An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V1) Cambridge International A Level Physics (9702) paper. Not affiliated with or reproduced from Cambridge.
Paper 11 Multiple Choice
Answer all forty questions on the multiple choice answer sheet. Each correct answer scores one mark.
40 Question · 40 marks
Question 1 · MCQ
1 marks
An astrophysical model suggests that the drag force \( F \) acting on a cosmic dust grain is given by the equation:
\[ F = \Gamma \frac{G M \rho}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of a nearby star, \( \rho \) is the density of the surrounding interstellar medium, and \( r \) is the distance from the star.
What are the SI base units of the constant \( \Gamma \)?
A.\( \text{m}^2 \)
B.\( \text{kg m}^2 \text{s}^{-2} \)
C.\( \text{kg}^{-1} \text{m}^5 \text{s}^{-2} \)
D.\( \text{kg}^2 \text{m}^{-2} \)
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Worked solution
To find the SI base units of the constant \( \Gamma \), we can express it in terms of the other quantities:
\[ \Gamma = \frac{F \cdot r}{G M \rho} \]
Let's determine the SI base units of each variable in the equation: - Force \( F \): \( \text{kg m s}^{-2} \) - Distance \( r \): \( \text{m} \) - Mass \( M \): \( \text{kg} \) - Density \( \rho \): \( \text{kg m}^{-3} \) - Gravitational constant \( G \): From the formula \( F = \frac{G M m}{r^2} \), we have \( [G] = \frac{[F][r^2]}{[M^2]} = \frac{\text{kg m s}^{-2} \cdot \text{m}^2}{\text{kg}^2} = \text{kg}^{-1} \text{m}^3 \text{s}^{-2} \).
Now we substitute these units into the expression for \( \Gamma \):
1 mark for the correct derivation of the SI base units of \( G \) and correctly rearranging the equation to solve for the base units of \( \Gamma \), leading to \( \text{m}^2 \).
Question 2 · MCQ
1 marks
A stone is thrown vertically upwards from the edge of a vertical cliff of height \( h \). The stone is thrown with an initial speed \( v \) and eventually lands on the ground at the base of the cliff. The total flight time of the stone is \( t \).
If air resistance is negligible, and \( g \) is the acceleration of free fall, which equation correctly relates \( h \), \( v \), \( t \) and \( g \)?
A.\( h = vt - \frac{1}{2}gt^2 \)
B.\( h = \frac{1}{2}gt^2 - vt \)
C.\( h = vt + \frac{1}{2}gt^2 \)
D.\( h = gt^2 - 2vt \)
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Worked solution
We use the SUVAT equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
Let us define the vertically upward direction as positive. - The initial velocity is upward, so \( u = +v \). - The acceleration is downwards due to gravity, so \( a = -g \). - The final position of the stone is on the ground, which is a distance \( h \) below the starting point, so the displacement is \( s = -h \).
Substituting these values into the SUVAT equation:
\[ -h = vt + \frac{1}{2}(-g)t^2 \]
\[ -h = vt - \frac{1}{2}gt^2 \]
Multiplying the entire equation by \(-1\) gives:
\[ h = \frac{1}{2}gt^2 - vt \]
Marking scheme
1 mark for correctly assigning vector signs to displacement, initial velocity, and acceleration of free fall, and substituting them into the kinematics equation to obtain \( h = \frac{1}{2}gt^2 - vt \).
Question 3 · MCQ
1 marks
A small object of weight \( W \) is suspended in equilibrium by two light, flexible strings. One string is at an angle of \( 35^\circ \) to the vertical, and the other string is horizontal.
What is the tension in the horizontal string?
A.\( W \sin 35^\circ \)
B.\( W \cos 35^\circ \)
C.\( W \tan 35^\circ \)
D.\( \frac{W}{\tan 35^\circ} \)
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Worked solution
Let \( T_1 \) be the tension in the angled string (at \( 35^\circ \) to the vertical), and \( T_2 \) be the tension in the horizontal string.
Since the object is in equilibrium, the forces must balance in both the vertical and horizontal directions:
Substitute the expression for \( T_1 \) into the horizontal equation: \[ T_2 = \left( \frac{W}{\cos 35^\circ} \right) \sin 35^\circ = W \tan 35^\circ \]
Marking scheme
1 mark for resolving forces vertically and horizontally to show that \( T_2 = W \tan 35^\circ \).
Question 4 · MCQ
1 marks
Two wires, X and Y, are made of the same metal. Wire X has length \( L \) and diameter \( d \). Wire Y has length \( 2L \) and diameter \( 2d \).
Both wires are subjected to the same tensile force \( F \), which causes them to stretch elastically.
What is the ratio of the tensile strain in wire X to the tensile strain in wire Y?
A.\( 1 : 2 \)
B.\( 1 : 1 \)
C.\( 2 : 1 \)
D.\( 4 : 1 \)
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Worked solution
By definition, Young modulus \( E \) is given by:
\[ E = \frac{\text{stress}}{\text{strain}} \implies \text{strain} = \frac{\text{stress}}{E} \]
Since both wires are made of the same metal, they have the same Young modulus \( E \). Thus, the strain is directly proportional to stress:
\[ \text{strain} \propto \text{stress} \]
Stress is defined as:
\[ \text{stress} = \frac{F}{A} \]
where \( A \) is the cross-sectional area. The area is proportional to the square of the diameter: \( A \propto d^2 \). Therefore:
\[ \text{stress} \propto \frac{F}{d^2} \]
Since both wires are subjected to the same force \( F \):
\[ \text{stress} \propto \frac{1}{d^2} \]
Let's compare wire X (diameter \( d \)) and wire Y (diameter \( 2d \)):
Therefore, the ratio of the tensile strain in wire X to that in wire Y is \( 4 : 1 \).
Marking scheme
1 mark for identifying that since Young's modulus is constant, the ratio of strains equals the ratio of stresses, which depends only on the cross-sectional area under equal force, leading to a ratio of \( 4 : 1 \).
Question 5 · MCQ
1 marks
A progressive transverse wave of frequency \( 25\text{ Hz} \) travels along a stretched horizontal string. The phase difference between two particles on the string that are separated by a horizontal distance of \( 0.15\text{ m} \) is \( \frac{\pi}{3}\text{ rad} \).
What is the speed of the wave?
A.\( 4.5\text{ m s}^{-1} \)
B.\( 22.5\text{ m s}^{-1} \)
C.\( 45\text{ m s}^{-1} \)
D.\( 90\text{ m s}^{-1} \)
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Worked solution
The relationship between phase difference \( \Delta \phi \) and path difference (distance between particles) \( \Delta x \) is given by:
\[ \Delta \phi = \frac{2\pi \Delta x}{\lambda} \]
Substituting \( \Delta \phi = \frac{\pi}{3}\text{ rad} \) and \( \Delta x = 0.15\text{ m} \):
Now, calculate the wave speed \( v \) using the wave equation:
\[ v = f \lambda = 25\text{ Hz} \times 0.90\text{ m} = 22.5\text{ m s}^{-1} \]
Marking scheme
1 mark for using the correct relation between phase difference and wavelength to find \( \lambda = 0.90\text{ m} \), and then applying \( v = f\lambda \) to get \( 22.5\text{ m s}^{-1} \).
Question 6 · MCQ
1 marks
A uniform metal wire of resistance \( R \) is stretched such that its length increases by \( 10\% \).
Assuming that both the density and the resistivity of the metal remain constant during stretching, what is the new resistance of the wire?
A.\( 1.10 R \)
B.\( 1.21 R \)
C.\( 1.33 R \)
D.\( 1.44 R \)
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Worked solution
The resistance \( R \) of a wire of length \( L \) and cross-sectional area \( A \) is given by:
\[ R = \rho \frac{L}{A} \]
Let the original length be \( L \) and the original cross-sectional area be \( A \). When the wire is stretched, its length increases by \( 10\% \), so the new length \( L' \) is:
\[ L' = 1.10 L \]
Since the density and mass are constant, the volume \( V \) of the wire remains constant during stretching:
\[ V = L A = L' A' \]
Substitute \( L' = 1.10 L \):
\[ L A = (1.10 L) A' \implies A' = \frac{A}{1.10} \]
1 mark for using the conservation of volume to determine that the cross-sectional area decreases by a factor of 1.10, and using the resistance formula to get \( R' = 1.21 R \).
Question 7 · MCQ
1 marks
A hypothetical meson has a net charge of \( +1e \) and contains a strange antiquark (\( \bar{\text{s}} \)).
Which of the following could be the other quark in the meson?
A.up (\( \text{u} \))
B.down (\( \text{d} \))
C.strange (\( \text{s} \))
D.anti-up (\( \bar{\text{u}} \))
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Worked solution
A meson consists of exactly one quark and one antiquark. We are given that this meson contains a strange antiquark (\( \bar{\text{s}} \)).
The charge of a strange quark (\( \text{s} \)) is \( -\frac{1}{3}e \), which means the charge of its antiparticle, the strange antiquark (\( \bar{\text{s}} \)), is \( +\frac{1}{3}e \).
