An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V2) Cambridge IGCSE Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
Paper 2 Multiple Choice (Extended)
There are forty questions on this paper. Answer all questions. Select one option among A, B, C, and D.
40 Question · 40 marks
Question 1 · multipleChoice
1 marks
A cyclist travels up a steep hill at a constant speed of \( 4.0\text{ m/s} \) for \( 120\text{ s} \). The cyclist then immediately turns around and rides down the same hill at a constant speed of \( 12.0\text{ m/s} \) for \( 40\text{ s} \).
What is the average speed of the cyclist for the entire journey?
A.\( 4.8\text{ m/s} \)
B.\( 6.0\text{ m/s} \)
C.\( 8.0\text{ m/s} \)
D.\( 16.0\text{ m/s} \)
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1. Calculate the distance travelled up the hill: \[ d_1 = \text{speed} \times \text{time} = 4.0\text{ m/s} \times 120\text{ s} = 480\text{ m} \]
2. Calculate the distance travelled down the hill: \[ d_2 = 12.0\text{ m/s} \times 40\text{ s} = 480\text{ m} \]
3. Find the total distance and total time: \[ d_{\text{total}} = d_1 + d_2 = 480\text{ m} + 480\text{ m} = 960\text{ m} \] \[ t_{\text{total}} = 120\text{ s} + 40\text{ s} = 160\text{ s} \]
4. Calculate the average speed: \[ \text{average speed} = \frac{960\text{ m}}{160\text{ s}} = 6.0\text{ m/s} \]
Therefore, the correct option is B.
Marking scheme
1 mark for the correct calculation of average speed, showing that total distance is divided by total time.
Question 2 · multipleChoice
1 marks
A ball of mass \( 0.15\text{ kg} \) hits a vertical wall horizontally with a speed of \( 8.0\text{ m/s} \) and rebounds horizontally with a speed of \( 6.0\text{ m/s} \).
The ball is in contact with the wall for \( 0.050\text{ s} \).
What is the average force exerted by the wall on the ball?
A.\( 6.0\text{ N} \)
B.\( 24\text{ N} \)
C.\( 42\text{ N} \)
D.\( 210\text{ N} \)
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Worked solution
To find the average force, we use the relation between force, change in momentum, and time:
\[ F = \frac{\Delta p}{\Delta t} \]
Let the direction of the initial velocity be positive. - Initial velocity, \( u = 8.0\text{ m/s} \) - Final velocity, \( v = -6.0\text{ m/s} \) (since it rebounds in the opposite direction)
Calculate the change in momentum (impulse): \[ \Delta p = m(v - u) = 0.15\text{ kg} \times (-6.0\text{ m/s} - 8.0\text{ m/s}) = 0.15\text{ kg} \times (-14.0\text{ m/s}) = -2.1\text{ N s} \]
The magnitude of the change in momentum is \( 2.1\text{ N s} \).
Calculate the average force: \[ F = \frac{2.1\text{ N s}}{0.050\text{ s}} = 42\text{ N} \]
Therefore, the correct option is C.
Marking scheme
1 mark for using the correct relationship between force, change in momentum, and contact time, taking into account the change in velocity direction.
Question 3 · multipleChoice
1 marks
A rectangular block of wood of mass \( 180\text{ g} \) has dimensions \( 5.0\text{ cm} \times 6.0\text{ cm} \times 8.0\text{ cm} \).
The block is placed in a beaker containing liquid X (density \( 0.80\text{ g/cm}^3 \)) and liquid Y (density \( 1.2\text{ g/cm}^3 \)), which do not mix.
Where does the block come to rest?
A.floating on top of liquid X
B.submerged entirely in liquid X, but above liquid Y
C.floating at the boundary between liquid X and liquid Y
D.sunk to the bottom of liquid Y
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Worked solution
First, calculate the volume of the block of wood: \[ V = 5.0\text{ cm} \times 6.0\text{ cm} \times 8.0\text{ cm} = 240\text{ cm}^3 \]
Next, calculate the density of the wood: \[ \rho = \frac{m}{V} = \frac{180\text{ g}}{240\text{ cm}^3} = 0.75\text{ g/cm}^3 \]
Now, compare the density of the wood block with the densities of the liquids: - Density of wood block = \( 0.75\text{ g/cm}^3 \) - Density of liquid X = \( 0.80\text{ g/cm}^3 \) - Density of liquid Y = \( 1.2\text{ g/cm}^3 \)
Since the density of the wood block (\( 0.75\text{ g/cm}^3 \)) is less than the density of liquid X (\( 0.80\text{ g/cm}^3 \)), the block will float on top of liquid X.
Therefore, the correct option is A.
Marking scheme
1 mark for calculating the density of the wood block and comparing it correctly to the liquid densities to determine its floating position.
Question 4 · multipleChoice
1 marks
An electric heater of power \( 60\text{ W} \) is used to heat a metal block of mass \( 1.5\text{ kg} \). The heater is switched on for \( 5.0\text{ minutes} \).
The temperature of the block rises from \( 20^\circ\text{C} \) to \( 44^\circ\text{C} \). No thermal energy is lost to the surroundings.
What is the specific heat capacity of the metal?
A.\( 5.0\text{ J / (kg }^\circ\text{C)} \)
B.\( 500\text{ J / (kg }^\circ\text{C)} \)
C.\( 3000\text{ J / (kg }^\circ\text{C)} \)
D.\( 30000\text{ J / (kg }^\circ\text{C)} \)
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Worked solution
First, calculate the total thermal energy supplied by the heater: \[ E = P \times t = 60\text{ W} \times (5.0 \times 60\text{ s}) = 60 \times 300 = 18\,000\text{ J} \]
Next, determine the temperature rise of the metal block: \[ \Delta T = 44^\circ\text{C} - 20^\circ\text{C} = 24^\circ\text{C} \]
Using the specific heat capacity formula: \[ E = m c \Delta T \] \[ 18\,000 = 1.5 \times c \times 24 \] \[ 18\,000 = 36 c \] \[ c = \frac{18\,000}{36} = 500\text{ J / (kg }^\circ\text{C)} \]
Therefore, the correct option is B.
Marking scheme
1 mark for calculating energy supplied and using the specific heat capacity equation to find the correct value.
Question 5 · multipleChoice
1 marks
A water wave travels from deep water into shallow water.
Which row correctly describes the changes, if any, to the speed, frequency, and wavelength of the wave?
A.speed: decreases | frequency: decreases | wavelength: stays the same
B.speed: decreases | frequency: stays the same | wavelength: decreases
C.speed: increases | frequency: stays the same | wavelength: increases
D.speed: stays the same | frequency: increases | wavelength: decreases
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Worked solution
When water waves enter a shallower region: 1. The speed of the wave decreases due to the interaction with the seabed. 2. The frequency of the wave remains unchanged because it depends solely on the source producing the wave. 3. Since \( v = f \lambda \), and the speed \( v \) decreases while the frequency \( f \) remains constant, the wavelength \( \lambda \) must also decrease.
Therefore, the correct option is B.
Marking scheme
1 mark for correctly identifying that wave frequency remains constant and speed and wavelength decrease when waves enter shallow water.
Question 6 · multipleChoice
1 marks
An electric kettle is connected to a \( 230\text{ V} \) mains supply. The current in the heating element of the kettle is \( 8.0\text{ A} \).
How much energy is transferred by the kettle to the water in \( 5.0\text{ minutes} \)?
A.\( 1.8\text{ kJ} \)
B.\( 9.2\text{ kJ} \)
C.\( 550\text{ kJ} \)
D.\( 5500\text{ kJ} \)
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Worked solution
To calculate the electrical energy transferred, we use the formula:
\[ E = V I t \]
Convert time to seconds: \[ t = 5.0\text{ minutes} = 5.0 \times 60\text{ s} = 300\text{ s} \]
Substitute the values into the formula: \[ E = 230\text{ V} \times 8.0\text{ A} \times 300\text{ s} = 1840\text{ W} \times 300\text{ s} = 552\,000\text{ J} = 552\text{ kJ} \]
Rounding to two significant figures gives \( 550\text{ kJ} \).
Therefore, the correct option is C.
Marking scheme
1 mark for using the electrical energy equation correctly and converting time to seconds.
Question 7 · multipleChoice
1 marks
The measured count rate of a sample containing a radioactive isotope is \( 430\text{ counts/minute} \) at the start of an experiment, which includes a constant background count rate of \( 30\text{ counts/minute} \).
After \( 6.0\text{ hours} \), the measured count rate is \( 80\text{ counts/minute} \).
What is the half-life of the radioactive isotope?
