An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V3) Cambridge International A Level Physics (9702) paper. Not affiliated with or reproduced from Cambridge.
Paper 13 (Multiple Choice)
Answer all forty questions on the separate multiple choice answer sheet. Each correct response scores one mark.
40 Question · 40 marks
Question 1 · multiple_choice
1 marks
A stone is thrown vertically upwards from the top of a cliff of height \(h\) with an initial speed of \(15\text{ m s}^{-1}\). The stone hits the ground at the base of the cliff after a total time of \(4.0\text{ s}\). Air resistance is negligible.
What is the height \(h\) of the cliff?
A.18 m
B.60 m
C.78 m
D.140 m
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Worked solution
Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with upwards as the positive direction: - Initial velocity \(u = +15\text{ m s}^{-1}\) - Acceleration due to gravity \(a = -9.81\text{ m s}^{-2}\) - Total time taken \(t = 4.0\text{ s}\)
The displacement \(s\) is \(-18.5\text{ m}\), which means the stone ends up \(18.5\text{ m}\) below its starting level. Therefore, the height \(h\) of the cliff is \(18\text{ m}\) (to two significant figures).
Marking scheme
A is the correct option. 1 mark for calculating the negative displacement and identifying the height as 18 m.
Question 2 · multiple_choice
1 marks
A helicopter of mass \(1200\text{ kg}\) hovers in a fixed position. The helicopter's rotors push a column of air vertically downwards at a constant speed of \(25\text{ m s}^{-1}\). Air resistance other than that causing lift is negligible.
What is the mass of air pushed downwards per second by the rotors?
A.48 kg s\(^{-1}\)
B.120 kg s\(^{-1}\)
C.470 kg s\(^{-1}\)
D.1.2 \times 10^{4} kg s\(^{-1}\)
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Worked solution
For the helicopter to hover, the upward force \(F\) from the air must equal its weight: \(F = mg = 1200\text{ kg} \times 9.81\text{ m s}^{-2} = 11772\text{ N}\)
By Newton's third law, the force exerted by the rotors on the air is also \(11772\text{ N}\). This force equals the rate of change of momentum of the air: \(F = \frac{\Delta p}{\Delta t} = \frac{\Delta (mv)}{\Delta t} = v \frac{\Delta m}{\Delta t}\)
where \(\frac{\Delta m}{\Delta t}\) is the mass of air pushed downwards per second and \(v = 25\text{ m s}^{-1}\).
C is the correct option. 1 mark for equating the upward lift force to the weight and applying the momentum formula to calculate mass per second.
Question 3 · multiple_choice
1 marks
A uniform wire of resistance \(R\) and length \(L\) has a uniform cross-sectional area. The wire is stretched so that its length increases to \(1.20 L\) while its volume and resistivity remain constant.
What is the resistance of the stretched wire?
A.0.83 R
B.1.20 R
C.1.44 R
D.1.73 R
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Worked solution
The volume \(V\) of the wire is \(V = A L\). Since volume is constant, if length increases to \(1.20 L\), the cross-sectional area must decrease to: \(A' = \frac{A}{1.20}\)
Resistance is given by \(R = \frac{\rho L}{A}\). The new resistance \(R'\) is: \(R' = \frac{\rho (1.20 L)}{A / 1.20} = 1.20^2 \times \frac{\rho L}{A} = 1.44 R\).
Marking scheme
C is the correct option. 1 mark for calculating the area reduction factor and using the formula to find the final resistance of 1.44 R.
Question 4 · multiple_choice
1 marks
A metal wire of length \(L\), diameter \(d\), and resistivity \(\rho\) is connected across a power supply of constant potential difference \(V\). The temperature of the wire is kept constant.
The wire is now replaced with another wire made of the same metal, but with length \(2L\) and diameter \(0.5d\).
What is the ratio \(\frac{\text{power dissipated in second wire}}{\text{power dissipated in first wire}}\)?
A.\frac{1}{8}
B.\frac{1}{4}
C.2
D.8
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Worked solution
The resistance of a wire is given by \(R = \frac{\rho L}{A} = \frac{4 \rho L}{\pi d^2}\). Thus, \(R \propto \frac{L}{d^2}\).