Let the charge of the other quark in the meson be \( Q_q \). The total charge of the meson is given as \( +1e \):
\[ Q_q + \left(+\frac{1}{3}e\right) = +1e \]
\[ Q_q = +1e - \frac{1}{3}e = +\frac{2}{3}e \]
Among the options, the quark that has a charge of \( +\frac{2}{3}e \) is the up (\( \text{u} \)) quark.
Marking scheme
1 mark for determining the charge of the strange antiquark, establishing the equation for the total charge, and identifying the up quark as having the required charge of \( +2/3 e \).
Question 8 · MCQ
1 marks
In a double-slit interference experiment, light of wavelength \( \lambda \) passes through two slits separated by a distance \( a \) to produce bright fringes of spacing \( x \) on a screen at a distance \( D \) from the slits.
The experiment is now modified: the slit separation is doubled, and the distance from the slits to the screen is halved.
What wavelength of light must be used to keep the fringe spacing \( x \) unchanged?
A.\( \frac{\lambda}{4} \)
B.\( \lambda \)
C.\( 2\lambda \)
D.\( 4\lambda \)
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Worked solution
The formula for the double-slit fringe spacing \( x \) is:
\[ x = \frac{\lambda D}{a} \]
Let the new wavelength of light used be \( \lambda' \). Under the new conditions: - New slit separation \( a' = 2a \) - New screen distance \( D' = \frac{D}{2} \)
We want the new fringe spacing \( x' \) to be equal to \( x \):
1 mark for using the double-slit formula, substituting the updated physical parameters, and solving for the new wavelength to show that it must be \( 4\lambda \).
Question 9 · MCQ
1 marks
A uniform picture frame of weight \(W\) and width \(d\) is suspended symmetrically from a smooth peg by a light inextensible string of total length \(L\), where \(L > d\). The ends of the string are attached to the top two corners of the frame. What is the tension \(T\) in the string?
A.\(\frac{W L}{2 \sqrt{L^2 - d^2}}\)
B.\(\frac{W d}{2 \sqrt{L^2 - d^2}}\)
C.\(\frac{W \sqrt{L^2 - d^2}}{2 L}\)
D.\(\frac{W L}{\sqrt{L^2 - d^2}}\)
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Worked solution
First, find the angle \(\theta\) that the string makes with the top horizontal edge of the frame. The string of length \(L\) is divided into two equal halves of length \(L/2\), meeting at the peg. The horizontal distance between the two corners is \(d\), so the horizontal distance from each corner to the vertical line through the peg is \(d/2\). Using trigonometry on one of the right-angled triangles formed: \(\cos\theta = \frac{d/2}{L/2} = \frac{d}{L}\). Thus, the sine of the angle is: \(\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{d^2}{L^2}} = \frac{\sqrt{L^2 - d^2}}{L}\). For vertical equilibrium, the vertical components of the tension in the two halves of the string must support the weight of the frame: \(2 T \sin\theta = W\). Substitute the expression for \(\sin\theta\): \(2 T \left(\frac{\sqrt{L^2 - d^2}}{L}\right) = W\). Solving for the tension \(T\): \(T = \frac{W L}{2 \sqrt{L^2 - d^2}}\).
Marking scheme
1 mark for identifying the correct vertical equilibrium equation and trigonometric substitution, resulting in A.
Question 10 · MCQ
1 marks
A particle of mass \(m\) undergoes simple harmonic motion with amplitude \(A\) and period \(T\). Which expression gives the maximum net force acting on the particle during its motion?
A.\(\frac{2\pi m A}{T}\)
B.\(\frac{4\pi^2 m A}{T^2}\)
C.\(\frac{4\pi^2 m A^2}{T^2}\)
D.\(\frac{2\pi^2 m A}{T^2}\)
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Worked solution
The angular frequency \(\omega\) of a simple harmonic oscillator is related to its period \(T\) by: \(\omega = \frac{2\pi}{T}\). The maximum acceleration \(a_{\text{max}}\) of the particle is given by: \(a_{\text{max}} = \omega^2 A = \left(\frac{2\pi}{T}\right)^2 A = \frac{4\pi^2 A}{T^2}\). According to Newton's second law, the maximum net force \(F_{\text{max}}\) is: \(F_{\text{max}} = m a_{\text{max}} = \frac{4\pi^2 m A}{T^2}\).
Marking scheme
1 mark for relating angular frequency to period, formulating maximum acceleration, and applying Newton's second law to get B.
Question 11 · MCQ
1 marks
A ball is thrown vertically upwards from the edge of a cliff with an initial speed of \(u\). After reaching its maximum height, it falls past the cliff edge and hits the ground at the base of the cliff with speed \(3u\). Air resistance is negligible. What is the height of the cliff?
A.\(\frac{2u^2}{g}\)
B.\(\frac{4u^2}{g}\)
C.\(\frac{8u^2}{g}\)
D.\(\frac{9u^2}{g}\)
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Worked solution
Using the equation of motion \(v^2 = u_i^2 + 2as\) where upward is positive. The initial velocity is \(u_i = +u\). The final velocity is \(v = -3u\) (downward). The acceleration is \(a = -g\). The displacement is \(s = -h\) where \(h\) is the height of the cliff. Substituting these into the equation: \((-3u)^2 = u^2 + 2(-g)(-h)\) yields \(9u^2 = u^2 + 2gh\). This simplifies to \(8u^2 = 2gh\), which gives \(h = \frac{4u^2}{g}\).
Marking scheme
1 mark for setting up the 1D kinematic equation with correct sign conventions to solve for h, yielding B.
Question 12 · MCQ
1 marks
A car of mass \(m\) travels up a slope angled at \(\theta\) to the horizontal at a constant speed \(v\). The resistive forces opposing the motion of the car are constant and equal to \(F\). What is the useful power output of the car’s engine?
A.\(m g v \sin\theta + F v\)
B.\(m g v \sin\theta - F v\)
C.\(m g v \cos\theta + F v\)
D.\((m g + F) v \sin\theta\)
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Worked solution
Since the car moves up the slope at a constant speed, the net force along the slope is zero. The driving force \(D\) exerted by the engine must balance the parallel component of the gravitational force down the slope and the resistive forces: \(D = mg \sin\theta + F\). Power \(P\) is the product of driving force and velocity: \(P = D v = (mg \sin\theta + F)v = mgv \sin\theta + Fv\).
Marking scheme
1 mark for calculating the driving force under equilibrium and using P = Fv, leading to A.
Question 13 · MCQ
1 marks
In a double-slit experiment, light of wavelength \(\lambda\) is incident on two parallel slits of separation \(d\). Fringes of spacing \(x\) are observed on a screen at a distance \(D\) from the slits. If the wavelength of the light is halved, the slit separation is doubled, and the distance to the screen is tripled, what is the new fringe spacing?
A.\(0.33 x\)
B.\(0.75 x\)
C.\(1.33 x\)
D.\(3.00 x\)
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Worked solution
The double-slit fringe spacing formula is \(x = \frac{\lambda D}{d}\). Substituting the modified variables \(\lambda' = 0.5\lambda\), \(d' = 2d\), and \(D' = 3D\) gives the new fringe spacing: \(x' = \frac{(0.5\lambda)(3D)}{2d} = 0.75 \frac{\lambda D}{d} = 0.75x\).
Marking scheme
1 mark for applying the double-slit equation and scaling factors correctly to obtain B.
Question 14 · MCQ
1 marks
A wire of length \(L\) and cross-sectional area \(A\) is made of a material of resistivity \(\rho\). The wire is stretched uniformly so that its length increases by \(10\%\) while its volume remains constant. What is the percentage increase in the electrical resistance of the wire?
A.\(10\%\)
B.\(21\%\)
C.\(44\%\)
D.\(121\%\)
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Worked solution
Initial resistance is \(R_1 = \rho \frac{L}{A}\). The new length is \(L_2 = 1.10 L\). Since volume \(V = A \times L\) is constant, the new area \(A_2\) must satisfy \(A_2 = \frac{A}{1.10}\). The new resistance is \(R_2 = \rho \frac{1.10 L}{A / 1.10} = 1.21 \rho \frac{L}{A} = 1.21 R_1\). The percentage increase is \(\frac{1.21 R_1 - R_1}{R_1} \times 100\% = 21\%\).
Marking scheme
1 mark for using constant volume to find area change and determining the resulting percentage change in resistance, yielding B.
Question 15 · MCQ
1 marks
A steel wire of length \(2.0\text{ m}\) and cross-sectional area \(1.5 \times 10^{-6}\text{ m}^2\) has a Young modulus of \(2.0 \times 10^{11}\text{ Pa}\). A load of \(150\text{ N}\) is suspended vertically from the wire, causing it to deform elastically. What is the strain energy stored in the wire?