A.\( 1.5\text{ hours} \)
B.\( 2.0\text{ hours} \)
C.\( 3.0\text{ hours} \)
D.\( 4.5\text{ hours} \)
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Worked solution
First, subtract the background radiation from both measurements to find the corrected count rates: - Corrected initial count rate: \( 430 - 30 = 400\text{ counts/minute} \) - Corrected final count rate: \( 80 - 30 = 50\text{ counts/minute} \)
Determine the fraction of the initial corrected count rate remaining: \[ \frac{50}{400} = \frac{1}{8} \]
Since \( \frac{1}{8} = \left(\frac{1}{2}\right)^3 \), the sample has undergone exactly \( 3 \) half-lives.
Calculate the duration of one half-life: \[ 3 \times t_{1/2} = 6.0\text{ hours} \] \[ t_{1/2} = \frac{6.0\text{ hours}}{3} = 2.0\text{ hours} \]
Therefore, the correct option is B.
Marking scheme
1 mark for calculating the corrected count rates, finding the number of half-lives elapsed, and determining the correct half-life.
Question 8 · multipleChoice
1 marks
Light from a distant galaxy is observed to have a longer wavelength than expected (redshift).
What does this redshift show about the motion of the galaxy, and what is the relationship between the galaxy's speed of recession and its distance from Earth?
A.motion of galaxy: moving away from Earth | relation: speed is directly proportional to distance
B.motion of galaxy: moving away from Earth | relation: speed is independent of distance
C.motion of galaxy: moving towards Earth | relation: speed is directly proportional to distance
D.motion of galaxy: moving towards Earth | relation: speed is independent of distance
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Worked solution
1. Redshift occurs when light waves from a source are stretched as the source moves away from the observer. This shows that distant galaxies are moving away from Earth. 2. According to Hubble's Law, the recessional speed of a galaxy is directly proportional to its distance from Earth (\( v = H_0 d \)).
Therefore, the correct option is A.
Marking scheme
1 mark for identifying that redshift shows galaxies are moving away and that speed of recession is proportional to distance.
Question 9 · multipleChoice
1 marks
An object is released from rest in a vacuum. It falls under gravity for a distance \(d\). Which graph represents how its kinetic energy \(E_k\) varies with the distance fallen \(d\)?
A.a straight line through the origin with a positive gradient
B.a curve starting from the origin with an increasing gradient
C.a curve starting from the origin with a decreasing gradient
D.a horizontal straight line
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Worked solution
Using conservation of energy, the gravitational potential energy lost is equal to the kinetic energy gained.
Since loss in G.P.E. = \(m g d\), where \(m\) is the mass of the object, \(g\) is the acceleration of free fall, and \(d\) is the distance fallen, we have: \(E_k = m g d\)
Since \(m\) and \(g\) are constant, \(E_k\) is directly proportional to \(d\). This relationship is represented by a straight line passing through the origin with a positive constant gradient.
Marking scheme
C1 for identifying that kinetic energy gained equals potential energy lost (\(E_k = m g d\)) A1 for selecting the correct proportional relationship (straight line through the origin)
Question 10 · multipleChoice
1 marks
A toy truck of mass 1.2 kg travelling at 2.5 m / s collides with a stationary toy car of mass 0.8 kg. After the collision, the truck and the car couple together and move off with a common velocity. What is the total kinetic energy lost during the collision?
A.0.60 J
B.1.50 J
C.2.25 J
D.3.75 J
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Worked solution
First, find the final common velocity \(v\) using conservation of momentum: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\) \((1.2 \times 2.5) + (0.8 \times 0) = (1.2 + 0.8) v\) \(3.0 = 2.0 v \implies v = 1.5 \text{ m / s}\)
Calculate the final kinetic energy \(E_{k,f}\): \(E_{k,f} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 2.0 \times 1.5^2 = 2.25 \text{ J}\)
Calculate the energy lost: \(\text{Energy lost} = 3.75 - 2.25 = 1.50 \text{ J}\)
Marking scheme
C1 for calculating final velocity using conservation of momentum (\(1.5 \text{ m / s}\)) C1 for calculating initial and final kinetic energies A1 for finding the correct difference in kinetic energy (\(1.50 \text{ J}\))
Question 11 · multipleChoice
1 marks
A gas is trapped in a cylinder by a piston. The volume of the gas is \(120 \text{ cm}^3\) and its pressure is \(1.5 \times 10^5 \text{ Pa}\). The piston is moved so that the volume decreases to \(90 \text{ cm}^3\) while the temperature is kept constant. What is the new pressure of the gas?
A.1.1 \times 10^5 \text{ Pa}
B.1.3 \times 10^5 \text{ Pa}
C.2.0 \times 10^5 \text{ Pa}
D.2.7 \times 10^5 \text{ Pa}
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Worked solution
For a constant mass of gas at constant temperature, Boyle's law applies: \(P_1 V_1 = P_2 V_2\)
Substitute the given values: \(1.5 \times 10^5 \times 120 = P_2 \times 90\)
C1 for using Boyle's Law formula (\(P_1 V_1 = P_2 V_2\)) A1 for calculating the correct final pressure (\(2.0 \times 10^5 \text{ Pa}\))
Question 12 · multipleChoice
1 marks
A ray of light is travelling inside a transparent plastic block towards the boundary with air. The refractive index of the plastic is 1.45. Which row correctly gives the critical angle for the plastic-air boundary and describes what happens to a ray incident at an angle of 45°?
A.The critical angle is 44° and the ray undergoes total internal reflection.
B.The critical angle is 44° and the ray is refracted into the air.
C.The critical angle is 46° and the ray undergoes total internal reflection.
D.The critical angle is 46° and the ray is refracted into the air.
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Worked solution
First, find the critical angle \(c\) using: \(\sin c = \frac{1}{n} = \frac{1}{1.45} \approx 0.6897\) \(c \approx 43.6^\circ \approx 44^\circ\)
Since the angle of incidence (45°) is greater than the critical angle (44°), the light cannot refract into the air and instead undergoes total internal reflection back into the plastic block.
Marking scheme
C1 for calculating the critical angle using \(\sin c = 1 / n\) (\(44^\circ\)) A1 for concluding that total internal reflection occurs since the angle of incidence is greater than the critical angle
Question 13 · multipleChoice
1 marks
A wire of length \(L\) and cross-sectional area \(A\) has a resistance of \(8.0 \ \Omega\). A second wire is made of the same metal but has a length of \(3L\) and a cross-sectional area of \(2A\). What is the resistance of the second wire?
A.5.3 \Omega
B.12 \Omega
C.16 \Omega
D.48 \Omega
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Worked solution
The resistance \(R\) of a wire is given by the formula: \(R = \rho \frac{\text{length}}{\text{area}}\)
For the first wire: \(R_1 = \rho \frac{L}{A} = 8.0 \ \Omega\)
For the second wire: \(R_2 = \rho \frac{3L}{2A} = 1.5 \times \left(\rho \frac{L}{A}\right) = 1.5 \times 8.0 = 12 \ \Omega\)
Marking scheme
C1 for relating resistance to length and cross-sectional area (\(R \propto L / A\)) A1 for calculating the correct resistance of the second wire (\(12 \ \Omega\))
Question 14 · multipleChoice
1 marks
An ideal step-down transformer has 400 turns on its primary coil and 100 turns on its secondary coil. An alternating current of 2.0 A is supplied to the primary coil at a voltage of 240 V. What are the secondary voltage and the secondary current?
A.secondary voltage = 60 V, secondary current = 0.50 A
B.secondary voltage = 60 V, secondary current = 8.0 A
C.secondary voltage = 960 V, secondary current = 0.50 A
D.secondary voltage = 960 V, secondary current = 8.0 A
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Worked solution
Using the transformer ratio equation: \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\) \(V_s = 240 \times \frac{100}{400} = 60 \text{ V}\)
For an ideal transformer, input power equals output power: \(V_p I_p = V_s I_s\) \(240 \times 2.0 = 60 \times I_s\) \(I_s = \frac{480}{60} = 8.0 \text{ A}\)
Marking scheme
C1 for calculating secondary voltage (60 V) C1 for using conservation of power (\(V_p I_p = V_s I_s\)) to find secondary current A1 for finding the correct combination: 60 V and 8.0 A
Question 15 · multipleChoice
1 marks
A detector is placed near a radioactive source. The background count rate is constant at 15 counts / minute. The initial reading on the detector is 255 counts / minute. After 12 hours, the reading on the detector is 45 counts / minute. What is the half-life of the radioactive source?
A.2.0 hours
B.3.0 hours
C.4.0 hours
D.6.0 hours
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Worked solution
First, subtract the background count rate to find the corrected count rates due to the source alone: \(\text{Initial corrected rate} = 255 - 15 = 240 \text{ counts / minute}\) \(\text{Final corrected rate} = 45 - 15 = 30 \text{ counts / minute}\)
Determine the fraction of active nuclei remaining: \(\text{Fraction remaining} = \frac{30}{240} = \frac{1}{8}\)
Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly 3 half-lives have elapsed.
