An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V3) Cambridge IGCSE Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
Paper 2 (Multiple Choice - Extended)
Answer all forty multiple choice questions. For each question, choose the one correct option (A, B, C or D).
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
A vehicle travels along a straight, horizontal track. From \( t = 0 \) to \( t = 4.0\text{ s} \), the vehicle accelerates uniformly from rest to a speed of \( 12\text{ m/s} \). From \( t = 4.0\text{ s} \) to \( t = 12.0\text{ s} \), it travels at a constant speed of \( 12\text{ m/s} \). From \( t = 12.0\text{ s} \) to \( t = 16.0\text{ s} \), it decelerates uniformly to rest. What is the average speed of the vehicle for the entire 16.0 s journey?
A.\( 6.0\text{ m/s} \)
B.\( 8.0\text{ m/s} \)
C.\( 9.0\text{ m/s} \)
D.\( 12\text{ m/s} \)
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Worked solution
To find the average speed, we must first calculate the total distance travelled, which is equal to the area under the speed-time graph (a trapezium):
\(\text{Total Distance} = \frac{1}{2} \times (a + b) \times h\) Where: - \( a = 16.0\text{ s} \) (the total time duration) - \( b = 12.0 - 4.0 = 8.0\text{ s} \) (the duration at constant speed) - \( h = 12\text{ m/s} \) (the maximum speed)
Now, calculate the average speed: \(\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{144\text{ m}}{16.0\text{ s}} = 9.0\text{ m/s} \)
Marking scheme
B1: for correctly calculating the total distance as 144 m B1: for dividing the distance by 16.0 s to get 9.0 m/s (C)
Question 2 · multiple-choice
1 marks
A tennis ball of mass \( 0.060\text{ kg} \) is moving horizontally at a speed of \( 25\text{ m/s} \) when it is struck by a racket. The racket applies a force in the opposite direction, causing the ball to rebound horizontally in that opposite direction at a speed of \( 35\text{ m/s} \). The ball is in contact with the racket for a time of \( 15\text{ ms} \). What is the average force exerted on the ball by the racket?
A.\( 40\text{ N} \)
B.\( 80\text{ N} \)
C.\( 240\text{ N} \)
D.\( 400\text{ N} \)
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Worked solution
Taking the direction of rebound as the positive direction, we have: - Initial velocity, \( u = -25\text{ m/s} \) - Final velocity, \( v = +35\text{ m/s} \) - Mass, \( m = 0.060\text{ kg} \) - Time interval, \( \Delta t = 15\text{ ms} = 0.015\text{ s} \)
Using the impulse-force equation: \( F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t} \) \( F = \frac{0.060 \times (35 - (-25))}{0.015} \) \( F = \frac{0.060 \times 60}{0.015} = \frac{3.6}{0.015} = 240\text{ N} \)
Marking scheme
B1: for correctly identifying the change in momentum as 3.6 Ns B1: for dividing by the time in seconds to obtain 240 N (C)
Question 3 · multiple-choice
1 marks
A diver is swimming at a depth of \( 25\text{ m} \) below the surface of a lake. The density of the fresh water in the lake is \( 1.0 \times 10^3\text{ kg/m}^3 \), the atmospheric pressure at the surface is \( 1.0 \times 10^5\text{ Pa} \), and the gravitational field strength \( g \) is \( 9.8\text{ N/kg} \). What is the total pressure experienced by the diver?
A.\( 2.5 \times 10^5\text{ Pa} \)
B.\( 3.0 \times 10^5\text{ Pa} \)
C.\( 3.5 \times 10^5\text{ Pa} \)
D.\( 3.9 \times 10^5\text{ Pa} \)
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Worked solution
The total pressure experienced by the diver is the sum of the atmospheric pressure at the surface and the hydrostatic pressure exerted by the water column:
\( P_{\text{total}} = P_{\text{atm}} + P_{\text{hydrostatic}} \) \( P_{\text{total}} = P_{\text{atm}} + \rho g h \)
B1: for correctly calculating the water pressure as 2.45 x 10^5 Pa B1: for adding atmospheric pressure to obtain the correct total pressure of 3.5 x 10^5 Pa (C)
Question 4 · multiple-choice
1 marks
An electric heater rated at \( 50\text{ W} \) is used to heat a metal block of mass \( 0.80\text{ kg} \). The heater is switched on for \( 4.0\text{ minutes} \) and the temperature of the metal block increases from \( 20^\circ\text{C} \) to \( 50^\circ\text{C} \). Assuming no thermal energy is lost to the surroundings, what is the specific heat capacity of the metal?
A.\( 100\text{ J / (kg }^\circ\text{C)} \)
B.\( 500\text{ J / (kg }^\circ\text{C)} \)
C.\( 1500\text{ J / (kg }^\circ\text{C)} \)
D.\( 3000\text{ J / (kg }^\circ\text{C)} \)
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Worked solution
Calculate the electrical energy supplied by the heater: \( E = P \times t \) Convert time to seconds: \( t = 4.0\text{ minutes} = 4.0 \times 60 = 240\text{ s} \) \( E = 50\text{ W} \times 240\text{ s} = 12\,000\text{ J} \)
Calculate the temperature change: \( \Delta \theta = 50^\circ\text{C} - 20^\circ\text{C} = 30^\circ\text{C} \)
Using the specific heat capacity formula: \( E = m c \Delta \theta \implies c = \frac{E}{m \Delta \theta} \) \( c = \frac{12\,000\text{ J}}{0.80\text{ kg} \times 30^\circ\text{C}} = \frac{12\,000}{24} = 500\text{ J / (kg }^\circ\text{C)} \)
Marking scheme
B1: for converting time to seconds and finding power energy as 12,000 J B1: for substituting values correctly into specific heat capacity equation to get 500 J / (kg °C) (B)
Question 5 · multiple-choice
1 marks
A monochromatic ray of light has a wavelength of \( 6.0 \times 10^{-7}\text{ m} \) in air. It enters a glass block of refractive index \( 1.5 \). The speed of light in air is \( 3.0 \times 10^8\text{ m/s} \). What is the frequency of the light inside the glass block?
A.\( 3.3 \times 10^{14}\text{ Hz} \)
B.\( 5.0 \times 10^{14}\text{ Hz} \)
C.\( 7.5 \times 10^{14}\text{ Hz} \)
D.\( 1.1 \times 10^{15}\text{ Hz} \)
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Worked solution
When a wave enters a different medium during refraction, its frequency remains unchanged. Therefore, the frequency of the light inside the glass block is exactly the same as its frequency in air.
Calculate the frequency in air: \( f = \frac{v}{\lambda} \) \( f = \frac{3.0 \times 10^8\text{ m/s}}{6.0 \times 10^{-7}\text{ m}} = 5.0 \times 10^{14}\text{ Hz} \)
Thus, the frequency in the glass is also \( 5.0 \times 10^{14}\text{ Hz} \).
Marking scheme
B1: for stating or using the fact that frequency is constant during refraction B1: for evaluating the frequency as 5.0 x 10^{14} Hz (B)
Question 6 · multiple-choice
1 marks
A circuit consists of a \( 12\text{ V} \) battery of negligible internal resistance connected to three resistors. A \( 6.0\ \Omega \) resistor is connected in series with a parallel combination of a \( 4.0\ \Omega \) resistor and a \( 12.0\ \Omega \) resistor. What is the potential difference across the \( 6.0\ \Omega \) resistor?
A.\( 4.0\text{ V} \)
B.\( 6.0\text{ V} \)
C.\( 8.0\text{ V} \)
D.\( 12\text{ V} \)
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Next, calculate the total resistance of the series circuit: \( R_{\text{total}} = R_{\text{series}} + R_p = 6.0\ \Omega + 3.0\ \Omega = 9.0\ \Omega \)
Using the potential divider formula, find the potential difference across the \( 6.0\ \Omega \) resistor: \( V_{6.0\ \Omega} = V_{\text{total}} \times \frac{6.0}{R_{\text{total}}} = 12\text{ V} \times \frac{6.0}{9.0} = 8.0\text{ V} \)
Marking scheme
B1: for finding equivalent parallel resistance of 3.0 ohms B1: for determining the voltage drop across the 6.0 ohm resistor as 8.0 V (C)
Question 7 · multiple-choice
1 marks
An ideal step-down transformer is connected to a \( 240\text{ V} \) a.c. mains supply. The primary coil has \( 1200\text{ turns} \) and the secondary coil has \( 60\text{ turns} \). The secondary coil is connected to a lamp rated at \( 12\text{ V}, 24\text{ W} \), which operates at its normal brightness. What is the current in the primary coil?
A.\( 0.05\text{ A} \)
B.\( 0.10\text{ A} \)
C.\( 2.0\text{ A} \)
D.\( 4.0\text{ A} \)
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Worked solution
For an ideal transformer, the power input to the primary coil is equal to the power output from the secondary coil (100% efficiency): \( P_{\text{primary}} = P_{\text{secondary}} \)
Given that the secondary load (lamp) operates at its normal rating: \( P_{\text{secondary}} = 24\text{ W} \) Therefore, \( P_{\text{primary}} = 24\text{ W} \).