A.\(7.5 \times 10^{-3}\text{ J}\)
B.\(3.8 \times 10^{-2}\text{ J}\)
C.\(7.5 \times 10^{-2}\text{ J}\)
D.\(1.5 \times 10^{-1}\text{ J}\)
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Worked solution
First find the extension: \(\Delta L = \frac{F L}{A E} = \frac{150 \times 2.0}{(1.5 \times 10^{-6}) \times (2.0 \times 10^{11})} = 1.0 \times 10^{-3}\text{ m}\). The strain energy stored is \(E_s = \frac{1}{2} F \Delta L = \frac{1}{2} \times 150 \times 1.0 \times 10^{-3} = 7.5 \times 10^{-2}\text{ J}\).
Marking scheme
1 mark for finding the correct extension and substituting it into the strain energy formula to get C.
Question 16 · MCQ
1 marks
A spacecraft of mass \(M\) is travelling in deep space at a constant velocity \(v\). An internal explosion splits the spacecraft into two fragments. One fragment of mass \(0.20 M\) is moving directly backwards with a speed of \(2.0 v\) relative to a stationary observer. What is the speed of the other fragment relative to the same stationary observer?
A.\(1.25 v\)
B.\(1.50 v\)
C.\(1.75 v\)
D.\(2.25 v\)
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Worked solution
By conservation of linear momentum: \(p_i = p_f\). Let the initial direction of motion be positive. \(M v = (0.20 M)(-2.0 v) + 0.80 M v_2\). Dividing by \(M\) gives \(v = -0.40 v + 0.80 v_2\). Solving for \(v_2\) gives \(1.40 v = 0.80 v_2\), which simplifies to \(v_2 = 1.75 v\).
Marking scheme
1 mark for applying conservation of momentum with correct signs to find the final velocity of the remaining fragment, yielding C.
Question 17 · MCQ
1 marks
The rate of heat flow \( P \) (in \(\text{W}\)) through a solid rod of length \( L \) and cross-sectional area \( A \) is given by the equation: \( P = \frac{k A \Delta T}{L} \), where \( \Delta T \) is the temperature difference across the rod (measured in kelvin, \(\text{K}\)) and \( k \) is the thermal conductivity of the material. What are the SI base units of thermal conductivity \( k \)?
A.\(\text{kg m s}^{-3} \text{K}^{-1}\)
B.\(\text{kg m}^2 \text{s}^{-2} \text{K}^{-1}\)
C.\(\text{kg s}^{-3} \text{K}^{-1}\)
D.\(\text{kg m}^2 \text{s}^{-3} \text{K}^{-1}\)
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Worked solution
We can rearrange the equation for \( k \): \( k = \frac{P L}{A \Delta T} \). The SI base units of the quantities are: \( [P] = \text{W} = \text{J s}^{-1} = \text{kg m}^2 \text{s}^{-3} \), \( [L] = \text{m} \), \( [A] = \text{m}^2 \), and \( [\Delta T] = \text{K} \). Substituting these into the formula for \( k \): \( [k] = \frac{(\text{kg m}^2 \text{s}^{-3}) \times \text{m}}{\text{m}^2 \times \text{K}} = \text{kg m s}^{-3} \text{K}^{-1} \).
Marking scheme
1 mark for the correct SI base unit derivation.
Question 18 · MCQ
1 marks
An object X is dropped from rest from a height of \( 80\text{ m} \) above the ground. At the same instant, object Y is thrown vertically upwards from ground level with an initial speed \( u \). The two objects pass each other at a height of \( 30\text{ m} \) above the ground. Air resistance is negligible. Take the acceleration of free fall \( g \) to be \( 9.81\text{ m s}^{-2} \). What is the initial speed \( u \) of object Y?
A.\(16.0\text{ m s}^{-1}\)
B.\(25.1\text{ m s}^{-1}\)
C.\(31.3\text{ m s}^{-1}\)
D.\(40.0\text{ m s}^{-1}\)
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Worked solution
First, find the time \( t \) taken for object X to fall from \( 80\text{ m} \) to \( 30\text{ m} \). The distance fallen is \( s_X = 80\text{ m} - 30\text{ m} = 50\text{ m} \). Using \( s = ut + \frac{1}{2}gt^2 \) with \( u = 0 \): \( 50 = \frac{1}{2}(9.81)t^2 \), which gives \( t = \sqrt{\frac{100}{9.81}} \approx 3.193\text{ s} \). Next, consider object Y which rises from the ground to \( 30\text{ m} \) in the same time \( t \). Using \( s_Y = ut - \frac{1}{2}gt^2 \): \( 30 = u(3.193) - \frac{1}{2}(9.81)(3.193)^2 \). Since \( \frac{1}{2}(9.81)(3.193)^2 = 50\text{ m} \), we have \( 30 = 3.193u - 50 \), so \( 3.193u = 80 \), giving \( u \approx 25.1\text{ m s}^{-1} \).
Marking scheme
1 mark for the correct calculated speed.
Question 19 · MCQ
1 marks
Two wires, P and Q, are made of the same material. Wire P has length \( L \) and diameter \( d \). Wire Q has length \( 2L \) and diameter \( 2d \). Wire P supports a load of weight \( F \), which produces an extension \( x \). What weight must be supported by wire Q to produce an extension of \( 2x \)? (Assume both wires behave elastically).
A.\(F\)
B.\(2F\)
C.\(4F\)
D.\(8F\)
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Worked solution
The Young modulus is \( E = \frac{F L}{A x} \), so extension \( x = \frac{F L}{A E} \). Since area \( A \propto d^2 \), we have \( x \propto \frac{F L}{d^2} \). For wire P: \( x = \frac{k F L}{d^2} \) where \( k \ is a constant. For wire Q: \) x_Q = \frac{k F_Q (2L)}{(2d)^2} = \frac{2 k F_Q L}{4 d^2} = \frac{k F_Q L}{2 d^2} \). We want \( x_Q = 2x \). Thus, \( \frac{k F_Q L}{2 d^2} = 2 \left( \frac{k F L}{d^2} \right) \), which simplifies to \( \frac{F_Q}{2} = 2F \), so \( F_Q = 4F \).
Marking scheme
1 mark for the correct factor for weight.
Question 20 · MCQ
1 marks
A sphere of mass \( m \) is moving to the right with speed \( 2v \). It undergoes a head-on, perfectly elastic collision with a sphere of mass \( 3m \) moving to the left with speed \( v \). What are the velocities of the two spheres after the collision?
A.Sphere of mass \( m \): \( 2.5v \) to the left; Sphere of mass \( 3m \): \( 0.5v \) to the right
B.Sphere of mass \( m \): \( 1.0v \) to the left; Sphere of mass \( 3m \): \( 0.5v \) to the right
C.Sphere of mass \( m \): \( 2.5v \) to the left; Sphere of mass \( 3m \): \( 1.5v \) to the right
D.Sphere of mass \( m \): \( 1.5v \) to the left; Sphere of mass \( 3m \): \( 1.0v \) to the right
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Worked solution
Let the right direction be positive. Initial velocities are \( u_1 = 2v \) and \( u_2 = -v \). For a perfectly elastic collision, the relative velocity of approach equals the relative velocity of separation: \( u_1 - u_2 = v_2 - v_1 \implies 2v - (-v) = v_2 - v_1 \implies v_2 - v_1 = 3v \). From conservation of linear momentum: \( m(2v) + 3m(-v) = m v_1 + 3m v_2 \implies -mv = m v_1 + 3m v_2 \implies v_1 + 3v_2 = -v \). Substituting \( v_1 = v_2 - 3v \) into this equation gives \( v_2 - 3v + 3v_2 = -v \implies 4v_2 = 2v \implies v_2 = +0.5v \). Therefore, \( v_1 = 0.5v - 3v = -2.5v \). Thus, the mass \( m \) moves to the left at \( 2.5v \) and the mass \( 3m \) moves to the right at \( 0.5v \).
Marking scheme
1 mark for the correct velocities of both spheres.
Question 21 · MCQ
1 marks
In a double-slit experiment, monochromatic light of wavelength \( 600\text{ nm} \) is incident on two slits separated by a distance \( a \). Interference fringes are observed on a screen at a distance \( D \) from the slits. The distance between adjacent bright fringes is \( 1.2\text{ mm} \). The light source is then replaced by another source of wavelength \( \lambda \), and the slit separation is halved, while the screen distance remains unchanged. The new fringe separation is \( 1.6\text{ mm} \). What is the wavelength \( \lambda \) of the second light source?
A.\(300\text{ nm}\)
B.\(400\text{ nm}\)
C.\(450\text{ nm}\)
D.\(900\text{ nm}\)
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Worked solution
Using the double-slit formula for fringe separation: \( x = \frac{\lambda D}{a} \). Initially, we have \( 1.2\text{ mm} = \frac{(600\text{ nm}) D}{a} \). Finally, the new fringe separation is \( 1.6\text{ mm} = \frac{\lambda D}{0.5a} = \frac{2\lambda D}{a} \). Dividing the second equation by the first: \( \frac{1.6}{1.2} = \frac{2\lambda}{600\text{ nm}} \implies \frac{4}{3} = \frac{\lambda}{300\text{ nm}} \implies \lambda = 400\text{ nm} \).
Marking scheme
1 mark for correct calculation of the second wavelength.
