An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V1) Cambridge IGCSE Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
Paper 2 (Extended Multiple Choice)
Answer all forty questions. For each question, choose the correct option from A, B, C, or D.
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 \(6.0\text{ s}\). It then travels at a constant speed for \(10\text{ s}\), and finally decelerates to rest in a further \(4.0\text{ s}\).
What is the average speed of the car for the entire journey?
A.\(6.0\text{ m/s}\)
B.\(9.0\text{ m/s}\)
C.\(10\text{ m/s}\)
D.\(12\text{ m/s}\)
Show answer & marking schemeHide answer & marking scheme
- Method to find maximum speed: \(v = 12\text{ m/s}\) - Method to calculate distances: \(d_1 = 36\text{ m}\), \(d_2 = 120\text{ m}\), \(d_3 = 24\text{ m}\) - Total distance: \(180\text{ m}\) - Average speed: \(180 / 20 = 9.0\text{ m/s}\)
Question 2 · multiple-choice
1 marks
A uniform beam of length \(1.2\text{ m}\) and weight \(20\text{ N}\) is suspended horizontally by two vertical strings, one at each end, P and Q.
A heavy block of weight \(60\text{ N}\) is placed on the beam at a distance of \(0.30\text{ m}\) from end P.
What is the tension in the string at end Q?
A.\(15\text{ N}\)
B.\(25\text{ N}\)
C.\(35\text{ N}\)
D.\(40\text{ N}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
To find the tension in the string at end Q (\(T_Q\)), we take moments about end P to eliminate the tension at P:
- The weight of the uniform beam (\(20\text{ N}\)) acts at its center of mass, which is at \(0.60\text{ m}\) from P. Clockwise moment of the beam = \(20\text{ N} \times 0.60\text{ m} = 12\text{ N m}\). - The block of weight \(60\text{ N}\) acts at \(0.30\text{ m}\) from P. Clockwise moment of the block = \(60\text{ N} \times 0.30\text{ m} = 18\text{ N m}\). - Total clockwise moment = \(12 + 18 = 30\text{ N m}\). - The tension \(T_Q\) acts upwards at end Q, which is \(1.2\text{ m}\) from P. Anticlockwise moment of \(T_Q\) = \(T_Q \times 1.2\text{ m}\).
Since the beam is in equilibrium: \[ T_Q \times 1.2 = 30 \] \[ T_Q = \frac{30}{1.2} = 25\text{ N} \]
Marking scheme
Award 1 mark for the correct option B.
- C1 for calculating clockwise moments correctly (e.g., beam moment = 12 N m, block moment = 18 N m) - A1 for the correct value of 25 N.
Question 3 · multiple-choice
1 marks
An object A of mass \(3.0\text{ kg}\) travels at a velocity of \(6.0\text{ m/s}\) to the right. It collides head-on with an object B of mass \(2.0\text{ kg}\) travelling at a velocity of \(4.0\text{ m/s}\) to the left.
After the collision, the two objects stick together and move with a common velocity.
What is this common velocity?
A.\(2.0\text{ m/s}\) to the left
B.\(2.0\text{ m/s}\) to the right
C.\(5.2\text{ m/s}\) to the left
D.\(5.2\text{ m/s}\) to the right
Show answer & marking schemeHide answer & marking scheme
Worked solution
Let the direction to the right be positive.
- Initial velocity of A: \(u_A = +6.0\text{ m/s}\) - Initial velocity of B: \(u_B = -4.0\text{ m/s}\) (since it moves to the left)
According to the principle of conservation of momentum: \[ m_A u_A + m_B u_B = (m_A + m_B) v \] \[ (3.0 \times 6.0) + (2.0 \times [-4.0]) = (3.0 + 2.0) v \] \[ 18.0 - 8.0 = 5.0 v \] \[ 10.0 = 5.0 v \implies v = +2.0\text{ m/s} \]
Since the velocity is positive, the combined object moves to the right at \(2.0\text{ m/s}\).
Marking scheme
Award 1 mark for the correct option B.
- C1 for setting up the momentum conservation equation with the correct sign for B's velocity (18 - 8 = 5v). - A1 for the correct velocity and direction.
Question 4 · multiple-choice
1 marks
An electric heater of power \(80\text{ W}\) is used to heat a \(2.0\text{ kg}\) block of metal. The heater is switched on for \(5.0\text{ minutes}\). The temperature of the block increases from \(20\text{ }^\circ\text{C}\) to \(50\text{ }^\circ\text{C}\). Assume there is no thermal energy loss to the surroundings.
What is the specific heat capacity of the metal?
A.\(6.7\text{ J / (kg }^\circ\text{C)}\)
B.\(200\text{ J / (kg }^\circ\text{C)}\)
C.\(400\text{ J / (kg }^\circ\text{C)}\)
D.\(800\text{ J / (kg }^\circ\text{C)}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. **Calculate the energy supplied by the heater:** - \(P = 80\text{ W}\) - \(t = 5.0\text{ minutes} = 5.0 \times 60 = 300\text{ s}\) - \(E = P \times t = 80 \times 300 = 24,000\text{ J}\)
2. **Calculate the temperature rise:** - \(\Delta\theta = 50 - 20 = 30\text{ }^\circ\text{C}\)
3. **Use the specific heat capacity formula:** - \(E = m c \Delta\theta\) - \(24,000 = 2.0 \times c \times 30\) - \(24,000 = 60 c \implies c = 400\text{ J / (kg }^\circ\text{C)}\)
Marking scheme
Award 1 mark for the correct option C.
- C1 for converting time to seconds (300 s) and calculating energy (24,000 J). - A1 for calculating \(c = 400\text{ J / (kg }^\circ\text{C)}\).
Question 5 · multiple-choice
1 marks
A ray of light travels from a glass block into air. The refractive index of the glass is \(1.60\).
What is the critical angle for this glass?
A.\(31^\circ\)
B.\(39^\circ\)
C.\(51^\circ\)
D.\(53^\circ\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
The relationship between refractive index \(n\) and critical angle \(c\) is: \[ \sin c = \frac{1}{n} \] \[ \sin c = \frac{1}{1.60} = 0.625 \] \[ c = \sin^{-1}(0.625) \approx 38.7^\circ \]
Rounding to the nearest degree, we get \(39^\circ\).
Marking scheme
Award 1 mark for the correct option B.
- C1 for using the formula \(\sin c = 1/n\) to find \(\sin c = 0.625\). - A1 for finding \(c \approx 39^\circ\).
Question 6 · multiple-choice
1 marks
A wire of length \(L\) and cross-sectional area \(A\) has a resistance of \(8.0\text{ }\Omega\).
A second wire, made of the same metal, has a length of \(2L\) and a cross-sectional area of \(4A\).
What is the resistance of the second wire?
A.\(1.0\text{ }\Omega\)
B.\(4.0\text{ }\Omega\)
C.\(16\text{ }\Omega\)
D.\(64\text{ }\Omega\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
The resistance of a wire is given by: \[ R = \rho \frac{\text{length}}{\text{area}} \]
For the first wire: \[ R_1 = \rho \frac{L}{A} = 8.0\text{ }\Omega \]
For the second wire: \[ R_2 = \rho \frac{2L}{4A} = \frac{2}{4} \left(\rho \frac{L}{A}\right) = 0.5 \times R_1 \] \[ R_2 = 0.5 \times 8.0 = 4.0\text{ }\Omega \]
Marking scheme
Award 1 mark for the correct option B.
- C1 for recognizing the relationship \(R \propto L/A\). - A1 for the correct value of 4.0 \(\Omega\).
Question 7 · multiple-choice
1 marks
An ideal transformer has \(400\) turns on its primary coil and \(100\) turns on its secondary coil. The primary coil is connected to an alternating current (a.c.) supply of voltage \(240\text{ V}\).
A resistor of resistance \(12\text{ }\Omega\) is connected across the secondary coil.
What is the current in the primary coil?
A.\(0.31\text{ A}\)
B.\(1.25\text{ A}\)
C.\(5.00\text{ A}\)
D.\(20.0\text{ A}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. **Calculate the secondary voltage (\(V_s\)):** \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = 240 \times \frac{100}{400} = 60\text{ V} \]
2. **Calculate the secondary current (\(I_s\)) using Ohm's Law:** \[ I_s = \frac{V_s}{R} = \frac{60\text{ V}}{12\text{ }\Omega} = 5.0\text{ A} \]
3. **Use the power conservation equation for an ideal transformer:** \[ I_p V_p = I_s V_s \] \[ I_p \times 240 = 5.0 \times 60 \] \[ 240 I_p = 300 \implies I_p = \frac{300}{240} = 1.25\text{ A} \]
Marking scheme
Award 1 mark for the correct option B.
- C1 for calculating \(V_s = 60\text{ V}\) and secondary current \(I_s = 5.0\text{ A}\). - A1 for applying \(I_p V_p = I_s V_s\) to get \(I_p = 1.25\text{ A}\).
Question 8 · multiple-choice
1 marks
A distant galaxy is measured to be at a distance of \(1.5 \times 10^{22}\text{ km}\) from Earth. Its speed of recession is determined to be \(3.3 \times 10^4\text{ km/s}\).
Using these data, what is the calculated value of the Hubble constant \(H_0\)?