Using the electrical power formula for the primary circuit: \( P_{\text{primary}} = V_{\text{primary}} \times I_{\text{primary}} \) \( 24\text{ W} = 240\text{ V} \times I_{\text{primary}} \) \( I_{\text{primary}} = \frac{24}{240} = 0.10\text{ A} \)
Marking scheme
B1: for using the principle of conservation of power (P_in = P_out = 24 W) for an ideal transformer B1: for calculating primary current as 0.10 A (B)
Question 8 · multiple-choice
1 marks
A Geiger-Müller (GM) tube and counter are used to measure the activity of a radioactive sample. The background count rate is constant at \( 24\text{ counts/minute} \). At \( t = 0 \), the total count rate registered is \( 344\text{ counts/minute} \). The half-life of the radioactive isotope is \( 4.0\text{ hours} \). What is the total count rate registered by the counter at \( t = 12\text{ hours} \)?
A.\( 40\text{ counts/minute} \)
B.\( 43\text{ counts/minute} \)
C.\( 64\text{ counts/minute} \)
D.\( 86\text{ counts/minute} \)
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Worked solution
First, calculate the initial corrected (source-only) count rate at \( t = 0 \): \( \text{Corrected Count Rate}_{t=0} = 344 - 24 = 320\text{ counts/minute} \)
Next, calculate the number of half-lives that elapse in \( 12\text{ hours} \): \( \text{Number of half-lives, } N = \frac{12\text{ hours}}{4.0\text{ hours}} = 3.0 \)
After 3 half-lives, the corrected count rate decays to: \( \text{Corrected Count Rate}_{t=12} = 320 \times \left(\frac{1}{2}\right)^3 = \frac{320}{8} = 40\text{ counts/minute} \)
Finally, add the background count rate back to find the total registered count rate: \( \text{Total Count Rate}_{t=12} = 40 + 24 = 64\text{ counts/minute} \)
Marking scheme
B1: for finding the initial corrected count rate of 320 counts/minute B1: for determining the final total count rate as 64 counts/minute after 3 half-lives (C)
Question 9 · multiple_choice
1 marks
A vehicle accelerates from rest with a constant acceleration of \(1.5\text{ m/s}^2\) for \(10\text{ s}\), then travels at a constant speed for \(15\text{ s}\), and finally decelerates uniformly to rest in \(5.0\text{ s}\). What is the total distance travelled by the vehicle during the entire journey?
A.300 m
B.338 m
C.375 m
D.450 m
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Worked solution
During the first phase (constant acceleration): \(d_1 = \frac{1}{2} a t^2 = \frac{1}{2} \times 1.5 \times 10^2 = 75\text{ m}\). The maximum speed reached is \(v = a t = 1.5 \times 10 = 15\text{ m/s}\). During the second phase (constant speed): \(d_2 = v \times t = 15 \times 15 = 225\text{ m}\). During the third phase (uniform deceleration to rest): \(d_3 = \frac{1}{2} v t = \frac{1}{2} \times 15 \times 5.0 = 37.5\text{ m}\). The total distance is: \(d = d_1 + d_2 + d_3 = 75 + 225 + 37.5 = 337.5\text{ m} \approx 338\text{ m}\).
Marking scheme
1 mark for the correct option B.
Question 10 · multiple_choice
1 marks
A ball of mass \(0.40\text{ kg}\) is attached to a string and whirled in a horizontal circle of radius \(0.50\text{ m}\) at a constant speed of \(3.0\text{ m/s}\). What is the direction and magnitude of the resultant force acting on the ball?
A.direction: away from the centre of the circle; magnitude: \(2.4\text{ N}\)
B.direction: along the tangent to the circle; magnitude: \(2.4\text{ N}\)
C.direction: towards the centre of the circle; magnitude: \(7.2\text{ N}\)
D.direction: towards the centre of the circle; magnitude: \(14.4\text{ N}\)
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Worked solution
An object moving in a circular path experiences a centripetal acceleration directed towards the centre of the circle. The magnitude of the centripetal force is given by: \(F = \frac{mv^2}{r}\). Substituting the given values: \(F = \frac{0.40 \times 3.0^2}{0.50} = \frac{0.40 \times 9.0}{0.50} = 7.2\text{ N}\). Therefore, the resultant force has a magnitude of \(7.2\text{ N}\) and is directed towards the centre of the circle.
Marking scheme
1 mark for the correct option C.
Question 11 · multiple_choice
1 marks
A car of mass \(1200\text{ kg}\) travelling at \(15\text{ m/s}\) collides with a stationary van of mass \(2000\text{ kg}\). After the collision, the two vehicles stick together and move off with a common velocity \(v\). What is the value of \(v\) and the change in the total kinetic energy of the system?
A.v = \(5.6\text{ m/s}\); change in kinetic energy = \(0\text{ kJ}\)
B.v = \(5.6\text{ m/s}\); change in kinetic energy = \(84\text{ kJ}\) decrease
C.v = \(9.4\text{ m/s}\); change in kinetic energy = \(51\text{ kJ}\) decrease
D.v = \(9.4\text{ m/s}\); change in kinetic energy = \(84\text{ kJ}\) increase
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An electric kettle containing \(0.80\text{ kg}\) of water at \(100\text{ }^\circ\text{C}\) is left switched on. The power of the heating element is \(2.4\text{ kW}\). The specific latent heat of vaporisation of water is \(2.3 \times 10^6\text{ J/kg}\). What mass of water is converted to steam if the kettle is left on for \(5.0\text{ minutes}\) at \(100\text{ }^\circ\text{C}\)?
A.0.0052 kg
B.0.31 kg
C.0.49 kg
D.0.80 kg
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Worked solution
The electrical energy supplied in \(5.0\text{ minutes}\) is: \(E = P \times t = 2400\text{ W} \times (5.0 \times 60)\text{ s} = 720\,000\text{ J}\). Using the formula for latent heat: \(E = m L\), \(m = \frac{E}{L} = \frac{720\,000}{2.3 \times 10^6} \approx 0.313\text{ kg} \approx 0.31\text{ kg}\). Since \(0.31\text{ kg}\) is less than the initial mass of water \(0.80\text{ kg}\), only this portion of water is converted to steam.
Marking scheme
1 mark for the correct option B.
Question 13 · multiple_choice
1 marks
A ray of light in air strikes the surface of a transparent polymer block at an angle of incidence of \(50^\circ\). The refractive index of the polymer is \(1.40\). What is the angle of refraction inside the polymer block and the critical angle for light inside the polymer block?
A.angle of refraction = \(33^\circ\); critical angle = \(46^\circ\)
B.angle of refraction = \(33^\circ\); critical angle = \(36^\circ\)
C.angle of refraction = \(36^\circ\); critical angle = \(46^\circ\)
D.angle of refraction = \(36^\circ\); critical angle = \(54^\circ\)
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Worked solution
From Snell's law: \(n = \frac{\sin i}{\sin r} \implies 1.40 = \frac{\sin 50^\circ}{\sin r} \implies \sin r = \frac{\sin 50^\circ}{1.40} \approx 0.5472 \implies r \approx 33.2^\circ\). For the critical angle \(c\): \(\sin c = \frac{1}{n} = \frac{1}{1.40} \approx 0.7143 \implies c \approx 45.6^\circ\). Rounding to the nearest degree gives \(r = 33^\circ\) and \(c = 46^\circ\).
Marking scheme
1 mark for the correct option A.
Question 14 · multiple_choice
1 marks
A wire of length \(L\) and diameter \(d\) has a resistance of \(8.0\text{ }\Omega\). A second wire is made of the same metal but has a length of \(3L\) and a diameter of \(2d\). What is the resistance of the second wire?
A.1.3 \(\Omega\)
B.6.0 \(\Omega\)
C.12 \(\Omega\)
D.24 \(\Omega\)
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Worked solution
The resistance of a wire is given by: \(R = \rho \frac{L}{A} = \rho \frac{L}{\pi (d/2)^2} \propto \frac{L}{d^2}\). For the second wire: \(R_2 \propto \frac{3L}{(2d)^2} = \frac{3L}{4d^2} \implies R_2 = \frac{3}{4} R\). Substituting the initial resistance: \(R_2 = \frac{3}{4} \times 8.0\text{ }\Omega = 6.0\text{ }\Omega\).
Marking scheme
1 mark for the correct option B.
Question 15 · multiple_choice
1 marks
A radioactive source has a half-life of \(4.0\text{ hours}\). A detector placed near the source registers a count rate of \(240\text{ counts/minute}\), which includes a constant background radiation of \(20\text{ counts/minute}\). What count rate will the detector register after \(12\text{ hours}\)?
A.27.5 counts/minute
B.30 counts/minute
C.47.5 counts/minute
D.50 counts/minute
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Worked solution
First, subtract the background count rate to find the initial count rate of the source alone: \(\text{Initial source count rate} = 240 - 20 = 220\text{ counts/minute}\). Next, find the number of half-lives in \(12\text{ hours}\): \(n = \frac{12}{4.0} = 3\text{ half-lives}\). Calculate the source count rate after 3 half-lives: \(\text{Final source count rate} = 220 \times \left(\frac{1}{2}\right)^3 = \frac{220}{8} = 27.5\text{ counts/minute}\). Finally, add the constant background radiation back to find the registered count rate: \(\text{Registered count rate} = 27.5 + 20 = 47.5\text{ counts/minute}\).
