2024 Cambridge IGCSE Physics (0625) Practice Paper with Answers

(a) Acceleration is the rate of change of velocity.

(b) \(v = u + at = 0 + (1.6 \times 5.0) = 8.0\text{ m/s}\).

(c) (i) The motion has three phases:
1. Acceleration phase (0 to 5.0 s): \(\text{distance}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5.0 \times 8.0 = 20\text{ m}\).
2. Constant speed phase (5.0 to 12.0 s): \(\text{time} = 7.0\text{ s}\), \(\text{distance}_2 = 7.0 \times 8.0 = 56\text{ m}\).
3. Deceleration phase (12.0 to 15.0 s): \(\text{time} = 3.0\text{ s}\), \(\text{distance}_3 = \frac{1}{2} \times 3.0 \times 8.0 = 12\text{ m}\).
Total distance = \(20 + 56 + 12 = 88\text{ m}\).

(ii) Speed is a scalar quantity which has magnitude only, whereas velocity is a vector quantity which has both magnitude and direction.

(a) rate of change of velocity OR change of velocity per unit time [1]

(b) \(v = u + at\) or \(1.6 \times 5.0\) [1]
\(8.0\text{ m/s}\) [1]

(c) (i) calculation of distance in any phase (e.g. \(20\text{ m}\) or \(56\text{ m}\) or \(12\text{ m}\)) [1]
summing of three phases [1]
correct calculation of total distance: \(88\text{ m}\) [1]

(ii) speed is scalar / magnitude only [1]
velocity is vector / has direction [1]

(a) In a closed system, the total momentum before a collision is equal to the total momentum after the collision.

(b) (i) \(p = m \times v = 0.80\text{ kg} \times 1.5\text{ m/s} = 1.2\text{ kg}\cdot\text{m/s}\) (or \(\text{N}\cdot\text{s}\)).
(ii) Total momentum before = total momentum after.
\(1.2 = (m_A + m_B) \times v_{\text{final}} = (0.80 + 1.2) \times v_{\text{final}} = 2.0 \times v_{\text{final}}\).
\(v_{\text{final}} = \frac{1.2}{2.0} = 0.60\text{ m/s}\).

(c) The collision is inelastic. Kinetic energy is not conserved during the collision. Before collision, \(E_k = \frac{1}{2} \times 0.80 \times 1.5^2 = 0.90\text{ J}\). After collision, \(E_k = \frac{1}{2} \times 2.0 \times 0.60^2 = 0.36\text{ J}\). Some kinetic energy is lost/converted to thermal energy.

(a) total momentum before collision = total momentum after collision [1]
in a closed system / if no external forces act [1]

(b) (i) \(p = m \times v\) in any form [1]
\(1.2\text{ kg}\cdot\text{m/s}\) (or \(\text{N}\cdot\text{s}\)) [1]

(ii) use of conservation of momentum equation [1]
substitution: \(1.2 = 2.0 \times v\) [1]
\(0.60\text{ m/s}\) [1]

(c) inelastic [1]
because kinetic energy is not conserved / kinetic energy decreases [1]

(a) Specific heat capacity is the energy required per unit mass to raise the temperature of a substance by one degree Celsius (or Kelvin).

(b) (i) \(E = P \times t = 150\text{ W} \times (4.0 \times 60\text{ s}) = 150 \times 240 = 36\ 000\text{ J}\) (or \(36\text{ kJ}\)).
(ii) \(c = \frac{E}{m \Delta \theta} = \frac{36\ 000}{2.0 \times 45.0} = \frac{36\ 000}{90} = 400\text{ J/(kg}\cdot^\circ\text{C)}\).

(c) Some of the thermal energy supplied by the heater is lost to the surroundings (or absorbed by the heater/container itself). This means the temperature rise of the block is lower than it would be if all energy was absorbed, leading to a calculated value of specific heat capacity that is higher than the actual value.

(a) energy / heat needed [1]
per unit mass per unit temperature rise (or \(1\text{ kg}\) by \(1^\circ\text{C}\)) [1]

(b) (i) \(E = P \times t\) or \(150 \times 4 \times 60\) [1]
\(36\ 000\text{ J}\) [1]

(ii) \(c = E / (m \Delta \theta)\) [1]
substitution: \(36\ 000 / (2.0 \times 45.0)\) [1]
\(400\text{ J/(kg}\cdot^\circ\text{C)}\) (accept \(\text{J/(kg}\cdot\text{K)}\)) [1]

(c) energy is lost to the surroundings / some thermal energy heats the heater itself [1]

(a) The critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is \(90^\circ\).

