An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V3) Cambridge International A Level Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
Section Multiple Choice (Extended)
Answer all forty questions. For each question there are four possible answers, A, B, C, and D. Choose the one you consider correct.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
A car accelerates from rest at a constant rate of \(2.0\text{ m/s}^2\) for \(5.0\text{ s}\), then travels at a constant speed for \(10\text{ s}\), and finally decelerates to rest at a constant rate of \(1.0\text{ m/s}^2\). What is the total distance traveled by the car?
A.\(125\text{ m}\)
B.\(150\text{ m}\)
C.\(175\text{ m}\)
D.\(225\text{ m}\)
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Worked solution
First phase (constant acceleration from rest): \(u = 0\text{ m/s}\), \(a = 2.0\text{ m/s}^2\), \(t = 5.0\text{ s}\). The final velocity of this phase is \(v = u + at = 0 + 2.0 \times 5.0 = 10\text{ m/s}\). The distance is \(s_1 = \frac{1}{2}(u + v)t = \frac{1}{2}(0 + 10) \times 5.0 = 25\text{ m}\). Second phase (constant speed): The car travels at \(10\text{ m/s}\) for \(10\text{ s}\). The distance is \(s_2 = v \times t = 10 \times 10 = 100\text{ m}\). Third phase (constant deceleration to rest): The car decelerates from \(10\text{ m/s}\) to \(0\text{ m/s}\) at a rate of \(1.0\text{ m/s}^2\). The time taken is \(t = \frac{10}{1.0} = 10\text{ s}\). The distance is \(s_3 = \frac{1}{2}(10 + 0) \times 10 = 50\text{ m}\). The total distance is \(s_1 + s_2 + s_3 = 25 + 100 + 50 = 175\text{ m}\).
Marking scheme
C is correct. 1 mark for calculating the distances of the three phases (25 m, 100 m, and 50 m) and summing them correctly to get 175 m.
Question 2 · multiple-choice
1 marks
An alloy is made by mixing metal X and metal Y. Metal X has a density of \(8.0\text{ g/cm}^3\) and a volume of \(30\text{ cm}^3\). Metal Y has a density of \(5.0\text{ g/cm}^3\) and a mass of \(150\text{ g}\). Assuming there is no change in total volume when they are mixed, what is the density of the alloy?
A.\(5.5\text{ g/cm}^3\)
B.\(6.5\text{ g/cm}^3\)
C.\(6.8\text{ g/cm}^3\)
D.\(7.0\text{ g/cm}^3\)
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Worked solution
First, calculate the mass of metal X: \(m_X = \text{density}_X \times \text{volume}_X = 8.0\text{ g/cm}^3 \times 30\text{ cm}^3 = 240\text{ g}\). Next, calculate the volume of metal Y: \(V_Y = \frac{m_Y}{\text{density}_Y} = \frac{150\text{ g}}{5.0\text{ g/cm}^3} = 30\text{ cm}^3\). The total mass of the alloy is \(m_{\text{total}} = 240\text{ g} + 150\text{ g} = 390\text{ g}\). The total volume of the alloy is \(V_{\text{total}} = 30\text{ cm}^3 + 30\text{ cm}^3 = 60\text{ cm}^3\). Finally, the density of the alloy is \(\rho_{\text{alloy}} = \frac{m_{\text{total}}}{V_{\text{total}}} = \frac{390\text{ g}}{60\text{ cm}^3} = 6.5\text{ g/cm}^3\).
Marking scheme
B is correct. 1 mark for finding the total mass (390 g) and total volume (60 cm3), and dividing them to find the correct density of 6.5 g/cm3.
Question 3 · multiple-choice
1 marks
An object of mass \(4.0\text{ kg}\) is acted upon by two forces at right angles to each other. One force is \(12\text{ N}\) acting north, and the other is \(9.0\text{ N}\) acting east. What is the magnitude of the acceleration of the object?
A.\(0.75\text{ m/s}^2\)
B.\(3.0\text{ m/s}^2\)
C.\(3.75\text{ m/s}^2\)
D.\(5.25\text{ m/s}^2\)
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Worked solution
Because the two forces act at right angles, the magnitude of the resultant force \(F\) can be found using Pythagoras' theorem: \(F = \sqrt{12^2 + 9.0^2} = \sqrt{144 + 81} = \sqrt{225} = 15\text{ N}\). Using Newton's second law, \(F = ma\), the magnitude of the acceleration \(a\) is: \(a = \frac{F}{m} = \frac{15\text{ N}}{4.0\text{ kg}} = 3.75\text{ m/s}^2\).
Marking scheme
C is correct. 1 mark for determining the resultant force is 15 N and then using Newton's second law to calculate an acceleration of 3.75 m/s^2.
Question 4 · multiple-choice
1 marks
A toy car of mass \(0.50\text{ kg}\) travels at \(4.0\text{ m/s}\) to the right. It collides head-on with a toy truck of mass \(1.5\text{ kg}\) traveling at \(2.0\text{ m/s}\) to the left. After the collision, the toy truck rebounds and travels at \(0.50\text{ m/s}\) to the right. What is the velocity of the toy car after the collision?
A.\(1.5\text{ m/s to the left}\)
B.\(1.5\text{ m/s to the right}\)
C.\(3.5\text{ m/s to the left}\)
D.\(3.5\text{ m/s to the right}\)
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Worked solution
Let rightwards be the positive direction. Initial momentum of the system: \(p_i = m_1u_1 + m_2u_2 = (0.50 \times 4.0) + (1.5 \times -2.0) = 2.0 - 3.0 = -1.0\text{ kg m/s}\). Final momentum of the system: \(p_f = m_1v_1 + m_2v_2 = 0.50v_1 + (1.5 \times 0.50) = 0.50v_1 + 0.75\). By conservation of momentum, \(p_i = p_f \implies -1.0 = 0.50v_1 + 0.75 \implies -1.75 = 0.50v_1 \implies v_1 = -3.5\text{ m/s}\). The negative sign indicates that the toy car is traveling at \(3.5\text{ m/s}\) to the left.
Marking scheme
C is correct. 1 mark for using the conservation of momentum with correct signs for direction, yielding a final velocity of 3.5 m/s to the left.
Question 5 · multiple-choice
1 marks
A crane lifts a load of mass \(600\text{ kg}\) vertically upwards through a height of \(12\text{ m}\) in a time of \(12\text{ s}\). The total electric power supplied to the crane's motor is \(9.8\text{ kW}\). Using \(g = 9.8\text{ N/kg}\), what is the efficiency of the crane system?
A.\(48\%\)
B.\(50\%\)
C.\(60\%\)
D.\(72\%\)
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Worked solution
The useful work done by the crane in lifting the load is equal to the increase in gravitational potential energy: \(\Delta E_p = mgh = 600\text{ kg} \times 9.8\text{ N/kg} \times 12\text{ m} = 70560\text{ J}\). The useful power output is: \(P_{\text{out}} = \frac{\text{Work}}{t} = \frac{70560\text{ J}}{12\text{ s}} = 5880\text{ W} = 5.88\text{ kW}\). The power input is \(9.8\text{ kW}\). The efficiency is: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{5.88\text{ kW}}{9.8\text{ kW}} \times 100\% = 60\%\).
Marking scheme
C is correct. 1 mark for calculating the useful output power as 5.88 kW and dividing by the input power of 9.8 kW to get an efficiency of 60%.
Question 6 · multiple-choice
1 marks
An electric heater of power \(800\text{ W}\) is used to heat a \(2.0\text{ kg}\) block of metal. The heater is switched on for \(3.0\text{ minutes}\). The temperature of the block rises from \(20^\circ\text{C}\) to \(110^\circ\text{C}\). Assuming no thermal energy is lost to the surroundings, what is the specific heat capacity of the metal?
A.\(130\text{ J / (kg }^\circ\text{C)}\)
B.\(400\text{ J / (kg }^\circ\text{C)}\)
C.\(800\text{ J / (kg }^\circ\text{C)}\)
D.\(1200\text{ J / (kg }^\circ\text{C)}\)
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Worked solution
The total thermal energy supplied by the heater is: \(E = P \times t = 800\text{ W} \times (3.0 \times 60\text{ s}) = 144,000\text{ J}\). The temperature rise is: \(\Delta T = 110^\circ\text{C} - 20^\circ\text{C} = 90^\circ\text{C}\). Using the formula \(E = mc\Delta T\), we rearrange for specific heat capacity \(c\): \(c = \frac{E}{m\Delta T} = \frac{144,000\text{ J}}{2.0\text{ kg} \times 90^\circ\text{C}} = 800\text{ J / (kg }^\circ\text{C)}\).
Marking scheme
C is correct. 1 mark for finding the energy supplied (144,000 J) and using it with the temperature change of 90 degrees Celsius to find a specific heat capacity of 800 J/(kg °C).
Question 7 · multiple-choice
1 marks
A wire has resistance \(R\), length \(L\), and cross-sectional area \(A\). A second wire made of the same material has twice the length and half the diameter of the first wire. What is the resistance of the second wire?
A.\(\frac{1}{2}R\)
B.R
C.4R
D.8R
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Worked solution
The resistance of a wire is given by \(R = \rho \frac{L}{A}\). The cross-sectional area is related to the diameter \(d\) by \(A = \pi \frac{d^2}{4}\). If the second wire has half the diameter, its cross-sectional area is \(A_2 = \pi \frac{(d/2)^2}{4} = \frac{A}{4}\). If it also has twice the length, its resistance is: \(R_2 = \rho \frac{2L}{A/4} = 8 \rho \frac{L}{A} = 8R\).
Marking scheme
D is correct. 1 mark for identifying that halving the diameter reduces the cross-sectional area by a factor of 4, which combined with doubling the length increases the resistance by a factor of 8.
Question 8 · multiple-choice
1 marks
A radioactive source is placed near a radiation detector. The detector measures a count rate of \(440\text{ counts/minute}\). The background count rate is constant at \(40\text{ counts/minute}\). The half-life of the source is \(3.0\text{ hours}\). What is the expected measured count rate after \(9.0\text{ hours}\)?