Marking scheme
1 mark for the correct option C.
Question 16 · multiple_choice
1 marks
Light from a distant galaxy is observed to have a fractional change in wavelength \(\frac{\Delta \lambda}{\lambda} = 0.050\). The Hubble constant \(H_0\) is \(2.2 \times 10^{-18}\text{ s}^{-1}\) and the speed of light \(c\) is \(3.0 \times 10^8\text{ m/s}\). Assuming the redshift is caused entirely by the expansion of space, what is the recessional speed of the galaxy and its approximate distance from Earth?
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Worked solution
The recessional speed \(v\) is given by: \(v = \frac{\Delta \lambda}{\lambda} c = 0.050 \times 3.0 \times 10^8\text{ m/s} = 1.5 \times 10^7\text{ m/s}\). Using Hubble's Law to find the distance \(d\): \(v = H_0 d \implies d = \frac{v}{H_0} = \frac{1.5 \times 10^7}{2.2 \times 10^{-18}} \approx 6.8 \times 10^{24}\text{ m}\).
Marking scheme
1 mark for the correct option A.
Question 17 · multiple-choice
1 marks
A car starts from rest and accelerates at a constant rate of 2.0 m/s^2 for 6.0 s. It then travels at a constant velocity for 10 s, and finally decelerates uniformly to rest in 4.0 s. What is the total distance travelled by the car?
A.120 m
B.144 m
C.180 m
D.204 m
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Worked solution
First phase (acceleration): The final velocity is v = u + at = 0 + (2.0 * 6.0) = 12 m/s. The distance is s1 = 0.5 * a * t^2 = 0.5 * 2.0 * 36 = 36 m. Second phase (constant velocity): The distance is s2 = v * t = 12 * 10 = 120 m. Third phase (deceleration): The distance is s3 = (v / 2) * t = (12 / 2) * 4.0 = 24 m. Total distance is 36 + 120 + 24 = 180 m.
Marking scheme
Award 1 mark for the correct option C.
Question 18 · multiple-choice
1 marks
A spring with a spring constant of 250 N/m has an unstretched length of 12.0 cm. When an object is hung from the spring, its total length becomes 16.4 cm. The limit of proportionality is not exceeded. What is the weight of the object?
A.1.1 N
B.11 N
C.110 N
D.1100 N
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Worked solution
Using Hooke's Law, F = k * x, where x is the extension. The extension is 16.4 cm - 12.0 cm = 4.4 cm = 0.044 m. The weight of the object is F = 250 N/m * 0.044 m = 11 N.
Marking scheme
Award 1 mark for the correct option B.
Question 19 · multiple-choice
1 marks
An object of mass 0.50 kg is thrown vertically upwards from the ground with an initial kinetic energy of 45 J. Air resistance is negligible. The acceleration of free fall g is 9.8 m/s^2. What is the kinetic energy of the object when it is at a height of 6.0 m above the ground?
A.15.0 J
B.15.6 J
C.29.4 J
D.45.0 J
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Worked solution
The total energy is conserved. The gravitational potential energy at 6.0 m is Ep = m * g * h = 0.50 kg * 9.8 m/s^2 * 6.0 m = 29.4 J. The remaining kinetic energy is Ek = E_total - Ep = 45 J - 29.4 J = 15.6 J.
Marking scheme
Award 1 mark for the correct option B.
Question 20 · multiple-choice
1 marks
A container holds a layer of oil floating on water. The oil has a density of 850 kg/m^3 and a depth of 0.20 m. The water has a density of 1000 kg/m^3 and a depth of 0.35 m. The acceleration of free fall g is 9.8 m/s^2. What is the pressure due to the liquids at the bottom of the container?
A.1666 Pa
B.3430 Pa
C.5096 Pa
D.5390 Pa
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Worked solution
The liquid pressure is the sum of the pressures of each layer. Oil pressure is p_oil = 850 kg/m^3 * 9.8 m/s^2 * 0.20 m = 1666 Pa. Water pressure is p_water = 1000 kg/m^3 * 9.8 m/s^2 * 0.35 m = 3430 Pa. Total pressure at the bottom is 1666 Pa + 3430 Pa = 5096 Pa.
Marking scheme
Award 1 mark for the correct option C.
Question 21 · multiple-choice
1 marks
A ray of light in air is incident on the flat surface of a semi-circular glass block at an angle of incidence of 40.0 degrees. The refractive index of the glass is 1.52. What is the angle of refraction inside the glass block, and what is the critical angle for the glass-air boundary?
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Worked solution
By Snell's Law, n = sin(i) / sin(r) so sin(r) = sin(40.0) / 1.52 = 0.4228, which gives r = 25.0 degrees. The critical angle c is given by sin(c) = 1 / n = 1 / 1.52 = 0.6579, which gives c = 41.1 degrees.
Marking scheme
Award 1 mark for the correct option A.
Question 22 · multiple-choice
1 marks
A wire of length L and cross-sectional area A has a resistance of 8.0 ohms. A second wire of the same metal has a length of 3L and a diameter that is twice that of the first wire. What is the resistance of the second wire?
A.1.5 ohms
B.6.0 ohms
C.12.0 ohms
D.24.0 ohms
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Worked solution
Resistance R is proportional to length / area. Since the diameter of the second wire is doubled, its cross-sectional area is quadrupled (4A). Thus, R2 = R1 * (3L / L) * (A / 4A) = 8.0 ohms * 3 / 4 = 6.0 ohms.
Marking scheme
Award 1 mark for the correct option B.
Question 23 · multiple-choice
1 marks
A radioactive sample has a half-life of 4.0 minutes. The initial count rate recorded by a detector near the sample is 420 counts/minute, which includes a constant background count rate of 20 counts/minute. What is the count rate recorded by the detector after 12.0 minutes?
A.50 counts/minute
B.70 counts/minute
C.72.5 counts/minute
D.125 counts/minute
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Worked solution
First, find the initial corrected count rate due to the sample alone: 420 - 20 = 400 counts/minute. The time of 12.0 minutes corresponds to 3 half-lives (12.0 / 4.0). The corrected count rate after 3 half-lives is 400 * (1/2)^3 = 50 counts/minute. Finally, add back the background count rate to get the recorded count rate: 50 + 20 = 70 counts/minute.
Marking scheme
Award 1 mark for the correct option B.
Question 24 · multiple-choice
1 marks
Light from a distant galaxy has a redshift that corresponds to a recession velocity of 1.5 * 10^4 km/s. Taking the Hubble constant H0 to be 2.2 * 10^-18 s^-1, what is the approximate distance from the Earth to this galaxy?
A.6.8 * 10^21 m
B.3.3 * 10^22 m
C.6.8 * 10^24 m
D.3.3 * 10^25 m
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Worked solution
Using Hubble's law, v = H0 * d. Convert velocity to m/s: v = 1.5 * 10^4 km/s = 1.5 * 10^7 m/s. Then d = v / H0 = (1.5 * 10^7 m/s) / (2.2 * 10^-18 s^-1) = 6.8 * 10^24 m.
Marking scheme
Award 1 mark for the correct option C.
Question 25 · multiple-choice
1 marks
A model rocket is launched vertically upwards from the ground. It accelerates uniformly from rest for 4.0 s to reach a speed of 48 m/s. The fuel is then immediately exhausted, and the rocket continues to move upwards under the influence of gravity alone (air resistance is negligible). Taking the acceleration of free fall as 9.8 m/s², what is the maximum height reached by the rocket?
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Worked solution
During the first phase of the journey, the rocket accelerates from rest to 48 m/s in 4.0 s. The distance travelled during this phase, \(s_1\), is given by: \(s_1 = \frac{u + v}{2} \times t = \frac{0 + 48}{2} \times 4.0 = 96\text{ m}\). In the second phase, the rocket decelerates under gravity from an initial velocity of 48 m/s to a final velocity of 0 m/s at the maximum height. Using the equation of motion: \(v^2 = u^2 + 2as_2\), we get: \(0^2 = 48^2 - 2 \times 9.8 \times s_2\) which simplifies to \(s_2 = \frac{2304}{19.6} \approx 117.6\text{ m}\). The maximum height reached is the sum of the distances in both phases: \(\text{Maximum Height} = s_1 + s_2 = 96\text{ m} + 117.6\text{ m} = 213.6\text{ m} \approx 214\text{ m}\).
Marking scheme
Award 1 mark for the correct option C. [1 mark for correct calculation of both phases of motion and summing them up]
Question 26 · multiple-choice
1 marks
A small box of mass 3.0 kg is pulled up a rough slope inclined at 30° to the horizontal. A constant pulling force of 25 N acts parallel to the slope. A constant frictional force of 4.0 N opposes the motion. The acceleration of free fall g is 9.8 m/s². What is the acceleration of the box up the slope?