(b) (i) \(n = \frac{c}{v} \Rightarrow v = \frac{c}{n} = \frac{3.0 \times 10^8}{1.52} \approx 1.97 \times 10^8\text{ m/s}\).
(ii) \(\sin(c) = \frac{1}{n} = \frac{1}{1.52} \approx 0.6579\).
\(c = \sin^{-1}(0.6579) \approx 41.1^\circ\).

(c) Since the angle of incidence (\(45^\circ\)) is greater than the critical angle (\(41.1^\circ\)), the light ray does not refract into the air. Instead, it undergoes total internal reflection, reflecting back into the glass block at an angle of reflection of \(45^\circ\).

(a) angle of incidence in more dense medium [1]
for which angle of refraction is \(90^\circ\) [1]

(b) (i) \(n = c/v\) in any form [1]
substitution: \(3.0 \times 10^8 / 1.52\) [1]
\(1.97 \times 10^8\text{ m/s}\) (or \(2.0 \times 10^8\text{ m/s}\)) [1]

(ii) \(\sin(c) = 1/n\) [1]
\(41.1^\circ\) (or \(41^\circ\)) [1]

(c) total internal reflection occurs [1]
because the angle of incidence is greater than the critical angle (\(45^\circ > 41.1^\circ\)) [1]

(a) (i) Frequency is the number of waves passing a point per second.
(ii) A wavefront is a line joining points on a wave that are in phase (e.g. all the crests).

(b) (i) \(v = f \lambda \Rightarrow f = \frac{v}{\lambda} = \frac{0.24\text{ m/s}}{0.030\text{ m}} = 8.0\text{ Hz}\).
(ii) The frequency of the waves remains constant as they enter shallow water.
(iii) \(\lambda = \frac{v}{f} = \frac{0.16\text{ m/s}}{8.0\text{ Hz}} = 0.020\text{ m}\) (or \(2.0\text{ cm}\)).

(c) The phenomenon is refraction. The change in medium density/depth changes the wave speed, causing a change in direction.

(a) (i) number of waves per second / unit time [1]
(ii) line joining points of same phase / peaks of waves [1]

(b) (i) \(v = f \lambda\) [1]
\(8.0\text{ Hz}\) [1]

(ii) stays the same / constant [1]

(iii) \(v = f \lambda\) or \(0.16 / 8.0\) [1]
\(0.020\text{ m}\) (or \(2.0\text{ cm}\)) [1]

(c) refraction [1]
speed changes, causing wavelength and direction to change [1]

(a) Electromotive force (e.m.f.) is the electrical work done by a source in moving a unit charge around a complete circuit.

(b) (i) \(R_{\text{total}} = R_1 + R_2 = 12 + 6.0 = 18\ \Omega\).
(ii) \(I = \frac{V}{R_{\text{total}}} = \frac{9.0\text{ V}}{18\ \Omega} = 0.50\text{ A}\).
(iii) \(V_{\text{resistor}} = I \times R = 0.50\text{ A} \times 12\ \Omega = 6.0\text{ V}\).

(c) When light intensity decreases, the resistance of the LDR increases. The LDR now takes a larger share of the total resistance, so the potential difference across it increases (as the circuit acts as a potential divider).

(a) work done / energy transferred [1]
per unit charge (around a complete circuit) [1]

(b) (i) \(18\ \Omega\) [1]

(ii) \(I = V / R\) [1]
\(0.50\text{ A}\) [1]

(iii) \(V = I \times R\) [1]
\(6.0\text{ V}\) [1]

(c) resistance of LDR increases [1]
p.d. across LDR increases (as it takes a larger fraction of total voltage) [1]

(a) Since a beta particle is an electron (\(^{0}_{-1}\beta\)), the nucleon number \(A\) remains unchanged, and the proton number \(Z\) increases by 1.
\(A = 14\)
\(Z = 7\).

(b) A beta particle is a fast-moving electron emitted from the nucleus of an atom.

(c) (i) Half-life is the time taken for half the radioactive nuclei in a sample to decay (or the time for the activity of a sample to halve).
(ii) Number of half-lives = \(\frac{17\ 190}{5730} = 3\).
After 1 half-life: \(800 / 2 = 400\text{ counts/s}\).
After 2 half-lives: \(400 / 2 = 200\text{ counts/s}\).
After 3 half-lives: \(200 / 2 = 100\text{ counts/s}\).