A.\(50\text{ counts/minute}\)
B.\(55\text{ counts/minute}\)
C.\(90\text{ counts/minute}\)
D.\(110\text{ counts/minute}\)
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Worked solution
First, subtract the background count rate to find the initial count rate due only to the source: \(440 - 40 = 400\text{ counts/minute}\). The time elapsed is \(9.0\text{ hours}\), which is exactly \(\frac{9.0}{3.0} = 3\) half-lives. After 3 half-lives, the count rate due to the source is halved 3 times: \(400 \times \left(\frac{1}{2}\right)^3 = 50\text{ counts/minute}\). Finally, add the background count rate back to find the expected total measured count rate: \(50 + 40 = 90\text{ counts/minute}\).
Marking scheme
C is correct. 1 mark for performing the background correction (subtracting 40 to get 400), decaying the source rate over 3 half-lives to get 50, and adding background back to find 90 counts/minute.
Question 9 · multiple-choice
1 marks
A toy car accelerates from rest at \(2.0 \text{ m/s}^2\) for \(4.0 \text{ s}\). It then travels at a constant speed for \(6.0 \text{ s}\). Finally, it decelerates uniformly to rest in \(5.0 \text{ s}\). What is the total distance traveled by the toy car?
A.\(64 \text{ m}\)
B.\(80 \text{ m}\)
C.\(84 \text{ m}\)
D.\(120 \text{ m}\)
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Worked solution
To find the total distance, we calculate the distance traveled in each of the three phases of motion: 1. First phase (constant acceleration): \(u = 0\), \(a = 2.0 \text{ m/s}^2\), \(t = 4.0 \text{ s}\). The final velocity is \(v = u + at = 0 + (2.0 \times 4.0) = 8.0 \text{ m/s}\). Distance \(d_1 = \frac{1}{2}(u + v)t = \frac{1}{2}(0 + 8.0) \times 4.0 = 16.0 \text{ m}\). 2. Second phase (constant speed): The car travels at \(8.0 \text{ m/s}\) for \(6.0 \text{ s}\). Distance \(d_2 = v \times t = 8.0 \times 6.0 = 48.0 \text{ m}\). 3. Third phase (constant deceleration to rest): The car slows from \(8.0 \text{ m/s}\) to \(0\) in \(5.0 \text{ s}\). Distance \(d_3 = \frac{1}{2}(8.0 + 0) \times 5.0 = 20.0 \text{ m}\). Adding these together, the total distance \(d = d_1 + d_2 + d_3 = 16.0 + 48.0 + 20.0 = 84.0 \text{ m}\).
Marking scheme
Award 1 mark for the correct answer C. Alternative method: Calculate area under a speed-time graph with a trapezium/triangles shape: \(16 \text{ m} + 48 \text{ m} + 20 \text{ m} = 84 \text{ m}\).
Question 10 · multiple-choice
1 marks
Two identical springs, each of spring constant \(400 \text{ N/m}\), are connected in parallel. A load of \(20 \text{ N}\) is suspended from this parallel combination. This parallel combination is then connected in series with a third identical spring of spring constant \(400 \text{ N/m}\) which supports the same total load. What is the total extension of this system of springs?
A.\(2.5 \text{ cm}\)
B.\(5.0 \text{ cm}\)
C.\(7.5 \text{ cm}\)
D.\(10.0 \text{ cm}\)
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Worked solution
Let's find the extension of each part of the system: 1. Parallel combination: Two identical springs of spring constant \(k = 400 \text{ N/m}\) in parallel behave as a single spring with an effective spring constant \(k_p = 2k = 800 \text{ N/m}\). Under a load of \(20 \text{ N}\), their extension is \(x_p = \frac{F}{k_p} = \frac{20 \text{ N}}{800 \text{ N/m}} = 0.025 \text{ m} = 2.5 \text{ cm}\). 2. Series spring: The third spring is in series, so it experiences the full load of \(20 \text{ N}\). Its extension is \(x_s = \frac{F}{k} = \frac{20 \text{ N}}{400 \text{ N/m}} = 0.050 \text{ m} = 5.0 \text{ cm}\). 3. Total extension: The total extension is the sum of the extensions of both parts: \(x_{\text{total}} = x_p + x_s = 2.5 \text{ cm} + 5.0 \text{ cm} = 7.5 \text{ cm}\).
Marking scheme
Award 1 mark for the correct option C. Reject A (extension of parallel combination only), reject B (extension of single spring only).
Question 11 · multiple-choice
1 marks
A block of mass \(3.0 \text{ kg}\) moving at \(4.0 \text{ m/s}\) collides with a stationary block of mass \(5.0 \text{ kg}\). After the collision, the \(3.0 \text{ kg}\) block rebounds at a speed of \(1.0 \text{ m/s}\) in the opposite direction. What is the speed of the \(5.0 \text{ kg}\) block after the collision?
A.\(1.8 \text{ m/s}\)
B.\(2.2 \text{ m/s}\)
C.\(3.0 \text{ m/s}\)
D.\(5.0 \text{ m/s}\)
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Worked solution
By the law of conservation of momentum, total momentum before collision equals total momentum after collision. Let the initial direction of the \(3.0 \text{ kg}\) block be positive. Initial momentum: \(p_{\text{initial}} = (3.0 \text{ kg} \times 4.0 \text{ m/s}) + (5.0 \text{ kg} \times 0 \text{ m/s}) = 12.0 \text{ kg m/s}\). Final momentum: The \(3.0 \text{ kg}\) block rebounds, so its velocity is \(-1.0 \text{ m/s}\). Let \(v\) be the final velocity of the \(5.0 \text{ kg}\) block. \(p_{\text{final}} = (3.0 \text{ kg} \times (-1.0 \text{ m/s})) + (5.0 \text{ kg} \times v) = -3.0 + 5.0v\). Setting \(p_{\text{initial}} = p_{\text{final}}\): \(12.0 = -3.0 + 5.0v\) \(15.0 = 5.0v \Rightarrow v = 3.0 \text{ m/s}\).
Marking scheme
Award 1 mark for the correct option C. Common error: forgetting the negative sign on the rebound velocity leads to \(12.0 = 3.0 + 5.0v \Rightarrow v = 1.8 \text{ m/s}\) (Option A), which is incorrect.
Question 12 · multiple-choice
1 marks
An electric heater of power \(50 \text{ W}\) is used to heat a \(0.20 \text{ kg}\) solid sample. The solid has a melting point of \(80^\circ\text{C}\) and is initially at a temperature of \(20^\circ\text{C}\). It takes \(4.0 \text{ minutes}\) for the heater to heat the sample to its melting point. Assuming no thermal energy is lost to the surroundings, what is the specific heat capacity of the solid?
A.\(17 \text{ J / (kg }^\circ\text{C)}\)
B.\(1000 \text{ J / (kg }^\circ\text{C)}\)
C.\(1200 \text{ J / (kg }^\circ\text{C)}\)
D.\(60\,000 \text{ J / (kg }^\circ\text{C)}\)
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Worked solution
Using the relation for electrical energy supplied and thermal energy absorbed: \(E = P \times t\) and \(E = m \times c \times \Delta T\). First, find the time in seconds: \(t = 4.0 \times 60 = 240 \text{ s}\). Energy supplied: \(E = 50 \text{ W} \times 240 \text{ s} = 12\,000 \text{ J}\). Temperature change: \(\Delta T = 80^\circ\text{C} - 20^\circ\text{C} = 60^\circ\text{C}\). Now calculate specific heat capacity \(c\): \(12\,000 = 0.20 \times c \times 60\) \(12\,000 = 12c\) \(c = 1000 \text{ J / (kg }^\circ\text{C)}\).
Marking scheme
Award 1 mark for the correct option B. Common distraction A comes from not converting minutes to seconds (\(t = 4.0\) leads to \(c = 16.7 \approx 17\)).
Question 13 · multiple-choice
1 marks
Monochromatic light travels from glass to air. The refractive index of this glass is \(1.5\). What is the critical angle for light travelling from this glass to air?
A.\(30.0^\circ\)
B.\(41.8^\circ\)
C.\(48.6^\circ\)
D.\(60.0^\circ\)
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Worked solution
The relationship between the critical angle \(c\) and the refractive index \(n\) is given by: \(\sin(c) = \frac{1}{n}\) Given \(n = 1.5\): \(\sin(c) = \frac{1}{1.5} = \frac{2}{3} \approx 0.667\) Therefore, \(c = \sin^{-1}(0.667) \approx 41.8^\circ\).
Marking scheme
Award 1 mark for the correct option B. Option A is \(\sin^{-1}(0.5)\), Option C is \(\sin^{-1}(0.75)\).
Question 14 · multiple-choice
1 marks
A potential divider circuit consists of a fixed resistor of resistance \(R\) and a light-dependent resistor (LDR) connected in series across a constant \(12 \text{ V}\) d.c. supply. The output voltage \(V_{\text{out}}\) is measured across the LDR. What happens to the resistance of the LDR and the value of \(V_{\text{out}}\) when the light intensity on the LDR increases?
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Worked solution
When light intensity on an LDR increases, its resistance decreases. Since the LDR is in series with a fixed resistor, a decrease in the resistance of the LDR means it takes a smaller share of the total \(12 \text{ V}\) potential difference. Therefore, the output voltage \(V_{\text{out}}\) measured across the LDR decreases.
Marking scheme
Award 1 mark for the correct option D.
Question 15 · multiple-choice
1 marks
The count rate from a radioactive source is measured near the source in a laboratory. The average background count rate is \(24 \text{ counts/minute}\). The initial reading of the detector is \(216 \text{ counts/minute}\). After \(6.0 \text{ hours}\), the detector reading is \(48 \text{ counts/minute}\). What is the half-life of the source?
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 count rate from the measured readings to obtain the corrected count rates: - Initial corrected count rate = \(216 - 24 = 192 \text{ counts/minute}\). - Final corrected count rate = \(48 - 24 = 24 \text{ counts/minute}\). Now, determine the fraction remaining: \(\frac{24}{192} = \frac{1}{8}\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly \(3\) half-lives have elapsed in \(6.0 \text{ hours}\). Therefore, the half-life \(t_{1/2} = \frac{6.0 \text{ hours}}{3} = 2.0 \text{ hours}\).
Marking scheme
Award 1 mark for the correct option B. Common mistake: not subtracting background yields \(48 / 216 \approx 0.222\), leading to a wrong calculation of around \(2.17\) half-lives, which leads to option A.