For the second wire: \(R_2 \propto \frac{2L}{(0.5d)^2} = \frac{2}{0.25} \frac{L}{d^2} = 8 R_1\).
The electrical power dissipated is \(P = \frac{V^2}{R}\). Since the potential difference \(V\) is constant: \(P \propto \frac{1}{R}\)
Therefore, the ratio of power is: \(\frac{P_2}{P_1} = \frac{R_1}{R_2} = \frac{1}{8}\).
Marking scheme
A is the correct option. 1 mark for determining that resistance increases by a factor of 8 and power decreases to 1/8 of the initial value.
Question 5 · multiple_choice
1 marks
A steel wire of initial length \(L\) and cross-sectional area \(A\) is stretched by a force \(F\), within its elastic limit, producing an extension \(x\).
A copper wire of the same initial length \(L\) but with twice the cross-sectional area (\(2A\)) is stretched by the same force \(F\), also within its elastic limit.
The Young modulus of steel is \(2.0 \times 10^{11}\text{ Pa}\) and the Young modulus of copper is \(1.1 \times 10^{11}\text{ Pa}\).
What is the extension of the copper wire?
A.0.28 x
B.0.55 x
C.0.91 x
D.1.8 x
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Worked solution
The Young modulus is \(E = \frac{F L}{A \Delta L}\), so the extension is \(\Delta L = \frac{F L}{A E}\).
Since \(F\) and \(L\) are the same for both wires, the extension is inversely proportional to the product \(A E\): \(\Delta L \propto \frac{1}{A E}\)
Let \(x\) be the extension of the steel wire. The extension of the copper wire \(x_c\) is: \(\frac{x_c}{x} = \frac{A_{\text{steel}} E_{\text{steel}}}{A_{\text{copper}} E_{\text{copper}}} = \frac{A \times (2.0 \times 10^{11})}{2A \times (1.1 \times 10^{11})} = \frac{2.0}{2.2} \approx 0.91\)
Thus, \(x_c = 0.91 x\).
Marking scheme
C is the correct option. 1 mark for formulating the relation between extension, area and Young modulus, and calculating the numerical ratio.
Question 6 · multiple_choice
1 marks
A uniform beam of length \(3.0\text{ m}\) and weight \(120\text{ N}\) is pivoted at one end. It is held horizontally by a cable attached to the beam at a distance of \(2.0\text{ m}\) from the pivot. The cable makes an angle of \(30^\circ\) with the horizontal beam.
What is the tension in the cable?
A.90 N
B.104 N
C.180 N
D.360 N
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Worked solution
Taking moments about the pivot in equilibrium: - The weight of the uniform beam acts at its center of gravity, which is at the midpoint of the beam: \(d_w = 1.5\text{ m}\). - Clockwise moment due to weight = \(120\text{ N} \times 1.5\text{ m} = 180\text{ N m}\). - The vertical component of the tension is \(T \sin(30^\circ)\), acting at a distance of \(2.0\text{ m}\) from the pivot. - Anticlockwise moment due to tension = \(T \sin(30^\circ) \times 2.0\text{ m} = T \times 0.5 \times 2.0 = T \times 1.0\text{ m}\).
C is the correct option. 1 mark for calculating the correct clockwise moment and equating it to the perpendicular component of tension multiplied by distance.
Question 7 · multiple_choice
1 marks
A \(\Sigma^0\) baryon has a quark composition of \(uds\). It decays into a \(\Lambda^0\) baryon (quark composition \(uds\)) and a photon:
\(\Sigma^0 \rightarrow \Lambda^0 + \gamma\)
In a different decay, a \(\Sigma^-\) baryon (quark composition \(dds\)) decays into a neutron (quark composition \(udd\)) and a \(\pi^-\) meson (quark composition \(d\bar{u}\)):
\(\Sigma^- \rightarrow n + \pi^-\)
Which row correctly identifies the interaction responsible for each of these two decays?
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Worked solution
- In the decay \(\Sigma^0 \rightarrow \Lambda^0 + \gamma\), a photon (\(\gamma\)) is emitted and there is no change in the flavor of the quarks. Therefore, this decay is mediated by the electromagnetic interaction. - In the decay \(\Sigma^- \rightarrow n + \pi^-\), a strange quark (\(s\)) in the \(\Sigma^-\) baryon is transformed into an up quark (\(u\)) (implied by the final state composition). Any decay involving a change in quark flavor is mediated by the weak interaction.