A.\(4.5 \times 10^{-19}\text{ s}^{-1}\)
B.\(2.2 \times 10^{-18}\text{ s}^{-1}\)
C.\(2.2 \times 10^{-15}\text{ s}^{-1}\)
D.\(4.5 \times 10^{17}\text{ s}^{-1}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
According to Hubble's Law: \[ v = H_0 d \implies H_0 = \frac{v}{d} \]
Substitute the given values (where the units of \(\text{km}\) cancel out): \[ H_0 = \frac{3.3 \times 10^4\text{ km/s}}{1.5 \times 10^{22}\text{ km}} \] \[ H_0 = \frac{3.3}{1.5} \times 10^{4 - 22}\text{ s}^{-1} \] \[ H_0 = 2.2 \times 10^{-18}\text{ s}^{-1} \]
Marking scheme
Award 1 mark for the correct option B.
- C1 for stating the formula \(H_0 = v/d\). - A1 for performing the calculation correctly to get \(2.2 \times 10^{-18}\text{ s}^{-1}\).
Question 9 · multiple-choice
1 marks
A car of mass 1200 kg travels along a straight road. It accelerates uniformly from rest to a speed of \(15\text{ m/s}\) in a time of \(6.0\text{ s}\). It then travels at a constant speed of \(15\text{ m/s}\) for \(10\text{ s}\), and finally decelerates uniformly to rest in a further \(4.0\text{ s}\). What is the average speed of the car for the entire journey?
A.\(7.50\text{ m/s}\)
B.\(11.3\text{ m/s}\)
C.\(12.5\text{ m/s}\)
D.\(15.0\text{ m/s}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
First, calculate the distance travelled in each of the three stages of motion: - Stage 1 (uniform acceleration from rest to \(15\text{ m/s}\) in \(6.0\text{ s}\)): \(d_1 = \frac{u + v}{2} \times t = \frac{0 + 15}{2} \times 6.0 = 45\text{ m}\) - Stage 2 (constant speed of \(15\text{ m/s}\) for \(10\text{ s}\)): \(d_2 = v \times t = 15 \times 10 = 150\text{ m}\) - Stage 3 (uniform deceleration to rest in \(4.0\text{ s}\)): \(d_3 = \frac{v + 0}{2} \times t = \frac{15 + 0}{2} \times 4.0 = 30\text{ m}\)
Now, calculate the total distance and total time: - Total distance \(D = 45 + 150 + 30 = 225\text{ m}\) - Total time \(T = 6.0 + 10 + 4.0 = 20.0\text{ s}\)
Finally, calculate the average speed: \(\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{225\text{ m}}{20.0\text{ s}} = 11.25\text{ m/s}\)
Rounding to 3 significant figures gives \(11.3\text{ m/s}\).
Marking scheme
1 mark for the correct option B. Show calculation of total distance (225 m) and division by total time (20 s).
Question 10 · multiple-choice
1 marks
A toy car of mass \(m\) travels at a constant speed \(v\) in a horizontal circular path of radius \(r\). The centripetal force acting on the car is \(F\). Which row describes the work done on the car by the centripetal force, and the magnitude of the change in momentum of the car, as it travels half-way around the circle?
A.work done: zero, magnitude of change in momentum: zero
B.work done: zero, magnitude of change in momentum: \(2mv\)
C.work done: \(\pi F r\), magnitude of change in momentum: zero
D.work done: \(\pi F r\), magnitude of change in momentum: \(2mv\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. Work done by a force is given by \(W = F d \cos(\theta)\). Since the centripetal force is always perpendicular to the direction of motion (displacement) at every instant, the angle \(\theta = 90^\circ\), which means the work done \(W = 0\). 2. Momentum is a vector quantity given by \(\vec{p} = m\vec{v}\). Initially, the car has a velocity \(\vec{v}\). After travelling half-way around the circle, its direction of motion is exactly reversed, so its final velocity is \(-\vec{v}\). The change in momentum is: \(\Delta \vec{p} = \vec{p}_f - \vec{p}_i = -m\vec{v} - m\vec{v} = -2m\vec{v}\) Thus, the magnitude of the change in momentum is \(2mv\).
Marking scheme
1 mark for the correct option B. Confirming work done is zero (force is perpendicular to displacement) and the magnitude of change in momentum is 2mv.
Question 11 · multiple-choice
1 marks
A \(500\text{ g}\) block of ice at \(-10^\circ\text{C}\) is heated at a constant rate of \(200\text{ W}\). The specific heat capacity of ice is \(2100\text{ J}/(\text{kg}\cdot^\circ\text{C})\). The specific latent heat of fusion of ice is \(3.3 \times 10^5\text{ J}/\text{kg}\). How long does it take for the ice to be completely converted into liquid water at \(0^\circ\text{C}\)?
A.\(52.5\text{ s}\)
B.\(825\text{ s}\)
C.\(878\text{ s}\)
D.\(1760\text{ s}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
The total energy required consists of two stages: 1. Heating the ice from \(-10^\circ\text{C}\) to \(0^\circ\text{C}\): \(Q_1 = m \cdot c_{\text{ice}} \cdot \Delta T = 0.500\text{ kg} \times 2100\text{ J}/(\text{kg}\cdot^\circ\text{C}) \times (0 - (-10))^\circ\text{C} = 10,500\text{ J}\)
2. Melting the ice at \(0^\circ\text{C}\): \(Q_2 = m \cdot L_f = 0.500\text{ kg} \times 3.3 \times 10^5\text{ J}/\text{kg} = 165,000\text{ J}\)
Total thermal energy required: \(Q_{\text{total}} = Q_1 + Q_2 = 10,500 + 165,000 = 175,500\text{ J}\)
Using the relation between power, energy, and time: \(t = \frac{Q_{\text{total}}}{P} = \frac{175,500\text{ J}}{200\text{ W}} = 877.5\text{ s}\)
Rounding to 3 significant figures gives \(878\text{ s}\).
Marking scheme
1 mark for the correct option C. Show sum of heating energy (10,500 J) and melting energy (165,000 J) divided by power (200 W).
Question 12 · multiple-choice
1 marks
A ray of light in air is incident on the surface of a transparent plastic block at an angle of \(40^\circ\) to the normal. The angle of refraction inside the block is \(25^\circ\). What is the critical angle for light travelling from the plastic block into air?
A.\(25^\circ\)
B.\(41^\circ\)
C.\(49^\circ\)
D.\(65^\circ\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. Find the refractive index \(n\) of the plastic block using Snell's Law: \(n = \frac{\sin(i)}{\sin(r)} = \frac{\sin(40^\circ)}{\sin(25^\circ)} = \frac{0.6428}{0.4226} \approx 1.521\)
2. Use the critical angle formula: \(\sin(c) = \frac{1}{n} = \frac{1}{1.521} \approx 0.6575\)
Rounding to 2 significant figures, we get \(41^\circ\).
Marking scheme
1 mark for the correct option B. Show calculation of refractive index first, then find critical angle using sin(c) = 1/n.
Question 13 · multiple-choice
1 marks
Two wires, \(X\) and \(Y\), are made of the same metal. Wire \(X\) has length \(L\) and radius \(r\). Wire \(Y\) has length \(2L\) and radius \(2r\). The two wires are connected in parallel across a power supply. What is the ratio of the current in wire \(X\) to the current in wire \(Y\), \(\frac{I_X}{I_Y}\)?
A.\(1 : 4\)
B.\(1 : 2\)
C.\(1 : 1\)
D.\(2 : 1\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
The resistance of a wire is given by \(R = \rho \frac{\text{length}}{\text{area}} = \rho \frac{\text{length}}{\pi \cdot \text{radius}^2}\).
Since the wires are connected in parallel, the potential difference \(V\) across both wires is the same. Using Ohm's law, the current \(I = \frac{V}{R}\): \(I_X = \frac{V}{R_X}\) \(I_Y = \frac{V}{R_Y} = \frac{V}{\frac{1}{2}R_X} = 2 I_X\)
Thus, the ratio of the currents is: \(\frac{I_X}{I_Y} = \frac{I_X}{2 I_X} = \frac{1}{2}\)
Marking scheme
1 mark for the correct option B. Finding the resistance of Y is half of X, so in parallel, the current in Y is twice that in X.
Question 14 · multiple-choice
1 marks
A detector is used to measure the count rate near a radioactive source. The background count rate is constant at \(24\text{ counts/minute}\). At time \(t = 0\), the detector records a count rate of \(408\text{ counts/minute}\). The half-life of the radioactive source is \(12\text{ minutes\dots}\) What count rate does the detector record at \(t = 36\text{ minutes}\)?
A.\(48\text{ counts/minute}\)
B.\(51\text{ counts/minute}\)
C.\(72\text{ counts/minute}\)
D.\(120\text{ counts/minute}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. At \(t = 0\), the total measured count rate is \(408\text{ counts/minute}\). Subtract the background to find the count rate due to the source alone: \(\text{Source count rate at } t=0 = 408 - 24 = 384\text{ counts/minute}\).
2. The time elapsed is \(36\text{ minutes}\), which corresponds to exactly 3 half-lives (since \(3 \times 12 = 36\)). After 3 half-lives, the source count rate decays to: \(\text{Source count rate at } t=36 = 384 \times \left(\frac{1}{2}\right)^3 = \frac{384}{8} = 48\text{ counts/minute}\).
3. The detector measures the source count rate plus the background count rate: \(\text{Total measured count rate} = 48 + 24 = 72\text{ counts/minute}\).
Marking scheme
1 mark for the correct option C. Show subtraction of background first, apply 3 half-lives, then add background back.