Question 16 · multiple-choice
1 marks
A specific spectral line of hydrogen has a wavelength of \(656 \text{ nm}\) in a laboratory on Earth. When analyzing light from a distant galaxy, the same line is observed at a wavelength of \(682 \text{ nm}\). What does this change in wavelength indicate about the motion of the galaxy, and what is its approximate recession speed? (Take the speed of light \(c = 3.0 \times 10^8 \text{ m/s}\))
A.Moving towards Earth at \(1.2 \times 10^7 \text{ m/s}\)
B.Moving towards Earth at \(1.3 \times 10^7 \text{ m/s}\)
C.Moving away from Earth at \(1.2 \times 10^7 \text{ m/s}\)
D.Moving away from Earth at \(1.3 \times 10^7 \text{ m/s}\)
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Worked solution
Since the observed wavelength (\(682 \text{ nm}\)) is greater than the laboratory wavelength (\(656 \text{ nm}\)), the light is redshifted. This indicates that the galaxy is moving away from Earth. To find the speed, we use the redshift formula: \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\) Where: \(\Delta \lambda = 682 - 656 = 26 \text{ nm}\) \(\lambda_0 = 656 \text{ nm}\) \(v = c \times \frac{\Delta \lambda}{\lambda_0} = (3.0 \times 10^8 \text{ m/s}) \times \frac{26}{656} \approx 1.19 \times 10^7 \text{ m/s} \approx 1.2 \times 10^7 \text{ m/s}\).
Marking scheme
Award 1 mark for the correct option C. Option D is calculated if \(\lambda_0\) is incorrectly taken as \(682 \text{ nm}\).
Question 17 · multiple-choice
1 marks
A trolley of mass \( 2.0\text{ kg} \) moving at a velocity of \( 6.0\text{ m/s} \) along a friction-free horizontal track collides with a stationary trolley of mass \( 4.0\text{ kg} \). After the collision, the two trolleys stick together and move with a common velocity.
What is the loss in total kinetic energy of the trolleys during the collision?
A.0 J
B.12 J
C.24 J
D.36 J
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Worked solution
First, apply the law of conservation of momentum to find the final velocity \( v_f \) of the joined trolleys: \( m_1 u_1 + m_2 u_2 = (m_1 + m_2) v_f \) \( (2.0 \times 6.0) + (4.0 \times 0) = (2.0 + 4.0) v_f \) \( 12 = 6.0 v_f \implies v_f = 2.0\text{ m/s} \)
Next, calculate the initial kinetic energy of the system: \( E_{k,\text{initial}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 2.0 \times 6.0^2 = 36\text{ J} \)
Calculate the final kinetic energy of the system: \( E_{k,\text{final}} = \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} \times 6.0 \times 2.0^2 = 12\text{ J} \)
The loss in kinetic energy is: \( \Delta E_k = 36\text{ J} - 12\text{ J} = 24\text{ J} \)
Marking scheme
1 mark for correct calculation of final velocity (2.0 m/s) and determining the difference between initial kinetic energy (36 J) and final kinetic energy (12 J) to obtain 24 J.
Question 18 · multiple-choice
1 marks
A ray of monochromatic light travels from a glass block into air. The refractive index of the glass is 1.60. The angle of incidence of the ray inside the glass is 35°. What is the angle of refraction in the air?
A.21°
B.48°
C.56°
D.67°
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Worked solution
According to Snell's Law, when light travels from a medium of higher refractive index (glass) to a medium of lower refractive index (air): \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \)
Where: \( n_1 = 1.60 \) (refractive index of glass) \( \theta_1 = 35^\circ \) (angle of incidence) \( n_2 = 1.00 \) (refractive index of air) \( \theta_2 \) is the angle of refraction.
1 mark for applying Snell's Law correctly, yielding \( \theta_2 \approx 67^\circ \). Reject calculation of critical angle or incorrect division by refractive index.
Question 19 · multiple-choice
1 marks
Two resistors of resistance \( 4.0\ \Omega \) and \( 12\ \Omega \) are connected in parallel. This parallel combination is then connected in series with a third resistor of resistance \( 5.0\ \Omega \) and a 24 V d.c. power supply.
What is the current in the \( 4.0\ \Omega \) resistor?
A.0.75 A
B.1.5 A
C.2.25 A
D.3.0 A
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Step 2: Calculate the total resistance of the circuit (\( R_{\text{total}} \)): \( R_{\text{total}} = R_p + R_3 = 3.0 + 5.0 = 8.0\ \Omega \)
Step 3: Calculate the total current (\( I_{\text{total}} \)) supplied by the 24 V source: \( I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{24}{8.0} = 3.0\text{ A} \)
Step 4: Calculate the potential difference across the parallel section (\( V_p \)): \( V_p = I_{\text{total}} \times R_p = 3.0\text{ A} \times 3.0\ \Omega = 9.0\text{ V} \)
Step 5: Calculate the current through the \( 4.0\ \Omega \) resistor (\( I_4 \)): \( I_4 = \frac{V_p}{4.0\ \Omega} = \frac{9.0}{4.0} = 2.25\text{ A} \)
Marking scheme
1 mark for determining the correct current through the branch using potential divider or current divider principles, arriving at 2.25 A.
Question 20 · multiple-choice
1 marks
Which statement correctly describes the orbital motion of planets in the Solar System?
A.The average orbital speed of a planet decreases as its distance from the Sun increases.
B.The orbital speed of a planet is constant throughout its entire elliptical path.
C.All planets move in perfectly circular orbits with the Sun at the exact geometric centre.
D.Planets further from the Sun have shorter orbital periods because they experience stronger gravitational fields.
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Worked solution
As the orbital distance from the Sun increases, the gravitational attraction becomes weaker, requiring a lower orbital speed to maintain a stable orbit. Therefore, the average orbital speed decreases as distance from the Sun increases (which is a core syllabus point). Option B is incorrect because orbital speed varies along an elliptical path. Option C is incorrect as orbits are elliptical. Option D is incorrect because further planets have longer orbital periods.
Marking scheme
1 mark for identifying that the average orbital speed decreases as distance from the Sun increases.
Question 21 · multiple-choice
1 marks
An ideal, 100% efficient step-down transformer has 1200 turns on its primary coil and 300 turns on its secondary coil. The primary coil is connected to a 240 V a.c. supply. The secondary coil is connected to a \( 5.0\ \Omega \) resistor.
What is the current in the primary coil?
A.0.75 A
B.3.0 A
C.12 A
D.48 A
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Worked solution
Step 1: Calculate the secondary voltage (\( V_s \)) using the transformer equation: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = 240 \times \frac{300}{1200} = 60\text{ V} \)
Step 2: Calculate the secondary current (\( I_s \)): \( I_s = \frac{V_s}{R} = \frac{60}{5.0} = 12\text{ A} \)
1 mark for applying the ideal power relation \( V_p I_p = V_s I_s \) (or using current turns ratio inversely) to find the primary current of 3.0 A.
Question 22 · multiple-choice
1 marks
An electric kettle containing 1.2 kg of water at 20 °C is switched on. The power rating of the kettle is 2.4 kW. The specific heat capacity of water is \( 4200\text{ J}/(\text{kg}\!\cdot\!{^\circ}\text{C}) \).
Assuming no thermal energy is lost to the surroundings, how much time does it take to heat the water to its boiling point of 100 °C?
A.42 s
B.140 s
C.168 s
D.210 s
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Worked solution
First, find the temperature rise: \( \Delta \theta = 100^\circ\text{C} - 20^\circ\text{C} = 80^\circ\text{C} \)
Calculate the thermal energy \( Q \) required: \( Q = m c \Delta \theta = 1.2 \times 4200 \times 80 = 403,200\text{ J} \)
Use the power formula \( P = \frac{Q}{t} \) to find the time \( t \): \( t = \frac{Q}{P} = \frac{403,200}{2400\text{ W}} = 168\text{ s} \)
Marking scheme
1 mark for calculating the correct heating time of 168 s using \( Q = mc\Delta\theta \) and \( t = Q/P \).
Question 23 · multiple-choice
1 marks
A radioactive isotope has a half-life of 6.0 hours. A sample of this isotope has an initial activity of 800 Bq.
What is the activity of the sample after 24 hours, and what percentage of the original nuclei have decayed?
A.Activity = 50 Bq; percentage decayed = 6.25%
B.Activity = 50 Bq; percentage decayed = 93.75%
C.Activity = 100 Bq; percentage decayed = 12.5%
D.Activity = 100 Bq; percentage decayed = 87.5%
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Worked solution
First, calculate the number of half-lives that have elapsed: \( N = \frac{24\text{ hours}}{6.0\text{ hours}} = 4\text{ half-lives} \)
This means \( \frac{1}{16} \) of the original radioactive nuclei remain. Expressed as a percentage: \( \text{Percentage remaining} = \frac{1}{16} \times 100\% = 6.25\% \)
Therefore, the percentage of nuclei that have decayed is: \( \text{Percentage decayed} = 100\% - 6.25\% = 93.75\% \)
Marking scheme
1 mark for finding both the correct final activity (50 Bq) and the correct decayed percentage (93.75%).
Question 24 · multiple-choice
1 marks
A uniform horizontal beam of length 4.0 m and weight 120 N is pivoted at one end. A vertical upward force \( F \) is applied at the other end to keep the beam in horizontal equilibrium. A load of 180 N is placed on the beam at a distance of 1.0 m from the pivot.
What is the magnitude of the force \( F \)?
A.45 N
B.75 N
C.105 N
D.165 N
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Worked solution
Since the beam is uniform, its weight of 120 N acts at its centre of mass (midpoint), which is 2.0 m from the pivot.
Take moments about the pivot for equilibrium: \( \text{Sum of anticlockwise moments} = \text{Sum of clockwise moments} \)
\( F \times 4.0\text{ m} = (120\text{ N} \times 2.0\text{ m}) + (180\text{ N} \times 1.0\text{ m}) \) \( 4.0 F = 240 + 180 \) \( 4.0 F = 420 \) \( F = 105\text{ N} \)
Marking scheme
1 mark for taking moments about the pivot, accounting correctly for the weight of the beam at 2.0 m, and finding \( F = 105\text{ N} \).
Question 25 · Multiple Choice (Extended)
1 marks
An object is released from rest in a uniform gravitational field where air resistance is not negligible. Which statement describes how its acceleration \( a \) varies with time \( t \) as it falls?
A.The acceleration starts at a maximum value and decreases at a decreasing rate towards zero.
B.The acceleration starts at a maximum value and decreases at a constant rate to zero.
C.The acceleration starts at zero and increases at a decreasing rate to a maximum constant value.