Hence, the correct row is A.
Marking scheme
A is the correct option. 1 mark for correctly identifying that photon emission corresponds to the electromagnetic interaction and quark flavor change corresponds to the weak interaction.
Question 8 · multiple_choice
1 marks
Two blocks, P of mass \(3.0\text{ kg}\) and Q of mass \(1.0\text{ kg}\), are placed in contact on a smooth horizontal surface. A constant horizontal force of \(12\text{ N}\) is applied to block P, pushing it against block Q.
What is the magnitude of the force exerted by block P on block Q?
A.3.0 N
B.4.0 N
C.9.0 N
D.12 N
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Worked solution
First, find the common acceleration \(a\) of the two blocks: \(F_{\text{total}} = (m_P + m_Q) a\) \(12\text{ N} = (3.0\text{ kg} + 1.0\text{ kg}) a \implies a = \frac{12}{4.0} = 3.0\text{ m s}^{-2}\)
Next, apply Newton's second law to block Q alone to find the contact force \(F_{PQ}\) exerted by P on Q: \(F_{PQ} = m_Q a = 1.0\text{ kg} \times 3.0\text{ m s}^{-2} = 3.0\text{ N}\).
Marking scheme
A is the correct option. 1 mark for finding the acceleration of the combined system and using it to find the horizontal force acting on block Q.
Question 9 · multiple-choice
1 marks
A stone is thrown vertically upwards from the edge of a cliff with an initial speed \(u\). At the same instant, a second stone is released from rest and falls vertically downwards from the same cliff edge. Both stones are still in flight.
What is the vertical distance between the two stones after a time \(t\) has elapsed?
A.\(u t\)
B.\(u t - g t^2\)
C.\(u t - \frac{1}{2} g t^2\)
D.\(u t + \frac{1}{2} g t^2\)
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Worked solution
Let upwards be the positive vertical direction.
For the first stone thrown upwards with initial speed \(u\): Its displacement from the cliff edge after time \(t\) is: \(s_1 = u t - \frac{1}{2} g t^2\)
For the second stone dropped from rest: Its displacement from the cliff edge after time \(t\) is: \(s_2 = -\frac{1}{2} g t^2\)
The distance \(d\) between the two stones is the magnitude of the difference between their displacements: \(d = |s_1 - s_2| = \left| \left( u t - \frac{1}{2} g t^2 \right) - \left( -\frac{1}{2} g t^2 \right) \right| = u t\)
Marking scheme
[1 mark] - Correctly identifies the equations of motion for both stones and subtracts them to obtain \(u t\). Award for selecting option A.
Question 10 · multiple-choice
1 marks
A glider of mass \(m\) is moving with speed \(3v\) to the right on a frictionless air track. It collides head-on with a glider of mass \(2m\) moving with speed \(v\) to the left. The two gliders stick together on impact.
What fraction of the initial total kinetic energy is lost during the collision?
A.\(\frac{1}{11}\)
B.\(\frac{2}{11}\)
C.\(\frac{30}{33}\)
D.\(\frac{32}{33}\)
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Worked solution
Let the direction to the right be positive.
1. Use conservation of linear momentum to find the final velocity \(v_f\): \(p_{\text{initial}} = m(3v) + 2m(-v) = 3mv - 2mv = mv\) \(p_{\text{final}} = (m + 2m)v_f = 3m v_f\) Since \(p_{\text{initial}} = p_{\text{final}}\): \(mv = 3m v_f \implies v_f = \frac{v}{3}\)
2. Calculate the initial total kinetic energy \(E_{ki}\): \(E_{ki} = \frac{1}{2}m(3v)^2 + \frac{1}{2}(2m)(-v)^2 = \frac{9}{2}mv^2 + mv^2 = \frac{11}{2}mv^2\)
3. Calculate the final total kinetic energy \(E_{kf}\): \(E_{kf} = \frac{1}{2}(3m)\left(\frac{v}{3}\right)^2 = \frac{3}{2}m \frac{v^2}{9} = \frac{1}{6}mv^2\)
4. Calculate the loss in kinetic energy \(\Delta E_k\): \(\Delta E_k = E_{ki} - E_{kf} = \frac{11}{2}mv^2 - \frac{1}{6}mv^2 = \left(\frac{33}{6} - \frac{1}{6}\right)mv^2 = \frac{32}{6}mv^2 = \frac{16}{3}mv^2\)
5. Find the fraction of initial kinetic energy lost: \(\text{Fraction} = \frac{\Delta E_k}{E_{ki}} = \frac{\frac{16}{3}mv^2}{\frac{11}{2}mv^2} = \frac{16}{3} \times \frac{2}{11} = \frac{32}{33}\)
Marking scheme
[1 mark] - Applies conservation of momentum to find final velocity, calculates initial and final kinetic energies, and computes the fraction lost. Award for selecting option D.