Question 15 · multiple-choice
1 marks
Light from a distant galaxy is observed to have a fractional increase in wavelength (redshift) of \(\frac{\Delta \lambda}{\lambda} = 0.050\). Using a value for the speed of light of \(3.0 \times 10^8\text{ m/s}\) and a Hubble constant of \(2.2 \times 10^{-18}\text{ s}^{-1}\), what is the estimated distance of this galaxy from the Earth?
A.\(1.5 \times 10^7\text{ m}\)
B.\(1.1 \times 10^{16\text{ m}}\)
C.\(6.8 \times 10^{24}\text{ m}\)
D.\(1.4 \times 10^{26}\text{ m}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. Find the speed of recession \(v\) of the galaxy using the redshift equation: \(\frac{\Delta \lambda}{\lambda} \approx \frac{v}{c}\) \(v = 0.050 \times (3.0 \times 10^8\text{ m/s}) = 1.5 \times 10^7\text{ m/s}\)
2. Use Hubble's law, \(v = H_0 d\), to calculate the distance \(d\): \(d = \frac{v}{H_0} = \frac{1.5 \times 10^7\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} \approx 6.8 \times 10^{24}\text{ m}\)
Marking scheme
1 mark for the correct option C. Show calculation of recession velocity (1.5 x 10^7 m/s) and dividing it by the Hubble constant.
Question 16 · multiple-choice
1 marks
An alternating current (a.c.) supply is connected to the primary coil of an ideal transformer. Which row describes the magnetic field in the soft iron core and the nature of the voltage induced across the secondary coil?
A.magnetic field in core: constant, voltage across secondary: direct voltage (d.c.)
B.magnetic field in core: constant, voltage across secondary: alternating voltage (a.c.)
C.magnetic field in core: changing, voltage across secondary: direct voltage (d.c.)
D.magnetic field in core: changing, voltage across secondary: alternating voltage (a.c.)
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. An alternating current in the primary coil flows back and forth, continuously changing in magnitude and direction. This produces a magnetic field in the soft iron core that is continually changing in both magnitude and direction. 2. According to Faraday's law of electromagnetic induction, a changing magnetic field linking a coil induces an electromotive force (voltage). Because the magnetic field in the core is alternating, the induced voltage across the secondary coil is also alternating (a.c.).
Marking scheme
1 mark for the correct option D. Recognizing that primary a.c. produces a changing magnetic field, which in turn induces an a.c. voltage in the secondary.
Question 17 · multiple-choice
1 marks
A car accelerates from rest at a constant rate of \(1.5\text{ m/s}^2\) for \(6.0\text{ s}\). It then travels at a constant speed for \(10\text{ s}\) before decelerating uniformly to rest in a further \(4.0\text{ s}\).
What is the total distance travelled by the car?
A.\(117\text{ m}\)
B.\(135\text{ m}\)
C.\(153\text{ m}\)
D.\(180\text{ m}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
To find the total distance travelled, we can calculate the distance for each of the three phases of motion:
A uniform beam of length \(1.2\text{ m}\) and weight \(60\text{ N}\) is pivoted at its centre. A block of weight \(40\text{ N}\) is placed \(0.20\text{ m}\) from the left-hand end of the beam.
At what distance from the pivot must a downward force of \(50\text{ N}\) be applied on the right-hand side to keep the beam in equilibrium?
A.\(0.16\text{ m}\)
B.\(0.32\text{ m}\)
C.\(0.40\text{ m}\)
D.\(0.48\text{ m}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
For the beam to be in equilibrium, the clockwise moments must equal the anticlockwise moments about the pivot.
1. **Anticlockwise moment (due to the 40 N weight on the left):** - The pivot is at the centre (\(0.60\text{ m}\) from either end). - The distance of the block from the pivot is \(0.60\text{ m} - 0.20\text{ m} = 0.40\text{ m}\). - Anticlockwise moment = \(40\text{ N} \times 0.40\text{ m} = 16\text{ N}\cdot\text{m}\).
2. **Clockwise moment (due to the 50 N force on the right):** - Let the distance of this force from the pivot be \(x\). - Clockwise moment = \(50\text{ N} \times x\).
Setting clockwise moment equal to anticlockwise moment: \(50 \times x = 16\) \(x = \frac{16}{50} = 0.32\text{ m}\).
Marking scheme
Award 1 mark for the correct option B.
Question 19 · multiple-choice
1 marks
A trolley of mass \(2.0\text{ kg}\) travels at a speed of \(6.0\text{ m/s}\) to the right. It 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.\(12\text{ J}\)
B.\(18\text{ J}\)
C.\(24\text{ J}\)
D.\(36\text{ J}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. **Find the common velocity \(v\) using conservation of momentum:** - Initial momentum: \(p_i = m_1 u_1 + m_2 u_2 = (2.0\text{ kg} \times 6.0\text{ m/s}) + (4.0\text{ kg} \times 0) = 12.0\text{ kg}\cdot\text{m/s}\). - Total mass after collision: \(m_{\text{total}} = 2.0\text{ kg} + 4.0\text{ kg} = 6.0\text{ kg}\). - Common velocity: \(v = \frac{12.0\text{ kg}\cdot\text{m/s}}{6.0\text{ kg}} = 2.0\text{ m/s}\).
3. **Calculate the loss in kinetic energy:** - \(\text{Loss} = E_{ki} - E_{kf} = 36.0\text{ J} - 12.0\text{ J} = 24.0\text{ J}\).
Marking scheme
Award 1 mark for the correct option C.
Question 20 · multiple-choice
1 marks
An electrical heater of power \(150\text{ W}\) is used to heat a metal block of mass \(2.0\text{ kg}\). The heater is switched on for \(4.0\text{ minutes}\) and the temperature of the block rises from \(20^\circ\text{C}\) to \(65^\circ\text{C}\).
What is the specific heat capacity of the metal, assuming no heat loss to the surroundings?
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. **Calculate the thermal energy supplied by the heater:** - Time \(t = 4.0\text{ minutes} = 4.0 \times 60 = 240\text{ s}\). - Energy supplied \(E = P \times t = 150\text{ W} \times 240\text{ s} = 36\,000\text{ J}\).
2. **Calculate the temperature rise:** - \(\Delta\theta = 65^\circ\text{C} - 20^\circ\text{C} = 45^\circ\text{C}\).
3. **Use the specific heat capacity formula:** - \(E = m c \Delta\theta\) - \(c = \frac{E}{m \Delta\theta} = \frac{36\,000\text{ J}}{2.0\text{ kg} \times 45^\circ\text{C}} = \frac{36\,000}{90} = 400\text{ J}/(\text{kg}\cdot^\circ\text{C})\).
Marking scheme
Award 1 mark for the correct option B.
Question 21 · multiple-choice
1 marks
A ray of light in air is incident on the surface of a glass block at an angle of incidence of \(45^\circ\). The angle of refraction inside the glass is \(28^\circ\).
What is the critical angle for light in this glass block when surrounded by air?
A.\(38^\circ\)
B.\(42^\circ\)
C.\(45^\circ\)
D.\(62^\circ\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. **Find the refractive index \(n\) of the glass:** - \(n = \frac{\sin(i)}{\sin(r)} = \frac{\sin(45^\circ)}{\sin(28^\circ)} \approx \frac{0.7071}{0.4695} \approx 1.506\).
*(Note: The time of \(5.0\text{ s}\) is not required since energy is directly calculated from the charge and potential difference.)*
Marking scheme
Award 1 mark for the correct option C.
Question 23 · multiple-choice
1 marks
An ideal (100% efficient) transformer has \(400\text{ turns}\) on its primary coil and \(100\text{ turns}\) on its secondary coil. An alternating current (a.c.) supply of \(240\text{ V}\) is connected to the primary coil. A resistor of resistance \(12\ \Omega\) is connected across the secondary coil.
What is the current in the primary coil?
A.\(0.31\text{ A}\)
B.\(1.25\text{ A}\)
C.\(5.0\text{ A}\)
D.\(20\text{ A}\)
Show answer & marking schemeHide answer & marking scheme
2. **Find the secondary current \(I_s\):** - \(I_s = \frac{V_s}{R} = \frac{60\text{ V}}{12\ \Omega} = 5.0\text{ A}\).
3. **Find the primary current \(I_p\) using the relationship for an ideal transformer:** - \(I_p V_p = I_s V_s\) - \(I_p \times 240\text{ V} = 5.0\text{ A} \times 60\text{ V}\) - \(I_p \times 240 = 300\) - \(I_p = \frac{300}{240} = 1.25\text{ A}\).
Marking scheme
Award 1 mark for the correct option B.
Question 24 · multiple-choice
1 marks
A distant galaxy is moving away from the Earth at a speed of \(3.3 \times 10^3\text{ km/s}\).
Using a value of \(2.2 \times 10^{-18}\text{ s}^{-1}\) for the Hubble constant \(H_0\), what is the estimated distance of this galaxy from the Earth?
A.\(1.5 \times 10^{18}\text{ km}\)
B.\(1.5 \times 10^{21}\text{ km}\)
C.\(7.3 \times 10^{21}\text{ km}\)
D.\(7.3 \times 10^{24}\text{ km}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. **Convert the speed \(v\) from km/s to m/s:** - \(v = 3.3 \times 10^3\text{ km/s} = 3.3 \times 10^6\text{ m/s}\).
2. **Use Hubble's law to find the distance \(d\) in meters:** - \(v = H_0 \times d \implies d = \frac{v}{H_0}\) - \(d = \frac{3.3 \times 10^6\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} = 1.5 \times 10^{24}\text{ m}\).