C1 for subtracting background to find corrected initial (240) and final (30) rates C1 for determining that 3 half-lives have elapsed A1 for finding the correct half-life (4.0 hours)
Question 16 · multipleChoice
1 marks
The light from a distant galaxy is observed to be redshifted. Which statement about the galaxy and the Universe is correct?
A.The galaxy is moving towards the Earth, showing that the Universe is contracting.
B.The galaxy is moving away from the Earth, showing that the Universe is contracting.
C.The galaxy is moving towards the Earth, showing that the Universe is expanding.
D.The galaxy is moving away from the Earth, showing that the Universe is expanding.
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Worked solution
Redshift is the increase in wavelength of electromagnetic radiation as it travels through expanding space. The observation of redshift in light from distant galaxies indicates that they are moving away from the Earth, which provides key evidence that the Universe is expanding.
Marking scheme
A1 for explaining that redshift indicates galaxies moving away and the expansion of the Universe
Question 17 · multipleChoice
1 marks
A skydiver of mass \(80\text{ kg}\) falls from rest. The upward air resistance force \(F\) acting on the skydiver is given by \(F = kv^2\), where \(v\) is the speed in \(\text{m/s}\) and \(k = 0.80\text{ N s}^2/\text{m}^2\). Take the acceleration of free fall \(g = 9.8\text{ m/s}^2\). What is the terminal velocity of the skydiver?
A.9.8 m/s
B.31 m/s
C.780 m/s
D.980 m/s
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Worked solution
At terminal velocity, the upward air resistance force equals the downward force of gravity (weight). Hence, \(F = W \Rightarrow kv^2 = mg\). Substituting the given values: \(0.80 \times v^2 = 80 \times 9.8\), which gives \(0.80 \times v^2 = 784\). Therefore, \(v^2 = 980\), giving \(v = \sqrt{980} \approx 31.3\text{ m/s}\). To two significant figures, this is \(31\text{ m/s}\).
Marking scheme
C1 for identifying that at terminal velocity, \(kv^2 = mg\). A1 for correct calculation of \(v = 31\text{ m/s}\).
Question 18 · multipleChoice
1 marks
An electric pump has an efficiency of \(65\%\). The pump lifts \(3.0\text{ kg}\) of water through a vertical height of \(8.0\text{ m}\) every second. What is the electrical power input to the pump? (Take the acceleration of free fall \(g = 9.8\text{ m/s}^2\).)
A.150 W
B.240 W
C.360 W
D.550 W
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Worked solution
First, find the useful power output: \(P_{\text{out}} = \frac{\text{work done}}{t} = \frac{mgh}{t} = \frac{3.0 \times 9.8 \times 8.0}{1.0} = 235.2\text{ W}\). Since the efficiency is \(65\%\), the electrical power input \(P_{\text{in}}\) is calculated as: \(P_{\text{in}} = \frac{P_{\text{out}}}{\text{efficiency}} = \frac{235.2}{0.65} \approx 361.8\text{ W}\). This is approximately \(360\text{ W}\).
Marking scheme
C1 for calculating the useful power output as \(235\text{ W}\). A1 for using efficiency to find the input power of \(360\text{ W}\).
Question 19 · multipleChoice
1 marks
A gas is trapped in a cylinder by a piston. The volume of the gas is \(120\text{ cm}^3\) and its pressure is \(1.0 \times 10^5\text{ Pa}\). The piston is pushed in slowly, at constant temperature, until the volume of the gas is reduced to \(40\text{ cm}^3\). What is the new pressure of the gas, and how does the average speed of the gas particles change?
A.New pressure: \(3.0 \times 10^5\text{ Pa}\); Average speed: increases
B.New pressure: \(3.0 \times 10^5\text{ Pa}\); Average speed: stays the same
C.New pressure: \(0.33 \times 10^5\text{ Pa}\); Average speed: decreases
D.New pressure: \(0.33 \times 10^5\text{ Pa}\); Average speed: stays the same
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Worked solution
By Boyle's Law at constant temperature, \(P_1 V_1 = P_2 V_2 \Rightarrow P_2 = \frac{1.0 \times 10^5 \times 120}{40} = 3.0 \times 10^5\text{ Pa}\). Since the temperature is kept constant, the average kinetic energy of the gas particles remains unchanged, meaning their average speed stays the same.
Marking scheme
C1 for calculating the new pressure as \(3.0 \times 10^5\text{ Pa}\) using Boyle's Law. A1 for identifying that the average speed stays the same because temperature is constant.
Question 20 · multipleChoice
1 marks
A ray of monochromatic light in a glass block reaches the boundary with air. The critical angle for glass to air is \(42^\circ\). What is the speed of light in this glass block? (The speed of light in air is \(3.0 \times 10^8\text{ m/s}\).)
A.1.3 \times 10^8 m/s
B.2.0 \times 10^8 m/s
C.3.0 \times 10^8 m/s
D.4.5 \times 10^8 m/s
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Worked solution
The relationship between the refractive index \(n\) and the critical angle \(c\) is given by: \(n = \frac{1}{\sin(c)} = \frac{1}{\sin(42^\circ)} \approx 1.494\). The speed of light in glass \(v\) is related to the speed of light in air \(c_0\) by: \(v = \frac{c_0}{n} = \frac{3.0 \times 10^8}{1.494} \approx 2.0 \times 10^8\text{ m/s}\).
Marking scheme
C1 for finding the refractive index \(n \approx 1.49\) using the critical angle formula. A1 for calculating the speed of light in glass as \(2.0 \times 10^8\text{ m/s}\).
Question 21 · multipleChoice
1 marks
A metal wire carries a current of \(0.40\text{ A}\) for a duration of \(5.0\text{ minutes}\). How much charge passes through any cross-section of the wire in this time, and what is the direction of the flow of electrons compared to the conventional current?
A.Charge: \(2.0\text{ C}\); Direction of electrons: opposite to the conventional current
B.Charge: \(2.0\text{ C}\); Direction of electrons: same as the conventional current
C.Charge: \(120\text{ C}\); Direction of electrons: opposite to the conventional current
D.Charge: \(120\text{ C}\); Direction of electrons: same as the conventional current
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Worked solution
First, convert time into seconds: \(t = 5.0 \times 60 = 300\text{ s}\). Use the formula \(Q = I \times t\) to calculate the charge: \(Q = 0.40 \times 300 = 120\text{ C}\). The flow of negatively charged electrons is always in the opposite direction to the conventional current, which represents the flow of positive charge.
Marking scheme
C1 for converting time to seconds and calculating charge as \(120\text{ C}\). A1 for stating that electron flow is opposite to conventional current.
Question 22 · multipleChoice
1 marks
A \(6.0\ \Omega\) resistor is connected in parallel with a \(3.0\ \Omega\) resistor. This parallel combination is then connected in series with a \(4.0\ \Omega\) resistor and a \(12\text{ V}\) power supply. What is the potential difference across the \(4.0\ \Omega\) resistor?
A.2.0 V
B.4.0 V
C.8.0 V
D.12 V
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Worked solution
Find the combined resistance of the parallel combination: \(R_p = \frac{6.0 \times 3.0}{6.0 + 3.0} = 2.0\ \Omega\). Find the total resistance of the circuit: \(R_{\text{total}} = R_p + 4.0 = 2.0 + 4.0 = 6.0\ \Omega\). Find the total current in the circuit: \(I = \frac{V}{R_{\text{total}}} = \frac{12}{6.0} = 2.0\text{ A}\). The potential difference across the \(4.0\ \Omega\) resistor is: \(V_4 = I \times 4.0 = 2.0 \times 4.0 = 8.0\text{ V}\).
Marking scheme
C1 for finding the equivalent parallel resistance as \(2.0\ \Omega\) and total circuit resistance as \(6.0\ \Omega\). A1 for calculating the voltage drop across the \(4.0\ \Omega\) resistor as \(8.0\text{ V}\).
Question 23 · multipleChoice
1 marks
A radioactive sample has a half-life of \(8.0\text{ days}\). Initially, the recorded count rate from the sample is \(500\text{ counts per minute}\), which includes a constant background radiation of \(20\text{ counts per minute}\). What will the recorded count rate be after \(24\text{ days}\)?
A.60 counts per minute
B.62.5 counts per minute
C.80 counts per minute
D.82.5 counts per minute
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Worked solution
The initial corrected count rate (from the source only) is: \(500 - 20 = 480\text{ counts/minute}\). The number of half-lives that elapse in \(24\text{ days}\) is: \(\frac{24}{8.0} = 3\text{ half-lives}\). After 3 half-lives, the corrected count rate of the source is: \(\frac{480}{2^3} = \frac{480}{8} = 60\text{ counts/minute}\). The recorded count rate (including background) will be: \(60 + 20 = 80\text{ counts/minute}\).
Marking scheme
C1 for subtracting the background radiation to get \(480\text{ cpm}\) and finding that 3 half-lives have elapsed. A1 for calculating the final recorded count rate as \(80\text{ cpm}\).