A.2.1 m/s²
B.3.5 m/s²
C.7.0 m/s²
D.8.3 m/s²
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Worked solution
The forces acting parallel to the slope are: 1) the pulling force, F_pull = 25 N (upwards); 2) the component of the weight down the slope, W_slope = m g sin(30°) = 3.0 kg × 9.8 m/s² × 0.5 = 14.7 N; 3) the frictional force, f = 4.0 N (downwards). The net force acting up the slope is: F_net = F_pull - W_slope - f = 25 - 14.7 - 4.0 = 6.3 N. Using Newton's second law, the acceleration a is: a = F_net / m = 6.3 N / 3.0 kg = 2.1 m/s².
Marking scheme
Award 1 mark for the correct option A. [1 mark for correct calculation of net force and acceleration]
Question 27 · multiple-choice
1 marks
An electric motor with an efficiency of 75% is used to lift a load of mass 80 kg vertically upwards through a height of 15 m in a time of 6.0 s. The acceleration of free fall g is 9.8 m/s². What is the electrical power input to the motor?
A.1.5 kW
B.2.0 kW
C.2.6 kW
D.12 kW
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Worked solution
First, calculate the useful work output done in lifting the mass: W_out = m g h = 80 kg × 9.8 m/s² × 15 m = 11760 J. Next, find the useful power output: P_out = W_out / t = 11760 J / 6.0 s = 1960 W. Using the efficiency formula, Power Input = P_out / efficiency = 1960 W / 0.75 ≈ 2613 W = 2.6 kW.
Marking scheme
Award 1 mark for the correct option C. [1 mark for correct calculation of power output and power input]
Question 28 · multiple-choice
1 marks
A cylindrical container has a base area of 0.12 m² and is filled with water of density 1000 kg/m³ to a depth of 1.5 m. The atmospheric pressure acting on the surface of the water is 1.0 × 10⁵ Pa. Taking the acceleration of free fall g as 9.8 m/s², what is the total force exerted by the water and the atmosphere on the bottom of the container?
A.1.8 × 10³ N
B.1.2 × 10⁴ N
C.1.4 × 10⁴ N
D.1.6 × 10⁴ N
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Worked solution
The pressure exerted by the water alone is given by: P_water = ρ g h = 1000 kg/m³ × 9.8 m/s² × 1.5 m = 14700 Pa. The total pressure at the bottom of the container is the sum of the atmospheric pressure and the water pressure: P_total = P_atm + P_water = 1.0 × 10⁵ Pa + 14700 Pa = 114700 Pa. The total force on the bottom is: F = P_total × Area = 114700 Pa × 0.12 m² = 13764 N ≈ 1.4 × 10⁴ N.
Marking scheme
Award 1 mark for the correct option C. [1 mark for correct calculation of total pressure and total force]
Question 29 · multiple-choice
1 marks
A ray of light in air strikes the flat surface of a transparent semicircular glass block at an angle of incidence of 50.0°. The refractive index of the glass is 1.52. What is the angle of refraction inside the glass block?
A.30.3°
B.32.9°
C.48.2°
D.76.0°
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Worked solution
According to Snell's Law: n_air × sin(θ_i) = n_glass × sin(θ_r). Substituting the values: 1.0 × sin(50.0°) = 1.52 × sin(θ_r). Since sin(50.0°) ≈ 0.7660, we have: sin(θ_r) = 0.7660 / 1.52 ≈ 0.5040. Taking the inverse sine: θ_r = arcsin(0.5040) ≈ 30.3°.
Marking scheme
Award 1 mark for the correct option A. [1 mark for correct application of Snell's Law]
Question 30 · multiple-choice
1 marks
A copper wire of length L and cross-sectional area A has a resistance of 8.0 Ω. A second copper wire, at the same temperature, has a length of 2.5L and a cross-sectional area of 0.50A. What is the resistance of the second wire?
A.1.6 Ω
B.10 Ω
C.16 Ω
D.40 Ω
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Worked solution
The resistance of a wire is given by R = ρ × (L / A). For the first wire: R_1 = ρ × (L / A) = 8.0 Ω. For the second wire: R_2 = ρ × (2.5L / 0.50A) = ρ × 5(L / A) = 5 × R_1 = 5 × 8.0 Ω = 40 Ω.
Marking scheme
Award 1 mark for the correct option D. [1 mark for correct ratio reasoning and calculation]
Question 31 · multiple-choice
1 marks
A radioactive isotope of carbon, Carbon-14 (¹⁴₆C), decays into nitrogen (N) by emitting a β⁻-particle. Which row correctly identifies the proton number (atomic number) and the nucleon number (mass number) of the nitrogen nuclide formed?
A.Proton number = 5, Nucleon number = 14
B.Proton number = 7, Nucleon number = 13
C.Proton number = 7, Nucleon number = 14
D.Proton number = 8, Nucleon number = 14
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Worked solution
During beta-minus (β⁻) decay, a neutron in the nucleus decays into a proton, an electron, and an antineutrino. This increases the proton number of the nucleus by 1 (from 6 to 7), while the total nucleon number remains unchanged at 14. Therefore, the product nucleus is Nitrogen-14 (¹⁴₇N), which has a proton number of 7 and a nucleon number of 14.
Marking scheme
Award 1 mark for the correct option C. [1 mark for correct understanding of beta decay nuclear change]
Question 32 · multiple-choice
1 marks
Light from a distant galaxy is observed to have a wavelength that is longer than the wavelength of the same light source in a laboratory on Earth. Which statement correctly explains this observation?
A.The galaxy is moving towards the Earth, causing a decrease in frequency (blueshift).
B.The galaxy is moving away from the Earth, causing an increase in wavelength (redshift).
C.The galaxy is rotating rapidly on its axis, causing its light to slow down.
D.The gravity of the galaxy is stretching the light as it escapes into space.
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Worked solution
An increase in observed wavelength of light from a celestial object is known as redshift. This shift towards the red end of the spectrum occurs because the galaxy is moving away from the Earth (receding), which stretches the electromagnetic waves as they travel through the expanding space.
Marking scheme
Award 1 mark for the correct option B. [1 mark for correct definition and explanation of redshift]
Question 33 · multiple-choice
1 marks
A motorcycle starts from rest and accelerates uniformly at \(1.5\text{ m/s}^2\) for \(12.0\text{ s}\). It then travels at a constant velocity for \(20.0\text{ s}\), and finally decelerates uniformly to rest in \(8.0\text{ s}\).
What is the total distance travelled by the motorcycle?
A.540 m
B.612 m
C.720 m
D.432 m
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C1: Calculate maximum speed reach during acceleration phase C1: Calculate distance in any one phase correctly A1: Correct total distance with units
Question 34 · multiple-choice
1 marks
A light spring has an unstretched length of \(8.0\text{ cm}\). When a load of \(24\text{ N}\) is hung from it, its length is \(12.0\text{ cm}\). Assume the limit of proportionality is not exceeded.
What load is required to stretch the spring to a total length of \(15.0\text{ cm}\)?
A.30 N
B.36 N
C.42 N
D.45 N
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2. Calculate the spring constant, \(k\): \(k = \frac{F_1}{\Delta x_1} = \frac{24\text{ N}}{4.0\text{ cm}} = 6.0\text{ N/cm}\)
3. Determine the required extension, \(\Delta x_2\), for a total length of \(15.0\text{ cm}\): \(\Delta x_2 = 15.0\text{ cm} - 8.0\text{ cm} = 7.0\text{ cm}\)
C1: Calculate initial extension or spring constant A1: Correct load required with appropriate steps shown
Question 35 · multiple-choice
1 marks
An electric motor is used to lift a crate of mass \(120\text{ kg}\) vertically upwards through a height of \(15.0\text{ m}\) in a time of \(18.0\text{ s}\). The efficiency of the motor system is \(65\%\).
Take the acceleration of free fall, \(g\), to be \(9.8\text{ m/s}^2\).
What is the electrical power input to the motor?
A.640 W
B.980 W
C.1500 W
D.2300 W
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Worked solution
1. Calculate the work done in lifting the crate (gaining gravitational potential energy): \(W = mgh = 120\text{ kg} \times 9.8\text{ m/s}^2 \times 15.0\text{ m} = 17640\text{ J}\)
2. Calculate the useful power output of the motor: \(P_{\text{out}} = \frac{W}{t} = \frac{17640\text{ J}}{18.0\text{ s}} = 980\text{ W}\)
3. Calculate the electrical power input using efficiency: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% \implies 0.65 = \frac{980}{P_{\text{in}}} \implies P_{\text{in}} = \frac{980}{0.65} \approx 1508\text{ W}\)
To two significant figures, this is \(1500\text{ W}\).
Marking scheme
C1: Use of mgh to find energy or power output C1: Division by efficiency to find power input A1: Final value rounded to appropriate significant figures (1500 W)
Question 36 · multiple-choice
1 marks
A diver is swimming at a depth of \(25.0\text{ m}\) in a freshwater lake.
The density of fresh water is \(1000\text{ kg/m}^3\). The atmospheric pressure is \(1.01 \times 10^5\text{ Pa}\).
Take the acceleration of free fall, \(g\), to be \(9.8\text{ m/s}^2\).
What is the total pressure experienced by the diver?