3. **Convert the distance back to kilometers (km):** - \(d = \frac{1.5 \times 10^{24}\text{ m}}{10^3} = 1.5 \times 10^{21}\text{ km}\).
Marking scheme
Award 1 mark for the correct option B.
Question 25 · multiple-choice
1 marks
A stone falls through the air from a high cliff. The air resistance acting on the stone increases as its speed increases. Which row correctly describes the acceleration and the speed of the stone before it reaches terminal velocity?
A.acceleration decreases, speed increases
B.acceleration decreases, speed decreases
C.acceleration increases, speed increases
D.acceleration remains constant, speed increases
Show answer & marking schemeHide answer & marking scheme
Worked solution
Before terminal velocity is reached, the downward gravitational force (weight) is greater than the upward air resistance force. Therefore, there is a net downward force, and the stone continues to accelerate, meaning its speed increases. However, as the speed increases, the air resistance increases. This reduces the net downward force (\(F = W - R\)), which in turn reduces the acceleration (\(a = F/m\)). Thus, the acceleration decreases.
Marking scheme
B1 for explaining that speed increases because there is a net downward force. B1 for explaining that acceleration decreases because air resistance increases with speed, reducing the net force. Select option A.
Question 26 · multiple-choice
1 marks
A toy car of mass \(0.50\text{ kg}\) travels at a speed of \(4.0\text{ m/s}\) to the right. It collides with a wall and rebounds at a speed of \(3.0\text{ m/s}\) to the left. The collision lasts for \(0.10\text{ s}\). What is the magnitude of the average force exerted by the wall on the car?
A.5.0 N
B.20 N
C.35 N
D.75 N
Show answer & marking schemeHide answer & marking scheme
Worked solution
Using the impulse-momentum equation: \(F \Delta t = \Delta p = m v_f - m v_i\). Let the direction to the right be positive. Then \(v_i = +4.0\text{ m/s}\) and \(v_f = -3.0\text{ m/s}\). Change in momentum \(\Delta p = 0.50 \times (-3.0) - 0.50 \times (+4.0) = -1.5 - 2.0 = -3.5\text{ kg m/s}\). The magnitude of the change in momentum is \(3.5\text{ kg m/s}\) (or \(\text{N s}\)). The average force is \(F = \frac{\Delta p}{\Delta t} = \frac{3.5}{0.10} = 35\text{ N}\).
Marking scheme
C1 for identifying the vector nature of velocity and calculating the correct change in momentum of \(3.5\text{ kg m/s}\). C1 for dividing by time interval of \(0.10\text{ s}\) to find force of \(35\text{ N}\). Select option C.
Question 27 · multiple-choice
1 marks
An electric motor is used to lift a load of weight \(120\text{ N}\) through a vertical height of \(5.0\text{ m}\) in \(4.0\text{ s}\). The electrical power input to the motor is \(250\text{ W}\). What is the efficiency of the motor?
A.24%
B.48%
C.60%
D.83%
Show answer & marking schemeHide answer & marking scheme
Worked solution
First, calculate the useful work done by the motor: \(W = F \times d = 120\text{ N} \times 5.0\text{ m} = 600\text{ J}\). Next, calculate the useful power output: \(P_{\text{out}} = \frac{W}{t} = \frac{600\text{ J}}{4.0\text{ s}} = 150\text{ W}\). Alternatively, the total electrical energy input is \(E_{\text{in}} = P_{\text{in}} \times t = 250\text{ W} \times 4.0\text{ s} = 1000\text{ J}\). Finally, calculate the efficiency: \(\text{efficiency} = \frac{\text{useful power output}}{\text{power input}} = \frac{150\text{ W}}{250\text{ W}} \times 100\% = 60\%\) (or \(\frac{600\text{ J}}{1000\text{ J}} \times 100\% = 60\%\)).
Marking scheme
C1 for calculating useful work done (\(600\text{ J}\)) or useful power output (\(150\text{ W}\)). C1 for applying the efficiency formula to obtain \(60\%\). Select option C.
Question 28 · multiple-choice
1 marks
A \(2.0\text{ kW}\) electric heater is used to heat a metal block of mass \(5.0\text{ kg}\). The heater is switched on for \(3.0\text{ minutes}\). The temperature of the block increases 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.240 J / (kg °C)
B.650 J / (kg °C)
C.800 J / (kg °C)
D.4000 J / (kg °C)
Show answer & marking schemeHide answer & marking scheme
Worked solution
Thermal energy supplied by the heater: \(E = P \times t = 2000\text{ W} \times (3.0 \times 60)\text{ s} = 360\,000\text{ J}\). Temperature rise of the block: \(\Delta T = 110^\circ\text{C} - 20^\circ\text{C} = 90^\circ\text{C}\). Using \(E = mc\Delta T\), we have \(360\,000 = 5.0 \times c \times 90\). Rearranging for \(c\): \(c = \frac{360\,000}{450} = 800\text{ J / (kg }^\circ\text{C)}\).
Marking scheme
C1 for calculating thermal energy supplied (\(360\,000\text{ J}\)) with correct unit conversions. C1 for calculating temperature change (\(90^\circ\text{C}\)) and applying the specific heat capacity formula. Select option C.
Question 29 · multiple-choice
1 marks
A ray of light in air is incident on the surface of a transparent glass block at an angle of \(40^\circ\) to the normal. The refractive index of the glass is \(1.5\). What is the angle of refraction inside the glass block, and what is the speed of light in the glass? (The speed of light in air is \(3.0 \times 10^8\text{ m/s}\).)
A.angle of refraction = 25°, speed of light in glass = 2.0 × 10⁸ m/s
B.angle of refraction = 25°, speed of light in glass = 4.5 × 10⁸ m/s
C.angle of refraction = 27°, speed of light in glass = 2.0 × 10⁸ m/s
D.angle of refraction = 27°, speed of light in glass = 4.5 × 10⁸ m/s
Show answer & marking schemeHide answer & marking scheme
Worked solution
According to Snell's law: \(n = \frac{\sin i}{\sin r}\). Substituting the values: \(1.5 = \frac{\sin 40^\circ}{\sin r} \implies \sin r = \frac{\sin 40^\circ}{1.5} \approx \frac{0.6428}{1.5} \approx 0.4285\). Therefore, \(r = \arcsin(0.4285) \approx 25.4^\circ\), which rounds to \(25^\circ\). The speed of light in glass is given by: \(v = \frac{c}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.5} = 2.0 \times 10^8\text{ m/s}\).
Marking scheme
C1 for correct application of Snell's law to find the angle of refraction as \(25^\regular\). C1 for calculating the speed of light in glass as \(2.0 \times 10^8\text{ m/s}\). Select option A.
Question 30 · multiple-choice
1 marks
A potential divider circuit consists of a \(12\text{ V}\) d.c. power supply, a fixed resistor of resistance \(4.0\text{ k}\Omega\) and a light-dependent resistor (LDR) connected in series. A voltmeter is connected across the LDR. In bright light, the resistance of the LDR is \(2.0\text{ k}\Omega\). What is the reading on the voltmeter in bright light?
A.2.0 V
B.4.0 V
C.6.0 V
D.8.0 V
Show answer & marking schemeHide answer & marking scheme
Worked solution
Using the potential divider formula: \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R_{\text{fixed}} + R_{\text{LDR}}}\). Substituting the given values: \(V_{\text{out}} = 12\text{ V} \times \frac{2.0\text{ k}\Omega}{4.0\text{ k}\Omega + 2.0\text{ k}\Omega} = 12 \times \frac{2.0}{6.0} = 4.0\text{ V}\).
Marking scheme
C1 for identifying the series combination and total resistance of \(6.0\text{ k}\Omega\). C1 for calculating the voltage across the LDR as \(4.0\text{ V}\). Select option B.
Question 31 · multiple-choice
1 marks
A sample of a radioactive isotope has an initial activity of \(800\text{ counts/s}\). After a time of \(24\text{ hours}\), the activity of the sample has decreased to \(100\text{ counts/s}\). What is the half-life of the radioactive isotope?
A.3.0 hours
B.6.0 hours
C.8.0 hours
D.12 hours
Show answer & marking schemeHide answer & marking scheme
Worked solution
The activity decreases from \(800\text{ counts/s}\) to \(100\text{ counts/s}\). Let's determine the number of half-lives that have passed: \(800 \xrightarrow{\text{1st half-life}} 400 \xrightarrow{\text{2nd half-life}} 200 \xrightarrow{\text{3rd half-life}} 100\). This represents exactly 3 half-lives. Since the total time is \(24\text{ hours}\), the length of one half-life is: \(T_{1/2} = \frac{24\text{ hours}}{3} = 8.0\text{ hours}\).
Marking scheme
C1 for determining that 3 half-lives have elapsed. C1 for dividing \(24\text{ hours}\) by 3 to get \(8.0\text{ hours}\). Select option C.
Question 32 · multiple-choice
1 marks
Astronomers observe light from a distant galaxy and determine that its spectral lines are redshifted. Which row correctly describes the direction of motion of the galaxy relative to Earth and how the observed wavelength of light compares to the emitted wavelength?