Question 22 · MCQ
1 marks
Two wires, X and Y, have the same mass and are made of metals with densities \( d_X \) and \( d_Y \) respectively. The resistivity of metal X is \( \rho_X \) and the resistivity of metal Y is \( \rho_Y \). Wire X has twice the length of wire Y. What is the ratio of the resistance of wire X to the resistance of wire Y, \( \frac{R_X}{R_Y} \)?
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Worked solution
Resistance is \( R = \frac{\rho L}{A} \). Mass is \( m = \text{volume} \times \text{density} = A L d \), so cross-sectional area \( A = \frac{m}{L d} \). Substituting this into the resistance equation gives \( R = \frac{\rho L^2 d}{m} \). Since both wires have the same mass \( m \), we can write the ratio of their resistances as: \( \frac{R_X}{R_Y} = \frac{\rho_X L_X^2 d_X}{\rho_Y L_Y^2 d_Y} \). Since \( L_X = 2L_Y \), we have \( L_X^2 = 4 L_Y^2 \). Thus, \( \frac{R_X}{R_Y} = 4 \left( \frac{\rho_X}{\rho_Y} \right) \left( \frac{d_X}{d_Y} \right) \).
Marking scheme
1 mark for identifying area relation from mass and density, and obtaining correct ratio.
Question 23 · MCQ
1 marks
A cell of electromotive force (e.m.f.) \( E \) and internal resistance \( r \) is connected to a variable external resistor of resistance \( R \). When \( R = 4.0\ \Omega \), the terminal potential difference across the cell is \( 3.0\text{ V} \). When \( R \) is increased to \( 9.0\ \Omega \), the terminal potential difference becomes \( 4.5\text{ V} \). What are the values of \( E \) and \( r \)?
A.\(E = 6.0\text{ V}, r = 4.0\ \Omega\)
B.\(E = 7.5\text{ V}, r = 6.0\ \Omega\)
C.\(E = 9.0\text{ V}, r = 8.0\ \Omega\)
D.\(E = 12.0\text{ V}, r = 12.0\ \Omega\)
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Worked solution
The relationship is \( E = V + I r = V \left(1 + \frac{r}{R}\right) \). For the first case: \( E = 3.0 \left(1 + \frac{r}{4.0}\right) \). For the second case: \( E = 4.5 \left(1 + \frac{r}{9.0}\right) \). Equating the two expressions for \( E \): \( 3.0 \left(1 + \frac{r}{4.0}\right) = 4.5 \left(1 + \frac{r}{9.0}\right) \). Dividing both sides by \( 1.5 \) gives \( 2 \left(1 + \frac{r}{4.0}\right) = 3 \left(1 + \frac{r}{9.0}\right) \implies 2 + \frac{r}{2} = 3 + \frac{r}{3} \implies \frac{r}{6} = 1 \implies r = 6.0\ \Omega \). Substituting \( r = 6.0\ \Omega \) back: \( E = 3.0 \left(1 + \frac{6.0}{4.0}\right) = 7.5\text{ V} \).
Marking scheme
1 mark for the correct system of equations and solving for E and r.
Question 24 · MCQ
1 marks
A nucleus of Carbon-11 (\( ^{11}_{6}\text{C} \)) decays by \( \beta^+ \) emission to a nucleus of Boron-11 (\( ^{11}_{5}\text{B} \)). Which row correctly describes the change in quark composition of the decaying nucleon and the leptons produced during this process?
A.Quark change: \( \text{u} \to \text{d} \); Leptons: positron and electron neutrino
B.Quark change: \( \text{d} \to \text{u} \); Leptons: electron and electron antineutrino
C.Quark change: \( \text{u} \to \text{d} \); Leptons: electron and electron antineutrino
D.Quark change: \( \text{d} \to \text{u} \); Leptons: positron and electron neutrino
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Worked solution
In \( \beta^+ \) decay, a proton decays into a neutron, a positron, and an electron neutrino. A proton has quark composition \( \text{uud} \) and a neutron has \( \text{udd} \), so an up quark (\( \text{u} \)) changes into a down quark (\( \text{d} \)). The leptons emitted are the positron (\( \text{e}^+ \)) and the electron neutrino (\( \nu_e \)).
Marking scheme
1 mark for identifying the quark change and correct leptons.
Question 25 · MCQ
1 marks
What is the SI base unit of electrical resistivity?
A.\text{kg m}^3 \text{s}^{-3} \text{A}^{-2}
B.\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}
C.\text{kg m}^3 \text{s}^{-2} \text{A}^{-1}
D.\text{kg m}^2 \text{s}^{-2} \text{A}^{-1}
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Worked solution
Electrical resistivity \(\rho\) is defined by the formula:
\rho = \frac{R A}{L}
where \(R\) is electrical resistance, \(A\) is cross-sectional area, and \(L\) is length.
First, find the SI base units of resistance \(R\). Resistance is defined as:
R = \frac{V}{I}
where \(V\) is potential difference and \(I\) is current (base unit: \(\text{A}\)).
Potential difference is work done per unit charge:
V = \frac{W}{Q} = \frac{W}{I t}
The SI base unit of work (energy) is \(\text{kg m}^2 \text{s}^{-2}\). Thus, the SI base unit of \(V\) is:
- Award 1 mark for showing the correct derivation from the base units of work, current, and length to get the correct option A. - Reject other options which do not simplify to the correct SI base unit combination.
Question 26 · MCQ
1 marks
A model rocket is launched vertically upwards from the ground. It accelerates uniformly from rest at \(4.0\text{ m s}^{-2}\) for \(5.0\text{ s}\). The fuel is then completely exhausted, and the rocket continues to move upwards under the influence of gravity alone (air resistance is negligible). What is the maximum height above the ground reached by the rocket?
A.50 m
B.70 m
C.91 m
D.120 m
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Worked solution
The motion of the rocket consists of two distinct phases.
**Phase 1: Powered ascent (with fuel)** The rocket starts from rest (\(u_1 = 0\)), accelerates at \(a_1 = 4.0\text{ m s}^{-2}\) for \(t_1 = 5.0\text{ s}\).
**Phase 2: Unpowered ascent (free fall)** The fuel is exhausted. The rocket has an initial upward velocity of \(u_2 = v_1 = 20\text{ m s}^{-1}\) and undergoes deceleration due to gravity \(a_2 = -g = -9.81\text{ m s}^{-2}\) until it momentarily stops at its maximum height (\(v_2 = 0\)).
Using \(v_2^2 = u_2^2 + 2 a_2 s_2\):
0 = 20^2 + 2(-9.81) s_2
0 = 400 - 19.62 s_2
s_2 = \frac{400}{19.62} \approx 20.4\text{ m}
**Total Height** The total height \(H\) above the ground is:
- Award 1 mark for calculating both the height of the powered stage (50 m) and the unpowered stage (approx. 20.4 m) and summing them up to obtain 70 m. - Reject 50 m (only powered stage) or 20 m (only unpowered stage).
Question 27 · MCQ
1 marks
A block of mass \(3.0\text{ kg}\) slides along a frictionless horizontal surface at \(4.0\text{ m s}^{-1}\) and collides with a second block of mass \(1.0\text{ kg}\) which is initially at rest. The collision is perfectly elastic.
What are the speeds of the two blocks after the collision?
A.1.0 m s^{-1} and 3.0 m s^{-1}
B.2.0 m s^{-1} and 2.0 m s^{-1}
C.2.0 m s^{-1} and 6.0 m s^{-1}
D.3.0 m s^{-1} and 9.0 m s^{-1}
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Worked solution
Let \(m_1 = 3.0\text{ kg}\) and \(u_1 = 4.0\text{ m s}^{-1}\). Let \(m_2 = 1.0\text{ kg}\) and \(u_2 = 0\). Let \(v_1\) and \(v_2\) be the final velocities of the two blocks.
By the principle of conservation of linear momentum:
Thus, the speeds are \(2.0\text{ m s}^{-1}\) and \(6.0\text{ m s}^{-1}\).
Marking scheme
1 mark for the correct answer C.
- Award 1 mark for setting up equations for conservation of momentum and the elastic collision relative speed relation, and solving them to get 2.0 m s^-1 and 6.0 m s^-1. - Reject other options that do not satisfy both conservation of momentum and conservation of kinetic energy.
Question 28 · MCQ
1 marks
A uniform rigid bar of length \(2.0\text{ m}\) and weight \(80\text{ N}\) is hinged at one end to a vertical wall. The bar is held horizontally by a cable attached to its other end. The cable makes an angle of \(35^\circ\) with the horizontal bar.
What is the tension in the cable?
A.40 N
B.70 N
C.98 N
D.140 N
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Worked solution
For the bar to remain in equilibrium, the sum of the moments about any point must be zero. We take moments about the hinge at the wall to eliminate the force exerted by the hinge.
The forces causing moments about the hinge are: 1. The weight of the uniform bar, \(W = 80\text{ N}\), which acts at its center of gravity (the midpoint of the bar, \(1.0\text{ m}\) from the hinge). This creates a clockwise moment. 2. The tension \(T\) in the cable, which acts at the far end (\(2.0\text{ m}\) from the hinge) at an angle of \(35^\circ\) above the horizontal. The perpendicular component of the tension is \(T \sin 35^\circ\), which creates an anticlockwise moment.