Question 24 · multipleChoice
1 marks
Light from a distant galaxy is observed to have a longer wavelength than the light emitted by the same elements in a laboratory on Earth. Which statement correctly explains this observation and its implication for the Universe?
A.The galaxy is moving away from the Earth because of redshift, indicating that the Universe is expanding.
B.The galaxy is moving towards the Earth because of redshift, indicating that the Universe is contracting.
C.The galaxy is moving away from the Earth because of blueshift, indicating that the Universe is contracting.
D.The galaxy is moving towards the Earth because of blueshift, indicating that the Universe is expanding.
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Worked solution
An increase in the wavelength of observed light is called redshift. This shift to longer wavelengths indicates that the light source (the galaxy) is moving away from the observer (Earth). Since almost all distant galaxies show redshift, this is major evidence that the Universe is expanding.
Marking scheme
C1 for associating the increase in wavelength with redshift. A1 for stating that redshift implies galaxies are moving away and the Universe is expanding.
Question 25 · multipleChoice
1 marks
A car accelerates from rest to a speed of \(20\text{ m/s}\) in \(10\text{ s}\). It then travels at this constant speed for \(15\text{ s}\), before decelerating uniformly to rest in a further \(5.0\text{ s}\). What is the average speed of the car for the entire journey?
A.\(11\text{ m/s}\)
B.\(15\text{ m/s}\)
C.\(17\text{ m/s}\)
D.\(20\text{ m/s}\)
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Worked solution
To find the average speed, we calculate the total distance travelled and divide it by the total time. 1. Acceleration phase (0 to 10 s): distance \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10\text{ s} \times 20\text{ m/s} = 100\text{ m}\). 2. Constant speed phase (10 to 25 s): distance \(d_2 = \text{speed} \times \text{time} = 20\text{ m/s} \times 15\text{ s} = 300\text{ m}\). 3. Deceleration phase (25 to 30 s): distance \(d_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5.0\text{ s} \times 20\text{ m/s} = 50\text{ m}\). Total distance \(d_{\text{total}} = 100\text{ m} + 300\text{ m} + 50\text{ m} = 450\text{ m}\). Total time \(t_{\text{total}} = 10\text{ s} + 15\text{ s} + 5.0\text{ s} = 30\text{ s}\). Average speed = \(\frac{450\text{ m}}{30\text{ s}} = 15\text{ m/s}\).
Marking scheme
B is the correct option. 1 mark for the correct calculation of average speed.
Question 26 · multipleChoice
1 marks
A ball of mass \(0.15\text{ kg}\) hits a vertical wall horizontally at a speed of \(12\text{ m/s}\). It rebounds horizontally in the opposite direction at a speed of \(8.0\text{ m/s}\). What is the magnitude of the impulse exerted on the ball by the wall?
A.\(0.60\text{ N s}\)
B.\(1.2\text{ N s}\)
C.\(1.8\text{ N s}\)
D.\(3.0\text{ N s}\)
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Worked solution
Impulse is equal to the change in momentum: \(\Delta p = m(v - u)\). Taking the direction towards the wall as positive, the initial velocity is \(u = +12\text{ m/s}\) and the rebound velocity is \(v = -8.0\text{ m/s}\). \(\text{Impulse} = 0.15\text{ kg} \times (-8.0\text{ m/s} - 12\text{ m/s}) = 0.15 \times (-20) = -3.0\text{ N s}\). The magnitude of the impulse is \(3.0\text{ N s}\).
Marking scheme
D is the correct option. 1 mark for the correct calculation of momentum change including the direction change.
Question 27 · multipleChoice
1 marks
An object floats in a beaker of liquid which has a density of \(1.2\text{ g/cm}^3\). Exactly \(60\%\) of the volume of the object is submerged beneath the surface of the liquid. What is the density of the object?
A.\(0.50\text{ g/cm}^3\)
B.\(0.72\text{ g/cm}^3\)
C.\(1.2\text{ g/cm}^3\)
D.\(2.0\text{ g/cm}^3\)
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Worked solution
For a floating object, the upthrust (buoyant force) equals the weight of the object: \(F_{\text{up}} = W\) \(\rho_{\text{liquid}} \times V_{\text{submerged}} \times g = \rho_{\text{object}} \times V_{\text{total}} \times g\) Since \(V_{\text{submerged}} = 0.60 \times V_{\text{total}}\), we have: \(\rho_{\text{object}} = 0.60 \times \rho_{\text{liquid}} = 0.60 \times 1.2\text{ g/cm}^3 = 0.72\text{ g/cm}^3\).
Marking scheme
B is the correct option. 1 mark for using the correct proportion of liquid density for floating equilibrium.
Question 28 · multipleChoice
1 marks
A portable lamp is connected to a battery. The potential difference across the lamp is \(6.0\text{ V}\). During a certain period, \(90\text{ J}\) of electrical energy is transferred to the lamp. How much charge passes through the lamp during this time?
A.\(0.067\text{ C}\)
B.\(15\text{ C}\)
C.\(540\text{ C}\)
D.\(3240\text{ C}\)
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Worked solution
The potential difference \(V\) is defined as the work done (energy transferred) \(E\) per unit charge \(Q\): \(V = \frac{E}{Q}\) Rearranging the equation to solve for charge gives: \(Q = \frac{E}{V} = \frac{90\text{ J}}{6.0\text{ V}} = 15\text{ C}\).
Marking scheme
B is the correct option. 1 mark for the correct calculation of charge using energy and voltage.
Question 29 · multipleChoice
1 marks
The North pole of a magnet is on the left and the South pole is on the right, creating a magnetic field pointing from left to right. A horizontal wire lies in this field and is moved vertically upwards, cutting the field lines. What is the direction of the induced electromotive force (e.m.f.)?
A.vertically downwards
B.horizontally to the right
C.into the page (away from the observer)
D.out of the page (towards the observer)
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Worked solution
According to Fleming's Right-Hand Rule: - The First finger (Field) points from left to right (North to South). - The Thumb (Motion) points vertically upwards. - Therefore, the Second finger (induced current / e.m.f.) points into the page, which is away from the observer.
Marking scheme
C is the correct option. 1 mark for correctly applying Fleming's Right-Hand Rule.
Question 30 · multipleChoice
1 marks
A fixed mass of gas is trapped inside a container of constant volume. The temperature of the gas is increased from \(20\text{ }^\circ\text{C}\) to \(100\text{ }^\circ\text{C}\). Which statement describes what happens to the gas particles and the pressure of the gas?
A.The gas particles move faster and hit the walls of the container with greater force, so the pressure increases.
B.The gas particles move slower and hit the walls of the container less frequently, so the pressure decreases.
C.The gas particles move faster, but because the volume is constant, the pressure remains the same.
D.The gas particles move at the same speed, but they collide with each other more frequently, so the pressure increases.
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Worked solution
An increase in temperature increases the average kinetic energy of the gas particles, making them move faster. Faster-moving particles collide with the walls of the container more frequently and with greater force, resulting in an increase in gas pressure.
Marking scheme
A is the correct option. 1 mark for explaining the link between kinetic energy, force of collisions, and pressure.
Question 31 · multipleChoice
1 marks
A radioactive source has a half-life of \(4.0\text{ hours}\). A detector measures a count rate of \(420\text{ counts/minute}\) near the source. The background count rate is constant at \(20\text{ counts/minute}\). What count rate does the detector measure after \(12\text{ hours}\)?
A.\(50\text{ counts/minute}\)
B.\(70\text{ counts/minute}\)
C.\(105\text{ counts/minute}\)
D.\(125\text{ counts/minute}\)
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Worked solution
1. Determine initial corrected count rate: \(420 - 20 = 400\text{ counts/minute}\). 2. Calculate number of half-lives in 12 hours: \(\frac{12}{4.0} = 3\text{ half-lives}\). 3. Calculate corrected count rate after 3 half-lives: \(400 \times \left(\frac{1}{2}\right)^3 = \frac{400}{8} = 50\text{ counts/minute}\). 4. Calculate measured count rate by adding the background count rate back: \(50 + 20 = 70\text{ counts/minute}\).
Marking scheme
B is the correct option. 1 mark for correctly accounting for background count rate before and after decay calculation.
Question 32 · multipleChoice
1 marks
A specific spectral line of hydrogen has a wavelength of \(486\text{ nm}\) when measured in a laboratory on Earth. In the light from a distant galaxy, the same spectral line is observed at a wavelength of \(513\text{ nm}\). Which statement about this galaxy is correct?
A.The galaxy is moving away from the Earth because the light has been blueshifted.
B.The galaxy is moving away from the Earth because the light has been redshifted.
C.The galaxy is moving towards the Earth because the light has been blueshifted.
D.The galaxy is moving towards the Earth because the light has been redshifted.