(a) \(A = 14\) [1]
\(Z = 7\) [1]

(b) electron [1]
fast-moving / emitted from nucleus [1]

(c) (i) time taken for the activity / count rate of a sample [1]
to decrease to half of its initial value [1]

(ii) calculation of number of half-lives: \(17\ 190 / 5730 = 3\) [1]
activity halved 3 times: \(800 / 2^3\) [1]
\(100\text{ counts/s}\) [1]

(a) Redshift is an increase in the observed wavelength of electromagnetic radiation emitted by a source that is moving away from the observer.

(b) Light from almost all distant galaxies is redshifted, meaning they are moving away from us. Galaxies that are further away show a greater redshift, meaning they are moving away faster. This indicates that the Universe is expanding. Extrapolating back in time suggests that all matter was once concentrated at a single, extremely dense, hot point (the Big Bang).

(c) (i) \(d = 2.5 \times 10^{21}\text{ km} \times 1000\text{ m/km} = 2.5 \times 10^{24}\text{ m}\).
(ii) \(v = H_0 \times d = 2.2 \times 10^{-18}\text{ s}^{-1} \times 2.5 \times 10^{24}\text{ m} = 5.5 \times 10^6\text{ m/s}\).

(a) increase in (observed) wavelength / decrease in frequency [1]
of light from galaxies moving away from us [1]

(b) redshift shows galaxies are moving away / expanding Universe [1]
further galaxies are moving faster [1]
suggests Universe started from a single point / origin [1]

(c) (i) \(2.5 \times 10^{24}\text{ m}\) [2] (1 mark for multiplying by 1000)

(ii) \(v = H_0 \times d\) [1]
\(5.5 \times 10^6\text{ m/s}\) [1]

(a) Momentum is defined as the product of mass and velocity (\(p = mv\)).

(b) (i) \(p = mv = 0.80\text{ kg} \times 3.0\text{ m/s} = 2.4\text{ kg m/s}\) (or \(\text{N s}\)).

(ii) Let the direction to the right be positive.
Initial total momentum: \(p_{\text{total, initial}} = p_{\text{car, initial}} + p_{\text{truck, initial}} = 2.4\text{ kg m/s} + 0 = 2.4\text{ kg m/s}\).
After the collision, the car rebounds to the left, so its velocity is \(-0.60\text{ m/s}\).
\(p_{\text{car, final}} = 0.80\text{ kg} \times (-0.60\text{ m/s}) = -0.48\text{ kg m/s}\).
By conservation of momentum:
\(p_{\text{total, initial}} = p_{\text{total, final}}\)
\(2.4 = -0.48 + 1.2 \times v_{\text{truck}}\)
\(2.88 = 1.2 \times v_{\text{truck}}\)
\(v_{\text{truck}} = 2.4\text{ m/s}\) (to the right).

(c) The impulse (change in momentum) on the toy car is:
\(\Delta p = p_{\text{final}} - p_{\text{initial}} = -0.48\text{ kg m/s} - 2.4\text{ kg m/s} = -2.88\text{ kg m/s}\).
Magnitude of change in momentum \(|\Delta p| = 2.88\text{ N s}\).
Average force: \(F = \frac{\Delta p}{\Delta t} = \frac{2.88\text{ N s}}{0.15\text{ s}} = 19.2\text{ N}\).

(a)
- product of mass and velocity [1 mark]

(b) (i)
- \(2.4\) [1 mark]
- \(\text{kg m/s}\) or \(\text{N s}\) [1 mark]

(b) (ii)
- use of conservation of momentum: \(m_1 u_1 + m_2 v_2 = m_1 v_1 + m_2 v_2\) (or statement of conservation) [1 mark]
- substitution showing correct sign for rebound car velocity: \(2.4 = -0.48 + 1.2 v\) [1 mark]
- value: \(2.4\text{ m/s}\) (accept 'to the right') [1 mark]

(c)
- calculation of change of momentum of car: \(2.88\text{ kg m/s}\) (or \(\text{N s}\)) [1 mark]
- use of \(F = \Delta p / t\) (e.g. \(2.88 / 0.15\)) [1 mark]
- value: \(19.2\text{ N}\) (or \(19\text{ N}\)) [1 mark]