A.direction of motion = moving away from Earth, observed wavelength = longer than emitted wavelength
B.direction of motion = moving away from Earth, observed wavelength = shorter than emitted wavelength
C.direction of motion = moving towards Earth, observed wavelength = longer than emitted wavelength
D.direction of motion = moving towards Earth, observed wavelength = shorter than emitted wavelength
Show answer & marking schemeHide answer & marking scheme
Worked solution
Redshift occurs when the spectral lines of light from a distant galaxy are shifted toward the red end of the electromagnetic spectrum (longer wavelengths / lower frequencies). This indicates that the galaxy is moving away from Earth. Therefore, the galaxy is moving away from Earth and the observed wavelength of light is longer than the emitted wavelength.
Marking scheme
B1 for stating that redshift implies motion away from Earth. B1 for stating that redshifted light has a longer observed wavelength. Select option A.
Question 33 · multiple-choice
1 marks
A tennis ball of mass 0.060 kg is moving horizontally at 20 m/s when it hits a wall. It rebounds horizontally in the opposite direction at 15 m/s. The contact time between the ball and the wall is 0.050 s. What is the magnitude of the average force exerted on the ball by the wall?
A.6.0 N
B.18 N
C.24 N
D.42 N
Show answer & marking schemeHide answer & marking scheme
Worked solution
The initial velocity is u = +20 m/s and the final velocity is v = -15 m/s. The change in momentum is delta p = m(v - u) = 0.060 * (-15 - 20) = -2.1 kg m/s. The magnitude of the impulse is 2.1 N s. The average force is calculated using F = delta p / delta t = 2.1 / 0.050 = 42 N.
Marking scheme
Award 1 mark for the correct option D. Method: calculate change in momentum including direction change, then divide by contact time.
Question 34 · multiple-choice
1 marks
A metal wire of length L and cross-sectional area A has a resistance R. A second wire of the same material has twice the length and half the diameter of the first wire. What is the resistance of the second wire in terms of R?
A.2R
B.4R
C.8R
D.16R
Show answer & marking schemeHide answer & marking scheme
Worked solution
The resistance of a wire is given by R = rho * L / A. The area A is proportional to the square of the diameter (d^2). The second wire has twice the length (2L) and half the diameter (d/2), which means its cross-sectional area is reduced by a factor of 4 (A/4). Therefore, the new resistance is R_2 = rho * (2L) / (A/4) = 8 * (rho * L / A) = 8R.
Marking scheme
Award 1 mark for the correct option C. Method: apply the resistivity formula and the relationship between area and diameter.
Question 35 · multiple-choice
1 marks
A specific spectral line of light emitted from a distant galaxy has a laboratory wavelength of 600 nm. When observed on Earth, this spectral line is redshifted to a wavelength of 612 nm. What is the recessional speed of this galaxy?
A.3.0 * 10^5 m/s
B.6.0 * 10^6 m/s
C.1.2 * 10^7 m/s
D.3.0 * 10^8 m/s
Show answer & marking schemeHide answer & marking scheme
Worked solution
The redshift z is given by z = delta lambda / lambda = (612 - 600) / 600 = 12 / 600 = 0.020. The recessional speed v is calculated using v = z * c, where c is the speed of light (3.0 * 10^8 m/s). This gives v = 0.020 * 3.0 * 10^8 = 6.0 * 10^6 m/s.
Marking scheme
Award 1 mark for the correct option B. Method: calculate the redshift parameter z, then multiply by the speed of light.
Question 36 · multiple-choice
1 marks
A ray of light in a glass block of refractive index 1.50 is incident on the boundary with air. The angle of incidence is 45°. Which statement correctly describes the path of the light ray?
A.It refracts into the air with an angle of refraction of 45°.
B.It refracts into the air with an angle of refraction of 70.7°.
C.It undergoes total internal reflection and reflects back into the glass with an angle of reflection of 45°.
D.It travels along the boundary between the glass and the air.
Show answer & marking schemeHide answer & marking scheme
Worked solution
The critical angle c for the glass-air boundary is calculated using sin(c) = 1 / n = 1 / 1.50, which gives c = 41.8°. Since the angle of incidence (45°) is greater than the critical angle (41.8°), the light ray undergoes total internal reflection and reflects back into the glass at an angle of 45°.
Marking scheme
Award 1 mark for the correct option C. Method: determine the critical angle and compare it to the angle of incidence to deduce that total internal reflection occurs.
Question 37 · multiple-choice
1 marks
An object moves in a horizontal circle at a constant speed. Which row correctly describes the direction of the resultant force on the object and the state of its velocity?
A.resultant force: towards the centre of the circle; velocity: constant
B.resultant force: towards the centre of the circle; velocity: changing
C.resultant force: tangent to the circle; velocity: constant
D.resultant force: tangent to the circle; velocity: changing
Show answer & marking schemeHide answer & marking scheme
Worked solution
An object moving in a circle has a centripetal resultant force acting on it, which always points towards the centre of the circle. Although the speed is constant, the direction of travel is continuously changing, which means the velocity (a vector quantity) is also continuously changing.
Marking scheme
Award 1 mark for the correct option B. Method: identify centripetal force direction and recognize that changing direction means changing velocity.
Question 38 · multiple-choice
1 marks
An electric crane lifts a load of mass 400 kg vertically upwards through a height of 12 m in a time of 6.0 s. The total electrical power input to the crane is 14 kW. What is the efficiency of the crane system?
A.34%
B.56%
C.78%
D.91%
Show answer & marking schemeHide answer & marking scheme
Worked solution
The useful work output is the gravitational potential energy gained: E_p = m * g * h = 400 * 9.8 * 12 = 47040 J. The useful power output is work / time = 47040 / 6.0 = 7840 W = 7.84 kW. The efficiency is (useful power output / total power input) * 100 = (7.84 / 14) * 100 = 56%.
Marking scheme
Award 1 mark for the correct option B. Method: calculate work output using mgh, convert to useful power output, and calculate efficiency as a percentage of the 14 kW input.
Question 39 · multiple-choice
1 marks
An electric heater of power 800 W is used to heat a 2.0 kg metal block. The heater is switched on for 5.0 minutes. The temperature of the block increases from 20 °C to 120 °C. Assuming no thermal energy is lost to the surroundings, what is the specific heat capacity of the metal?
A.40 J / (kg °C)
B.600 J / (kg °C)
C.1200 J / (kg °C)
D.2400 J / (kg °C)
Show answer & marking schemeHide answer & marking scheme
Worked solution
The thermal energy supplied is E = P * t = 800 W * (5.0 * 60) s = 240,000 J. The temperature rise is delta theta = 120 - 20 = 100 °C. The specific heat capacity is c = E / (m * delta theta) = 240,000 / (2.0 * 100) = 1200 J / (kg °C).
Marking scheme
Award 1 mark for the correct option C. Method: calculate total energy input in Joules, calculate temperature change, and use the specific heat capacity formula.
Question 40 · multiple-choice
1 marks
A nucleus of Uranium-238 (superscript 238, subscript 92, U) decays through a series of steps to stable Lead-206 (superscript 206, subscript 82, Pb). How many alpha-particles and beta-minus particles are emitted in this decay chain?
A.6 alpha-particles and 6 beta-particles
B.8 alpha-particles and 6 beta-particles
C.8 alpha-particles and 8 beta-particles
D.32 alpha-particles and 10 beta-particles
Show answer & marking schemeHide answer & marking scheme
Worked solution
Each alpha-decay reduces the nucleon (mass) number A by 4 and the proton number Z by 2. The total change in A is 238 - 206 = 32, which corresponds to 32 / 4 = 8 alpha-decays. These 8 alpha-decays would reduce Z by 8 * 2 = 16, resulting in a proton number of 92 - 16 = 76. Since the final proton number of Lead is 82, there must be 82 - 76 = 6 beta-minus decays to increase Z back to 82.
Marking scheme
Award 1 mark for the correct option B. Method: determine the number of alpha decays from the mass number change, then calculate the number of beta decays needed to match the final proton number.
Ready to test yourself?
Turn these notes into exam-style practice. Get unlimited AI questions on this topic with instant marking and explanations.
Answer all nine structured theoretical questions. Show all working, state correct units, and express answers to 2 or 3 significant figures.
9 Question · 74 marks
Question 1 · structured
8 marks
A model rocket is fired vertically upwards. It accelerates upwards under constant thrust from rest, reaching a speed of \(45\text{ m/s}\) at time \(t = 3.0\text{ s}\). The engine then cuts out, and the rocket continues to rise until it reaches its maximum height. Air resistance is negligible. Take \(g = 9.8\text{ m/s}^2\).\ \ (a) Calculate the acceleration of the rocket during the first 3.0 s. [2]\ \ (b) Calculate the height reached by the rocket at \(t = 3.0\text{ s}\). [2]\ \ (c) State the magnitude and direction of the acceleration of the rocket immediately after the engine cuts out. [2]\ \ (d) Calculate the further height the rocket rises after the engine cuts out. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) acceleration \(a = \frac{v - u}{t} = \frac{45 - 0}{3.0} = 15\text{ m/s}^2\).\ \ (b) distance \(s = \text{average speed} \times t = \frac{0 + 45}{2} \times 3.0 = 22.5 \times 3.0 = 67.5\text{ m}\).\ \ (c) The rocket is now in free fall, so its acceleration is \(9.8\text{ m/s}^2\) directed vertically downwards.\ \ (d) Using \(v^2 = u^2 + 2as\) where \(v = 0\), \(u = 45\text{ m/s}\), \(a = -9.8\text{ m/s}^2\): \(0 = 45^2 - 2 \times 9.8 \times s \implies 19.6 s = 2025 \implies s = \frac{2025}{19.6} \approx 103\text{ m}\).
