2024 Cambridge IGCSE Physics (0625) Practice Paper with Answers

(a) The graph starts at the origin (0,0) and rises with a constant positive gradient to (3.0 s, 6.0 m/s). It then continues as a horizontal line at 6.0 m/s from 3.0 s to 7.0 s. Finally, it decreases with a constant negative gradient from (7.0 s, 6.0 m/s) to (9.0 s, 0 m/s). (b) Acceleration \(a = \frac{v - u}{t} = \frac{6.0 - 0}{3.0} = 2.0\text{ m/s}^2\). (c) Distance is the area under the graph. Area = area of trapezium = \(\frac{1}{2} \times (a + b) \times h = \frac{1}{2} \times (4.0 + 9.0) \times 6.0 = 39\text{ m}\). (d) Average speed = \(\frac{\text{Total distance}}{\text{Total time}} = \frac{39\text{ m}}{9.0\text{ s}} = 4.33\text{ m/s}\).

(a) Description of three distinct regions: constant positive gradient, constant zero gradient (horizontal), and constant negative gradient [3 marks]. (b) Use of \(a = \Delta v / t\) [1 mark], correct answer \(2.0\text{ m/s}^2\) [1 mark]. (c) Calculation of area under graph [2 marks], correct answer \(39\text{ m}\) [1 mark]. (d) Use of \(\text{average speed} = \text{total distance} / \text{total time}\) and correct calculation of \(4.33\text{ m/s}\) [0.88 marks].

(a) Momentum is defined as the product of mass and velocity of an object: \(p = mv\). Its SI unit is \(\text{kg}\cdot\text{m/s}\). (b) Total initial momentum \(p_{\text{initial}} = m_1 v_1 + m_2 v_2 = (120 \times 4.5) + (80 \times 0) = 540\text{ kg}\cdot\text{m/s}\). (c) By the law of conservation of momentum, \(p_{\text{final}} = p_{\text{initial}} = 540\text{ kg}\cdot\text{m/s}\). \((120 + 80) v = 540 \implies 200 v = 540 \implies v = 2.7\text{ m/s}\). (d) The change in momentum of the second trolley is \(\Delta p_2 = m_2 v_2^f - m_2 v_2^i = 80 \times 2.7 - 0 = 216\text{ kg}\cdot\text{m/s}\). The average force \(F = \frac{\Delta p}{\Delta t} = \frac{216}{0.15} = 1440\text{ N}\).

(a) Definition of momentum as mass \(\times\) velocity [1 mark] and unit \(\text{kg}\cdot\text{m/s}\) [1 mark]. (b) Calculation of initial momentum: \(120 \times 4.5 = 540\text{ kg}\cdot\text{m/s}\) [2 marks]. (c) Equating initial and final momentum [1 mark], and finding velocity \(2.7\text{ m/s}\) [1.88 marks]. (d) Use of \(F = \Delta p / \Delta t\) [1 mark], and correct calculation of \(1440\text{ N}\) [1 mark].

(a) Thermal energy supplied \(E = P \times t = 120\text{ W} \times (5.0 \times 60)\text{ s} = 36000\text{ J}\). (b) Heat gained by copper block \(Q = m c \Delta \theta\). Assuming no heat loss, \(Q = E = 36000\text{ J}\). Temperature rise \(\Delta \theta = 115.0 - 21.0 = 94.0^\circ\text{C}\). \(c = \frac{Q}{m \Delta \theta} = \frac{36000}{0.80 \times 94.0} \approx 479\text{ J}/(\text{kg}\cdot^\circ\text{C})\). (c) In reality, some energy is lost to the surroundings. Therefore, more energy must be supplied to achieve the same temperature rise, which makes the calculated value of \(c\) higher than the actual value. To reduce this, we can wrap the copper block in insulating material (lagging).

(a) Formula \(E = P \times t\) [1 mark], correct calculation \(36000\text{ J}\) [1.88 marks]. (b) Formula \(c = Q / (m \Delta \theta)\) [1 mark], correct temperature change \(94.0^\circ\text{C}\) [1 mark], correct calculated specific heat capacity \(479\text{ J}/(\text{kg}\cdot^\circ\text{C})\) [1 mark]. (c) Explanation that thermal energy is lost to surroundings so supplied heat \(Q\) is overestimate [2 marks], suggestion of lagging or using an insulating cover [1 mark].