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Worked solution
The wavelength of the light has increased from \(486\text{ nm}\) to \(513\text{ nm}\). An increase in wavelength corresponds to a shift towards the red end of the spectrum (redshift). Redshift indicates that the galaxy is moving away from the observer on Earth.
Marking scheme
B is the correct option. 1 mark for linking increased wavelength to redshift and recessional motion.
Question 33 · multipleChoice
1 marks
An athlete runs a race. A speed–time graph represents their motion.
The athlete accelerates from rest at a constant rate for \(4.0\text{ s}\) until reaching a speed of \(9.0\text{ m/s}\). They then run at this constant speed of \(9.0\text{ m/s}\) for \(6.0\text{ s}\).
What is the average speed of the athlete during these \(10.0\text{ s}\)?
A.\(4.5\text{ m/s}\)
B.\(7.2\text{ m/s}\)
C.\(8.1\text{ m/s}\)
D.\(9.0\text{ m/s}\)
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Worked solution
First, calculate the distance travelled in each section of the journey: - During acceleration (first \(4.0\text{ s}\)), distance \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 9.0\text{ m/s} = 18.0\text{ m}\). - During constant speed (next \(6.0\text{ s}\)), distance \(d_2 = \text{speed} \times \text{time} = 9.0\text{ m/s} \times 6.0\text{ s} = 54.0\text{ m}\).
Calculate total distance and total time: - Total distance \(d = d_1 + d_2 = 18.0\text{ m} + 54.0\text{ m} = 72.0\text{ m}\). - Total time \(t = 10.0\text{ s}\).
The alloy contains \(30\text{ cm}^3\) of metal X of density \(8.0\text{ g/cm}^3\) and \(70\text{ cm}^3\) of metal Y of density \(10.0\text{ g/cm}^3\).
No volume change occurs when the metals are mixed.
What is the average density of the alloy?
A.\(8.6\text{ g/cm}^3\)
B.\(9.0\text{ g/cm}^3\)
C.\(9.4\text{ g/cm}^3\)
D.\(9.6\text{ g/cm}^3\)
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Worked solution
First, calculate the total mass of the alloy: - Mass of metal X \(m_{\text{X}} = \text{density} \times \text{volume} = 8.0\text{ g/cm}^3 \times 30\text{ cm}^3 = 240\text{ g}\). - Mass of metal Y \(m_{\text{Y}} = 10.0\text{ g/cm}^3 \times 70\text{ cm}^3 = 700\text{ g}\). - Total mass \(m = m_{\text{X}} + m_{\text{Y}} = 240\text{ g} + 700\text{ g} = 940\text{ g}\).
Next, determine total volume: - Total volume \(V = 30\text{ cm}^3 + 70\text{ cm}^3 = 100\text{ cm}^3\).
A trolley X of mass \(2.0\text{ kg}\) travels at \(4.0\text{ m/s}\) and collides with a stationary trolley Y of mass \(3.0\text{ kg}\).
After the collision, trolley X bounces back with a speed of \(0.5\text{ m/s}\) in the opposite direction.
What is the magnitude of the impulse exerted on trolley Y during the collision?
A.\(1.0\text{ N s}\)
B.\(7.0\text{ N s}\)
C.\(9.0\text{ N s}\)
D.\(10.0\text{ N s}\)
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Worked solution
By Newton's third law and the conservation of momentum, the impulse exerted on trolley Y is equal in magnitude to the change in momentum of trolley X.
Let the initial direction of trolley X be positive. - Initial velocity of X, \(u_{\text{X}} = +4.0\text{ m/s}\). - Final velocity of X, \(v_{\text{X}} = -0.5\text{ m/s}\).
Calculate the change in momentum (impulse) of trolley X: - \(\Delta p_{\text{X}} = m_{\text{X}} (v_{\text{X}} - u_{\text{X}})\) - \(\Delta p_{\text{X}} = 2.0\text{ kg} \times (-0.5\text{ m/s} - 4.0\text{ m/s}) = 2.0 \times (-4.5) = -9.0\text{ N s}\).
Therefore, the magnitude of the impulse exerted on trolley Y is \(9.0\text{ N s}\).
Marking scheme
Award 1 mark for the correct option C.
Question 36 · multipleChoice
1 marks
The temperature of a gas in a sealed, rigid container of fixed volume is increased.
How does this affect the average kinetic energy of the gas molecules and the frequency of their collisions with the container walls?
A.Average kinetic energy: increases / Frequency of collisions: increases
B.Average kinetic energy: increases / Frequency of collisions: stays the same
C.Average kinetic energy: stays the same / Frequency of collisions: increases
D.Average kinetic energy: stays the same / Frequency of collisions: stays the same
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Worked solution
Temperature is a measure of the average kinetic energy of the gas molecules. As temperature increases, the average kinetic energy of the molecules increases.
Because the molecules are moving faster on average in a container of fixed volume, they will collide with the walls of the container more frequently. Therefore, both variables increase.
Marking scheme
Award 1 mark for the correct option A.
Question 37 · multipleChoice
1 marks
A water wave in a ripple tank has a wavelength of \(1.5\text{ cm}\) and a frequency of \(8.0\text{ Hz}\).
As the wave enters a shallower region of water, its speed decreases to half of its initial speed.
What is the wavelength of the wave in the shallower water?
A.\(0.75\text{ cm}\)
B.\(1.5\text{ cm}\)
C.\(3.0\text{ cm}\)
D.\(12\text{ cm}\)
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Worked solution
When a wave enters a different medium (or shallower water), its frequency remains constant because the frequency is determined entirely by the source.
Using the wave equation \(v = f \lambda\), since frequency \(f\) is constant, the wavelength \(\lambda\) is directly proportional to the speed \(v\).
Since the speed decreases to half, the wavelength also decreases to half: - \(\lambda_{\text{new}} = \frac{1}{2} \times 1.5\text{ cm} = 0.75\text{ cm}\).
Marking scheme
Award 1 mark for the correct option A.
Question 38 · multipleChoice
1 marks
A charge of \(30\text{ C}\) passes through a lamp in \(2.0\text{ minutes}\). The potential difference (p.d.) across the lamp is \(6.0\text{ V}\).
What is the current in the lamp and how much energy is transferred to the lamp during this time?
A.Current: \(0.25\text{ A}\) / Energy transferred: \(180\text{ J}\)
B.Current: \(0.25\text{ A}\) / Energy transferred: \(15\text{ J}\)
C.Current: \(15\text{ A}\) / Energy transferred: \(180\text{ J}\)
D.Current: \(15\text{ A}\) / Energy transferred: \(1080\text{ J}\)
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Worked solution
First, convert time into seconds: - \(t = 2.0\text{ minutes} = 120\text{ s}\).
Calculate energy transferred: - \(E = V Q = 6.0\text{ V} \times 30\text{ C} = 180\text{ J}\).
Marking scheme
Award 1 mark for the correct option A.
Question 39 · multipleChoice
1 marks
A circuit consists of a \(6.0\ \Omega\) resistor connected in parallel with a \(12\ \Omega\) resistor. This parallel combination is connected in series with a \(4.0\ \Omega\) resistor across a \(12\text{ V}\) d.c. supply.
What is the current drawn from the d.c. supply?
A.\(0.55\text{ A}\)
B.\(1.0\text{ A}\)
C.\(1.5\text{ A}\)
D.\(3.0\text{ A}\)
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Worked solution
First, calculate the resistance of the parallel combination, \(R_{\text{p}}\): - \(\frac{1}{R_{\text{p}}} = \frac{1}{6.0\ \Omega} + \frac{1}{12\ \Omega} = \frac{2 + 1}{12} = \frac{3}{12}\) - \(R_{\text{p}} = 4.0\ \Omega\).
Next, calculate the total resistance of the circuit, \(R_{\text{total}}\): - \(R_{\text{total}} = R_{\text{p}} + R_{\text{series}} = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega\).
Finally, calculate the total current drawn using Ohm's law: - \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\).
Marking scheme
Award 1 mark for the correct option C.
Question 40 · multipleChoice
1 marks
A detector is used to measure the activity of a radioactive source. The background count rate is constant at \(20\text{ counts/minute}\).
Initially, the detector measures a total count rate of \(340\text{ counts/minute}\). After \(6.0\text{ hours}\), the detector measures a total count rate of \(60\text{ counts/minute}\).
What is the half-life of the radioactive source?
A.\(1.2\text{ hours}\)
B.\(2.0\text{ hours}\)
C.\(3.0\text{ hours}\)
D.\(6.0\text{ hours}\)
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Worked solution
Subtract background radiation to find the corrected count rates of the source: - Initial corrected count rate \(= 340 - 20 = 320\text{ counts/minute}\). - Final corrected count rate \(= 60 - 20 = 40\text{ counts/minute}\).
Find the ratio of final to initial corrected count rates: - \(\frac{40}{320} = \frac{1}{8} = \left(\frac{1}{2}\right)^3\).
Answer all questions. Show all working clearly and include appropriate units.