D.The acceleration remains constant at a non-zero value throughout the fall.
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Worked solution
At the start of the fall, speed \( v = 0 \), so air resistance is zero and the acceleration is equal to the gravitational field strength \( g \). As the object speeds up, air resistance increases, which opposes the downward force of gravity and reduces the net force. Consequently, the acceleration decreases. Because the acceleration is decreasing, the rate of increase of speed also slows down, which in turn means the air resistance increases at a slower rate. Therefore, the rate at which the acceleration decreases also reduces over time, making the acceleration asymptotically approach zero. This corresponds to a curve that starts at a maximum value and decreases with a decreasing rate of change towards zero.
Marking scheme
1 mark: Correctly identify that the acceleration starts at a maximum and decreases at a decreasing rate to zero.
Question 26 · Multiple Choice (Extended)
1 marks
An electric motor with an efficiency of 60% is used to lift a mass of \( 40\text{ kg} \) vertically upwards through a height of \( 12\text{ m} \). The lift takes \( 8.0\text{ s} \). The gravitational field strength \( g \) is \( 10\text{ N/kg} \). What is the electrical power input to the motor?
A.\( 360\text{ W} \)
B.\( 600\text{ W} \)
C.\( 1000\text{ W} \)
D.\( 1500\text{ W} \)
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Worked solution
First, find the useful work done against gravity: \( W = mgh = 40\text{ kg} \times 10\text{ N/kg} \times 12\text{ m} = 4800\text{ J} \). Next, calculate the useful power output: \( P_{\text{out}} = \frac{W}{t} = \frac{4800\text{ J}}{8.0\text{ s}} = 600\text{ W} \). Finally, use the efficiency to find the input electrical power: \( \text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies 0.60 = \frac{600\text{ W}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{600}{0.60} = 1000\text{ W} \).
Marking scheme
1 mark: Correctly calculate the input power as 1000 W.
Question 27 · Multiple Choice (Extended)
1 marks
A uniform metal beam of mass \( 12\text{ kg} \) and length \( 2.0\text{ m} \) is suspended horizontally in equilibrium by two identical vertical springs, one at each end. Each spring has a spring constant of \( 400\text{ N/m} \). The gravitational field strength \( g \) is \( 10\text{ N/kg} \). What is the extension of each spring?
A.\( 0.15\text{ m} \)
B.\( 0.30\text{ m} \)
C.\( 0.60\text{ m} \)
D.\( 1.2\text{ m} \)
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Worked solution
The total downward force is the weight of the beam: \( W = mg = 12\text{ kg} \times 10\text{ N/kg} = 120\text{ N} \). Since the beam is uniform and suspended symmetrically in equilibrium, each spring supports exactly half of the weight: \( F = \frac{120\text{ N}}{2} = 60\text{ N} \). Using Hooke's Law \( F = kx \), the extension \( x \) of each spring is \( x = \frac{F}{k} = \frac{60\text{ N}}{400\text{ N/m}} = 0.15\text{ m} \).
Marking scheme
1 mark: Correctly calculate the extension of each spring as 0.15 m.
Question 28 · Multiple Choice (Extended)
1 marks
A rubber ball of mass \( 0.20\text{ kg} \) hits a vertical wall horizontally with a speed of \( 15\text{ m/s} \). It rebounds horizontally in the opposite direction with a speed of \( 10\text{ m/s} \). 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 during the collision?
A.\( 20\text{ N} \)
B.\( 50\text{ N} \)
C.\( 100\text{ N} \)
D.\( 250\text{ N} \)
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Worked solution
Define the direction towards the wall as positive. The initial momentum is \( p_i = m v_i = 0.20\text{ kg} \times 15\text{ m/s} = 3.0\text{ kg m/s} \). The final momentum is \( p_f = m v_f = 0.20\text{ kg} \times (-10\text{ m/s}) = -2.0\text{ kg m/s} \). The change in momentum (impulse) is \( \Delta p = p_f - p_i = -2.0 - 3.0 = -5.0\text{ kg m/s} \). The magnitude of the change in momentum is \( 5.0\text{ N s} \). Using the relationship \( F = \frac{\Delta p}{\Delta t} \), the average force is \( F = \frac{5.0\text{ N s}}{0.050\text{ s}} = 100\text{ N} \).
Marking scheme
1 mark: Correctly calculate the average force as 100 N.
Question 29 · Multiple Choice (Extended)
1 marks
A battery drives a current of \( 2.5\text{ A} \) through a resistor for \( 4.0\text{ minutes} \). During this time, the total electrical energy transferred to the resistor is \( 7.2\text{ kJ} \). What is the potential difference across the resistor?
A.\( 0.72\text{ V} \)
B.\( 12\text{ V} \)
C.\( 48\text{ V} \)
D.\( 720\text{ V} \)
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Worked solution
First, convert the time to seconds: \( t = 4.0\text{ minutes} = 4.0 \times 60 = 240\text{ s} \). Next, convert the energy to joules: \( E = 7.2\text{ kJ} = 7200\text{ J} \). Using the relationship for electrical energy \( E = V I t \), rearrange to find potential difference \( V \): \( V = \frac{E}{I t} = \frac{7200\text{ J}}{2.5\text{ A} \times 240\text{ s}} = \frac{7200}{600} = 12\text{ V} \).
Marking scheme
1 mark: Correctly calculate the potential difference as 12 V.
Question 30 · Multiple Choice (Extended)
1 marks
A circuit contains three resistors connected to a \( 12\text{ V} \) d.c. power supply of negligible internal resistance. A \( 6.0\ \Omega \) resistor is in series with a parallel combination of a \( 4.0\ \Omega \) resistor and an unknown resistor \( R \). The total current supplied by the power supply is \( 1.5\text{ A} \). What is the resistance of the unknown resistor \( R \)?
A.\( 2.0\ \Omega \)
B.\( 4.0\ \Omega \)
C.\( 8.0\ \Omega \)
D.\( 12\ \Omega \)
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Worked solution
First, find the total resistance of the circuit: \( R_{\text{total}} = \frac{V}{I} = \frac{12\text{ V}}{1.5\text{ A}} = 8.0\ \Omega \). Since the \( 6.0\ \Omega \) resistor is in series with the parallel pair, the combined resistance of the parallel pair \( R_p \) must be: \( R_{\text{total}} = 6.0 + R_p \implies 8.0 = 6.0 + R_p \implies R_p = 2.0\ \Omega \). For the parallel pair: \( \frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{R} \implies \frac{1}{2.0} = \frac{1}{4.0} + \frac{1}{R} \implies \frac{1}{R} = \frac{1}{2.0} - \frac{1}{4.0} = \frac{1}{4.0} \implies R = 4.0\ \Omega \).
Marking scheme
1 mark: Correctly calculate the resistance of the unknown resistor as 4.0 ohms.
Question 31 · Multiple Choice (Extended)
1 marks
A ray of light in a glass block is incident on the glass-air boundary. The refractive index of the glass is 1.6. Which row correctly describes the behavior of the light ray for the given angles of incidence?
A.At \( 35^\circ \), the ray is refracted into the air; at \( 45^\circ \), the ray is totally internally reflected.
B.At \( 35^\circ \), the ray is totally internally reflected; at \( 45^\circ \), the ray is refracted into the air.
C.At both \( 35^\circ \) and \( 45^\circ \), the ray is refracted into the air.
D.At both \( 35^\circ \) and \( 45^\circ \), the ray is totally internally reflected.
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Worked solution
Calculate the critical angle \( c \) for the boundary: \( \sin c = \frac{1}{n} = \frac{1}{1.6} = 0.625 \implies c = \sin^{-1}(0.625) \approx 38.7^\circ \). If the angle of incidence is less than the critical angle, the ray is refracted out of the glass into the air. If the angle of incidence is greater than the critical angle, total internal reflection occurs. Thus, at an angle of incidence of \( 35^\circ \) (which is less than \( 38.7^\circ \)), the light is refracted into the air, and at an angle of incidence of \( 45^\circ \) (which is greater than \( 38.7^\circ \)), total internal reflection occurs.
Marking scheme
1 mark: Correctly identify the behavior at both angles of incidence based on the critical angle.
Question 32 · Multiple Choice (Extended)
1 marks
Light from a distant galaxy is analyzed. The spectrum of light shows a redshift compared to a laboratory source on Earth. Which statement about this galaxy and the Universe is correct?
A.The galaxy is moving away from Earth, and this redshift provides evidence that the Universe is expanding.
B.The galaxy is moving towards Earth, and this redshift provides evidence that the Universe is contracting.
C.The galaxy is moving away from Earth, and its speed of recession is the same for all galaxies regardless of distance.
D.The galaxy is moving towards Earth, and its speed of approach is proportional to its distance from Earth.
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Worked solution
Redshift means the observed wavelength of the electromagnetic radiation is longer than the emitted wavelength. This occurs because the source of light (the galaxy) is moving away from the observer (Earth). The redshift observed in light from almost all distant galaxies is a key piece of evidence showing that the galaxies are receding from us, which supports the theory that the Universe is expanding.
Marking scheme
1 mark: Correctly identify that redshift indicates the galaxy is moving away and provides evidence for the expansion of the Universe.
Question 33 · multiple-choice
1 marks
A toy car travels along a straight line. From rest, it accelerates uniformly to a speed of \(8.0\text{ m/s}\) in \(4.0\text{ s}\). It then travels at this constant speed of \(8.0\text{ m/s}\) for a time \(T\). Finally, it decelerates uniformly to rest in \(3.0\text{ s}\). The total distance travelled by the car during the whole journey is \(80\text{ m}\). What is the total time of the journey?
A.\(10.0\text{ s}\)
B.\(13.5\text{ s}\)
C.\(15.5\text{ s}\)
D.\(17.0\text{ s}\)
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Worked solution
The motion can be divided into three parts. 1) Uniform acceleration: Distance \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 8.0\text{ m/s} = 16\text{ m}\). 2) Constant velocity: Distance \(d_2 = 8.0 \times T\). 3) Uniform deceleration: Distance \(d_3 = \frac{1}{2} \times 3.0\text{ s} \times 8.0\text{ m/s} = 12\text{ m}\). The total distance is given as \(80\text{ m}\), so \(d_1 + d_2 + d_3 = 80\text{ m}\), which means \(16 + 8.0T + 12 = 80\), leading to \(28 + 8.0T = 80\), then \(8.0T = 52\), giving \(T = 6.5\text{ s}\). The total time of the journey is \(4.0\text{ s} + T + 3.0\text{ s} = 4.0 + 6.5 + 3.0 = 13.5\text{ s}\).