(a) Using Snell's law: \(n = \frac{\sin i}{\sin r} \implies 1.52 = \frac{\sin(43.0^\circ)}{\sin r} \implies \sin r \approx 0.4487 \implies r \approx 26.7^\circ\). (b) Critical angle \(c\) is given by \(\sin c = \frac{1}{n} = \frac{1}{1.52} \implies c \approx 41.1^\circ\). (c) Since the angle of incidence inside the glass is \(45.0^\circ\), which is greater than the critical angle of \(41.1^\circ\), the ray cannot refract into the air. Instead, it undergoes total internal reflection, reflecting back into the glass at an angle of reflection of \(45.0^\circ\).

(a) Snell's Law stated or used [1 mark], correct substitution [1 mark], final answer \(26.7^\circ\) [1 mark]. (b) Equation \(\sin c = 1/n\) [1 mark], calculation of \(c = 41.1^\circ\) [1.88 marks]. (c) Statement that total internal reflection occurs [1 mark], reason given that angle of incidence is greater than critical angle [1 mark], statement that angle of reflection is \(45.0^\circ\) [1 mark].

(a) Frequency \(f = \frac{v}{\lambda} = \frac{0.28\text{ m/s}}{0.035\text{ m}} = 8.0\text{ Hz}\). (b) The frequency remains constant because it is determined solely by the source of the waves. (c) In shallow water, \(v = f \lambda \implies 0.18 = 8.0 \times \lambda \implies \lambda = 0.0225\text{ m} = 2.25\text{ cm}\). (d) The wave direction bends towards the normal because the wave speed decreases in the shallow water, causing the wavefronts to slow down and crowd closer together.

(a) Formula \(v = f \lambda\) used [1 mark], conversion of units \(3.5\text{ cm} = 0.035\text{ m}\) [1 mark], correct calculation of \(8.0\text{ Hz}\) [1 mark]. (b) State that frequency is unchanged/constant [1.88 marks]. (c) Use of \(v = f \lambda\) with \(f = 8.0\text{ Hz}\) [1 mark], correct answer \(2.25\text{ cm}\) [1 mark]. (d) Bends towards the normal [1 mark], reason given that wave speed is slower in shallow water [1 mark].

(a) Ohm's law states that the current through a conductor is directly proportional to the potential difference across it, provided physical conditions (such as temperature) remain constant. (b) Current \(I = \frac{V}{R} = \frac{4.5\text{ V}}{3.0\ \Omega} = 1.5\text{ A}\). (c) Charge \(Q = I \times t = 1.5\text{ A} \times (2.0 \times 60)\text{ s} = 180\text{ C}\). (d) Resistance \(R = \rho \frac{L}{A}\). For the second wire, \(L_2 = 2 L_1\) and \(A_2 = 0.5 A_1\). Thus, \(R_2 = \rho \frac{2 L_1}{0.5 A_1} = 4 \times R_1 = 4 \times 3.0 = 12.0\ \Omega\).

(a) Stating proportional relationship between \(V\) and \(I\) [1 mark], and condition of constant temperature [1 mark]. (b) Formula \(I = V / R\) [1 mark], correct calculation of \(1.5\text{ A}\) [1 mark]. (c) Formula \(Q = I t\) [1 mark], correct conversion of time and calculation of \(180\text{ C}\) [1.88 marks]. (d) Identifying that resistance is proportional to length and inversely proportional to area [1 mark], correct calculation of \(12.0\ \Omega\) [1 mark].

(a) Since the components are in series, the total resistance \(R_{\text{total}} = R_1 + R_{\text{thermistor}} = 40.0 + 20.0 = 60.0\ \Omega\). The current \(I = \frac{V}{R_{\text{total}}} = \frac{12.0}{60.0} = 0.20\text{ A}\). (b) The potential difference across the thermistor \(V_{\text{thermistor}} = I \times R_{\text{thermistor}} = 0.20 \times 20.0 = 4.0\text{ V}\). (c) When temperature increases, the resistance of the thermistor decreases. This reduces the total resistance of the circuit, which increases the current in the circuit. Since \(V_{R_1} = I \times R_1\) and \(R_1\) is constant, the increase in current causes the potential difference across \(R_1\) to increase.