11 Question · 78 marks
Question 1 · structured
7 marks
A drone of mass \(1.5\text{ kg}\) starts from rest and accelerates uniformly vertically upwards at \(2.4\text{ m/s}^2\) for a duration of \(5.0\text{ s}\).
(a) (i) Calculate the velocity of the drone at \(t = 5.0\text{ s}\).
(ii) Calculate the height the drone reaches during these \(5.0\text{ s}\).
(b) After \(5.0\text{ s}\), the motors are adjusted so that the drone continues to ascend at a constant velocity. Explain, in terms of the forces acting on the drone, why its acceleration is now zero.
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Worked solution
(a) (i) Using \(v = u + at\): \(v = 0 + (2.4 \times 5.0) = 12\text{ m/s}\)
(ii) Using \(s = ut + \frac{1}{2}at^2\): \(s = 0 + \frac{1}{2} \times 2.4 \times (5.0)^2 = 1.2 \times 25 = 30\text{ m}\)
(b) When ascending at constant velocity, the acceleration is zero. According to Newton's first law, this means there is no resultant force. The upward thrust force produced by the rotors must be equal in magnitude and opposite in direction to the sum of the downward forces (the weight of the drone and any air resistance/drag).
Marking scheme
(a) (i) - \(v = u + at\) or \(2.4 \times 5.0\) [1] - \(12\text{ m/s}\) [1]
(ii) - \(s = ut + \frac{1}{2}at^2\) or \(0.5 \times 2.4 \times 25\) [1] - \(30\text{ m}\) [1]
(b) - constant velocity means zero resultant force / balanced forces [1] - upward force / thrust is equal to downward force(s) [1] - downward forces are weight and air resistance / drag [1]
Question 2 · structured
7 marks
(a) Define momentum in words.
(b) Bumper car A of mass \(240\text{ kg}\) (including its driver) is travelling at \(3.2\text{ m/s}\). It collides with bumper car B of mass \(200\text{ kg}\) which is travelling in the same direction at \(1.5\text{ m/s}\). After the collision, car A has a velocity of \(1.8\text{ m/s}\) in the same direction.
(i) Calculate the velocity of car B immediately after the collision.
(ii) The duration of the collision is \(0.25\text{ s}\). Calculate the average force exerted on car A during the collision.
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Worked solution
(a) Momentum is the product of mass and velocity (\(p = mv\)).
(ii) - formula for impulse or force: \(F = \frac{m\Delta v}{t}\) [1] - correct calculation of change in momentum of car A: \(-336\text{ kg m/s}\) (or \(336\text{ kg m/s}\)) [1] - \(-1344\text{ N}\) (or magnitude \(1340\text{ N}\) / \(1300\text{ N}\)) [1]
Question 3 · structured
7 marks
A solar panel is mounted on the roof of a house to heat domestic water.
(a) State the name of the main process of thermal energy transfer by which energy is transferred from the Sun to the solar panel.
(b) The panel has a dull black surface covered with a transparent sheet of glass.
(i) Explain how the dull black surface helps to heat the water inside the panel.
(ii) Explain how the glass cover reduces thermal energy loss from the panel by convection.
(c) On a clear day, the water in the panel absorbs \(4.2 \times 10^6\text{ J}\) of thermal energy over a period of \(2.0\text{ hours}\). Calculate the average useful power output of the panel.
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Worked solution
(a) Radiation (or infrared radiation).
(b) (i) Dull black surfaces are excellent absorbers of thermal radiation, ensuring that the maximum possible amount of energy from the Sun is absorbed to heat the water.
(ii) The glass cover traps a layer of air inside the panel, which stops warm air from escaping and prevents wind from setting up convection currents that would carry heat away from the hot pipes.
(b) The critical angle is the angle of incidence in the optically denser medium that produces an angle of refraction of \(90^\circ\). Total internal reflection occurs when: 1. Light travels from an optically denser medium to a less dense medium. 2. The angle of incidence is greater than the critical angle.
(b) - critical angle is angle of incidence for which the angle of refraction is \(90^\circ\) [1] - TIR condition 1: travelling from optically denser to less dense medium [1] - TIR condition 2: angle of incidence is greater than the critical angle [1]
Question 5 · structured
7 marks
(a) Define electromotive force (e.m.f.) in terms of energy and charge.
(b) An electric toaster is connected to a \(230\text{ V}\) mains supply. The current in its heating element is \(4.5\text{ A}\).
(i) Calculate the resistance of the heating element.
(ii) Calculate the electrical energy transferred in the element when the toaster is operated for \(3.0\text{ minutes}\).
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Worked solution
(a) Electromotive force (e.m.f.) is the electrical work done per unit charge by a source (such as a battery or generator) in driving charge around a complete circuit.
(ii) - conversion of minutes to seconds: \(180\text{ s}\) [1] - \(E = VIt\) or \(E = P \times t\) [1] - \(1.9 \times 10^5\text{ J}\) (or \(1.86 \times 10^5\text{ J}\)) [1]
Question 6 · structured
7 marks
A step-down transformer has a primary coil with \(600\text{ turns}\) and a secondary coil with \(150\text{ turns}\).
(a) The primary coil is connected to a \(240\text{ V}\) alternating current (a.c.) supply.
(i) Calculate the output voltage across the secondary coil.
(ii) Explain why the transformer does not produce an output voltage if the primary coil is connected to a \(12\text{ V}\) direct current (d.c.) battery.
(b) State one reason why a real transformer is not 100% efficient, and suggest a practical design feature used to minimize this loss.
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(ii) A d.c. battery produces a constant current, which creates a constant, unchanging magnetic field in the core. A voltage is only induced in the secondary coil when there is a changing magnetic field cutting through it (electromagnetic induction requires a change in magnetic flux linkage).
(b) One reason is the electrical resistance of the copper wire coils, which generates thermal energy. This is minimized by using thicker wires with lower resistance. (Another reason is eddy currents in the iron core, minimized by using a laminated core).
(ii) - d.c. produces a constant current / steady magnetic field [1] - induction requires a changing magnetic field / flux [1] - no rate of change of flux linkage means no induced e.m.f. [1]
(b) - identify cause: resistance of coils OR eddy currents in core OR magnetic hysteresis [1] - identify solution: thick copper wire / low resistance OR laminated core OR soft iron core [1]
Question 7 · structured
7 marks
(a) The radioactive isotope iodine-131 (\(^{131}_{\ \ 53}\text{I}\)) decays to xenon-131 (\(^{131}_{\ \ 54}\text{Xe}\)) by emitting a \(\beta\)-particle.
(i) State the number of protons and the number of neutrons in a nucleus of iodine-131.
(ii) Write down the nuclide notation for a \(\beta\)-particle.
(b) Iodine-131 has a half-life of \(8.0\text{ days}\). A sample initially has an activity of \(1.6 \times 10^5\text{ Bq}\).
(i) Calculate the activity of the sample after \(24\text{ days}\).
(ii) Explain why radioactive waste with a very short half-life (e.g., several hours) is considered less of a long-term environmental hazard than waste with a very long half-life.
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Worked solution
(a) (i) Protons = 53 Neutrons = \(131 - 53 = 78\)
(ii) Nuclide notation: \(^{\ 0}_{-1}\text{e}\) or \(^{\ 0}_{-1}\beta\)
(b) (i) Number of half-lives = \(\frac{24\text{ days}}{8.0\text{ days}} = 3\) Activity after 3 half-lives = \(\frac{1.6 \times 10^5}{2^3} = \frac{1.6 \times 10^5}{8} = 2.0 \times 10^4\text{ Bq}\)
(ii) Waste with a very short half-life decays very rapidly. Within a few days or weeks, its activity will decrease to safe, negligible levels. This means it does not require complex, secure long-term geological storage for thousands of years like waste with long half-lives.
Marking scheme
(a) (i) - Protons = 53 AND Neutrons = 78 [1]
(ii) - \(^{\ 0}_{-1}\text{e}\) or \(^{\ 0}_{-1}\beta\) [1]
(ii) - short half-life means high rate of decay / decays quickly [1] - activity drops to safe / background levels in a short time [1] - no need for long-term secure containment [1]
Question 8 · structured
7 marks
Mars orbits the Sun at an average distance of \(2.28 \times 10^{11}\text{ m}\). It takes \(687\text{ Earth days}\) to complete one orbital revolution.
(a) (i) Show that the orbital period of Mars is approximately \(5.94 \times 10^7\text{ s}\).
(ii) Calculate the average orbital speed of Mars in \(\text{m/s}\).
(b) State the name of the force that keeps Mars in its orbit and explain how this force depends on the mass of Mars and its distance from the Sun.
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(b) The force is gravity (or gravitational force). This force is directly proportional to the mass of Mars, and it decreases as the distance between Mars and the Sun increases (specifically, it is inversely proportional to the square of the distance).