Question 11 · multiple-choice
1 marks
A uniform plank of length \(4.0\text{ m}\) and weight \(240\text{ N}\) is supported horizontally by two pivots. One pivot is at the left-hand end and the other pivot is at a distance of \(1.0\text{ m}\) from the right-hand end. A person of weight \(600\text{ N}\) stands on the plank at a distance of \(1.5\text{ m}\) from the left-hand end.
What is the upward force exerted by the right-hand pivot on the plank?
A.380 N
B.460 N
C.480 N
D.620 N
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Worked solution
Let the plank lie along the x-axis from \(x = 0\) to \(x = 4.0\text{ m}\). - The left pivot is at \(x = 0\text{ m}\). - The right pivot is at \(x = 4.0 - 1.0 = 3.0\text{ m}\). - The weight of the uniform plank (\(240\text{ N}\)) acts at its center of gravity, \(x = 2.0\text{ m}\). - The person's weight (\(600\text{ N}\)) acts at \(x = 1.5\text{ m}\).
Taking moments about the left pivot (\(x = 0\)): \(\text{Sum of clockwise moments} = \text{Sum of counter-clockwise moments}\) \((600\text{ N} \times 1.5\text{ m}) + (240\text{ N} \times 2.0\text{ m}) = F_R \times 3.0\text{ m}\) \(900 + 480 = 3.0 F_R\) \(1380 = 3.0 F_R\) \(F_R = 460\text{ N}\)
Marking scheme
[1 mark] - Correctly sets up the principle of moments about the left pivot to solve for the right pivot force. Award for selecting option B.
Question 12 · multiple-choice
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 suspended vertically from a ceiling and support identical loads at their lower ends. Both wires deform elastically.
What is the ratio \(\frac{\text{strain energy stored in wire X}}{\text{strain energy stored in wire Y}}\)?
A.0.5
B.1
C.2
D.4
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Worked solution
The strain energy \(E_s\) stored in a wire under load \(F\) is given by: \(E_s = \frac{1}{2} F \Delta L\)
From Young's modulus \(E_Y = \frac{F L}{A \Delta L}\), the extension is \(\Delta L = \frac{F L}{A E_Y}\). Substituting this into the energy formula: \(E_s = \frac{F^2 L}{2 A E_Y}\)
Since \(F\) and \(E_Y\) are the same for both wires, the strain energy is proportional to: \(E_s \propto \frac{L}{A} \propto \frac{L}{d^2}\)
For wire X: \(E_X \propto \frac{L}{d^2}\)
For wire Y: \(E_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\)
Therefore, the ratio is: \(\frac{E_X}{E_Y} = \frac{L/d^2}{L/(2d^2)} = 2\)
Marking scheme
[1 mark] - Derives the relationship between strain energy, length, and diameter for a constant load, and calculates the correct ratio. Award for selecting option C.
Question 13 · multiple-choice
1 marks
A uniform copper wire of length \(L\) and cross-sectional area \(A\) has a resistance \(R\).
Another wire made of the same copper material consists of two segments connected end-to-end: a first segment of length \(0.4L\) and cross-sectional area \(A\), and a second segment of length \(0.6L\) and cross-sectional area \(2A\).
What is the total resistance of this second wire in terms of \(R\)?