(a) Addition of series resistances [1 mark], formula \(I = V / R\) [1 mark], correct total resistance and current [1.88 marks]. (b) Use of \(V = I R\) or potential divider equation [1 mark], correct calculation of \(4.0\text{ V}\) [1 mark]. (c) Stating thermistor resistance decreases with temperature [1 mark], total resistance decreases and current increases [1 mark], concluding that voltage across \(R_1\) increases [1 mark].

(a) Redshift is the observed increase in the wavelength of light from astronomical objects moving away from us. It indicates that the galaxy is moving away from the Earth (receding). (b) Change in wavelength \(\Delta \lambda = \lambda_{\text{observed}} - \lambda_{\text{emitted}} = 512\text{ nm} - 486\text{ nm} = 26\text{ nm}\). (c) Hubble's law states that the recessional velocity \(v\) of a distant galaxy is directly proportional to its distance \(d\) from Earth (\(v = H_0 d\)). The observation of redshift shows that almost all distant galaxies are moving away from us, and those further away are moving faster, indicating that the Universe is expanding. Extrapolating back in time, all matter must have started from a single, hot, dense point (the Big Bang).

(a) Definition of redshift as increase in observed wavelength [1 mark], indicating galaxy is moving away [1 mark]. (b) Correct calculation of change in wavelength: \(512 - 486 = 26\text{ nm}\) [1.88 marks]. (c) Statement of Hubble's law [2 marks], explanation that redshift shows expansion [2 marks], linking expansion back to a common starting point (Big Bang) [1 mark].

\( \textbf{(a)} \)
Using the momentum formula:
\( p = m \times v \)
\( p = 0.40 \text{ kg} \times 1.5 \text{ m/s} = 0.60 \text{ kg m/s} \) (or \( \text{N s} \))

\( \textbf{(b)} \)
By the principle of conservation of momentum:
\( \text{Total momentum before} = \text{Total momentum after} \)
\( m_A v_A + m_B v_B = (m_A + m_B) v \)
\( 0.60 \text{ kg m/s} + 0 = (0.40 \text{ kg} + 0.60 \text{ kg}) \times v \)
\( 0.60 = 1.00 \times v \)
\( v = 0.60 \text{ m/s} \)

\( \textbf{(c)} \)
Calculate the change in momentum of truck \( B \):
\( \Delta p_B = m_B v - m_B u_B = 0.60 \text{ kg} \times 0.60 \text{ m/s} - 0 = 0.36 \text{ kg m/s} \) (or \( \text{N s} \))

Now calculate the average force exerted on truck \( B \):
\( F = \frac{\Delta p}{\Delta t} \)
\( F = \frac{0.36 \text{ N s}}{0.12 \text{ s}} = 3.0 \text{ N} \)

**(a)** [Total: 2 marks]
- \( 1 \text{ mark} \): Correct use of \( p = mv \) with substitution \( 0.40 \times 1.5 \)
- \( 1 \text{ mark} \): Correct calculation of \( 0.60 \) with appropriate unit (\( \text{kg m/s} \) or \( \text{N s} \))

**(b)** [Total: 3 marks]
- \( 1 \text{ mark} \): Correct application of momentum conservation principle (total momentum before = total momentum after)
- \( 1 \text{ mark} \): Correct substitution into conservation equation: \( 0.60 = 1.0 \times v \)
- \( 1 \text{ mark} \): Correct calculation of \( v = 0.60 \text{ m/s} \)

**(c)** [Total: 3 marks]
- \( 1 \text{ mark} \): Correct calculation of change in momentum of truck B, \( \Delta p = 0.36 \text{ kg m/s} \) (or correct acceleration \( a = \frac{0.60}{0.12} = 5.0 \text{ m/s}^2 \))
- \( 1 \text{ mark} \): Correct force formula used: \( F = \frac{\Delta p}{t} \) (or \( F = ma \)) and substitution
- \( 1 \text{ mark} \): Correct calculation of \( 3.0 \text{ N} \)