(ii) - \(v = \frac{2\pi r}{T}\) [1] - correct substitution of \(r\) and \(T\) [1] - \(2.41 \times 10^4\text{ m/s}\) (accept \(2.4 \times 10^4\text{ m/s}\)) [1]
(b) - gravity / gravitational force [1] - increases with mass of Mars AND decreases with distance [1]
Question 9 · structured
7 marks
(a) A ray of light travels from a glass block into air.
(i) Explain what is meant by the term *critical angle*. [2]
(ii) The refractive index of the glass is 1.54. Calculate the critical angle for light traveling from this glass into air. [2]
(b) A ray of light is incident normally on one of the shorter faces of a right-angled isosceles glass prism (with angles \(45^\circ\), \(45^\circ\), and \(90^\circ\)). The glass has a refractive index of 1.54.
State and explain what happens to the ray when it strikes the hypotenuse face of the prism. [3]
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Worked solution
(a) (i) The critical angle is the angle of incidence in the optically denser medium that results in an angle of refraction of \(90^\circ\) in the less dense medium.
(ii) Using the formula: \[\sin(c) = \frac{1}{n}\] \[\sin(c) = \frac{1}{1.54} = 0.6494\] \[c = \sin^{-1}(0.6494) \approx 40.5^\circ \text{ (or } 41^\circ\)\]
(b) Since the ray enters normally, it passes into the prism without bending and strikes the hypotenuse face at an angle of incidence of \(45^\circ\).
Since this angle of incidence (\(45^\circ\)) is greater than the critical angle (\(40.5^\circ\)), the light cannot refract out of the glass. Instead, it undergoes total internal reflection and reflects at \(45^\circ\) inside the prism.
Marking scheme
(a) (i) - Angle of incidence in the denser medium [1] - Resulting in an angle of refraction of 90 degrees [1]
(ii) - Recall/use of \(\sin(c) = 1/n\) [1] - Correct calculation to give \(40.5^\circ\) or \(41^\circ\) [1]
(b) - Identifies that the angle of incidence at the hypotenuse face is \(45^\circ\) [1] - States that \(45^\circ > \text{critical angle}\) (or \(40.5^\circ\)) [1] - Concludes that total internal reflection occurs [1]
Question 10 · structured
8 marks
(a) Bismuth-212 (\(^{212}_{83}\text{Bi}\)) is a radioactive isotope that decays to polonium-212 (\(^{212}_{84}\text{Po}\)).
(i) Identify the type of radiation emitted during this decay. [1]
(ii) Write a complete nuclear equation for this decay. [3]
(b) Alternatively, a nucleus of bismuth-212 can decay by emitting an alpha particle (\(\alpha\)) to form a nucleus of thallium (\(\text{Tl}\)).
State the proton number (atomic number) and the nucleon number (mass number) of the thallium nucleus formed in this decay. [2]
(c) The half-life of bismuth-212 is 60.5 minutes. Calculate the fraction of bismuth-212 remaining in a sample after 242 minutes. [2]
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Worked solution
(a) (i) Since the proton number increases by 1 (from 83 to 84) while the nucleon number remains the same (212), the decay must be a beta-minus (\(\beta^-\)) decay.
(ii) The nuclear decay equation is: \[^{212}_{83}\text{Bi} \rightarrow\ ^{212}_{84}\text{Po} +\ ^{0}_{-1}\text{e}\] (or using \(\beta\) in place of \(\text{e}\)).
(b) An alpha particle is a helium nucleus, \(^{4}_{2}\text{He}\). Thus, the nucleon number decreases by 4 and the proton number decreases by 2: - Proton number: \(83 - 2 = 81\) - Nucleon number: \(212 - 4 = 208\)
(c) First, determine the number of half-lives that have elapsed: \[\text{Number of half-lives} = \frac{242 \text{ minutes}}{60.5 \text{ minutes}} = 4.0\]
(ii) - Left-hand side correct: \(^{212}_{83}\text{Bi}\) [1] - Right-hand side has \(^{212}_{84}\text{Po}\) [1] - Right-hand side has \(^{0}_{-1}\text{e}\) or \(^{0}_{-1}\beta\) [1]
(c) - Calculates number of half-lives as 4 [1] - Calculates fraction remaining as 1/16 or 0.0625 [1]
Question 11 · structured
7 marks
(a) Define *resistance*. [1]
(b) A metallic wire of length \(L\) and uniform cross-sectional area \(A\) has a resistance of \(3.0\ \Omega\).
(i) A second wire is made of the same metal but has a length of \(2.5L\) and a cross-sectional area of \(0.50A\). Calculate the resistance of this second wire. [2]
(ii) The original \(3.0\ \Omega\) wire is connected across the terminals of a battery with an electromotive force (e.m.f.) of \(9.0\text{ V}\). The current in the circuit is \(2.5\text{ A}\). Calculate the internal resistance of the battery. [2]
(iii) Calculate the rate at which electrical energy is converted into thermal energy in the \(3.0\ \Omega\) wire. [2]
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Worked solution
(a) Resistance is defined as the ratio of potential difference (p.d.) across a component to the current flowing through it: \[R = \frac{V}{I}\]
(b) (i) Resistance is proportional to length and inversely proportional to cross-sectional area: \[R \propto \frac{L}{A}\] For the second wire: \[R_2 = R_1 \times \frac{L_2}{L_1} \times \frac{A_1}{A_2} = 3.0 \times 2.5 \times \frac{1}{0.50} = 3.0 \times 5.0 = 15\ \Omega\]
(ii) The total resistance of the circuit (including the internal resistance \(r\) of the battery) is: \[R_{\text{total}} = \frac{E}{I} = \frac{9.0\text{ V}}{2.5\text{ A}} = 3.6\ \Omega\] Since the external wire has a resistance of \(3.0\ \Omega\): \[r = R_{\text{total}} - R = 3.6\ \Omega - 3.0\ \Omega = 0.60\ \Omega\]
(iii) The rate of thermal energy dissipation (power) in the external wire is: \[P = I^2 R = (2.5\text{ A})^2 \times 3.0\ \Omega = 6.25 \times 3.0 = 18.75\text{ W}\] (or \(19\text{ W}\) to 2 significant figures).
Marking scheme
(a) - ratio of potential difference / voltage to current (or \(R = V/I\) with terms defined) [1]
(ii) - Calculates total circuit resistance as \(9.0 / 2.5 = 3.6\ \Omega\) [1] - Subtracts external resistance to get internal resistance \(r = 0.60\ \Omega\) [1]
(iii) - Use of \(P = I^2 R\) (or equivalent, e.g., \(V = I \times R = 7.5\text{ V}\) then \(P = V \times I\)) [1] - Correct calculation to give \(18.75\text{ W}\) or \(19\text{ W}\) [1]
Paper 6 Alternative to Practical
Answer all questions. Show all working and identify experimental control parameters.
4 Question · 40 marks
Question 1 · practical
10 marks
A student investigates the mass of a uniform meter rule using a balancing method.
Fig. 1.1 shows the experimental setup where a pivot is placed at the center of the rule (the \(50.0\text{ cm}\) mark). A fixed mass \(M = 100\text{ g}\) is positioned at a distance \(d\) to the left of the pivot. A suspended mass \(m = 50\text{ g}\) is positioned at a distance \(x\) to the right of the pivot and adjusted until the rule is balanced.
(a) In one test, the mass \(M\) is centered at the \(20.0\text{ cm}\) mark. Calculate the distance \(d\) from the pivot.
\(d = \) .................... \( \text{cm} \)
(b) The mass \(m = 50\text{ g}\) is positioned to balance the ruler. The center of mass \(m\) is aligned with the \(85.4\text{ cm}\) mark. Calculate the distance \(x\) from the pivot.
\(x = \) .................... \( \text{cm} \)
(c) The student repeats the experiment for several values of \(d\) and records the balancing distances in Table 1.1.
(c) Plot the points \((15.0, 30.2)\), \((20.0, 40.1)\), \((25.0, 49.8)\), \((30.0, 60.3)\), and \((35.0, 70.0)\). Draw a straight line of best fit that passes closely through all points.
(d) (i) Choose two distant points on the line, e.g., \((15.0, 30.0)\) and \((35.0, 70.0)\). \(G = \frac{70.0 - 30.0}{35.0 - 15.0} = \frac{40.0}{20.0} = 2.0\).
(ii) Using the formula: \(m_{\text{rule}} = 50 \times 2.0 = 100\text{ g}\).
(e) Ensure that the mass is hung from a thin loop of thread so that the exact position of its center of gravity aligns with the ruler's scale, or ensure the experiment is performed in a draft-free environment to avoid external forces disturbing the balance.