A.2.45 * 10^5 Pa
B.3.46 * 10^5 Pa
C.3.55 * 10^5 Pa
D.4.47 * 10^5 Pa
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Worked solution
1. Calculate the hydrostatic pressure exerted by the fresh water: \(p_{\text{water}} = \rho g h = 1000\text{ kg/m}^3 \times 9.8\text{ m/s}^2 \times 25.0\text{ m} = 245000\text{ Pa} = 2.45 \times 10^5\text{ Pa}\)
2. Add atmospheric pressure to find the total pressure: \(p_{\text{total}} = p_{\text{water}} + p_{\text{atm}} = 2.45 \times 10^5\text{ Pa} + 1.01 \times 10^5\text{ Pa} = 3.46 \times 10^5\text{ Pa}\)
Marking scheme
C1: Correct use of \(p = \rho g h\) to find pressure of the water A1: Addition of atmospheric pressure to yield correct total pressure
Question 37 · multiple-choice
1 marks
A block of copper of mass \(2.5\text{ kg}\) is heated by a \(150\text{ W}\) heater for \(4.0\text{ minutes}\). The temperature of the copper block rises from \(20^\circ\text{C}\) to \(57^\circ\text{C}\).
Assuming no thermal energy is lost to the surroundings, what is the specific heat capacity of copper?
A.150 J/(kg K)
B.390 J/(kg K)
C.420 J/(kg K)
D.970 J/(kg K)
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Worked solution
1. Calculate the energy supplied by the heater: \(E = P \times t = 150\text{ W} \times (4.0 \times 60\text{ s}) = 36000\text{ J}\)
2. Calculate the temperature change: \(\Delta \theta = 57^\circ\text{C} - 20^\circ\text{C} = 37\text{ K}\)
To two significant figures, this is \(390\text{ J/(kg K)}\).
Marking scheme
C1: Correctly calculate the energy supplied in joules (converting minutes to seconds) C1: Correct temperature rise A1: Calculation of specific heat capacity matching 390 J/(kg K)
Question 38 · multiple-choice
1 marks
A ray of light in air is incident on a flat boundary with a glass block. The angle of incidence in air is \(48^\circ\) and the angle of refraction in the glass is \(29^\circ\).
What is the critical angle for total internal reflection in this glass?
A.29°
B.41°
C.45°
D.49°
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Worked solution
1. Calculate the refractive index \(n\) of the glass using Snell's law: \(n = \frac{\sin i}{\sin r} = \frac{\sin 48^\circ}{\sin 29^\circ} \approx \frac{0.7431}{0.4848} \approx 1.533\)
2. Use the critical angle formula: \(\sin c = \frac{1}{n} = \frac{1}{1.533} \approx 0.652\)
C1: Use of Snell's law to calculate the refractive index C1: Rearranging the critical angle formula A1: Correct angle
Question 39 · multiple-choice
1 marks
A potential divider circuit consists of a \(12\text{ V}\) power supply connected in series with a fixed resistor of resistance \(8.0\ \Omega\) and a thermistor. At room temperature, the resistance of the thermistor is \(16\ \Omega\).
What is the potential difference across the thermistor?
A.4.0 V
B.6.0 V
C.8.0 V
D.12 V
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Worked solution
Using the potential divider formula to find the voltage across the thermistor: \(V_{\text{thermistor}} = V_{\text{supply}} \times \left( \frac{R_{\text{thermistor}}}{R_{\text{fixed}} + R_{\text{thermistor}}} \right)\)
Answer all structured and quantitative questions. Show all your working and state units where appropriate.
10 Question · 80 marks
Question 1 · Structured
8 marks
A prototype electric car of mass \(1200\text{ kg}\) accelerates uniformly from rest to a speed of \(30\text{ m/s}\) in a time of \(12\text{ s}\).
(a) Calculate the acceleration of the car during the first \(12\text{ s}\). [2]
(b) The car then continues at a constant speed of \(30\text{ m/s}\) for a further \(15\text{ s}\). It then decelerates uniformly to rest in a time of \(8.0\text{ s}\).
(i) Show that the deceleration of the car is \(3.75\text{ m/s}^2\). [2]
(ii) Calculate the total distance travelled by the car during the entire journey. [4]
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Worked solution
(a) Use the definition of acceleration: \(a = \frac{v - u}{t} = \frac{30\text{ m/s} - 0}{12\text{ s}} = 2.5\text{ m/s}^2\).
(b)(i) Use \(a = \frac{v - u}{t} = \frac{0 - 30\text{ m/s}}{8.0\text{ s}} = -3.75\text{ m/s}^2\). The magnitude of the deceleration is \(3.75\text{ m/s}^2\).
(b)(ii) The total distance is the sum of the distances in the three stages: - Stage 1 (acceleration): \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12\text{ s} \times 30\text{ m/s} = 180\text{ m}\). - Stage 2 (constant speed): \(d_2 = \text{speed} \times \text{time} = 30\text{ m/s} \times 15\text{ s} = 450\text{ m}\). - Stage 3 (deceleration): \(d_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8.0\text{ s} \times 30\text{ m/s} = 120\text{ m}\).
(a) [2 marks] - C1 for formula: \(a = \frac{\Delta v}{t}\) or \(\frac{30}{12}\) - A1 for correct value with unit: \(2.5\text{ m/s}^2\)
(b)(i) [2 marks] - C1 for formula showing substitution: \(a = \frac{0 - 30}{8.0}\) or \(30 / 8.0\) - A1 for clear statement that deceleration is \(3.75\text{ m/s}^2\)
(b)(ii) [4 marks] - C1 for calculating distance during acceleration: \(180\text{ m}\) - C1 for calculating distance during constant speed: \(450\text{ m}\) - C1 for calculating distance during deceleration: \(120\text{ m}\) - A1 for total distance: \(750\text{ m}\)
Question 2 · Structured
8 marks
An electric heater of power \(150\text{ W}\) is placed inside a cavity in a block of ice of mass \(0.50\text{ kg}\) at a temperature of \(-10^\circ\text{C}\).
(a) The specific heat capacity of ice is \(2100\text{ J}/(\text{kg}\cdot^\circ\text{C})\).
(i) Calculate the energy required to raise the temperature of the ice to \(0^\circ\text{C}\). [2]
(ii) Calculate the time taken to raise the temperature of the ice to \(0^\circ\text{C}\). [2]
(b) When the ice reaches \(0^\circ\text{C}\), it begins to melt. The specific latent heat of fusion of ice is \(3.3 \times 10^5\text{ J}/\text{kg}\).
Calculate the additional time required for the entire block of ice to melt. [4]
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(b) Energy required to melt ice: \(Q = mL_f = 0.50\text{ kg} \times 3.3 \times 10^5\text{ J}/\text{kg} = 1.65 \times 10^5\text{ J}\). Time taken to melt: \(t = \frac{Q}{P} = \frac{1.65 \times 10^5\text{ J}}{150\text{ W}} = 1100\text{ s}\).
Marking scheme
(a)(i) [2 marks] - C1 for formula: \(E = mc\Delta \theta\) or \(0.50 \times 2100 \times 10\) - A1 for: \(10500\text{ J}\) or \(1.05 \times 10^4\text{ J}\)
(a)(ii) [2 marks] - C1 for formula: \(t = E/P\) or \(10500 / 150\) (allow ecf from a(i)) - A1 for: \(70\text{ s}\)
(b) [4 marks] - C1 for formula: \(Q = mL_f\) - C1 for calculating energy: \(1.65 \times 10^5\text{ J}\) - C1 for expression for time: \(1.65 \times 10^5 / 150\) - A1 for: \(1100\text{ s}\) or \(1.1 \times 10^3\text{ s}\)
Question 3 · Structured
8 marks
A student investigates the stretching of a steel spring. The unstretched length of the spring is \(15.0\text{ cm}\).
(a) When a load of \(6.0\text{ N}\) is suspended from the spring, its new length becomes \(19.0\text{ cm}\).
(i) Calculate the spring constant \( k \) of the spring. [3]
(ii) Calculate the length of the spring when a load of \(15.0\text{ N}\) is suspended from it, assuming the limit of proportionality is not exceeded. [2]
(b) State what is meant by the *limit of proportionality* of a spring, and describe what happens to the relation between load and extension after this limit is exceeded. [3]
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(a)(ii) Extension for a \(15.0\text{ N}\) load: \(x = \frac{F}{k} = \frac{15.0\text{ N}}{1.5\text{ N/cm}} = 10.0\text{ cm}\). New length = \(15.0\text{ cm} + 10.0\text{ cm} = 25.0\text{ cm}\).
(b) The limit of proportionality is the point beyond which extension is no longer directly proportional to the applied load. Once exceeded, Hooke's law is no longer obeyed, and a further small increase in load will cause a much larger, non-linear increase in extension (or permanent plastic deformation).
