = 1.56 x 10^-4 m^3; density = 0.081 / 1.56 x 10^-4 = about 520 kg/m^3.
   Examiner tips

   • Track powers of ten carefully when multiplying small dimensions.

3. Question 3

   1 marksConservation of momentum in a collision
   Momentum is conserved. Take rightward as positive. Total momentum before = 0.45 x 0.34 + 0.21 x (-0.12) = 0.153 - 0.0252 = 0.1278 kg m/s. Total mass after = 0.45 + 0.21 = 0.66 kg. Velocity after = 0.1278 / 0.66 = about 0.19 m/s, positive, so to the right.
   Method:

   Total momentum 0.1278 kg m/s; total mass 0.66 kg; velocity = 0.1278 / 0.66 = about 0.19 m/s right.
   Examiner tips

   • Always assign a positive direction first when applying conservation of momentum.

4. Question 4

   1 marksEmission of thermal radiation
   At a given temperature, a dull black surface is a much better emitter of infrared radiation than a shiny white surface. Container A therefore radiates thermal energy away faster than container B, and so its water cools more quickly.
   Method:

   Container A cools fastest because its dull black surface emits infrared faster.
   Examiner tips

   • Dull black = best emitter and absorber; shiny white = worst.

5. Question 5

   1 marksWavelength of microwaves
   Use the wave equation: speed = frequency times wavelength, so wavelength = speed / frequency. Convert 2.48 GHz to 2.48 x 10^9 Hz. Then wavelength = (3.0 x 10^8) / (2.48 x 10^9) = about 0.12 m.
   Method:

   Divide the speed of light by the frequency in Hz: 3.0 x 10^8 / 2.48 x 10^9 = about 0.12 m.
   Examiner tips

   • Convert prefixes before substituting: G means x 10^9.

6. Question 6

   1 marksImage formed by a converging lens at 2F
   Placing the object at twice the focal length (2F) of a converging lens produces an image at 2F on the opposite side. The image is real, inverted, and the same size as the object. This is a standard case of the converging lens.
   Method:

   Recognise the object is at 2F; the image is real, inverted and the same size.
   Examiner tips

   • Know the four standard converging-lens cases by object distance.

7. Question 7

   1 marksEnergy in kilowatt-hours
   One kilowatt-hour is the energy used by a 1 kW device in 1 hour. The lamp's power is 25 W = 0.025 kW. Energy = power x time = 0.025 kW x 220 h = 5.5 kW h.
   Method:

   Convert 25 W to 0.025 kW, multiply by 220 h to get 5.5 kW h.
   Examiner tips

   • Use kilowatts and hours when working in kilowatt-hours.

8. Question 8

   1 marksResistance from voltage and current
   Resistance = potential difference divided by current. So R = V / I = 5.0 / 2.4 = about 2.1 ohms.
   Method:

   Divide voltage by current: 5.0 / 2.4 = about 2.1 ohms.
   Examiner tips

   • Resistance is always voltage divided by current.

9. Question 9

   1 marksAlpha decay of radium-223
   An alpha particle has proton number 2 and nucleon number 4. In alpha decay the parent nucleus loses two protons and two neutrons, so the proton number drops by 2 (from 88 to 86) and the nucleon number drops by 4 (from 223 to 219). The product is therefore radon with proton number 86 and nucleon number 219.
   Method:

   Subtract 2 from the proton number and 4 from the nucleon number to get radon-219.
   Examiner tips

   • Alpha decay: proton number -2, nucleon number -4.

10. Question 10

    1 marksLife cycle of a massive star
    After its stable phase, a massive star swells into a red supergiant. When it can no longer support itself by fusion it explodes as a supernova. The core left behind collapses to form a neutron star or, if massive enough, a black hole.
    Method:

    Apply the massive-star sequence: red supergiant, supernova, neutron star or black hole.
    Examiner tips

    • Distinguish low-mass and high-mass star end stages.