Marking scheme
Award 1 mark for the correct option B. 1 mark for calculating the distances during acceleration (16 m) and deceleration (12 m), finding the constant velocity time T (6.5 s), and obtaining the total time of 13.5 s.
Question 34 · multiple-choice
1 marks
An empty flask has a mass of \(120\text{ g}\). When completely filled with liquid X, the total mass is \(360\text{ g}\). When the same flask is completely filled with liquid Y, the total mass is \(300\text{ g}\). If the density of liquid X is \(1.2\text{ g/cm}^3\), what is the density of liquid Y?
A.\(0.60\text{ g/cm}^3\)
B.\(0.75\text{ g/cm}^3\)
C.\(0.90\text{ g/cm}^3\)
D.\(1.00\text{ g/cm}^3\)
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Worked solution
First, find the mass of liquid X: \(m_X = 360\text{ g} - 120\text{ g} = 240\text{ g}\). Since the density of liquid X is \(\rho_X = 1.2\text{ g/cm}^3\), the volume of the flask \(V\) is: \(V = \frac{m_X}{\rho_X} = \frac{240\text{ g}}{1.2\text{ g/cm}^3} = 200\text{ cm}^3\). Next, find the mass of liquid Y: \(m_Y = 300\text{ g} - 120\text{ g} = 180\text{ g}\). The volume of liquid Y is also equal to the volume of the flask, \(200\text{ cm}^3\). The density of liquid Y is: \(\rho_Y = \frac{m_Y}{V} = \frac{180\text{ g}}{200\text{ cm}^3} = 0.90\text{ g/cm}^3\).
Marking scheme
Award 1 mark for the correct option C. 1 mark for finding the volume of the flask (200 cm³) and using it to calculate the density of liquid Y (0.90 g/cm³).
Question 35 · multiple-choice
1 marks
A uniform wooden beam of length \(2.0\text{ m}\) and weight \(120\text{ N}\) is supported by a pivot at a distance of \(0.50\text{ m}\) from its left-hand end. To keep the beam horizontal, a vertical upward force \(F\) is applied at the right-hand end of the beam. What is the magnitude of force \(F\)?
A.\(30\text{ N}\)
B.\(40\text{ N}\)
C.\(60\text{ N}\)
D.\(80\text{ N}\)
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Worked solution
Since the beam is uniform, its weight of \(120\text{ N}\) acts at its midpoint, which is \(1.0\text{ m}\) from either end. The pivot is located \(0.50\text{ m}\) from the left end, so the weight acts at a distance of \(1.0\text{ m} - 0.50\text{ m} = 0.50\text{ m}\) to the right of the pivot. Taking moments about the pivot: clockwise moment due to weight = \(120\text{ N} \times 0.50\text{ m} = 60\text{ N m}\). The vertical upward force \(F\) is at the right end, which is at a distance of \(2.0\text{ m} - 0.50\text{ m} = 1.50\text{ m}\) to the right of the pivot. This force creates an anti-clockwise moment equal to \(F \times 1.50\text{ m}\). For rotational equilibrium, the sum of clockwise moments equals the sum of anti-clockwise moments: \(1.50 \times F = 60\), which yields \(F = \frac{60}{1.50} = 40\text{ N}\).
Marking scheme
Award 1 mark for the correct option B. 1 mark for identifying the correct distances from pivot to center of mass (0.50 m) and to force F (1.50 m), and solving the principle of moments equation.
Question 36 · multiple-choice
1 marks
A trolley A of mass \(3.0\text{ kg}\) moving at a speed of \(4.0\text{ m/s}\) to the right collides with a stationary trolley B of mass \(2.0\text{ kg}\). After the collision, the two trolleys stick together and move as a single combined body. What is the loss of kinetic energy during this collision?
A.\(0\text{ J}\)
B.\(9.6\text{ J}\)
C.\(14.4\text{ J}\)
D.\(24.0\text{ J}\)
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Worked solution
First, use the principle of conservation of momentum to find the final velocity \(v\) of the combined trolleys: \(m_A u_A + m_B u_B = (m_A + m_B) v\), which is \((3.0\text{ kg} \times 4.0\text{ m/s}) + 0 = (3.0\text{ kg} + 2.0\text{ kg}) \times v\), so \(12 = 5.0 \times v\) leading to \(v = 2.4\text{ m/s}\). Now calculate the initial and final kinetic energies: Initial kinetic energy is \(E_{\text{ki}} = \frac{1}{2} m_A u_A^2 = \frac{1}{2} \times 3.0 \times (4.0)^2 = 24.0\text{ J}\). Final kinetic energy is \(E_{\text{kf}} = \frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2} \times 5.0 \times (2.4)^2 = 14.4\text{ J}\). The loss of kinetic energy is: \(\Delta E_{\text{k}} = 24.0\text{ J} - 14.4\text{ J} = 9.6\text{ J}\).
Marking scheme
Award 1 mark for the correct option B. 1 mark for calculating final velocity (2.4 m/s), computing initial (24 J) and final (14.4 J) kinetic energies, and finding the difference of 9.6 J.
Question 37 · multiple-choice
1 marks
An electric motor is used to lift a load of \(250\text{ N}\) vertically through a height of \(12\text{ m}\) in a time of \(6.0\text{ s}\). The electrical power input to the motor is \(800\text{ W}\). What is the efficiency of the motor?
A.\(31.25\%\)
B.\(37.5\%\)
C.\(62.5\%\)
D.\(80.0\%\)
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Worked solution
First, calculate the useful work done by the motor in lifting the load: \(W = F \times d = 250\text{ N} \times 12\text{ m} = 3000\text{ J}\). Next, calculate the useful power output of the motor: \(P_{\text{out}} = \frac{W}{t} = \frac{3000\text{ J}}{6.0\text{ s}} = 500\text{ W}\). Finally, calculate the efficiency of the motor: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{500\text{ W}}{800\text{ W}} \times 100\% = 62.5\%\).
Marking scheme
Award 1 mark for the correct option C. 1 mark for finding the useful power output (500 W) and the efficiency (62.5%).
Question 38 · multiple-choice
1 marks
A simple mercury barometer has a vertical column of mercury of height \(h\) above the reservoir level. The atmosphere exerts a pressure of \(1.01 \times 10^5\text{ Pa}\). The density of mercury is \(1.36 \times 10^4\text{ kg/m}^3\) and the gravitational field strength \(g = 9.8\text{ N/kg}\). What is the height \(h\) of the mercury column?
A.\(0.076\text{ m}\)
B.\(0.76\text{ m}\)
C.\(1.36\text{ m}\)
D.\(7.6\text{ m}\)
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Worked solution
The atmospheric pressure \(P\) is balanced by the pressure exerted by the column of mercury: \(P = \rho g h\). Where \(P = 1.01 \times 10^5\text{ Pa}\), \(\rho = 1.36 \times 10^4\text{ kg/m}^3\), and \(g = 9.8\text{ N/kg}\). Rearranging the formula to solve for \(h\): \(h = \frac{P}{\rho g} = \frac{1.01 \times 10^5}{1.36 \times 10^4 \times 9.8} = \frac{101000}{133280} \approx 0.758\text{ m}\). Rounded to two significant figures, this is \(0.76\text{ m}\).
Marking scheme
Award 1 mark for the correct option B. 1 mark for rearranging the liquid pressure formula correctly and finding the height as approximately 0.76 m.
Question 39 · multiple-choice
1 marks
A \(2.0\text{ kW}\) electric heater is placed in a well-insulated container containing \(1.5\text{ kg}\) of water at \(20\text{ }^\circ\text{C}\). The heater is switched on for \(3.0\text{ minutes}\). The specific heat capacity of water is \(4200\text{ J/(kg }^\circ\text{C)}
eef. What is the final temperature of the water?
A.\(38\text{ }^\circ\text{C}\)
B.\(57\text{ }^\circ\text{C}\)
C.\(77\text{ }^\circ\text{C}\)
D.\(97\text{ }^\circ\text{C}\)
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Worked solution
First, calculate the total thermal energy supplied by the heater: \(E = P \times t = 2000\text{ W} \times (3.0 \times 60\text{ s}) = 2000 \times 180 = 360,000\text{ J}\). Using the formula for heat energy transfer: \(E = m c \Delta T\), which gives \(360,000 = 1.5 \times 4200 \times \Delta T\), so \(360,000 = 6300 \times \Delta T\), and thus \(\Delta T = \frac{360,000}{6300} \approx 57.1\text{ }^\circ\text{C}\). The final temperature of the water is \(T_{\text{final}} = T_{\text{initial}} + \Delta T = 20\text{ }^\circ\text{C} + 57.1\text{ }^\circ\text{C} = 77.1\text{ }^\circ\text{C}\), which is rounded to \(77\text{ }^\circ\text{C}\).
Marking scheme
Award 1 mark for the correct option C. 1 mark for calculating the energy supplied (360,000 J), the temperature change (57.1 °C), and adding it to the initial temperature to get 77 °C.
Question 40 · multiple-choice
1 marks
A uniform metal wire of resistance \(R\) and length \(L\) has a cross-sectional area \(A\). The wire is stretched uniformly until its length becomes \(2L\) and its volume remains unchanged. What is the new resistance of the wire?
A.\(0.5R\)
B.\(R\)
C.\(2R\)
D.\(4R\)
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Worked solution
The resistance of a wire of resistivity \(\rho\), length \(L\), and cross-sectional area \(A\) is given by: \(R = \rho \frac{L}{A}\). Since the volume \(V = A \times L\) of the wire remains constant during stretching, if the length doubles to \(2L\), the cross-sectional area must halve to \(\frac{A}{2}\). Substitute the new values into the resistance formula: \(R_{\text{new}} = \rho \frac{2L}{\frac{A}{2}} = 4 \left(\rho \frac{L}{A}\right) = 4R\).
Marking scheme
Award 1 mark for the correct option D. 1 mark for identifying that area halves when length doubles (to keep volume constant), leading to a fourfold increase in resistance.
Section Theory (Extended)
Answer all questions. Show your working clearly. Give your final answers to 2 or 3 significant figures with units.