T = \frac{80}{1.147} \approx 69.7\text{ N} \approx 70\text{ N}
Marking scheme
1 mark for the correct answer B.
- Award 1 mark for using the correct moments equation, taking the weight at the center of gravity (1.0 m from the pivot) and the vertical component of the tension at 2.0 m from the pivot, yielding approx. 70 N. - Reject 140 N (which forgets that the weight acts at the midpoint).
Question 29 · MCQ
1 marks
A wire of length \(L\) and cross-sectional area \(A\) is made of a metal of Young modulus \(E\). Under a tensile force \(F\), the wire extends by \(x\).
A second wire made of the same metal has length \(2L\) and a diameter twice that of the first wire.
What is the extension of this second wire when subjected to the same tensile force \(F\)?
A.0.25x
B.0.5x
C.x
D.2x
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Worked solution
The Young modulus \(E\) of a wire is given by:
E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}
Rearranging for the extension \(x\):
x = \frac{FL}{AE}
For the second wire: - The material is the same, so the Young modulus is still \(E\). - The tensile force is the same, \(F\). - The length is \(L' = 2L\). - The diameter is doubled, so the radius is also doubled. Since area is proportional to the square of the diameter (\(A = \frac{\pi d^2}{4}\)), the new cross-sectional area is \(A' = 4A\).
Substituting these values into the formula for the new extension \(x\)' :
- Award 1 mark for setting up the ratio of extensions correctly, recognizing that doubling the diameter quadruples the cross-sectional area, resulting in an extension of 0.5x. - Reject 2x (which fails to account for the change in area) or other options.
Question 30 · MCQ
1 marks
In a double-slit interference experiment using red laser light of wavelength \(650\text{ nm}\), the interference pattern is observed on a screen placed at a distance of \(1.5\text{ m}\) from the slits. The distance between the centers of the two slits is \(0.30\text{ mm}\).
What is the distance between adjacent bright fringes on the screen?
A.0.33 mm
B.1.3 mm
C.3.3 mm
D.13 mm
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Worked solution
The formula for the fringe spacing \(x\) in a double-slit interference experiment is:
x = \frac{\lambda D}{a}
where: - \(\lambda\) is the wavelength of the light = \(650\text{ nm} = 6.50 \times 10^{-7}\text{ m}\) - \(D\) is the distance from the slits to the screen = \(1.5\text{ m}\) - \(a\) is the slit separation = \(0.30\text{ mm} = 3.0 \times 10^{-4}\text{ m}\)
To two significant figures, this is \(3.3\text{ mm}\).
Marking scheme
1 mark for the correct answer C.
- Award 1 mark for correct unit conversion of all quantities to meters and substituting them correctly into \(x = \frac{\lambda D}{a}\) to get 3.3 mm. - Reject options with incorrect powers of ten due to incorrect unit conversions (such as A, B, and D).
Question 31 · MCQ
1 marks
A potential difference of \(6.0\text{ V}\) is applied across a uniform resistance wire of length \(2.0\text{ m}\) and cross-sectional area \(1.5 \times 10^{-7}\text{ m}^2\). The current in the wire is \(0.80\text{ A}\).
What is the resistivity of the metal from which the wire is made?
A.3.6 \times 10^{-7} \ m
B.4.5 \times 10^{-7} \ m
C.5.6 \times 10^{-7} \ m
D.7.5 \times 10^{-7} \ m
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Worked solution
First, find the resistance \(R\) of the wire using Ohm's law:
R = \frac{V}{I} = \frac{6.0\text{ V}}{0.80\text{ A}} = 7.5\ \Omega
The relationship between resistance \(R\) and resistivity \(\rho\) is given by:
- Award 1 mark for first calculating resistance (7.5 ohms) and then correctly applying the resistivity formula to get \(5.6 \times 10^{-7}\ \Omega\text{ m}\). - Reject B (calculated using voltage instead of resistance) or other options.
Question 32 · MCQ
1 marks
A baryon has a net charge of \(+e\) and a strangeness of \(-1\).
Knowing that the strange quark (\(s\)) has a charge of \(-\frac{1}{3}e\) and a strangeness of \(-1\), and that the baryon contains only up (\(u\)), down (\(d\)), and strange (\(s\)) quarks, what is the quark composition of this baryon?
A.uds
B.uus
C.uss
D.udd
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Worked solution
A baryon is a hadron made of three quarks.
1. The baryon has a strangeness of \(-1\). Since only the strange (\(s\)) quark carries a strangeness of \(-1\) (and \(u\) and \(d\) quarks have strangeness 0), the baryon must contain exactly one \(s\) quark.
2. Let the baryon have the composition \(q_1 q_2 s\), where \(q_1\) and \(q_2\) are either \(u\) or \(d\) quarks.
3. The total charge of the baryon is \(+1e\). The charge of the \(s\) quark is \(-\frac{1}{3}e\).
Therefore, the sum of the charges of \(q_1\) and \(q_2\) must satisfy:
Since the charge of an up quark (\(u\)) is \(+\frac{2}{3}e\) and the charge of a down quark (\(d\)) is \(-\frac{1}{3}e\), we can determine \(q_1\) and \(q_2\):
\frac{2}{3}e + \frac{2}{3}e = +\frac{4}{3}e
Thus, both \(q_1\) and \(q_2\) must be up quarks (\(u\)).
The quark composition is therefore \(uus\).
Marking scheme
1 mark for the correct answer B.
- Award 1 mark for analyzing that a strangeness of -1 requires exactly one s quark, and that to get a net charge of +1e with 3 quarks, the remaining two quarks must both be up quarks, giving 'uus'. - Reject other combinations which have different net charges or strangeness.
Question 33 · MCQ
1 marks
A stone is projected vertically upwards with speed \(v\) from a cliff of height \(H\). It hits the ground at the base of the cliff with speed \(3v\). Air resistance is negligible. What is the maximum height reached by the stone above its starting point?
A.\(\frac{H}{8}\)
B.\(\frac{H}{4}\)
C.\(\frac{H}{3}\)
D.\(\frac{H}{2}\)
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Worked solution
Let the upward direction be positive. Using the equation of motion \(v_f^2 = v_i^2 + 2as\) from the launch point to the ground:
\((-3v)^2 = v^2 + 2(-g)(-H)\)
\(9v^2 = v^2 + 2gH\)
\(8v^2 = 2gH \implies v^2 = \frac{gH}{4}\)
At the maximum height \(h\) above the launch point, the velocity of the stone is zero:
\(0 = v^2 - 2gh \implies h = \frac{v^2}{2g}\)
Substituting the expression for \(v^2\) into this equation:
\(h = \frac{gH/4}{2g} = \frac{H}{8}\).
Marking scheme
1 mark for the correct option A.
Question 34 · MCQ
1 marks
A uniform rigid beam of weight \(W\) and length \(L\) is held horizontally by a hinge at one end and a light cable attached to the other end. The cable makes an angle of \(30^\circ\) with the horizontal beam. What is the magnitude of the horizontal component of the force exerted by the hinge on the beam?
A.\(0.29 W\)
B.\(0.50 W\)
C.\(0.87 W\)
D.\(1.00 W\)
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Worked solution
Let the hinge be at \(x=0\) and the cable at \(x=L\). The weight \(W\) of the uniform beam acts downwards at its midpoint \(L/2\). Taking moments about the hinge:
\(T \sin(30^\circ) \times L = W \times \frac{L}{2}\)
\(T (0.5) = 0.5 W \implies T = W\)
For horizontal equilibrium, the horizontal force exerted by the hinge \(F_x\) must balance the horizontal component of the tension force:
\(F_x = T \cos(30^\circ) = W \cos(30^\circ) \approx 0.87 W\).
Marking scheme
1 mark for the correct option C.
Question 35 · MCQ
1 marks
Two wires \(P\) and \(Q\) are made of the same metal. Wire \(P\) has twice the diameter and half the length of wire \(Q\). Both wires are suspended vertically and support identical loads, which stretch them elastically. What is the ratio \(\frac{\text{strain in } P}{\text{strain in } Q}\)?
A.\(0.125\)
B.\(0.25\)
C.\(0.50\)
D.\(2.0\)
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Worked solution
The Young modulus \(E\) of the material is given by:
Since both wires are made of the same metal, \(E\) is identical. Since they support identical loads, \(F\) is also identical. Therefore, strain is inversely proportional to the cross-sectional area \(A\), and hence inversely proportional to the square of the diameter \(d^2\):
\(\text{strain} \propto \frac{1}{d^2}\)
\(\frac{\text{strain in } P}{\text{strain in } Q} = \left(\frac{d_Q}{d_P}\right)^2 = \left(\frac{1}{2}\right)^2 = 0.25\).
Note that the length of the wires does not affect the strain under these conditions (though it does affect the overall extension).
Marking scheme
1 mark for the correct option B.