Marking scheme
(a)\ C1: formula \(a = \Delta v / t\) or substitution \(45 / 3.0\)\ A1: \(15\text{ m/s}^2\)\ \ (b)\ C1: formula \(s = \frac{1}{2}(u+v)t\) or \(s = \frac{1}{2}at^2\) or substitution \(\frac{1}{2} \times 45 \times 3.0\)\ A1: \(67.5\text{ m}\)\ \ (c)\ B1: \(9.8\text{ m/s}^2\) (or \(g\))\ B1: downwards / towards Earth\ \ (d)\ C1: formula \(v^2 = u^2 + 2as\) or \(E_k = E_p\) or substitution \(45^2 / (2 \times 9.8)\) \ A1: \(103\text{ m}\)
Question 2 · structured
8 marks
A wooden crate of mass \(12\text{ kg}\) is pulled along a flat, rough horizontal floor by a constant horizontal pulling force \(F\). The crate starts from rest and accelerates uniformly at \(1.5\text{ m/s}^2\). The constant frictional force opposing the motion is \(18\text{ N}\).\ \ (a) Calculate the resultant force acting on the crate. [2]\ \ (b) Calculate the magnitude of the pulling force \(F\). [2]\ \ (c) State and explain how the frictional force would change, if at all, if the same pulling force was applied at an angle upwards from the horizontal. [2]\ \ (d) The crate reaches a speed of \(6.0\text{ m/s}\). Calculate its kinetic energy at this speed. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Resultant force \(F_{\text{res}} = m \times a = 12 \times 1.5 = 18\text{ N}\).\ \ (b) Since \(F_{\text{res}} = F - F_{\text{friction}} \implies 18 = F - 18 \implies F = 36\text{ N}\).\ \ (c) The upward component of the force reduces the downward force on the floor, thereby reducing the normal contact force. Since friction depends on the normal contact force, the frictional force decreases.\ \ (d) \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 12 \times 6.0^2 = 6 \times 36 = 216\text{ J}\).
Marking scheme
(a)\ C1: \(F = ma\) or substitution \(12 \times 1.5\)\ A1: \(18\text{ N}\)\ \ (b)\ C1: \(F_{\text{res}} = F - F_{\text{f}}\) or \(F = 18 + 18\)\ A1: \(36\text{ N}\)\ \ (c)\ B1: frictional force decreases\ B1: because the normal contact force decreases due to the vertical component of the pulling force\ \ (d)\ C1: \(E_k = \frac{1}{2}mv^2\) or substitution \(\frac{1}{2} \times 12 \times 6.0^2\)\ A1: \(216\text{ J}\)
Question 3 · structured
8 marks
A toy railway car A of mass \(0.50\text{ kg}\) moves at a velocity of \(2.4\text{ m/s}\) to the right. It collides with a stationary railway car B of mass \(0.30\text{ kg}\). After the collision, the two cars couple together and move off together.\ \ (a) Define momentum in terms of mass and velocity. [1]\ \ (b) Calculate the momentum of car A before the collision. [2]\ \ (c) Calculate the common velocity of the coupled cars after the collision. [3]\ \ (d) Show, by calculating the total kinetic energy before and after the collision, that the collision is inelastic. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Momentum \(p = \text{mass } m \times \text{velocity } v\).\ \ (b) \(p = 0.50 \times 2.4 = 1.2\text{ kg m/s}\).\ \ (c) By conservation of momentum: \(m_A v_A + m_B v_B = (m_A + m_B) v \implies 1.2 = (0.50 + 0.30) v \implies 1.2 = 0.80 v \implies v = 1.5\text{ m/s}\).\ \ (d) Total KE before = \(\frac{1}{2} m_A v_A^2 = \frac{1}{2} \times 0.50 \times 2.4^2 = 1.44\text{ J}\).\ Total KE after = \(\frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2} \times 0.80 \times 1.5^2 = 0.90\text{ J}\).\ Since the kinetic energy decreases (is not conserved), the collision is inelastic.
Marking scheme
(a)\ B1: mass \(\times\) velocity\ \ (b)\ C1: \(p = mv\) or substitution \(0.50 \times 2.4\)\ A1: \(1.2\text{ kg m/s}\)\ \ (c)\ C1: state conservation of momentum or formula \(m_A v_A = (m_A + m_B)v\)\ C1: substitution \(1.2 = 0.80 v\)\ A1: \(1.5\text{ m/s}\)\ \ (d)\ C1: calculation of KE before (1.44 J) AND KE after (0.90 J)\ A1: statement that KE is not conserved, therefore inelastic
Question 4 · structured
8 marks
An electrical immersion heater of power \(80\text{ W}\) is placed inside a block of copper of mass \(1.5\text{ kg}\). The heater is switched on for \(5.0\text{ minutes}\). The temperature of the copper block rises from \(20^\circ\text{C}\) to \(56^\circ\text{C}\).\ \ (a) Calculate the electrical energy supplied to the heater in this time. [2]\ \ (b) Calculate the specific heat capacity of copper from these measurements. [3]\ \ (c) Suggest one reason why the value calculated in (b) is likely to be larger than the actual accepted specific heat capacity of copper. [1]\ \ (d) State two modifications that could be made to the experimental setup to obtain a more accurate value for the specific heat capacity. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) \(E = P \times t = 80 \times (5.0 \times 60) = 80 \times 300 = 24\\ 000\text{ J}\).\ \ (b) \(\Delta \theta = 56 - 20 = 36^\circ\text{C}\).\ Using \(E = m c \Delta \theta \implies 24\\ 000 = 1.5 \times c \times 36 \implies 24\\ 000 = 54 c \implies c = \frac{24\\ 000}{54} \approx 444\text{ J/(kg }^\circ\text{C)}\), which rounds to \(440\text{ J/(kg }^\circ\text{C)}\) to 2 s.f.\ \ (c) Some thermal energy is lost to the surroundings, meaning less energy is actually transferred to the copper block than was supplied, which overestimates the calculated specific heat capacity.\ \ (d) 1. Add lagging/insulation around the copper block. 2. Put oil in the thermometer hole to ensure good thermal contact.
Marking scheme
(a)\ C1: \(E = Pt\) or substitution \(80 \times 300\)\ A1: \(24\\ 000\text{ J}\)\ \ (b)\ C1: \(\Delta\theta = 36^\circ\text{C}\)\ C1: \(c = E / (m \Delta\theta)\) or substitution \(24\\ 000 / (1.5 \times 36)\) \ A1: \(440\text{ J/(kg }^\circ\text{C)}\)\ \ (c)\ B1: thermal energy lost to surroundings / container\ \ (d)\ B1: lag/insulate the block\ B1: use oil in the thermometer hole to improve thermal contact
Question 5 · structured
9 marks
A water wave in a ripple tank has a wavelength of \(4.5\text{ cm}\) in deep water. The frequency of the wave is \(12\text{ Hz}\).\ \ (a) Calculate the speed of the water wave in deep water. Give your answer in \(\text{m/s}\). [2]\ \ (b) The wave passes from deep water into shallow water. The speed of the wave decreases.\ (i) State what happens to the frequency of the wave as it enters shallow water. [1]\ (ii) State and explain what happens to the wavelength of the wave. [2]\ \ (c) Describe the motion of water molecules as this transverse wave passes. [2]\ \ (d) Explain how a longitudinal wave differs from a transverse wave. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) \(v = f \lambda = 12 \times 0.045 = 0.54\text{ m/s}\).\ \ (b)(i) The frequency remains constant (unchanged) because it is determined solely by the source.\ (b)(ii) Since \(v = f \lambda\) and \(f\) is constant, if the speed decreases, the wavelength must also decrease.\ \ (c) The water molecules vibrate up and down (oscillate) in a direction perpendicular to the direction of wave travel (energy transfer).\ \ (d) In a longitudinal wave, the particles vibrate parallel to the direction of wave travel (energy transfer), whereas in a transverse wave, they vibrate perpendicular to it.
Marking scheme
(a)\ C1: \(v = f\lambda\) or substitution \(12 \times 0.045\)\ A1: \(0.54\text{ m/s}\)\ \ (b)(i)\ B1: remains constant / unchanged\ \ (b)(ii)\ B1: wavelength decreases\ B1: because speed decreases (and frequency is constant) / \(v \propto \lambda\)\ \ (c)\ B1: oscillate/vibrate up and down / vertically\ B1: perpendicular to direction of wave propagation / travel\ \ (d)\ B1: longitudinal: vibrations/oscillations parallel to direction of wave travel\ B1: transverse: vibrations/oscillations perpendicular to direction of wave travel
Question 6 · structured
8 marks
A battery of electromotive force (e.m.f.) \(9.0\text{ V}\) is connected in a complete circuit to a resistor of resistance \(15\\ \Omega\).\ \ (a) Define electromotive force (e.m.f.) in terms of energy and charge. [2]\ \ (b) Calculate the current in the resistor. [2]\ \ (c) Calculate the total charge passing through the resistor in \(4.0\text{ minutes}\). [2]\ \ (d) Calculate the energy transferred by the battery to the charge carriers in this time. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Electromotive force is the energy supplied by a source per unit charge in driving charge around a complete circuit.\ \ (b) \(I = \frac{V}{R} = \frac{9.0}{15} = 0.60\text{ A}\).\ \ (c) \(Q = I \times t = 0.60 \times (4.0 \times 60) = 0.60 \times 240 = 144\text{ C}\).\ \ (d) \(E = V \times Q = 9.0 \times 144 = 1296\text{ J}\), which is approximately \(1300\text{ J}\) (or \(1.3\text{ kJ}\)). Alternatively, \(E = V I t = 9.0 \times 0.60 \times 240 = 1296\text{ J}\).