11 Question · 79.96999999999997 marks
Question 1 · structured-theory
7.27 marks
A small steel ball of mass 0.15 kg is released from rest in a tall cylinder of oil. At a particular instant during its fall, the acceleration of the ball is 4.0 m/s^2 downwards. (Take the acceleration of free fall g = 9.8 m/s^2).
(a) Calculate the weight of the steel ball. (b) Determine the upward drag force acting on the ball at this instant. (c) Describe the motion of the ball after a very long period of time has passed, explaining your answer in terms of the forces acting on it.
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Worked solution
(a) Weight: \(W = m \times g = 0.15 \times 9.8 = 1.47\text{ N}\). (b) Resultant force: \(F_{\text{net}} = m \times a = 0.15 \times 4.0 = 0.60\text{ N}\). Using Newton\'s second law: \(F_{\text{net}} = W - F_{\text{drag}} \Rightarrow 0.60 = 1.47 - F_{\text{drag}} \Rightarrow F_{\text{drag}} = 1.47 - 0.60 = 0.87\text{ N}\). (c) As velocity increases, drag increases until \(F_{\text{drag}} = W\). At this point, the resultant force is zero, so the acceleration becomes zero, and the ball falls at a constant terminal velocity.
Marking scheme
Part (a) [2 marks]: - 1 mark for formula \(W = mg\) - 1 mark for correct calculation \(1.47\text{ N}\) (accept \(1.5\text{ N}\) if \(g=10\text{ m/s}^2\) used)
Part (b) [3 marks]: - 1 mark for calculating resultant force \(F = 0.60\text{ N}\) - 1 mark for setting up equation \(F = W - F_{\text{drag}}\) - 1 mark for final answer \(0.87\text{ N}\) (or \(0.90\text{ N}\) if using \(g=10\text{ m/s}^2\))
Part (c) [2.27 marks]: - 1 mark for stating constant speed / terminal velocity is reached - 1.27 marks for explaining that drag force increases to equal weight, resulting in zero net force / zero acceleration
Question 2 · structured-theory
7.27 marks
An electric pump is used to raise 6.0 kg of water per second through a vertical height of 8.0 m. (Take g = 9.8 m/s^2).
(a) Calculate the useful power output of the pump. (b) The electrical power input to the pump is 720 W. Calculate the efficiency of the pump. (c) State one reason why the efficiency of the pump is less than 100%.
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Worked solution
(a) Useful work done per second (power): \(P_{\text{out}} = \frac{\Delta E_p}{t} = \frac{mgh}{t} = 6.0 \times 9.8 \times 8.0 = 470.4\text{ W}\) (approx 470 W). (b) Efficiency: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100 = \frac{470.4}{720} \times 100 \approx 65.3\%\) (approx 65%). (c) Energy is lost as heat to the surroundings due to friction in the pump\'s moving parts or electrical resistance in the motor.
Marking scheme
Part (a) [3 marks]: - 1 mark for \(E_p = mgh\) or \(P = \frac{mgh}{t}\) - 1 mark for substitution: \(6.0 \times 9.8 \times 8.0\) - 1 mark for correct calculation: \(470\text{ W}\) or \(470.4\text{ W}\)
Part (b) [3 marks]: - 1 mark for efficiency formula: \(\frac{\text{Power out}}{\text{Power in}}\) - 1 mark for substitution: \(\frac{470.4}{720}\) - 1 mark for correct calculation: \(65\%\) (or \(65.3\%\))
Part (c) [1.27 marks]: - 1.27 marks for identifying work done against friction / heat losses in motor windings
Question 3 · structured-theory
7.27 marks
A ray of monochromatic light is directed from air into a flat semi-circular glass block. The angle of incidence in air is 45 degrees, and the angle of refraction in the glass is 28 degrees.
(a) Calculate the refractive index of the glass. (b) Determine the critical angle for light travelling inside this glass block towards the air boundary. (c) Describe what happens to the ray of light if it meets the boundary from inside the glass at an angle of incidence of 50 degrees.
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Worked solution
(a) Refractive index \(n = \frac{\sin(i)}{\sin(r)} = \frac{\sin(45^\circ)}{\sin(28^\circ)} = \frac{0.7071}{0.4695} \approx 1.51\). (b) Critical angle \(c\): \(\sin(c) = \frac{1}{n} = \frac{1}{1.51} \approx 0.662\), which gives \(c = \arcsin(0.662) \approx 41.5^\circ\). (c) Since the angle of incidence (50 degrees) is greater than the critical angle (41.5 degrees), total internal reflection occurs, and all light is reflected back inside the glass.
Marking scheme
Part (a) [3 marks]: - 1 mark for formula \(n = \frac{\sin i}{\sin r}\) - 1 mark for substitution: \(\frac{\sin 45}{\sin 28}\) - 1 mark for correct calculation: \(1.51\) (accept 1.5)
Part (b) [2.27 marks]: - 1 mark for formula \(\sin c = \frac{1}{n}\) - 1.27 marks for correct calculation: \(41.5^\circ\) (accept \(41^\circ\) to \(42^\circ\) depending on rounding)
Part (c) [2 marks]: - 1 mark for stating that total internal reflection occurs - 1 mark for explaining that the angle of incidence is greater than the critical angle
Question 4 · structured-theory
7.27 marks
An electrical immersion heater rated at 12 V, 3.0 A is placed inside a 0.45 kg block of metal. The heater is turned on for 5.0 minutes. The temperature of the block increases from 21 degrees Celsius to 39 degrees Celsius.
(a) Calculate the electrical energy supplied to the heater. (b) Calculate the specific heat capacity of the metal block, assuming no energy is lost to the surroundings. (c) State and explain how the calculated value of the specific heat capacity would change if some thermal energy was lost to the surroundings during the experiment.
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Worked solution
(a) Energy \(E = V \times I \times t = 12 \times 3.0 \times (5.0 \times 60) = 36 \times 300 = 10,800\text{ J}\). (b) Temperature change \(\Delta \theta = 39 - 21 = 18^\circ\text{C}\). Using \(E = m c \Delta \theta\): \(c = \frac{E}{m \Delta \theta} = \frac{10800}{0.45 \times 18} = \frac{10800}{8.1} \approx 1333 \approx 1300\text{ J/(kg}\cdot^\circ\text{C)}\). (c) If heat was lost, the actual energy absorbed by the block is less than the calculated electrical energy supplied. This makes the calculated specific heat capacity higher than the actual value because the formula assumes all supplied energy was kept in the block.
Marking scheme
Part (a) [2 marks]: - 1 mark for converting minutes to seconds (300 s) - 1 mark for formula and correct calculation: \(10,800\text{ J}\) (or \(10.8\text{ kJ}\))
Part (b) [3 marks]: - 1 mark for calculating temperature change (18 degrees) - 1 mark for rearranging formula \(c = \frac{E}{m \Delta T}\) - 1 mark for correct calculation: \(1300\text{ J/(kg }^\circ\text{C)}\) (accept \(1333\text{ J/(kg }^\circ\text{C)}\))
Part (c) [2.27 marks]: - 1 mark for stating the calculated value would be higher / larger - 1.27 marks for explaining that more electrical energy is assumed to heat the metal than actually did, or that less temperature rise occurs for the same input energy
Question 5 · structured-theory
7.27 marks
A 12.0 V battery of negligible internal resistance is connected to a circuit. The circuit consists of a 6.0 ohm resistor connected in series with a parallel combination of an 8.0 ohm resistor and a 12.0 ohm resistor.
(a) Determine the combined resistance of the parallel combination. (b) Calculate the total current flowing from the battery. (c) Calculate the potential difference across the parallel combination.
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Worked solution
(a) For parallel resistors: \(\frac{1}{R_p} = \frac{1}{8.0} + \frac{1}{12.0} = \frac{3+2}{24} = \frac{5}{24}\), so \(R_p = \frac{24}{5} = 4.8\ \Omega\). (b) Total resistance \(R_{\text{total}} = 6.0 + 4.8 = 10.8\ \Omega\). Total current \(I = \frac{V}{R_{\text{total}}} = \frac{12.0}{10.8} \approx 1.11\text{ A}\). (c) Potential difference across the parallel combination: \(V_p = I \times R_p = 1.11 \times 4.8 \approx 5.33\text{ V}\).
Marking scheme
Part (a) [2 marks]: - 1 mark for parallel formula \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\) or product-over-sum - 1 mark for correct calculation: \(4.8\ \Omega\)
Part (b) [3 marks]: - 1 mark for adding the series resistor to find total resistance \(10.8\ \Omega\) - 1 mark for \(I = \frac{V}{R}\) - 1 mark for correct calculation: \(1.11\text{ A}\) (accept \(1.1\text{ A}\))
Part (c) [2.27 marks]: - 1 mark for formula \(V = I R_p\) using their current from (b) - 1.27 marks for correct calculation: \(5.33\text{ V}\) (accept \(5.3\text{ V}\))
Question 6 · structured-theory
7.27 marks
A ball A of mass 0.40 kg is rolling along a frictionless track at a velocity of 5.0 m/s. It collides head-on with a stationary ball B of mass 0.60 kg. After the collision, ball A rebounds in the opposite direction at a speed of 1.0 m/s.
(a) State the principle of conservation of momentum. (b) Calculate the velocity of ball B after the collision. (c) Determine the impulse exerted on ball B during the collision.
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Worked solution
(a) The total momentum of a closed system remains constant, provided no external forces act on it. (b) Using conservation of momentum (taking initial direction of A as positive): \(m_A u_A + m_B u_B = m_A v_A + m_B v_B\) \((0.40 \times 5.0) + (0.60 \times 0) = (0.40 \times (-1.0)) + (0.60 \times v_B)\) \(2.0 = -0.40 + 0.60 v_B\) \(2.40 = 0.60 v_B \Rightarrow v_B = 4.0\text{ m/s}\). (c) Impulse on B = change in momentum of B = \(m_B (v_B - u_B) = 0.60 \times (4.0 - 0) = 2.4\text{ N}\cdot\text{s}\) (or \(2.4\text{ kg}\cdot\text{m/s}\)).