Question 36 · MCQ
1 marks
A cell of electromotive force (e.m.f.) \(E\) and internal resistance \(r\) is connected to a variable resistor of resistance \(R\). A voltmeter is connected across the terminals of the cell. As the resistance \(R\) is increased from a very low value to a very high value, how do the current \(I\) in the circuit and the terminal potential difference \(V\) measured by the voltmeter change?
A.current decreases, terminal p.d. increases
B.current decreases, terminal p.d. decreases
C.current increases, terminal p.d. increases
D.current increases, terminal p.d. decreases
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Worked solution
The total resistance of the circuit is \(R + r\). As \(R\) increases, the total resistance increases, which causes the current \(I = \frac{E}{R + r}\) to decrease.
The terminal potential difference is given by \(V = E - Ir\). As the current \(I\) decreases, the 'lost volts' term \(Ir\) decreases, which causes the terminal potential difference \(V\) to increase, eventually approaching \(E\) as \(R\) becomes very large.
Marking scheme
1 mark for the correct option A.
Question 37 · MCQ
1 marks
In a double-slit interference experiment using light of wavelength \(600\text{ nm}\), the fringe separation on a screen placed at a fixed distance from the slits is \(1.2\text{ mm}\). If the light source is replaced by one of wavelength \(450\text{ nm}\) and the slit separation is halved, what is the new fringe separation?
A.\(0.45\text{ mm}\)
B.\(0.90\text{ mm}\)
C.\(1.6\text{ mm}\)
D.\(1.8\text{ mm}\)
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Worked solution
The formula for double-slit fringe separation is \(x = \frac{\lambda D}{a}\).
Two balls, X and Y, travel along a frictionless horizontal track. Ball X of mass \(2m\) moving with velocity \(u\) collides head-on with ball Y of mass \(m\) moving in the opposite direction with velocity \(2u\). The collision is perfectly elastic. What are the velocities of X and Y after the collision (taking the direction of the initial velocity of X as positive)?
A.velocity of X is \(u\); velocity of Y is \(-2u\)
B.velocity of X is \(-u\); velocity of Y is \(2u\)
C.velocity of X is \(-2u\); velocity of Y is \(u\)
D.both balls have zero velocity (they come to rest)
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Worked solution
Let the direction of the initial velocity of X be positive. Therefore, \(u_X = u\) and \(u_Y = -2u\).
Since the collision is perfectly elastic, the relative speed of approach equals the relative speed of separation:
\(u_X - u_Y = v_Y - v_X\)
\(u - (-2u) = v_Y - v_X \implies 3u = v_Y - v_X\)
Substituting \(v_Y = -2 v_X\) into this equation:
\(3u = -2 v_X - v_X = -3 v_X \implies v_X = -u\)
Using this to find \(v_Y\):
\(v_Y = -2(-u) = 2u\).
Thus, the velocity of X after the collision is \(-u\) and the velocity of Y is \(2u\).
Marking scheme
1 mark for the correct option B.
Question 39 · MCQ
1 marks
A metal wire of resistance \(R\) is stretched such that its length increases by \(10\%\) while its volume remains constant. What is the percentage increase in the resistance of the wire?
A.\(10\%\)
B.\(20\%\)
C.\(21\%\)
D.\(44\%\)
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Worked solution
The resistance of a wire is given by \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area.
Since the volume \(V = A L\) is constant, we can express the area as \(A = \frac{V}{L}\).
Substituting this into the resistance formula gives:
\(R = \rho \frac{L^2}{V}\)
Since \(\rho\) and \(V\) remain constant, the resistance is proportional to the square of the length: \(R \propto L^2\).
With a \(10\%\) increase in length, the new length is \(L' = 1.10 L\).
A particle known as a kaon has a quark composition of \(u\bar{s}\), where \(u\) represents an up quark and \(\bar{s}\) represents an antistrange quark. What are the electric charge of this kaon (in terms of the elementary charge \(e\)) and its hadron classification?
Since the particle consists of one quark and one antiquark (a quark-antiquark pair), it is classified as a meson.
Marking scheme
1 mark for the correct option D.
Paper 21 Structured Questions
Answer all questions in the spaces provided on the question paper. Show all working and appropriate units.
7 Question · 59.99 marks
Question 1 · Structured
8.57 marks
A uniform metal rod of length 1.5 m and weight 45 N is pivoted at one end to a vertical wall. The rod is held horizontally by a wire attached to the other end of the rod. The wire makes an angle of \(30^\circ\) with the horizontal rod. (a) Define center of gravity. (b) By taking moments about the pivot, calculate the tension in the wire. (c) Calculate the vertical component of the force exerted on the rod by the pivot.
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Worked solution
(a) The center of gravity is the point at which the entire weight of a body may be considered to act. (b) For the rod in equilibrium, take moments about the pivot. Clockwise moment = Weight \(\times\) distance to center of gravity = \(45\text{ N} \times 0.75\text{ m} = 33.75\text{ N m}\). Anticlockwise moment = Tension vertical component \(\times\) length = \(T \sin(30^\circ) \times 1.5\text{ m} = 0.75 T\). Equating moments: \(0.75 T = 33.75\), which gives \(T = 45\text{ N}\). (c) For vertical equilibrium: Pivot vertical force \(F_y\) + \(T \sin(30^\circ) = \text{Weight}\). \(F_y + 45 \sin(30^\circ) = 45 \implies F_y + 22.5 = 45 \implies F_y = 22.5\text{ N}\).
Marking scheme
(a) 1 mark for stating center of gravity is the point where the weight of the object is considered to act. (b) 1 mark for calculating clockwise moment of 33.75 N m. 1 mark for setting up correct moment equation: \(T \sin(30^\circ) \times 1.5 = 45 \times 0.75\). 1 mark for correct calculation of tension \(T = 45\text{ N}\). (c) 1 mark for vertical force balance equation: \(F_y + T \sin(30^\circ) = W\). 1 mark for final vertical force of 22.5 N.
Question 2 · Structured
8.57 marks
An object of mass 0.20 kg undergoes simple harmonic motion. The displacement x of the object varies with time t according to the equation \(x = 0.050 \cos(8.0 t)\), where x is in meters and t is in seconds. (a) State the two conditions required for a system to undergo simple harmonic motion. (b) Calculate the maximum kinetic energy of the object. (c) Determine the magnitude of the acceleration of the object when its displacement is 0.030 m.
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Worked solution
(a) The conditions are: 1. Acceleration is directly proportional to displacement. 2. Acceleration is always directed towards a fixed point (or opposite in direction to displacement). (b) Comparing with \(x = x_0 \cos(\omega t)\), the angular frequency \(\omega = 8.0\text{ rad s}^{-1}\) and the amplitude \(x_0 = 0.050\text{ m}\). Maximum speed \(v_0 = \omega x_0 = 8.0 \times 0.050 = 0.40\text{ m s}^{-1}\). Maximum kinetic energy \(E_k = \frac{1}{2} m v_0^2 = 0.5 \times 0.20 \times (0.40)^2 = 0.016\text{ J}\). (c) Magnitude of acceleration \(a = \omega^2 x = (8.0)^2 \times 0.030 = 64 \times 0.030 = 1.92\text{ m s}^{-2}\).
Marking scheme
(a) 1 mark for stating acceleration is proportional to displacement. 1 mark for stating acceleration is directed towards a fixed point. (b) 1 mark for identifying angular frequency as 8.0 rad/s. 1 mark for calculating maximum speed as 0.40 m/s. 1 mark for maximum kinetic energy of 0.016 J. (c) 1 mark for using formula \(a = \omega^2 x\). 1 mark for final acceleration value of 1.92 m/s^2.
Question 3 · Structured
8.57 marks
A ball is projected horizontally from the top of a cliff of height 45 m with an initial horizontal velocity of \(12\text{ m s}^{-1}\). Air resistance is negligible. (a) Calculate the time taken for the ball to reach the ground. (b) Calculate the horizontal distance traveled by the ball before it hits the ground. (c) Determine the magnitude and direction of the velocity of the ball just as it strikes the ground.
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Worked solution
(a) Vertical motion: initial vertical velocity \(u_y = 0\), vertical acceleration \(a_y = 9.81\text{ m s}^{-2}\), vertical displacement \(s_y = 45\text{ m}\). Using \(s_y = u_y t + \frac{1}{2} g t^2 \implies 45 = 0 + 0.5 \times 9.81 \times t^2 \implies t = \sqrt{90 / 9.81} = 3.03\text{ s}\). (b) Horizontal motion: horizontal velocity remains constant at \(12\text{ m s}^{-1}\). Horizontal distance \(s_x = v_x \times t = 12 \times 3.03 = 36.4\text{ m}\). (c) Just before hitting the ground: \(v_x = 12\text{ m s}^{-1}\), \(v_y = u_y + g t = 9.81 \times 3.03 = 29.7\text{ m s}^{-1}\). Magnitude of velocity \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 29.7^2} = 32.0\text{ m s}^{-1}\). Direction \(\theta = \tan^{-1}(v_y / v_x) = \tan^{-1}(29.7 / 12) = 68^\circ\) below the horizontal.