Marking scheme
(a)\ B1: energy supplied per unit charge\ B1: in driving charge around a complete circuit\ \ (b)\ C1: \(I = V/R\) or substitution \(9.0 / 15\)\ A1: \(0.60\text{ A}\)\ \ (c)\ C1: \(Q = It\) or substitution \(0.60 \times 240\)\ A1: \(144\text{ C}\)\ \ (d)\ C1: \(E = VQ\) or \(E = VIt\) or substitution \(9.0 \times 144\)\ A1: \(1300\text{ J}\)
Question 7 · structured
8 marks
The isotope bismuth-210 (\(^{210}_{83}\text{Bi}\)) decays to polonium-210 (\(^{210}_{84}\text{Po}\)) by emitting a beta-minus (\(\beta^-\)) particle.\ \ (a) State the nature of a beta-minus particle. [1]\ \ (b) Complete the nuclear decay equation for the decay of bismuth-210:\ \[ ^{210}_{83}\text{Bi} \rightarrow \text{Po} + \beta \]\ Ensure you include all nucleon (mass) and proton (atomic) numbers. [3]\ \ (c) The half-life of bismuth-210 is 5.0 days. A sample initially contains bismuth-210 with an activity of \(360\text{ Bq}\).\ (i) Explain what is meant by the term half-life. [1]\ (ii) Calculate the activity of the sample after 15.0 days. [3]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) A beta-minus particle is a high-speed electron emitted from the nucleus.\ \ (b) \(^{210}_{83}\text{Bi} \rightarrow ^{210}_{84}\text{Po} + ^{0}_{-1}\text{e}\).\ \ (c)(i) Half-life is the time taken for the activity of a radioactive sample to decrease to half of its initial value.\ \ (c)(ii) Number of half-lives \(n = \frac{15.0}{5.0} = 3\). After 3 half-lives, the activity is: \(360 \xrightarrow{\text{1st}} 180 \xrightarrow{\text{2nd}} 90 \xrightarrow{\text{3rd}} 45\text{ Bq}\).
Marking scheme
(a)\ B1: high-speed electron\ \ (b)\ B1: polonium symbol with correct numbers: \(^{210}_{84}\text{Po}\)\ B1: beta particle with correct numbers: \(^{0}_{-1}\text{e}\) (or \(\beta\))\ B1: fully balanced equation\ \ (c)(i)\ B1: time taken for the activity / count-rate / number of radioactive nuclei to halve\ \ (c)(ii)\ C1: identify 3 half-lives have passed\ C1: successive halving \(360 \rightarrow 180 \rightarrow 90 \rightarrow 45\)\ A1: \(45\text{ Bq}\)
Question 8 · structured
8 marks
A galaxy is at a distance of \(3.8 \times 10^{22}\text{ km}\) from Earth. Light from a specific element in this galaxy is shifted towards the red end of the spectrum (redshift).\ \ (a) State what redshift tells us about the motion of this galaxy relative to Earth. [1]\ \ (b) Show that the distance of the galaxy in meters is \(3.8 \times 10^{25}\text{ m}\) . [1]\ \ (c) Use the Hubble constant \(H_0 = 2.2 \times 10^{-18}\text{ s}^{-1}\) to calculate the speed of recession of this galaxy in \(\text{m/s}\). [3]\ \ (d) Calculate an estimate for the age of the Universe in years. Take \(1\text{ year} \approx 3.16 \times 10^7\text{ s}\). [3]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Redshift indicates that the galaxy is moving away from the Earth (receding).\ \ (b) \(d = 3.8 \times 10^{22}\text{ km} = 3.8 \times 10^{22} \times 10^3\text{ m} = 3.8 \times 10^{25}\text{ m}\).\ \ (c) \(v = H_0 d = (2.2 \times 10^{-18}\text{ s}^{-1}) \times (3.8 \times 10^{25}\text{ m}) = 8.36 \times 10^7\text{ m/s} \approx 8.4 \times 10^7\text{ m/s}\).\ \ (d) \(T = \frac{1}{H_0} = \frac{1}{2.2 \times 10^{-18}\text{ s}^{-1}} \approx 4.545 \times 10^{17}\text{ s}\). In years, this is: \(T_{\text{years}} = \frac{4.545 \times 10^{17}}{3.16 \times 10^7} \approx 1.44 \times 10^{10}\text{ years} \approx 1.4 \times 10^{10}\text{ years}\) (or 14 billion years).
A toy railway car A of mass 0.80 kg travels at a speed of 3.0 m/s along a frictionless horizontal track. It collides with a stationary railway car B of mass 1.2 kg. The two cars couple together during the collision and move off together.
(b) During the collision, the interaction lasts for a time of 0.15 s. Calculate the average force exerted by car A on car B.
average force = ..................................................... [3]
(c) State and explain whether the collision is elastic or inelastic. Support your answer with calculations of the total kinetic energy before and after the collision.
**(b)** Using the impulse-momentum relationship on car B: $$\text{Impulse} = \Delta p_B = m_B \times v_f - 0 = 1.2\text{ kg} \times 1.2\text{ m/s} = 1.44\text{ N}\cdot\text{s}$$ Using $$F = \frac{\Delta p}{\Delta t}$$: $$F = \frac{1.44\text{ N}\cdot\text{s}}{0.15\text{ s}} = 9.6\text{ N}$$
**(c)** Calculating the initial kinetic energy: $$E_{k,i} = \frac{1}{2} m_A v_A^2 = 0.5 \times 0.80 \times (3.0)^2 = 3.6\text{ J}$$ Calculating the final kinetic energy: $$E_{k,f} = \frac{1}{2} (m_A + m_B) v_f^2 = 0.5 \times 2.0 \times (1.2)^2 = 1.44\text{ J}$$ Since the total kinetic energy decreases (is not conserved), the collision is **inelastic**.
Marking scheme
**(a) (i)** - $p = m \times v$ OR $0.80 \times 3.0$ [C1] - $2.4\text{ kg}\cdot\text{m/s}$ (or $\text{N}\cdot\text{s}$) [A1]
**(ii)** - Use of conservation of momentum: $m_A v_A = (m_A + m_B) v_f$ OR $2.4 = 2.0 \times v_f$ [C1] - $1.2\text{ m/s}$ [A1]
**(b)** - Change of momentum of B (or A) calculated: $1.2 \times 1.2 = 1.44\text{ kg}\cdot\text{m/s}$ [C1] - $F = \Delta p / \Delta t$ OR $1.44 / 0.15$ [C1] - $9.6\text{ N}$ [A1]
**(c)** - Calculation of total KE before ($3.6\text{ J}$) AND total KE after ($1.44\text{ J}$) [B1] - States 'inelastic' because total kinetic energy decreases / is not conserved [B1]
Paper 6 (Alternative to Practical)
Answer all four experimental and data analysis questions. Complete tables, plot graphs, and detail experimental improvements.
4 Question · 40 marks
Question 1 · structured
10 marks
A student investigates the vertical oscillations of a wooden ruler clamped to a bench with a mass of 100 g attached to its free end (a cantilever setup).
Fig. 1.1 shows the projecting length of the ruler, $L$, beyond the edge of the clamp. The student measures $L$ using a metre ruler.
(a) On Fig. 1.1, the scale is 1:5. The measured distance on the diagram for $L$ is $6.0\text{ cm}$. Calculate the actual projecting length $L$ in $\text{cm}$.
(b) The student displaces the free end of the ruler vertically and releases it so that it oscillates. The student records the time $t$ for 15 complete oscillations for two different lengths: - For $L = 60.0\text{ cm}$, $t = 12.3\text{ s}$. - For $L = 40.0\text{ cm}$, $t = 6.8\text{ s}$.
(i) Calculate the period $T$ of oscillation (the time for one complete oscillation) for each length. (ii) Calculate $T^2$ for each length. (iii) State the correct units for the column headings if these data were to be entered into a table with columns: $L$, $t$, $T$, and $T^2$.
(c) Explain why timing 15 oscillations gives a more accurate result for the period $T$ than timing 1 oscillation.
(d) State one safety precaution the student should take when performing this experiment with a heavy mass at the end of the vibrating ruler.
(e) The student wants to plot a graph of $T^2$ against $L$. Suggest three other suitable actual values of $L$ to obtain sufficient data to plot a reliable graph.
(f) Describe one technique to avoid parallax error when measuring the actual length $L$ with a metre ruler.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Using the scale 1:5, the actual length is $L = 6.0\text{ cm} \times 5 = 30.0\text{ cm}$. (b)(i) Period $T = t / 15$. For $L = 60.0\text{ cm}$, $T = 12.3 / 15 = 0.82\text{ s}$. For $L = 40.0\text{ cm}$, $T = 6.8 / 15 = 0.45\text{ s}$. (ii) For $L = 60.0\text{ cm}$, $T^2 = 0.82^2 = 0.67\text{ s}^2$. For $L = 40.0\text{ cm}$, $T^2 = 0.45^2 = 0.20\text{ s}^2$. (iii) Column headings: $L/\text{cm}$, $t/\text{s}$, $T/\text{s}$, $T^2/\text{s}^2$. (c) Human reaction time error at the start and stop of timing is constant. Spreading this error over 15 oscillations significantly reduces its effect on the calculated period for a single oscillation. (d) Ensure the G-clamp is secured tightly to the bench so the ruler does not slip, and keep eyes/face at a safe distance from the oscillating end. (e) Suitable intermediate values between $30.0\text{ cm}$ and $60.0\text{ cm}$ such as $50.0\text{ cm}$, $45.0\text{ cm}$, and $35.0\text{ cm}$. (f) Position the eye directly vertically above the mark being read on the ruler to avoid a parallax offset.