Marking scheme
Part (a) [1.27 marks]: - 1.27 marks for definition mentioning total momentum remains constant in a closed/isolated system (no external forces)
Part (b) [4 marks]: - 1 mark for momentum formula: \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\) - 1 mark for correct handling of the negative sign for the rebounding ball \((-1.0\text{ m/s})\) - 1 mark for substitution: \(2.0 = -0.40 + 0.60 v_B\) - 1 mark for correct calculation of velocity: \(4.0\text{ m/s}\)
Part (c) [2 marks]: - 1 mark for impulse formula: \(\Delta p = m \Delta v\) - 1 mark for correct calculation: \(2.4\text{ N s}\) (or \(2.4\text{ kg m/s}\))
Question 7 · structured-theory
7.27 marks
A radioactive sample contains Radium-226, which decays by emitting alpha particles.
(a) Radium-226 decays into Radon-222 according to the following equation: $$\text{Ra}_{88}^{226} \rightarrow \text{Rn}_{86}^{222} + \alpha_{Z}^{A}$$ State the values of A and Z for the alpha particle.
(b) In a laboratory test, a sample of Radon-222 has an initial count rate of 320 counts per second. After 11.4 days, the count rate drops to 40 counts per second. Calculate the half-life of Radon-222.
(c) Define the term "half-life".
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Worked solution
(a) An alpha particle is a helium nucleus. Mass number (nucleon number) \(A = 4\), proton number \(Z = 2\). (b) Find the number of half-lives that have elapsed: \(320 \rightarrow 160 \rightarrow 80 \rightarrow 40\) (which is 3 half-lives). Therefore, 3 half-lives = 11.4 days. One half-life = \(\frac{11.4}{3} = 3.8\text{ days}\). (c) Half-life is the time taken for the activity (or the number of undecayed nuclei) of a radioactive source to decrease to half of its initial value.
Marking scheme
Part (a) [2 marks]: - 1 mark for \(A = 4\) - 1 mark for \(Z = 2\)
Part (b) [3.27 marks]: - 1 mark for identifying that the count rate has halved 3 times (3 half-lives) - 1 mark for setting up \(3 \times t_{1/2} = 11.4\) - 1.27 marks for correct calculation: \(3.8\text{ days}\)
Part (c) [2 marks]: - 1 mark for 'time taken' - 1 mark for 'for activity / count rate / mass / number of nuclei to halve'
Question 8 · structured-theory
7.27 marks
Astronomers observe the light from a distant galaxy and determine its recession speed using the redshift of its spectral lines.
(a) Explain what is meant by the redshift of light from a galaxy. (b) The galaxy is located at a distance of \(1.5 \times 10^{24}\text{ m}\) from Earth. The Hubble constant \(H_0\) is \(2.2 \times 10^{-18}\text{ s}^{-1}\). Calculate the recession speed of the galaxy in m/s. (c) State what the redshift of light from distant galaxies suggests about the history of the Universe.
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Worked solution
(a) Redshift means light from a galaxy moving away from Earth is shifted towards longer wavelengths (towards the red end of the spectrum). (b) Using Hubble\'s Law: \(v = H_0 \times d\) \(v = (2.2 \times 10^{-18}\text{ s}^{-1}) \times (1.5 \times 10^{24}\text{ m}) = 3.3 \times 10^6\text{ m/s}\). (c) The redshift of distant galaxies suggests that the Universe is expanding. Extrapolating back in time, it suggests that the Universe originated from a single, hot, dense point (the Big Bang).
Marking scheme
Part (a) [2 marks]: - 1 mark for stating that the observed wavelength of light increases - 1 mark for relating this to galaxies moving away from us
Part (b) [3.27 marks]: - 1 mark for formula \(v = H_0 d\) - 1 mark for correct substitution: \(2.2 \times 10^{-18} \times 1.5 \times 10^{24}\) - 1.27 marks for correct calculation with unit: \(3.3 \times 10^6\text{ m/s}\)
Part (c) [2 marks]: - 1 mark for stating that the Universe is expanding - 1 mark for stating that the Universe originated from a single point / Big Bang theory
Question 9 · structured-theory
7.27 marks
A toy cart of mass \( 0.80 \text{ kg} \) is moving along a frictionless horizontal track at a speed of \( 2.5 \text{ m/s} \). It collides with and couples to a second cart of mass \( 1.2 \text{ kg} \) which is initially stationary.
(a) Calculate the momentum of the first cart before the collision. Include the unit.
(b) State the total momentum of both carts after the collision.
(c) Calculate the velocity of the coupled carts after the collision.
(d) Determine by calculation whether this collision is elastic or inelastic.
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Worked solution
(a) Momentum of the first cart: \( p = m_1 \times v_1 = 0.80 \text{ kg} \times 2.5 \text{ m/s} = 2.0 \text{ kg m/s} \) (or \( \text{N s} \)).
(b) According to the principle of conservation of momentum, total momentum is conserved in the absence of external forces. Total momentum after collision = \( 2.0 \text{ kg m/s} \).
(c) Let \( v \) be the final velocity of the coupled carts. \( p_{\text{total}} = (m_1 + m_2) \times v \) \( 2.0 \text{ kg m/s} = (0.80 \text{ kg} + 1.2 \text{ kg}) \times v \) \( 2.0 = 2.0 \times v \) \( v = 1.0 \text{ m/s} \).
Since the final kinetic energy (\( 1.0 \text{ J} \)) is less than the initial kinetic energy (\( 2.5 \text{ J} \)), kinetic energy is not conserved, meaning the collision is inelastic.
(b) [1 mark] - \(2.0 \text{ kg m/s}\) (allow follow-through from part a) [1 mark]
(c) [2 marks] - Use of conservation of momentum: \(v = \frac{2.0}{0.80 + 1.2}\) or \(2.0 = 2.0 \times v\) [1 mark] - \(1.0 \text{ m/s}\) [1 mark]
(d) [2 marks] - Calculation of initial \(E_k\) (\(2.5 \text{ J}\)) and final \(E_k\) (\(1.0 \text{ J}\)) [1 mark] - Conclusion that collision is inelastic because kinetic energy is lost / not conserved [1 mark]
Question 10 · structured-theory
7.27 marks
A ray of monochromatic light traveling in air is incident on the flat surface of a glass block. The angle of incidence is \( 42.0^\circ \) and the angle of refraction inside the glass is \( 26.0^\circ \).
(a) Calculate the refractive index of the glass.
(b) Define the term 'critical angle' as it applies to light at a boundary between two optical media.
(c) Calculate the critical angle for a ray of light traveling from this glass block into air.
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Worked solution
(a) Refractive index \( n \) is given by Snell's Law: \( n = \frac{\sin i}{\sin r} \) \( n = \frac{\sin(42.0^\circ)}{\sin(26.0^\circ)} \approx \frac{0.6691}{0.4384} \approx 1.526 \approx 1.53 \).
(b) The critical angle is defined as the angle of incidence in the optically denser medium (medium with higher refractive index) that results in an angle of refraction of \( 90^\circ \) in the less dense medium (medium with lower refractive index).
(c) The critical angle \( c \) is calculated using the formula: \( \sin c = \frac{1}{n} \) \( \sin c = \frac{1}{1.526} \approx 0.6553 \) \( c = \sin^{-1}(0.6553) \approx 40.9^\circ \).
(b) [2 marks] - Angle of incidence in the denser medium [1 mark] - For which the angle of refraction is \(90^\circ\) or light travels along the boundary [1 mark]
(c) [3 marks] - Formula: \(\sin c = \frac{1}{n}\) [1 mark] - Substitution: \(\sin c = \frac{1}{1.53}\) (accept follow-through from part a) [1 mark] - \(40.9^\circ\) (accept range \(40.8^\circ\) to \(41.0^\circ\)) [1 mark]
Question 11 · structured-theory
7.27 marks
An electrical circuit contains a \( 9.0\text{ V} \) 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 \( 12\ \Omega \) resistor.
(a) Calculate the combined resistance of the \( 4.0\ \Omega \) and \( 12\ \Omega \) resistors connected in parallel.
(b) State the total resistance of the entire circuit.
(c) Calculate the total current flowing from the battery.
(d) Calculate the potential difference across the parallel combination.
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(b) The total resistance of the circuit \( R_{\text{total}} \) is the series sum: \( R_{\text{total}} = 6.0\ \Omega + R_p = 6.0 + 3.0 = 9.0\ \Omega \).
(c) Using Ohm's Law for the whole circuit: \( I = \frac{V}{R_{\text{total}}} = \frac{9.0\text{ V}}{9.0\ \Omega} = 1.0\text{ A} \).
(d) The potential difference across the parallel combination \( V_p \) is: \( V_p = I \times R_p = 1.0\text{ A} \times 3.0\ \Omega = 3.0\text{ V} \). (Alternatively, \( V_p = 9.0\text{ V} - I \times 6.0\ \Omega = 9.0 - (1.0 \times 6.0) = 3.0\text{ V} \)).
(d) [2 marks] - \(V_p = I \times R_p\) or \(9.0 - (I \times 6.0)\) or potential divider formula [1 mark] - \(3.0\text{ V}\) [1 mark]
Section Alternative to Practical
Answer all questions. Use your knowledge of practical physics techniques, graphing, and experimental design.
4 Question · 40 marks
Question 1 · practical-structured
10 marks
A student is investigating the rate of cooling of water under different conditions.
A thermometer scale is used to measure the initial temperature of the water at time \(t = 0\text{ s}\). The level of the liquid column is at the mark exactly one small division below the 80.0 °C mark, where each small division represents 1.0 °C.
(a) Record the initial temperature \(\theta_0\) shown on the thermometer.
(b) The student records the temperature of the water in an uninsulated beaker A every 30 s for 180 s. The readings are: - At \(t = 0\text{ s}\): \(\theta = \theta_0\) - At \(t = 30\text{ s}\): \(\theta = 74.0^\circ\text{C}\) - At \(t = 60\text{ s}\): \(\theta = 69.5^\circ\text{C}\) - At \(t = 90\text{ s}\): \(\theta = 66.0^\circ\text{C}\) - At \(t = 120\text{ s}\): \(\theta = 63.0^\circ\text{C}\) - At \(t = 150\text{ s}\): \(\theta = 60.5^\circ\text{C}\) - At \(t = 180\text{ s}\): \(\theta = 58.5^\circ\text{C}\)
Calculate the temperature fall \(\Delta \theta_A\) for beaker A over the 180 s.
(c) The experiment is repeated using beaker B, which is wrapped in a layer of cotton wool. The initial temperature of the water in beaker B is the same as in beaker A. At \(t = 180\text{ s}\), the temperature of beaker B is 64.0 °C.
Calculate the temperature fall \(\Delta \theta_B\) for beaker B over the 180 s.