Marking scheme
(a) 1 mark for choosing correct vertical equation: \(45 = 0.5 \times 9.81 \times t^2\). 1 mark for calculating time \(t = 3.03\text{ s}\) (accept 3.0 s). (b) 1 mark for horizontal distance formula \(s = v t\). 1 mark for calculating distance \(s_x = 36.4\text{ m}\) (or 36 m). (c) 1 mark for calculating final vertical velocity component \(v_y = 29.7\text{ m s}^{-1}\). 1 mark for using Pythagoras' theorem. 1 mark for final velocity magnitude \(32\text{ m s}^{-1}\). 1 mark for angle \(68^\circ\) below the horizontal.
Question 4 · Structured
8.57 marks
An electric motor is used to lift a load of mass 80 kg vertically upwards through a height of 15 m in a time of 6.0 s. The electrical power input to the motor is 2.5 kW. (a) Calculate the useful work done in lifting the load. (b) Calculate the useful output power of the motor. (c) Determine the efficiency of the motor.
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Worked solution
(a) The useful work done is equal to the gain in gravitational potential energy of the load: \(W_y = m g h = 80\text{ kg} \times 9.81\text{ m s}^{-2} \times 15\text{ m} = 11772\text{ J} \approx 1.18 \times 10^4\text{ J}\). (b) The useful output power of the motor is: \(P_{\text{out}} = \frac{\text{Work done}}{\text{time}} = \frac{11772\text{ J}}{6.0\text{ s}} = 1962\text{ W} \approx 1.96\text{ kW}\). (c) Electrical power input is \(P_{\text{in}} = 2.5\text{ kW} = 2500\text{ W}\). Efficiency \(\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{1962}{2500} \times 100\% = 78.5\%\).
Marking scheme
(a) 1 mark for formula \(W = m g h\). 1 mark for correct calculation of work \(1.18 \times 10^4\text{ J}\) (or \(1.2 \times 10^4\text{ J}\)). (b) 1 mark for formula \(P = W / t\). 1 mark for correct calculation of power \(1.96\text{ kW}\) (or \(2.0\text{ kW}\)). (c) 1 mark for efficiency equation \(P_{\text{out}} / P_{\text{in}}\). 1 mark for correct calculation of efficiency \(78.5\%\) (or \(79\%\)).
Question 5 · Structured
8.57 marks
Laser light of wavelength 632 nm is incident normally on a double slit. The slit separation is 0.25 mm. Interference fringes are observed on a screen placed parallel to the double slit at a distance of 1.8 m. (a) State what is meant by coherent light sources. (b) Calculate the fringe separation observed on the screen. (c) Describe the effect on the fringe separation if the wavelength of the light is increased to 700 nm while keeping all other parameters constant.
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Worked solution
(a) Coherent light sources have a constant phase difference between them. (b) Using the double-slit interference formula \(x = \frac{\lambda D}{a}\), where \(\lambda = 632 \times 10^{-9}\text{ m}\), \(D = 1.8\text{ m}\), and \(a = 0.25 \times 10^{-3}\text{ m}\). \(x = \frac{632 \times 10^{-9} \times 1.8}{0.25 \times 10^{-3}} = 4.55 \times 10^{-3}\text{ m} \approx 4.6\text{ mm}\). (c) Since the fringe separation \(x\) is directly proportional to the wavelength \(\lambda\) (from \(x = \frac{\lambda D}{a}\)), increasing the wavelength from 632 nm to 700 nm will increase the fringe separation on the screen.
Marking scheme
(a) 1 mark for stating constant phase difference. (b) 1 mark for selecting the correct formula \(x = \lambda D / a\). 1 mark for correct conversion of units (nm and mm). 1 mark for calculating fringe separation \(4.6\text{ mm}\) (or \(4.55\text{ mm}\)). (c) 1 mark for stating that the fringe separation increases. 1 mark for explaining that fringe separation is directly proportional to wavelength.
Question 6 · Structured
8.57 marks
A uniform copper wire of length 2.0 m and cross-sectional area \(1.2 \times 10^{-7}\text{ m}^2\) has a resistance of \(0.28\ \Omega\). (a) Calculate the resistivity of copper. (b) The wire is stretched uniformly until its length becomes 3.0 m. Assuming the volume and resistivity of the wire remain constant, calculate the new resistance of the wire.
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Worked solution
(a) Using the resistance formula \(R = \frac{\rho L}{A}\), where \(R = 0.28\ \Omega\), \(L = 2.0\text{ m}\), and \(A = 1.2 \times 10^{-7}\text{ m}^2\). Resistivity \(\rho = \frac{R A}{L} = \frac{0.28 \times 1.2 \times 10^{-7}}{2.0} = 1.68 \times 10^{-8}\ \Omega\text{ m}\). (b) Since the volume \(V = A L\) is constant, and the length increases from 2.0 m to 3.0 m (by a factor of 1.5), the area must decrease by a factor of 1.5, so the new area \(A' = \frac{A}{1.5}\). The new resistance \(R' = \frac{\rho L'}{A'} = \frac{\rho \times 1.5 L}{A / 1.5} = 2.25 \times \frac{\rho L}{A} = 2.25 R\). \(R' = 2.25 \times 0.28 = 0.63\ \Omega\).
Marking scheme
(a) 1 mark for using \(R = \rho L / A\). 1 mark for calculating resistivity \(1.68 \times 10^{-8}\ \Omega\text{ m}\). (b) 1 mark for stating volume is constant, so \(A_1 L_1 = A_2 L_2\). 1 mark for finding the new cross-sectional area \(8.0 \times 10^{-8}\text{ m}^2\). 1 mark for substituting new dimensions back into the resistance formula. 1 mark for calculating new resistance \(0.63\ \Omega\).
Question 7 · Structured
8.57 marks
A free neutron decays into a proton, an electron, and an electron antineutrino. (a) Write down the balanced nuclear equation representing this decay, including the nucleon numbers and proton numbers for all particles. (b) State the quark composition of a neutron and a proton. (c) Describe the change in quark composition that occurs during this decay.
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Worked solution
(a) The balanced nuclear equation is: \(^{1}_{0}\text{n} \rightarrow\ ^{1}_{1}\text{p} + \ ^{0}_{-1}\text{e} + \bar{\nu}_e\). (b) The quark composition of a neutron is udd (one up quark, two down quarks). The quark composition of a proton is uud (two up quarks, one down quark). (c) During the beta-minus decay, a down quark (d) inside the neutron changes into an up quark (u), converting the neutron into a proton.
Marking scheme
(a) 1 mark for correct reactant \(^{1}_{0}\text{n}\) and product \(^{1}_{1}\text{p}\). 1 mark for adding electron \(^{0}_{-1}\text{e}\). 1 mark for adding electron antineutrino \(\bar{\nu}_e\). (b) 1 mark for identifying neutron as udd. 1 mark for identifying proton as uud. (c) 1 mark for stating that a down quark changes into an up quark.
Paper 31 Practical Skills
Perform the two practical experiments as directed, collecting data, plotting graphs, and evaluating limitations.
**An investigation into the vertical oscillations of a pivoted wooden rod.**
In this experiment, you will investigate how the period of oscillation of a wooden rod depends on the position of its supporting spring.
**Apparatus:** - Retort stand, boss, and clamp - G-clamp or heavy weight to secure the stand - A second retort stand, boss, and clamp - A wooden rod (e.g., a half-metre rule) with a small hole drilled near the 0.0 cm mark to act as a pivot - A nail or pin to act as a pivot axis through the hole, clamped securely - A spring (spring constant approximately \( 25 \text{ N m}^{-1} \)) - A 100 g mass, taped securely to the free end of the rod (at the 50.0 cm mark) - A stopwatch - A metre rule
**Procedure:** 1. Set up the apparatus. Clamp one end of the wooden rod using the pivot pin so that the rod can rotate freely in a vertical plane. Connect the spring between the second clamp and a loop of thread situated at a distance \( d \) from the pivot. Adjust the height of the second clamp so that the wooden rod is horizontal. Ensure the 100 g mass is firmly taped to the end of the rod.
2. Measure and record the distance \( d \) from the pivot to the spring attachment point. The value of \( d \) should initially be approximately 15.0 cm.
3. Displace the free end of the rod vertically downwards by a small distance and release it so that it performs vertical oscillations. Measure and record the time \( t \) for 10 complete oscillations. Repeat this measurement to obtain a second reading for \( t \), and calculate the mean time \( t_{\text{mean}} \). Determine the period \( T \) of the oscillations.
4. Change \( d \) by moving the spring attachment loop along the rod. Adjust the height of the second clamp to keep the rod horizontal. Repeat step 3 for five further values of \( d \) in the range \( 15.0 \text{ cm} \le d \le 45.0 \text{ cm} \).
5. Record all six sets of values for \( d \), \( t_1 \), \( t_2 \), \( t_{\text{mean}} \), and \( T \) in a table. Include columns for \( d^2 \) and \( T^{-2} \).