Marking scheme
(a)(i) [3 marks] - C1 for calculating extension: \(19.0 - 15.0 = 4.0\text{ cm}\) - C1 for Hooke's Law formula: \(k = F/x\) or \(6.0 / 4.0\) - A1 for spring constant: \(1.5\text{ N/cm}\) (or \(150\text{ N/m}\) with correct units)
(a)(ii) [2 marks] - C1 for calculating new extension: \(15.0 / 1.5 = 10.0\text{ cm}\) (allow ecf from a(i)) - A1 for new length: \(25.0\text{ cm}\) (or \(0.25\text{ m}\) with correct units)
(b) [3 marks] - B1 for definition: limit up to which extension is proportional to force/load - B1 for behavior change: extension and load are no longer proportional - B1 for physical result: spring undergoes non-linear stretching / permanent deformation
Question 4 · Structured
8 marks
A ray of monochromatic light is incident on the flat face of a semi-circular glass block at an angle of incidence of \(42^\circ\). The refractive index of the glass is \(1.52\).
(a) Calculate the angle of refraction inside the glass block. [3]
(b) Calculate the critical angle \( c \) for this glass-to-air boundary. [3]
(c) State and explain what happens to the ray of light if the angle of incidence on the flat face is increased to \(48^\circ\). [2]
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(b) The formula for critical angle is: \(\sin c = \frac{1}{n} = \frac{1}{1.52} = 0.6579\). \(c = \sin^{-1}(0.6579) = 41.1^\circ\).
(c) If the angle of incidence is increased to \(48^\circ\), which is greater than the critical angle of \(41.1^\circ\), the light cannot refract out of the block. Instead, total internal reflection occurs, and all of the light is reflected back inside the glass block.
Marking scheme
(a) [3 marks] - C1 for formula: \(n = \frac{\sin i}{\sin r}\) - C1 for rearrangement: \(\sin r = \frac{\sin 42}{1.52}\) - A1 for angle of refraction: \(26.1^\circ\) (accept range \(26.0^\circ - 26.2^\circ\))
(b) [3 marks] - C1 for formula: \(\sin c = \frac{1}{n}\) - C1 for substitution: \(\sin c = \frac{1}{1.52}\) - A1 for critical angle: \(41.1^\circ\) (accept \(41^\circ\))
(c) [2 marks] - B1 for stating: total internal reflection occurs - B1 for explanation: because the angle of incidence (\(48^\circ\)) is greater than the critical angle (\(41.1^\circ\))
Question 5 · Structured
8 marks
A portable water heater is rated at \(230\text{ V}, 800\text{ W}\).
(a) Calculate:
(i) the current in the heating element when it is operating at its rated voltage. [2]
(ii) the resistance of the heating element. [2]
(b) Calculate the total charge that passes through the heater when it is switched on for \(15\text{ minutes}\). [2]
(c) The heater is connected using a thinner copper wire of the same length. Explain how this change affects the current in the heater, assuming the supply voltage remains \(230\text{ V}\). [2]
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Worked solution
(a)(i) Using \(P = VI \Rightarrow I = \frac{P}{V} = \frac{800\text{ W}}{230\text{ V}} = 3.48\text{ A}\).
(b) First, convert time to seconds: \(t = 15 \times 60 = 900\text{ s}\). Using \(Q = It = 3.48\text{ A} \times 900\text{ s} = 3130\text{ C}\).
(c) A thinner wire has a greater resistance. This increases the total resistance of the circuit, which decreases the current flowing through the heater.
(a)(ii) [2 marks] - C1 for formula: \(R = V/I\) or \(R = V^2/P\) - A1 for: \(66.1\Omega\) (accept range \(66.0\Omega - 66.2\Omega\))
(b) [2 marks] - C1 for converting time to seconds: \(900\text{ s}\) AND formula \(Q = It\) - A1 for charge: \(3130\text{ C}\) (or \(3100\text{ C}\) from \(3.5\text{ A}\))
(c) [2 marks] - B1 for: thinner wire has a higher resistance - B1 for: total circuit resistance increases so current decreases
Question 6 · Structured
8 marks
A step-up transformer is used at a wind farm to increase the output voltage of a generator from \(600\text{ V}\) to \(24\text{ kV}\) for transmission.
(a) The primary coil of the transformer has \(150\text{ turns}\). Calculate the number of turns on the secondary coil. [3]
(b) The generator supplies a power of \(2.0\text{ MW}\) to the primary coil. Assuming the transformer is \(100\%\) efficient:
(i) Calculate the current in the transmission cables connected to the secondary coil. [3]
(ii) State and explain one advantage of transmitting electrical energy at a high voltage. [2]
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Worked solution
(a) Using the transformer turn ratio formula: \(\frac{V_p}{V_s} = \frac{N_p}{N_s} \Rightarrow N_s = N_p \times \frac{V_s}{V_p}\). \(N_s = 150 \times \frac{24000\text{ V}}{600\text{ V}} = 150 \times 40 = 6000\text{ turns}\).
(b)(i) Power on the secondary is equal to the power on the primary (\(2.0\text{ MW} = 2.0 \times 10^6\text{ W}\)). Using \(P = V_s I_s \Rightarrow I_s = \frac{P}{V_s} = \frac{2.0 \times 10^6\text{ W}}{24000\text{ V}} = 83.3\text{ A}\).
(b)(ii) High voltage reduces transmission current. Since power loss in cables is given by \(P_{\text{loss}} = I^2 R\), reducing the current significantly reduces energy losses as heat in the transmission lines, making transmission more efficient.
Marking scheme
(a) [3 marks] - C1 for turn ratio formula: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) - C1 for converting voltage: \(24\text{ kV} = 24000\text{ V}\) - A1 for turns: \(6000\text{ turns}\)
(b)(i) [3 marks] - C1 for converting power: \(2.0\text{ MW} = 2.0 \times 10^6\text{ W}\) - C1 for formula: \(I = P/V\) or \(2.0 \times 10^6 / 24000\) - A1 for current: \(83.3\text{ A}\) (accept \(83\text{ A}\))
(b)(ii) [2 marks] - B1 for: high voltage results in a lower current - B1 for: less energy/heat lost in cables because of lower current / power loss is proportional to \(I^2\)
Question 7 · Structured
8 marks
A radioactive source contains a nuclide of sodium, \(^{24}_{11}\text{Na}\). It decays by emitting a \(\beta\)-particle to form an isotope of magnesium (\(\text{Mg}\)).
(a) Write a complete nuclide equation for the decay of \(^{24}_{11}\text{Na}\). [3]
(b) The half-life of \(^{24}_{11}\text{Na}\) is \(15\text{ hours}\). A sample originally has an activity of \(800\text{ counts/s}\).
(i) Calculate the activity of the sample after \(45\text{ hours}\). [2]
(ii) Calculate the time taken for the activity of the sample to decrease to \(25\text{ counts/s}\). [3]
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Worked solution
(a) The decay equation is: \(^{24}_{11}\text{Na} \rightarrow ^{24}_{12}\text{Mg} + \,^{0}_{-1}\beta\) (or \(^{0}_{-1}\text{e}\))
(b)(i) Number of half-lives in 45 hours: \(N = \frac{45}{15} = 3\). After 3 half-lives, the activity is: \(A = \frac{A_0}{2^3} = \frac{800}{8} = 100\text{ counts/s}\).
(b)(ii) The activity needs to drop from \(800\) to \(25\text{ counts/s}\). \(\frac{25}{800} = \frac{1}{32} = \left(\frac{1}{2}\right)^5\). This corresponds to 5 half-lives. Time taken = \(5 \times 15\text{ hours} = 75\text{ hours}\).
Marking scheme
(a) [3 marks] - B1 for correct Magnesium symbol with correct nucleon number 24: \(^{24}\text{Mg}\) - B1 for correct Magnesium proton number 12: \(_{12}\text{Mg}\) - B1 for correct beta particle symbol: \(^{0}_{-1}\beta\) or \(^{0}_{-1}\text{e}\)
(b)(i) [2 marks] - C1 for calculating number of half-lives: \(45 / 15 = 3\) - A1 for: \(100\text{ counts/s}\)
(b)(ii) [3 marks] - C1 for identifying fraction of activity remaining: \(25 / 800 = 1/32\) - C1 for determining number of half-lives: \(32 = 2^5\) so 5 half-lives - A1 for time: \(75\text{ hours}\)
Question 8 · Structured
8 marks
(a) State what is meant by the term *redshift* and explain how it provides evidence for the expansion of the Universe. [3]
(b) Light from a distant galaxy is observed to have a redshift that indicates it is moving away from Earth at a speed of \(1.8 \times 10^7\text{ m/s}\).
(i) Using a Hubble constant value of \( H_0 = 2.2 \times 10^{-18}\text{ s}^{-1} \), calculate the distance \( d \) to this galaxy in meters. [3]
(ii) Estimate the age of the Universe in years, based on this value of the Hubble constant. [2]
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Worked solution
(a) Redshift is an increase in the observed wavelength of light from distant galaxies. Because the light is shifted towards the red end of the spectrum, it shows that the source is moving away from us. Since almost all distant galaxies show redshift, they are all moving away, proving the Universe is expanding.
(b)(i) Using Hubble's law: \(v = H_0 d \Rightarrow d = \frac{v}{H_0}\). \(d = \frac{1.8 \times 10^7\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} = 8.18 \times 10^{24}\text{ m}\).