Marking scheme
Part (a): 1 mark for actual length of 30.0 cm. Part (b)(i) & (ii): 1 mark for both T values correct, 1 mark for both T^2 values correct with appropriate decimal places. Part (b)(iii): 1 mark for all correct units (cm, s, s, s^2). Part (c): 1 mark for identifying that human reaction time / timing error is spread over more oscillations, reducing percentage uncertainty. Part (d): 1 mark for a valid safety precaution (e.g., place a cushion underneath, ensure G-clamp is secure, stand back). Part (e): 1 mark for suggesting three reasonable values of L spanning the range (e.g. between 25 cm and 60 cm). Part (f): 2 marks for clear description of placing ruler close to the object and viewing perpendicular to the scale.
Question 2 · structured
10 marks
A student investigates the thermal insulation of different materials wrapped around a metal beaker containing hot water.
(a) Fig. 2.1 shows a thermometer measuring the initial room temperature, $\theta_R$. State the value of $\theta_R$ shown on the thermometer scale.
(b) The student pours $150\text{ cm}^3$ of hot water into a beaker wrapped with Insulation X. The temperature $\theta$ is recorded every $30\text{ s}$. Complete the column headings for the table of results:
(c) Plot a graph of temperature $\theta$ on the y-axis against time $t$ on the x-axis. You do not need to start the y-axis at 0.
(d) With reference to the shape of your graph line, state how the rate of cooling of the water changes as time increases. Explain your answer.
(e) State two variables that must be kept constant to ensure a fair comparison when the experiment is repeated using Insulation Y.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Reading the thermometer scale: $\theta_R = 21.5^\circ\text{C}$. (b) The units of measurement are seconds for time ($t/\text{s}$) and degrees Celsius for temperature ($\theta/^\circ\text{C}$). (c) The graph should have axes labeled correctly with quantities and units. Scales should be chosen so the plotted points cover more than half the grid. Points must be plotted accurately within half a small square, and a smooth best-fit curve drawn. (d) The rate of cooling decreases. This is shown by the gradient (slope) of the curve becoming flatter / less steep as time increases, because the temperature of the water approaches room temperature. (e) Control variables for a fair comparison: volume of hot water ($150\text{ cm}^3$), initial temperature of the water, same beaker size/shape, same thickness of insulation, same room temperature.
Marking scheme
Part (a): 1 mark for reading 21.5 C (accept 21.5 to 22.0 C). Part (b): 1 mark for correct units (s and C or degrees Celsius). Part (c): 4 marks total: - 1 mark for correct axes labeled with units and right way round. - 1 mark for suitable scale (points cover at least half of the grid). - 1 mark for accurate plotting of all points to within half a small square. - 1 mark for a smooth, single, continuous best-fit curve. Part (d): 2 marks total: - 1 mark for stating that rate of cooling decreases. - 1 mark for explaining with reference to the decreasing gradient of the graph. Part (e): 2 marks for any two valid control variables (volume of water, initial temperature, thickness of insulation, same beaker shape/material).
Question 3 · structured
10 marks
A student investigates the refraction of a ray of light through a rectangular glass block to determine its refractive index.
Fig. 3.1 shows the ray-trace sheet obtained by the student. The line $NN'$ is the normal to the surface of the block.
(a) On Fig. 3.1: (i) Measure the angle of incidence $i$ between the incident ray and the normal $NN'$. (ii) Measure the angle of refraction $r$ between the refracted ray and the normal $NN'$ inside the block.
(b) Calculate the refractive index $n_1$ of the glass block using the equation:
$$n_1 = \frac{\sin i}{\sin r}$$
Give your answer to three significant figures.
(c) Describe two precautions that the student should take when using optical pins to trace the light rays accurately.
(d) The student repeats the experiment using a different angle of incidence and obtains a second value of refractive index, $n_2 = 1.48$. State and explain whether the two values ($n_1$ and $n_2$) are equal within the limits of experimental accuracy. (Assume $i = 45.0^\circ$ and $r = 28.0^\circ$ are measured in part a).
(e) Suggest one reason why the path of the light ray marked on the paper might not perfectly represent the true path of the light inside the glass block.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a)(i) Measuring the angle of incidence: $i = 45.0^\circ$ (accept $44^\circ - 46^\circ$). (ii) Measuring the angle of refraction: $r = 28.0^\circ$ (accept $27^\circ - 29^\circ$). (b) Using the formula: $n_1 = \sin(45.0^\circ) / \sin(28.0^\circ) = 0.7071 / 0.4695 = 1.506 \approx 1.51$ (to 3 sig figs). (c) Precautions when using pins: 1. Space the pins as far apart as possible (at least $5\text{ cm}$ apart) to define the ray direction more accurately. 2. View the bases of the pins through the glass block to ensure alignment, avoiding errors from bent pins. (d) Calculate percentage difference: $\text{diff} = \frac{1.51 - 1.48}{1.51} \times 100\% \approx 2.0\%$. Since $2.0\% < 10\%$, the values are equal within the limits of experimental accuracy. (e) Possible reasons: The pencil lines drawn have a finite thickness; the glass block may have shifted slightly during the experiment; or there is difficulty in aligning the pins exactly with the edge of the block.
Marking scheme
Part (a): 2 marks (1 mark for i = 45.0 +/- 1 degree, 1 mark for r = 28.0 +/- 1 degree). Part (b): 2 marks (1 mark for correct calculation of n = 1.51, 1 mark for giving the answer to exactly 3 significant figures). Part (c): 2 marks (1 mark each for any two valid precautions: spacing pins > 5 cm, pins vertical, viewing at the base, sharp pencil, etc.). Part (d): 2 marks (1 mark for stating 'Yes, they are equal within limits of accuracy', 1 mark for justification showing percentage difference is less than 10%). Part (e): 2 marks for suggesting a valid source of experimental error (pencil lines have thickness, alignment shifts, or difficulty placing the block exactly on the outline).
Question 4 · structured
10 marks
A student wants to investigate how the electrical resistance of a metallic constantan wire depends on its length.
The following apparatus is available to the student: - a d.c. power supply (battery) - an ammeter - a voltmeter - a switch - a metre ruler - a length of constantan wire mounted on a wooden board - connecting wires and crocodile clips.
Plan an experiment to investigate how the resistance of the wire depends on its length.
You should: 1. Draw a complete circuit diagram showing how the apparatus is connected to measure the current in the wire and the potential difference across a measured length of the wire. 2. Explain briefly how to carry out the investigation, detailing the measurements to be taken. 3. State the key variables that must be kept constant. 4. Draw a table, with column headings and units, to display the readings (you are not required to enter any data). 5. Explain how you would use your results to reach a conclusion.
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. **Circuit Diagram**: Draw a series circuit consisting of the battery, switch, ammeter, and the test wire. Connect the voltmeter in parallel across the two crocodile clips attached to the test wire.
2. **Method**: Connect the crocodile clips to a short length of wire (e.g., $10.0\text{ cm}$). Close the switch and record the current $I$ from the ammeter and the potential difference $V$ from the voltmeter. Open the switch to prevent heating. Move one crocodile clip to increase the length of wire in the circuit (e.g., to $20.0\text{ cm}, 30.0\text{ cm}, 40.0\text{ cm}, 50.0\text{ cm}$). Repeat the readings of $V$ and $I$ for at least 5 different lengths.
3. **Control Variables**: Keep the diameter/cross-sectional area of the wire constant (use the same wire). Keep the temperature of the wire constant by keeping current low and switching off between readings.
4. **Table of Results**: $$\begin{array}{|c|c|c|c|}\hline \text{Length } L / \text{cm} & \text{Voltage } V / \text{V} & \text{Current } I / \text{A} & \text{Resistance } R / \Omega \\hline & & & \\hline\end{array}$$
5. **Conclusion**: Calculate resistance for each length using $R = V / I$. Plot a graph of Resistance $R$ (y-axis) against Length $L$ (x-axis). If $R$ is directly proportional to $L$, the graph will be a straight line passing through the origin.
Marking scheme
1. Circuit diagram: 2 marks (1 mark for series circuit with battery, switch, ammeter, test wire; 1 mark for voltmeter correctly placed in parallel across the test wire). 2. Method: 2 marks (1 mark for explaining how to vary length L and measure V and I; 1 mark for repeating for at least 5 different lengths). 3. Control variables: 2 marks (1 mark for stating thickness/diameter of wire must be kept constant; 1 mark for keeping temperature constant / switching off between readings). 4. Table: 2 marks (1 mark for clear columns; 1 mark for including all correct units: L/cm or m, V/V, I/A, R/ohms). 5. Conclusion: 2 marks (1 mark for calculating R = V/I; 1 mark for plotting a graph of R against L and stating that a straight line through the origin indicates direct proportionality).
Wondering how well you actually know this?
thinka is an AI practice app for IGCSE & IB students: unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.