(d) Calculate the average rate of cooling \(R\) for each beaker during the 180 s using the equation \(R = \frac{\Delta \theta}{t}\). (i) Calculate the rate of cooling for beaker A, \(R_A\). (ii) Calculate the rate of cooling for beaker B, \(R_B\). (iii) State the standard unit of the rate of cooling in this experiment.
(e) Suggest two variables that must be kept constant to ensure a fair comparison between the rate of cooling of water in beaker A and beaker B.
(f) State one precaution that the student should take when reading the thermometer to ensure the temperature readings are accurate.
(g) Explain, in terms of heat transfer, why wrapping the beaker with cotton wool reduces the rate of cooling.
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Worked solution
(a) Since the meniscus is one division below 80.0 °C, \(\theta_0 = 79.0^\circ\text{C}\).
(b) Temperature fall for beaker A: \(\Delta \theta_A = 79.0^\circ\text{C} - 58.5^\circ\text{C} = 20.5^\circ\text{C}\).
(c) Temperature fall for beaker B: \(\Delta \theta_B = 79.0^\circ\text{C} - 64.0^\circ\text{C} = 15.0^\circ\text{C}\).
(d) (i) \(R_A = \frac{20.5}{180} = 0.114^\circ\text{C/s}\) (or 0.11). (ii) \(R_B = \frac{15.0}{180} = 0.083^\circ\text{C/s}\). (iii) The unit of rate of cooling is \(^\circ\text{C/s}\) (or \(^\circ\text{C }\text{s}^{-1}\)).
(e) Any two from: volume of water, initial temperature of water, shape/material of the beaker, room temperature / presence of draughts.
(f) Read the scale perpendicular to the thermometer tube at eye level to avoid parallax error; stir the water before taking a reading; make sure the bulb is fully immersed and not touching the bottom/sides of the beaker.
(g) Cotton wool is a poor thermal conductor / insulator (or traps pockets of air which are poor conductors), reducing the rate of thermal energy transfer from the beaker to the surroundings by conduction and convection.
Marking scheme
(a) 79.0 °C [1] (accept 79) (b) 20.5 °C [1] (c) 15.0 °C [1] (d)(i) 0.114 °C/s (accept 0.11) [1] (d)(ii) 0.083 °C/s [1] (d)(iii) °C/s (or °C s⁻¹) [1] (e) Any two variables: e.g. volume of water, room temperature, beaker size/material [2] (f) Read perpendicular to scale/at eye level to avoid parallax OR stir before reading OR keep bulb fully immersed and away from walls [1] (g) Insulation/poor conductor/trapped air reduces rate of heat transfer by conduction/convection [1]
Question 2 · practical-structured
10 marks
A student is investigating the refraction of light through a transparent rectangular glass block using optical pins.
She places the block on a sheet of paper and draws its outline ABCD. An incident ray is drawn meeting the side AB at point P. The normal NN' is perpendicular to AB at P.
(a) The student measures the angle of incidence \(i\) using a protractor. The incident ray is at an angle of 45.0° to the normal. State the value of \(i\).
(b) The student places two pins, \(P_1\) and \(P_2\), on the incident ray. (i) State where the pins \(P_1\) and \(P_2\) should be positioned relative to each other on the incident ray to ensure high accuracy. (ii) State a practical reason why the distance between the two pins should be at least 5.0 cm.
(c) The student looks through the block from the opposite side CD and places two pins, \(P_3\) and \(P_4\), so that they appear to line up with the images of \(P_1\) and \(P_2\). State how the student must view the pins to ensure they are aligned correctly.
(d) After removing the block, the student draws the path of the refracted ray inside the block and measures the angle of refraction to be \(r = 28.0^\circ\). (i) Calculate the value of \(\sin i\). (ii) Calculate the value of \(\sin r\). (iii) Calculate the refractive index \(n\) of the glass block using the equation \(n = \frac{\sin i}{\sin r}\).
(e) The student repeats the experiment for several different angles of incidence and plans to plot a graph of \(\sin i\) (y-axis) against \(\sin r\) (x-axis). (i) Describe the expected shape of the graph. (ii) Explain how the gradient of this graph can be used to determine the refractive index \(n\). (iii) Suggest one precaution, other than pin separation, to obtain accurate drawings or line alignments in this pin experiment.
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Worked solution
(a) The angle of incidence is the angle between the incident ray and the normal, so \(i = 45.0^\circ\).
(b) (i) The pins should be placed as far apart as possible on the incident ray. (ii) Placing pins far apart reduces the percentage error / uncertainty in drawing the straight line through the pin marks.
(c) The student should view the pins from their bases / at eye level, ensuring that the bottom of the pins appear in a single straight line.
(e) (i) A straight line passing through the origin. (ii) Since \(\sin i = n \sin r\), the gradient of the graph of \(\sin i\) against \(\sin r\) is equal to \(n\). (iii) Ensure pins are perpendicular/vertical; use thin/sharp pins; use a sharp pencil to draw lines; place the block back in exactly the same outline if it moves.
Marking scheme
(a) 45.0° (or 45°) [1] (b)(i) Far apart [1] (b)(ii) Reduces percentage error in alignment / direction of the ray [1] (c) View the bases/bottoms of the pins at eye level [1] (d)(i) 0.707 (accept 0.71) [1] (d)(ii) 0.469 (accept 0.47) [1] (d)(iii) 1.51 (or 1.50 to 1.52, no units) [1] (e)(i) Straight line passing through the origin [1] (e)(ii) Gradient is equal to the refractive index n [1] (e)(iii) Any one from: ensure pins are vertical, use very thin pins, use a sharp pencil [1]
Question 3 · practical-structured
10 marks
A student investigates the density of a metal key.
First, she determines the mass \(M\) of a uniform metre rule by balancing it on a pivot. She places a known mass \(m = 100.0\text{ g}\) on the rule. The rule balances when the pivot is at the 40.0 cm mark. The known mass \(m\) is centered at the 15.0 cm mark. The center of gravity of the uniform metre rule is at the 50.0 cm mark.
(a) (i) Determine the distance \(d_1\) from the center of mass \(m\) to the pivot. (ii) Determine the distance \(d_2\) from the pivot to the center of gravity of the metre rule.
(b) Calculate the mass \(M\) of the metre rule using the principle of moments equation: \(M \times d_2 = m \times d_1\).
(c) State one precaution the student should take to find the balance point as accurately as possible.
(d) Explain how the student can check that the metre rule is uniform before starting the experiment.
(e) Next, the student uses a measuring cylinder to determine the volume \(V\) of the metal key. - The initial volume of water in the measuring cylinder is at the graduation mark exactly at the 50.0 cm³ mark. - When the key is fully submerged, the water level rises to two small divisions above the 50.0 cm³ mark, where each small division represents 2.0 cm³.
(i) State the initial water volume \(V_1\). (ii) State the water volume \(V_2\) with the submerged key. (iii) Calculate the volume \(V\) of the key.
(f) The mass of the key is determined to be 31.2 g. Calculate the density \(\rho\) of the metal key using the equation \(\rho = \frac{\text{mass}}{V}\). State the unit.
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(a)(i) d1 = 25.0 cm [1] (a)(ii) d2 = 10.0 cm [1] (b) M = 250 g [1] (c) Move mass/rule slowly/gently OR ensure pivot is perpendicular to rule [1] (d) Balance rule alone on the pivot to see if it balances at 50.0 cm [1] (e)(i) V1 = 50.0 cm³ [1] (e)(ii) V2 = 54.0 cm³ [1] (e)(iii) V = 4.0 cm³ [1] (f) rho = 7.8 [1], unit = g/cm³ [1]
Question 4 · practical-structured
10 marks
A student is investigating how the resistance of a wire depends on its length.
(a) The student sets up a circuit to measure the current in the wire and the potential difference across it. The circuit contains a power supply, a switch, an ammeter, a voltmeter, and a length of resistance wire connected between two terminals P and Q. (i) Explain how the ammeter must be connected relative to the resistance wire PQ in the circuit. (ii) Explain how the voltmeter must be connected relative to the resistance wire PQ in the circuit.
(b) The student closes the switch and records the current \(I\) and potential difference \(V\). - On the ammeter, the pointer is exactly halfway between 0.40 A and 0.50 A. - On the voltmeter, the pointer is at the second graduation mark past the 2.5 V mark, where each graduation mark represents 0.1 V.
(i) Record the current \(I\). (ii) Record the potential difference \(V\). (iii) Calculate the resistance \(R\) of the wire PQ using the equation \(R = \frac{V}{I}\).
(c) State one reason why it is important to open the switch between taking readings.
(d) The student wants to extend the experiment to investigate how the resistance of the wire depends on its length \(L\). (i) Describe how the student would vary the length of the wire in the circuit. (ii) State two other variables that must be kept constant during this investigation. (iii) Predict how the resistance of the wire changes as its length increases.
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Worked solution
(a) (i) The ammeter must be connected in series with the resistance wire PQ to measure the current flowing through it. (ii) The voltmeter must be connected in parallel across the resistance wire PQ to measure the potential difference across it.
(b) (i) Halfway between 0.40 A and 0.50 A gives \(I = 0.45\text{ A}\). (ii) Two divisions past 2.5 V where each division is 0.1 V gives \(V = 2.5 + 0.2 = 2.7\text{ V}\). (iii) \(R = \frac{2.7}{0.45} = 6.0\text{ }\Omega\).
(c) Opening the switch prevents current flowing when not taking readings, which prevents the wire from heating up. Heating changes the resistance of the wire.
(d) (i) Use a sliding contact, crocodile clip, or jockey to connect to the wire at different measured lengths. (ii) Maintain constant: material of the wire, cross-sectional area (diameter/thickness) of the wire, temperature of the room/environment. (iii) The resistance increases as length increases (resistance is directly proportional to length).
Marking scheme
(a)(i) Ammeter connected in series [1] (a)(ii) Voltmeter connected in parallel [1] (b)(i) I = 0.45 A [1] (b)(ii) V = 2.7 V [1] (b)(iii) R = 6.0 Ω (or 6 Ω) [1] (c) Prevent heating of the wire / temperature rise (which changes resistance) [1] (d)(i) Use sliding contact / jockey / crocodile clip to connect at different points [1] (d)(ii) Any two of: material of wire, diameter/thickness/cross-sectional area, temperature [2] (d)(iii) Resistance increases / is directly proportional [1]
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