(b)(ii) The age of the Universe can be estimated as \(T = \frac{1}{H_0}\): \(T = \frac{1}{2.2 \times 10^{-18}\text{ s}^{-1}} = 4.545 \times 10^{17}\text{ s}\). Convert seconds into years: \(T_{\text{years}} = \frac{4.545 \times 10^{17}\text{ s}}{365 \times 24 \times 3600\text{ s/year}} \approx 1.44 \times 10^{10}\text{ years}\) (or \(1.4 \times 10^{10}\text{ years}\)).
Marking scheme
(a) [3 marks] - B1 for: definition of redshift as increase in observed wavelength of light / light shifted to red end of spectrum - B1 for: indicating that redshift is caused by galaxies moving away from us - B1 for: linking the movement of almost all galaxies away from us to the expansion of space / the Universe
(b)(i) [3 marks] - C1 for formula: \(v = H_0 d\) or \(d = v / H_0\) - C1 for substitution: \(1.8 \times 10^7 / 2.2 \times 10^{-18}\) - A1 for distance: \(8.2 \times 10^{24}\text{ m}\) (accept range \(8.1 \times 10^{24}\text{ m} - 8.3 \times 10^{24}\text{ m}\))
(b)(ii) [2 marks] - C1 for formula \(t = 1/H_0\) and attempting conversion of units: \(\frac{1}{2.2 \times 10^{-18} \times 3.15 \times 10^7}\) - A1 for age: \(1.4 \times 10^{10}\text{ years}\) (accept range \(1.4 \times 10^{10} - 1.5 \times 10^{10}\text{ years}\))
Question 9 · Structured
8 marks
A ball of mass 0.40 kg is released from rest from a height of 1.8 m above the ground. (a) Calculate the gravitational potential energy of the ball before it is released. Take the acceleration of free fall $g = 9.8\text{ m/s}^2$. [2] (b) Show that the speed of the ball just before it hits the ground is approximately $5.9\text{ m/s}$. Air resistance is negligible. [2] (c) On bouncing, the ball is in contact with the ground for 0.15 s and rebounds vertically with an initial speed of 4.5 m/s. (i) Calculate the change in momentum (impulse) of the ball during the collision with the ground. [2] (ii) Calculate the average force exerted on the ball by the ground during the collision. [2]
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Worked solution
(a) Gravitational potential energy: $E_p = mgh = 0.40\text{ kg} \times 9.8\text{ m/s}^2 \times 1.8\text{ m} = 7.056\text{ J} \approx 7.1\text{ J}$. (b) Conservation of energy: $E_k = E_p \implies \frac{1}{2}mv^2 = mgh \implies v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.8} = \sqrt{35.28} = 5.939\text{ m/s} \approx 5.9\text{ m/s}$. (c)(i) Define upwards as the positive direction: Initial velocity just before collision, $u = -5.94\text{ m/s}$ (downwards). Final velocity just after collision, $v = +4.5\text{ m/s}$ (upwards). Change in momentum: $\Delta p = m(v - u) = 0.40\text{ kg} \times (4.5\text{ m/s} - (-5.94\text{ m/s})) = 0.40 \times 10.44 = 4.176\text{ kg m/s} \approx 4.2\text{ N s}$. (c)(ii) Average force: $F = \frac{\Delta p}{\Delta t} = \frac{4.176\text{ N s}}{0.15\text{ s}} = 27.84\text{ N} \approx 28\text{ N}$.
Marking scheme
Part (a): - Formula $E_p = mgh$ or correct substitution [1 mark] - Correct calculation of $7.1\text{ J}$ (or $7.06\text{ J}$) with correct unit [1 mark] Part (b): - Equating GPE lost to KE gained or using $v^2 = u^2 + 2as$ [1 mark] - Correct calculation leading to $5.9\text{ m/s}$ with intermediate steps shown [1 mark] Part (c)(i): - Use of $\Delta p = m\Delta v$ with correct sign consideration (giving velocity change $\approx 10.4\text{ m/s}$) [1 mark] - Correct value $4.2\text{ N s}$ (or $4.2\text{ kg m/s}$) with correct unit [1 mark] Part (c)(ii): - Formula $F = \Delta p / t$ or substitution [1 mark] - Correct calculation of $28\text{ N}$ (allow $27.7\text{ N}$ to $28.0\text{ N}$ depending on rounding of velocity) [1 mark]
Question 10 · Structured
8 marks
A ray of monochromatic light travels inside a transparent plastic block towards its flat boundary with air. (a) State what is meant by monochromatic light. [1] (b) The refractive index of the plastic block is 1.62. (i) Calculate the speed of light in the plastic. The speed of light in air is $3.0 \times 10^8\text{ m/s}$. [2] (ii) Calculate the critical angle for light travelling from the plastic into air. [2] (c) The angle of incidence of the ray at the plastic-air boundary is $42^\circ$. State and explain what happens to the light ray at the boundary. [3]
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Worked solution
(a) Monochromatic light is light consisting of a single frequency or a single wavelength. (b)(i) Speed of light in plastic: $n = \frac{c}{v} \implies v = \frac{c}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.62} = 1.85 \times 10^8\text{ m/s} \approx 1.9 \times 10^8\text{ m/s}$. (b)(ii) Critical angle: $\sin c = \frac{1}{n} = \frac{1}{1.62} = 0.6173 \implies c = \sin^{-1}(0.6173) \approx 38.1^\circ \approx 38^\circ$. (c) The angle of incidence ($42^\circ$) is greater than the critical angle ($38^\circ$). Because the ray is traveling inside the denser medium (plastic) towards the less dense medium (air), total internal reflection occurs, meaning all of the light is reflected back inside the block.
Marking scheme
Part (a): - Mention of single frequency / single wavelength [1 mark] Part (b)(i): - Formula $n = c/v$ or $v = c/n$ [1 mark] - Correct calculation of $1.9 \times 10^8\text{ m/s}$ (or $1.85 \times 10^8\text{ m/s}$) with unit [1 mark] Part (b)(ii): - Formula $\sin c = 1/n$ [1 mark] - Correct calculation of $38^\circ$ (or $38.1^\circ$) [1 mark] Part (c): - State that total internal reflection occurs / ray is reflected back inside the block [1 mark] - Explanation that the angle of incidence ($42^\circ$) is greater than the critical angle ($38^\circ$) [1 mark] - Explanation that light is traveling from a more optically dense to a less optically dense medium (or plastic to air) [1 mark]
Paper 6 (Alternative to Practical)
Answer all questions. Use your knowledge of practical physics to plot graphs, identify sources of error, and design experiments.
4 Question · 40 marks
Question 1 · open_ended
10 marks
A student investigates the rate of cooling of hot water in a beaker under two different conditions: **Beaker A** (with no lid) and **Beaker B** (with a plastic lid).
The thermometer in Fig. 1.1 shows the initial temperature \(\theta_0\) of the hot water as \(85.0\text{ }^\circ\text{C}\).
(a) (i) Record the temperature \(\theta_0\) shown on the thermometer. \(\theta_0\) = .................... \(^\circ\text{C}\) [1]
(ii) State one precaution that should be taken when reading a thermometer to ensure that the temperature is measured accurately. [1]
(b) The student records the temperature \(\theta\) of the water in each beaker every 30 s. The readings are shown in Table 1.1.
Table 1.1: | Time \(t\) / s | Temperature \(\theta_A\) in Beaker A / \(^\circ\text{C}\) | Temperature \(\theta_B\) in Beaker B / \(^\circ\text{C}\) | |---|---|---| | 0 | 85.0 | 85.0 | | 30 | 79.5 | 81.5 | | 60 | 75.0 | 78.5 | | 90 | 71.0 | 76.0 | | 120 | 67.5 | 74.0 | | 150 | 64.5 | 72.0 | | 180 | 62.0 | 70.5 |
(i) Calculate the total temperature drop \(\Delta\theta_A\) of the water in Beaker A and \(\Delta\theta_B\) of the water in Beaker B over the 180 s period. \(\Delta\theta_A\) = .................... \(^\circ\text{C}\) \(\Delta\theta_B\) = .................... \(^\circ\text{C}\) [2]
(c) Describe how the student would display the results on a graph of temperature against time, specifying which variable goes on each axis. [2]
(d) Under another condition, Beaker C has a reflective foil lid. Suggest why a shiny foil lid is better at reducing heat loss than a dark plastic lid. [2]
(e) State two variables that should be controlled (kept constant) to ensure a fair comparison of cooling rates between Beaker A and Beaker B. [2]
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Worked solution
(a) (i) The liquid level on the thermometer indicates \(85.0\text{ }^\circ\text{C}\). (ii) To avoid parallax error, the thermometer must be read perpendicular to the scale at eye level. Alternatively, ensure the thermometer bulb does not touch the sides or bottom of the beaker.
(c) Temperature is the dependent variable and is plotted on the vertical y-axis. Time is the independent variable and is plotted on the horizontal x-axis.
(d) Shiny silver foil reflects infrared radiation back into the beaker and is a poor emitter of thermal radiation, whereas dark surfaces are excellent emitters of heat radiation.