6. Plot a graph of \( T^{-2} \) on the y-axis against \( d^2 \) on the x-axis. Draw the straight line of best fit.
7. Determine the gradient and y-intercept of this line.
8. The quantities \( T \) and \( d \) are related by the equation: \( \frac{1}{T^2} = p d^2 + q \) where \( p \) and \( q \) are constants. Use your answers from part 7 to determine the values of \( p \) and \( q \). Include appropriate units for both constants.
9. The spring constant \( k \) of the spring is related to the constant \( p \) by the formula: \( p = \frac{k}{4 \pi^2 m} \) where \( m = 0.100 \text{ kg} \) is the mass attached to the end of the rod. Calculate the value of \( k \), giving its unit.
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### Graph and Analysis - Plotting \( T^{-2} \) against \( d^2 \) yields a straight line with a positive gradient and negligible y-intercept. - Gradient \( p = \frac{7.19 - 0.796}{0.2025 - 0.0225} = \frac{6.394}{0.180} \approx 35.5 \text{ s}^{-2} \text{ m}^{-2} \) (or \( 0.00355 \text{ s}^{-2} \text{ cm}^{-2} \)). - y-intercept \( q \approx 0.0 \text{ s}^{-2} \).
### Calculation of spring constant \( k \) Using \( p = \frac{k}{4 \pi^2 m} \): \( k = 4 \pi^2 m p = 4 \pi^2 \times 0.100 \times 35.5 \approx 140 \text{ N m}^{-1} \).
Marking scheme
**Data Collection (5 marks)** - [1] Six or more sets of readings of \( d \) and \( t \) taken with different values of \( d \). - [1] Range of \( d \): \( d_{\text{max}} - d_{\text{min}} \ge 25.0 \text{ cm} \). - [1] Precision of raw \( d \) values: all values recorded to the nearest millimetre. - [1] Precision of raw times \( t_1, t_2 \): recorded to 0.01 s. - [1] Quality: points on graph are close to the straight line.
**Table of Results (4 marks)** - [1] Column headings: must contain name of quantity and unit separated by a solidus (e.g., \( d^2 / \text{cm}^2 \)). - [1] Consistency: raw values of \( t_1 \) and \( t_2 \) recorded to the same decimal place. - [1] Significant figures: calculated values of \( d^2 \) have same or one more significant figure than raw \( d \). - [1] Arithmetic: \( T^{-2} \) calculated correctly.
**Graph (4 marks)** - [1] Axes: scales must be linear, with no awkward factors. Points must occupy at least half the grid in both directions. - [1] Plotting: points plotted correctly to within half a small square. - [1] Best-fit line: balanced about the points, single thin line drawn with a ruler. - [1] Quality of data: no anomalous points, maximum deviation from the line is less than 5% of the range.
**Gradient and Intercept (2 marks)** - [1] Gradient: calculated using a triangle with hypotenuse length at least half the length of the line. - [1] Intercept: read directly from the y-axis if \( x=0 \), or calculated using \( y = m x + c \) with a point from the line.
**Constants (5 marks)** - [1] Value of \( p \) set equal to gradient. - [1] Value of \( q \) set equal to intercept. - [1] Correct units: \( p \) in \( \text{s}^{-2}\text{m}^{-2} \) (or \( \text{s}^{-2}\text{cm}^{-2} \)) and \( q \) in \( \text{s}^{-2} \). - [1] Correct calculation of \( k = 4 \pi^2 m p \) with correct units. - [1] Value of \( k \) in the range \( 15 \text{ to } 150 \text{ N m}^{-1} \) given to 2 or 3 significant figures.
**An investigation into the deflection of a loaded wooden rule supported by a spring.**
In this experiment, you will investigate how the deflection of a horizontal wooden rule depends on the position of a load suspended from it.
**Apparatus:** - Metre rule (to be used as the cantilever) - Stand, boss, and clamp to clamp one end of the metre rule horizontally (at the 0.0 cm mark) - A second stand, boss, and clamp to support the top of a spring - A spring (spring constant approximately \( 10 \text{ to } 15 \text{ N m}^{-1} \)) - A 100 g mass hanger or mass with a hook - A half-metre rule (for vertical height measurements) - A set-square to ensure vertical measurements
**Procedure:** 1. Set up the apparatus. Clamp the metre rule horizontally at its 0.0 cm mark. Attach the spring vertically between the second clamp and a loop of thread placed at the 40.0 cm mark of the metre rule. Adjust the height of the second clamp so that the rule is horizontal.
2. (a) Measure and record the unstretched length \( L_0 \) of the spring (the length of the coiled section) to the nearest millimetre.
(b) Measure and record the vertical height \( h_0 \) of the 50.0 cm mark of the horizontal metre rule above the surface of the bench.
3. (c) Suspend a 100 g mass from the metre rule at a distance \( x = 45.0 \text{ cm} \) from the clamped end (the pivot). Measure and record the new vertical height \( h \) of the 50.0 cm mark of the rule above the bench. Calculate the vertical deflection \( z = h_0 - h \).
4. (d) Estimate the percentage uncertainty in your value of \( z \). Show your working.
5. (e) Move the 100 g mass to a position \( x = 25.0 \text{ cm} \) from the clamped end. Measure and record the new height \( h \) of the 50.0 cm mark above the bench. Calculate the new vertical deflection \( z \).
6. (f) It is suggested that the deflection \( z \) and position \( x \) are related by the equation: \( z = C x \) where \( C \) is a constant. (i) Calculate two values of \( C \) using your measurements. (ii) Explain whether your results support the suggested relationship. State a clear criterion for your decision and show your calculations. (iii) The spring constant \( k \) of the spring can be estimated from the relationship: \( k = \frac{m g L}{C d^2} \) where \( m = 0.100 \text{ kg} \), \( g = 9.81 \text{ m s}^{-2} \), \( L = 0.500 \text{ m} \), and \( d = 0.400 \text{ m} \). Using your second value of \( C \) (with \( x \) converted to metres), calculate the spring constant \( k \). Include its unit.
7. (g) Describe the limitations and sources of error in this experiment. Suggest improvements that could be made to improve the accuracy. You should list four distinct limitations and four corresponding improvements.
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Worked solution
### Representative Measurements and Calculations - Unstretched spring length: \( L_0 = 3.2 \text{ cm} \) - Unloaded height of 50.0 cm mark: \( h_0 = 25.4 \text{ cm} \)
### Comparison and Evaluation - Percentage difference: \( \frac{|C_1 - C_2|}{(C_1 + C_2)/2} \times 100\% = \frac{|0.256 - 0.252|}{0.254} \times 100\% \approx 1.6\% \). - Criterion: Since the percentage difference is less than 10%, the suggested relationship is supported.
### Estimation of Spring Constant \( k \) Using \( C_2 = 0.252 \) (converting unit: \( C_2 = \frac{0.063 \text{ m}}{0.250 \text{ m}} = 0.252 \)): \( k = \frac{m g L}{C_2 d^2} = \frac{0.100 \times 9.81 \times 0.500}{0.252 \times 0.400^2} = \frac{0.4905}{0.04032} \approx 12.2 \text{ N m}^{-1} \).
Marking scheme
**Data Collection and Calculations (9 marks)** - [1] Measurement of \( L_0 \) to the nearest millimetre with correct unit. - [1] Measurement of \( h_0 \) with unit, recorded to nearest millimetre. - [1] Measurement of \( h \) for first setup with correct calculation of \( z \). - [1] Percentage uncertainty in \( z \): showing correct absolute uncertainty (usually 0.2 cm) and correct percentage calculation. - [2] Measurement of second height \( h \) and calculation of second \( z \) with correct trend (smaller \( x \) gives smaller \( z \)). - [1] Calculation of two values of \( C \) with correct units (or dimensionless if ratio of same units used). - [2] Comparison and Conclusion: correct calculation of percentage difference between \( C \) values, comparing with a stated limit (e.g. 10%) and stating whether relationship is supported.
**Estimation of Spring Constant (3 marks)** - [1] Correct equation substitution using \( C_2 \) and converted metre units. - [1] Calculated value of \( k \) in range \( 5.0 \text{ to } 25.0 \text{ N m}^{-1} \). - [1] Correct unit \( \text{N m}^{-1} \) (or equivalent) and 2 or 3 significant figures.
**Limitations and Improvements (8 marks - 4 pairs, 1 mark per limit, 1 mark per improvement)** - [1] Limit 1: Difficult to measure \( h \) vertically due to tilt / ruler twisting. - [1] Improvement 1: Clamp a guide or use a plumb line to ensure ruler remains vertically aligned. - [1] Limit 2: Parallax error during height reading with ruler. - [1] Improvement 2: Attach a pointer to the horizontal rule and read against a closely placed vertical scale. - [1] Limit 3: Friction at the clamp pivot restricts free vertical deflection. - [1] Improvement 3: Use a low-friction hinge or pin-joint at the pivot end. - [1] Limit 4: Difficulty ensuring the rule is perfectly horizontal when unloaded. - [1] Improvement 4: Use a spirit level resting on top of the rule.
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