October/November 2025 Paper 41 Worked Answers (IGCSE Physics 0625 Extended)

10 questions · 80 marks · 75 minutes

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1. Question 1

   1 marksStopping distance from velocity and deceleration
   Use v squared = u squared - 2 a s with v = 0, so s = u squared / (2 a) = 56 squared / (2 x 1.2) = 3136 / 2.4 = about 1300 m.
   Method:

   Set v to zero and rearrange to s = u squared / (2 a) = 1300 m.
   Examiner tips

   • Use the suvat equation that connects v, u, a and s directly.

2. Question 2

   1 marksWeight on a different planet
   Mass is the same everywhere. Mass = weight on Earth divided by g on Earth = 42 / 9.8 = 4.29 kg. Weight on Mars = mass times g on Mars = 4.29 x 3.7 = 15.9, which is about 16 N.
   Method:

   Compute mass on Earth (4.29 kg), then weight on Mars = 4.29 x 3.7 = 16 N.
   Examiner tips

   • Mass is a constant property; weight changes with gravitational field strength.

3. Question 3

   1 marksSolar cell efficiency and output
   Useful output power = efficiency as a fraction times the input power = 0.75 x 0.72 = 0.54 kW.
   Method:

   Multiply the input power by 0.75 to get 0.54 kW.
   Examiner tips

   • Efficiency as a decimal multiplies the input to give the useful output.

4. Question 4

   1 marksEnergy to heat water
   Energy supplied = mass times specific heat capacity times change in temperature = 0.15 x 4200 x (58 - 20) = 0.15 x 4200 x 38 = 23 940 J, which is about 24 000 J.
   Method:

   Substitute m = 0.15, c = 4200, change in T = 38 into E = m c change in T.
   Examiner tips

   • Always use the temperature change, not the final temperature.

5. Question 5

   1 marksLoudness, pitch and the audible range
   Loudness depends on amplitude: a larger amplitude gives a louder sound. Pitch depends on frequency: a higher frequency gives a higher pitch. Wave X has both a larger amplitude and a higher frequency, so it is louder and has a higher pitch.
   Method:

   Match the amplitude to loudness and the frequency to pitch.
   Examiner tips

   • Amplitude controls loudness; frequency controls pitch.

6. Question 6

   1 marksCritical angle for an optical fibre
   Using the relation that the sine of the critical angle equals 1 divided by the refractive index: the sine of c = 1 / 1.4 = 0.714. The critical angle is therefore about 46 degrees (45.6 degrees more precisely).
   Method:

   Compute sine c = 0.714 then take the inverse sine to get c about 46 degrees.
   Examiner tips

   • A denser medium gives a smaller critical angle for total internal reflection.

7. Question 7

   1 marksSeries resistance from current and voltage
   Total resistance = supply voltage divided by current = 6.0 / 0.080 = 75 ohms. Since two identical resistors are in series, each has resistance = 75 / 2 = 37.5 ohms, about 38 ohms.
   Method:

   R_total = 6.0 / 0.080 = 75 ohms; each = 37.5 ohms.
   Examiner tips

   • Equal series resistors share the total resistance equally.

8. Question 8

   1 marksField strength inside a solenoid
   The magnetic field of a solenoid is strongest inside the coil, where the field lines run almost parallel and very close together. Close spacing of field lines indicates a strong magnetic field.
   Method:

   Identify the inside of the solenoid; close field lines indicate a strong field.
   Examiner tips

   • Inside a solenoid the field is strong and almost uniform.

9. Question 9

   1 marksStructure of a helium-4 atom
   The proton number is 2, so the nucleus contains 2 protons. The nucleon number is 4, so it has 4 - 2 = 2 neutrons. A neutral atom has the same number of electrons as protons, so there are 2 electrons in shells around the nucleus.
   Method:

   Use the proton and nucleon numbers to find protons, neutrons and electrons.
   Examiner tips

   • For a neutral atom: electrons = protons; neutrons = nucleon number - protons.

10. Question 10

    1 marksComet orbital shape
    Planets follow nearly circular (slightly elliptical) orbits around the Sun. Comets follow highly elliptical orbits, coming very close to the Sun at one end and travelling far out into the Solar System at the other end. Their orbital speed is high near the Sun and low far away.
    Method:

    Recall that comets have highly elliptical orbits around the Sun.
    Examiner tips

    • Planets: nearly circular orbits. Comets: highly elliptical orbits.