Marking scheme
*(a)* [1 mark] \(30.0\text{ cm}\) (must have units or unit in header). *(b)* [1 mark] \(35.4\text{ cm}\). *(c)* [3 marks total]: - Axes correctly labeled with quantity and unit (x-axis: \(d / \text{cm}\), y-axis: \(x / \text{cm}\)) [1] - At least 4 points plotted correctly to within half a small square [1] - Best-fit straight line drawn with a thin, clear, continuous line [1] *(d)* [3 marks total]: - (i) Triangle method shown on graph covering at least half of the line [1] - Gradient \(G\) calculated in the range \(1.95 - 2.05\) [1] - (ii) Correct calculation of \(m_{\text{rule}}\) using the candidate's gradient value, rounded to 2 or 3 sig. figs. with unit (\(\text{g}\)) [1] *(e)* [2 marks total]: - Any valid precaution, e.g., hang masses using thin thread / look perpendicular to the scale to avoid parallax error / perform in draft-free area [1] - Explanation of how it improves accuracy [1]
Question 2 · practical
10 marks
A student investigates the thermal insulation of different materials by measuring the rate of cooling of hot water in two beakers.
Beaker A is uninsulated. Beaker B is wrapped in a layer of thick wool.
(a) Fig. 2.1 shows a thermometer indicating the room temperature before the start of the experiment. Read and record the room temperature \(\theta_R\).
(b) The student pours \(150\text{ cm}^3\) of hot water into each beaker and records the temperature of the water every \(30\text{ seconds}\) for \(180\text{ seconds}\). Table 2.1 shows the results.
(c) State, with reference to the data in Table 2.1, which beaker has the more effective insulation. Explain your reasoning.
(d) Calculate the average rate of cooling \(R_A\) of the water in Beaker A during the first \(60\text{ seconds}\). Use the equation: \(R_A = \frac{\theta_{\text{start}} - \theta_{60}}{60}\) Include the unit.
\(R_A = \) ....................
(e) Identify two variables that the student must keep constant to ensure a fair comparison between the cooling rates of the two beakers.
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Worked solution
(a) Reading from the thermometer diagram shows a level halfway between 21 and 22, hence \(\theta_R = 21.5\text{ °C}\).
(c) Beaker B is wrapped in thick wool and has a lower temperature drop (\(14.5\text{ °C}\) vs \(21.0\text{ °C}\)), demonstrating that the wool insulation restricts thermal energy transfer more effectively than no insulation.
(e) Control variables are essential to maintain validity. These include using the same volume of water (\(150\text{ cm}^3\)), starting at the same initial temperature (\(85.0\text{ °C}\)), and using identical beakers in a draft-free environment.
Marking scheme
*(a)* [1 mark] \(21.5\text{ °C}\) (allow \(21.0\text{ °C} - 22.0\text{ °C}\)). *(b)* [2 marks total]: - (i) \(21.0\text{ °C}\) [1] - (ii) \(14.5\text{ °C}\) [1] *(c)* [2 marks total]: - Beaker B identified as having better insulation [1] - Quantitative reason given comparing temperature drops (e.g., \(14.5\text{ °C} < 21.0\text{ °C}\)) [1] *(d)* [2 marks total]: - Correct numerical calculation: \(0.17\) or \(0.167\) [1] - Correct unit: \(\text{°C / s}\) or \(\text{°C / min}\) if calculated appropriately [1] *(e)* [3 marks total]: - Volume of hot water [1] - Initial temperature of water [1] - Same room temperature / thickness of beaker glass / use of lid on both [1] (any two)
Question 3 · practical
10 marks
A student investigates the resistance of a constantan wire of different lengths.
Fig. 3.1 shows the experimental circuit used by the student.
(a) Fig. 3.2 shows the ammeter and voltmeter readings when a wire of length \(L = 40.0\text{ cm}\) is connected in the circuit.
(i) Read and record the current \(I\) from the ammeter.
\(I = \) .................... \(\text{A}\)
(ii) Read and record the potential difference \(V\) from the voltmeter.
\(V = \) .................... \(\text{V}\)
(b) Calculate the resistance \(R\) of the \(40.0\text{ cm}\) length of wire using the equation: \(R = \frac{V}{I}\) Include the unit.
\(R = \) ....................
(c) The student records the resistance for other lengths \(L\) in Table 3.1.
(c) (i) Enter \(3.6\) in Table 3.1. (ii) The relationship is direct proportionality. This is justified because as length is doubled from \(20.0\text{ cm}\) to \(40.0\text{ cm}\), resistance doubles from \(1.8\text{ }\Omega\) to \(3.6\text{ }\Omega\), keeping the ratio \(R/L\) constant.
(d) The switch is used to control current flow. Opening it between readings prevents continuous current, which would cause the wire to heat up, changing its resistivity and introducing error into the measurements.
(e) Draw a standard rectangular resistor symbol with a diagonal arrow pointing upwards and to the right.
Marking scheme
*(a)* [2 marks total]: - (i) \(0.40\text{ A}\) (accept \(0.4\text{ A}\)) [1] - (ii) \(1.44\text{ V}\) (accept \(1.4\text{ V} - 1.5\text{ V}\)) [1] *(b)* [2 marks total]: - Correct numerical value based on readings: \(3.6\) [1] - Correct unit: \(\Omega\) or Ohm [1] *(c)* [3 marks total]: - (i) Table completed with value from (b) [1] - (ii) Statement: Directly proportional (or equivalent) [1] - Justification: Proof that doubling length doubles resistance within experimental error [1] *(d)* [2 marks total]: - To prevent the wire from heating up / temperature rising [1] - Resistance of the wire changes with temperature, so keeping temperature constant ensures a fair test [1] *(e)* [1 mark]: - Correct symbol for variable resistor (standard rectangle with diagonal arrow).
Question 4 · practical
10 marks
A student wants to investigate how the terminal velocity of a falling object depends on its mass.
A single lightweight cupcake paper case is dropped from a height, and it quickly reaches terminal velocity. The mass can be increased by nesting identical paper cases inside each other, which increases the total mass while keeping the size and shape the same.
Plan an experiment to investigate this relationship.
You are provided with: - a supply of identical cupcake paper cases - a digital balance - a meter rule
You may use other common laboratory apparatus.
In your plan, you should: 1. list any additional apparatus needed 2. describe the method for carrying out the investigation, including how you ensure the paper cases have reached terminal velocity before measurement starts 3. state the key variables to control 4. present a suitable table with column headings and units to show how the readings will be recorded (no data is required) 5. explain how you would process the results to draw a conclusion.
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Worked solution
1. **Additional Apparatus**: - A stopwatch (to measure fall time). - Sticky tape or markers to mark start and finish levels on a wall or vertical stand. - A vertical height line marker (such as a plumb line or second meter rule).
2. **Method**: - Measure the mass \(M\) of a single paper case using the balance. - Set up a clear vertical drop zone. Mark a starting timing line at least \(1.0\text{ m}\) below the release point. This ensure the case has accelerated and achieved constant (terminal) velocity before timing begins. - Mark a finish line at a distance \(d = 1.00\text{ m}\) below the start line. - Drop the case and measure the time \(t\) taken to fall between the start and finish lines. - Repeat the drop 3 times and calculate the average time. - Increase the mass by nesting a second case inside the first. Repeat the measurements. - Continue for stacks of up to 5 cases.
3. **Control Variables**: - The shape and surface area of the falling object (by using identical cases nested tightly). - The timing distance \(d\) between the marks. - Draft-free environment (doors and windows closed to prevent air currents).
4. **Results Table**:
| Number of cases \(N\) | Mass \(m / \text{g}\) | Distance \(d / \text{m}\) | Time \(t_1 / \text{s}\) | Time \(t_2 / \text{s}\) | Average Time \(t_{\text{avg}} / \text{s}\) | Terminal Velocity \(v / \text{m/s}\) | |---|---|---|---|---|---|---|
5. **Processing Results**: - Calculate the terminal velocity using the equation \(v = \frac{d}{t_{\text{avg}}}\). - Plot a graph of terminal velocity \(v\) (y-axis) against mass \(m\) (x-axis) to analyze how terminal velocity depends on mass.
Marking scheme
*(Planning Question - 10 Marks total)*: - **Apparatus** [2 marks]: - Identifies a stopwatch / timing device [1] - Identifies tape/markers to mark the start/end lines of the timing zone [1] - **Method** [3 marks]: - Clear description of releasing paper case and timing it over a fixed distance [1] - Explains that the start line must be below the release point (e.g., \(\ge 1.0\text{ m}\)) to allow terminal velocity to be reached [1] - Mentions repeating each drop and taking an average time [1] - **Control Variables** [2 marks]: - States that the shape/surface area of the cases must be kept constant [1] - Mentions using the same timing distance \(d\) or ensuring a draft-free room [1] - **Results Table** [2 marks]: - Includes columns for number of cases / mass with correct units [1] - Includes columns for time and calculated terminal velocity with units [1] - **Analysis** [1 mark]: - States the formula used to calculate velocity (\(v = d/t\)) and describes plotting a graph of \(v\) against \(m\) to find the relationship [1]
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