(e) Variables to control: same initial temperature (\(85.0\text{ }^\circ\text{C}\)), same volume of water in both beakers, same room temperature, and same material/thickness of the beakers.
Marking scheme
(a) (i) 85.0 °C [1] (ii) Look at eye level / perpendicular to the scale OR do not let bulb touch the bottom/sides [1] (b) (i) Delta theta_A = 23.0 °C AND Delta theta_B = 14.5 °C (both correct for 2 marks, 1 mark if only one is correct) [2] (c) Temperature on y-axis, time on x-axis [1]; appropriate labels with units (°C and s) [1] (d) Reference to poor emission / good reflection of thermal radiation by shiny/silver foil [1]; compared to plastic/dark plastic which is a better emitter [1] (e) Any two correct control variables: same volume of water, same initial temperature, same room temperature, or same beaker type [2]
Question 2 · open_ended
10 marks
A student investigates the extension of a spring when different loads are suspended from it.
(a) Fig. 2.1 shows the spring with no load. A ruler next to the spring is graduated in millimetres.
The top of the spring is aligned with the 0.0 cm mark.
(i) State the unstretched length \(l_0\) of the spring shown in Fig. 2.1, where the pointer is aligned with 42 mm. \(l_0\) = .................... mm [1]
(ii) State one precaution to take when using a ruler to measure the length of the spring to ensure the reading is accurate. [1]
(b) Suspended loads \(L\) of 1.0 N, 2.0 N, 3.0 N, 4.0 N, and 5.0 N are added in turn.
Table 2.1 shows the lengths \(l\) of the spring under these loads.
(i) Calculate the extension \(e\) for a load \(L = 4.0\text{ N}\). \(e\) = .................... mm [1]
(ii) Describe the relationship between the load \(L\) and the extension \(e\) for loads from 0.0 N to 4.0 N. [2]
(iii) Explain whether Hooke's Law is still obeyed at a load of 5.0 N. Justify your answer with reference to the values in Table 2.1. [2]
(c) Suggest one reason why it is important not to suspend excessively heavy loads from the spring during this experiment. [1]
(d) State how the student can check that the spring has not been permanently damaged (permanently stretched) after removing the 5.0 N load. [2]
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Worked solution
(a) (i) \(l_0 = 42\text{ mm}\) as indicated. (ii) Keep the eye level with the pointer to avoid parallax error, or ensure the ruler is parallel to the spring.
(b) (i) \(e = 106 - 42 = 64\text{ mm}\). (ii) Since the extension increases by exactly \(16\text{ mm}\) for every \(1.0\text{ N}\) of load added (up to \(4.0\text{ N}\)), the extension is directly proportional to the load. (iii) At \(5.0\text{ N}\), the extension is \(83\text{ mm}\). If Hooke's Law were obeyed, it should be \(5 \times 16 = 80\text{ mm}\). Since the extension is larger than expected, the limit of proportionality has been exceeded and Hooke's Law is no longer obeyed.
(c) Excessively heavy loads can permanently deform the spring so that it does not return to its original length, ruining the elastic property.
(d) Unhook the load and check if the pointer returns exactly to the initial \(42\text{ mm}\) mark.
Marking scheme
(a) (i) 42 mm (accept 4.2 cm) [1] (ii) Read perpendicular to the scale / keep ruler parallel to the spring [1] (b) (i) 64 mm [1] (ii) Extension is directly proportional to the load [1]; because there is a constant increase of 16 mm per 1.0 N [1] (iii) Hooke's law is not obeyed [1]; because the extension increases more rapidly than expected / the extension is 83 mm instead of 80 mm [1] (c) To avoid exceeding the limit of proportionality / permanent stretching [1] (d) Remove all loads [1]; check that the length returns to the original unstretched length (42 mm) [1]
Question 3 · open_ended
10 marks
A student investigates the refraction of a ray of light passing through a rectangular glass block.
Fig. 3.1 shows the student's ray-trace sheet. The line \(NM\) is the normal to the surface of the block at point \(B\). The incident ray is represented by the line \(AB\), and the refracted ray inside the block is represented by the line \(BC\).
For this online simulation, the angles measured are as follows: - The angle of incidence \(i\) between \(AB\) and \(NM\) is \(30^\circ\). - The angle of refraction \(r\) between \(BC\) and \(NM\) is \(19^\circ\).
(a) (i) State the angle of incidence \(i\). \(i\) = .................... \(^\circ\) [1]
(ii) State the angle of refraction \(r\). \(r\) = .................... \(^\circ\) [1]
(b) The student calculates the refractive index \(n\) of the glass using the formula:
\[n = \frac{\sin i}{\sin r}\]
(i) Calculate the refractive index \(n\). Give your answer to an appropriate number of significant figures. \(n\) = .................... [2]
(ii) Suggest why the refractive index \(n\) has no unit. [1]
(c) In this type of experiment, the student uses optical pins to locate the path of the light rays.
(i) State a suitable minimum distance between the pins used to trace the incident ray \(AB\). [1]
(ii) Explain why placing the pins too close together reduces the accuracy of the ray-trace. [2]
(d) When drawing the outline of the glass block on the ray-trace sheet, the student must use a sharp pencil.
Explain why using a blunt pencil would lead to inaccurate results in measuring angles and calculating the refractive index. [2]
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Worked solution
(a) (i) \(i = 30^\circ\). (ii) \(r = 19^\circ\).
(b) (i) \(n = \frac{\sin 30^\circ}{\sin 19^\circ} = \frac{0.500}{0.3256} \approx 1.54\). (ii) Since \(n\) is the ratio of two sine values, which are dimensionless numbers, the units cancel out, leaving \(n\) unitless.
(c) (i) The minimum distance between tracking pins should be at least \(5.0\text{ cm}\) (or \(50\text{ mm}\)). (ii) If pins are too close, any slight misalignment in placing the ruler along the pins results in a significantly larger uncertainty in the direction of the traced line.
(d) A thick pencil line introduces uncertainty in the exact position of the block's edge. This uncertainty propagates into the measurement of the angles \(i\) and \(r\), resulting in an inaccurate value for \(n\).
Marking scheme
(a) (i) 30(°) [1] (ii) 19(°) [1] (b) (i) sin 30 / sin 19 seen [1]; correct value 1.54 (accept 1.5) [1] (ii) Ratio of two sines / same units cancel out [1] (c) (i) Any value from 5.0 cm to 10.0 cm (or 50 mm to 100 mm) [1] (ii) Larger angle of uncertainty / harder to align ruler accurately with pins if they are close [2] (d) Thick lines make it hard to define exact contact points / normals [1]; leading to errors in protractor measurements [1]
Question 4 · open_ended
10 marks
A student investigates the resistance of combinations of resistors.
Fig. 4.1 shows three circuits: Circuit P (one resistor \(R_1\)), Circuit Q (two identical resistors in series), and Circuit R (two identical resistors in parallel).
(a) The student measures the current \(I\) and the potential difference \(V\) for Circuit P: - The voltmeter reading is exactly \(3.6\text{ V}\). - The ammeter reading is exactly \(0.30\text{ A}\).
(i) Record the potential difference \(V_P\) for Circuit P. \(V_P\) = .................... V [1]
(ii) Record the current \(I_P\) for Circuit P. \(I_P\) = .................... A [1]
(b) The readings for Circuit Q are \(V_Q = 3.6\text{ V}\) and \(I_Q = 0.15\text{ A}\).
The readings for Circuit R are \(V_R = 3.6\text{ V}\) and \(I_R = 0.60\text{ A}\).
Calculate the resistance for Circuit P (\(R_P\)), Circuit Q (\(R_Q\)), and Circuit R (\(R_R\)) using the formula:
\[R = \frac{V}{I}\]
\(R_P\) = .................... \(\Omega\)
\(R_Q\) = .................... \(\Omega\)
\(R_R\) = .................... \(\Omega\) [3]
(c) State, with reference to your calculations in (b), how the total resistance of the circuit changes when:
(i) a second identical resistor is connected in series with the first resistor. [1]
(ii) a second identical resistor is connected in parallel with the first resistor. [1]
(d) (i) Draw the standard circuit symbol for a variable resistor (rheostat). [1]
(ii) Suggest why a variable resistor is included in these circuits during experimental testing. [2]
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(c) (i) In series, resistance increases (doubles from 12 \(\Omega\) to 24 \(\Omega\)). (ii) In parallel, resistance decreases (halves from 12 \(\Omega\) to 6.0 \(\Omega\)).
(d) (i) The standard symbol is a rectangle with a diagonal arrow pointing up and to the right. (ii) A variable resistor allows the student to change the current and potential difference in the circuit, allowing them to take a series of readings to confirm Ohm's Law or find average resistance values.
Marking scheme
(a) (i) 3.6 V [1] (ii) 0.30 A [1] (b) R_P = 12 (Ω) [1]; R_Q = 24 (Ω) [1]; R_R = 6.0 (Ω) [1] (c) (i) Resistance doubles / increases to 24 Ω [1] (ii) Resistance halves / decreases to 6.0 Ω [1] (d) (i) Correct symbol drawn (rectangle with diagonal arrow) [1] (ii) To vary current or potential difference [1]; to get multiple sets of readings [1]
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