A.0.5 R
B.0.7 R
C.1.0 R
D.1.4 R
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Worked solution
The resistance of a wire is given by \(R = \frac{\rho L}{A}\).
The two segments of the second wire are in series, so the total resistance is the sum of their individual resistances: \(R_{\text{total}} = R_1 + R_2\)
For the first segment: \(R_1 = \frac{\rho (0.4L)}{A} = 0.4 \left(\frac{\rho L}{A}\right) = 0.4 R\)
For the second segment: \(R_2 = \frac{\rho (0.6L)}{2A} = 0.3 \left(\frac{\rho L}{A}\right) = 0.3 R\)
Therefore, the total resistance is: \(R_{\text{total}} = 0.4 R + 0.3 R = 0.7 R\)
Marking scheme
[1 mark] - Correctly uses the formula for resistivity to express the resistance of each segment in series and finds their sum. Award for selecting option B.
Question 14 · multiple-choice
1 marks
A sodium-22 nucleus (\(^{22}_{11}\text{Na}\)) undergoes \(\beta^+\) decay to form a neon-22 nucleus (\(^{22}_{10}\text{Ne}\)).
Which change in quark flavor occurs in a nucleon during this decay?
A.A down quark changes to an up quark.
B.An up quark changes to a down quark.
C.A down quark changes to a strange quark.
D.A strange quark changes to an up quark.
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Worked solution
During \(\beta^+\) decay, a proton decays into a neutron, a positron, and an electron neutrino: \(p \to n + e^+ + \nu_e\)
The quark composition of a proton is \(uud\) and the quark composition of a neutron is \(udd\).
Therefore, one up (\(u\)) quark in the proton changes into a down (\(d\)) quark to form the neutron: \(u \to d\)
Marking scheme
[1 mark] - Identifies \(\beta^+\) decay as a proton-to-neutron transition and correctly determines the corresponding quark change. Award for selecting option B.
Question 15 · multiple-choice
1 marks
A tennis ball of mass \(0.060\text{ kg}\) travels horizontally with a speed of \(25\text{ m s}^{-1}\) when it hits a wall. It rebounds horizontally in the opposite direction with a speed of \(15\text{ m s}^{-1}\). The collision lasts for a time interval of \(8.0\text{ ms}\).
What is the average force exerted by the wall on the ball during the collision?
A.75 N
B.110 N
C.190 N
D.300 N
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Worked solution
Let the initial direction of motion towards the wall be positive. - Initial velocity, \(u = +25\text{ m s}^{-1}\) - Final velocity, \(v = -15\text{ m s}^{-1}\) (rebound direction)
The change in momentum \(\Delta p\) of the ball is: \(\Delta p = m(v - u) = 0.060\text{ kg} \times (-15 - 25)\text{ m s}^{-1} = 0.060 \times (-40) = -2.4\text{ N s}\)
The magnitude of the change in momentum is \(2.4\text{ N s}\).
Using Newton's second law, the average force \(F\) is: \(F = \frac{\Delta p}{\Delta t} = \frac{2.4\text{ N s}}{8.0 \times 10^{-3}\text{ s}} = 300\text{ N}\)
Marking scheme
[1 mark] - Correctly calculates the change in momentum taking into account the directions of the velocities, and divides by the collision time. Award for selecting option D.
Question 16 · multiple-choice
1 marks
A copper wire of cross-sectional area \(A\) carries a current \(I\). The average drift velocity of the free electrons in this wire is \(v\).
Another copper wire has a cross-sectional area of \(\frac{1}{2}A\) and carries a current of \(2I\).
What is the average drift velocity of the free electrons in the second wire?
A.\(\frac{1}{2}v\)
B.\(v\)
C.\(2v\)
D.\(4v\)
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Worked solution
The relationship between electric current \(I\) and drift velocity \(v\) is given by: \(I = n A v e\) where: - \(n\) is the number density of free electrons (which is constant for the same material, copper), - \(A\) is the cross-sectional area, - \(e\) is the elementary charge.
Rearranging for drift velocity: \(v = \frac{I}{n A e} \implies v \propto \frac{I}{A}\)
For the second wire with current \(I_2 = 2I\) and area \(A_2 = \frac{1}{2}A\): \(v_2 \propto \frac{2I}{\frac{1}{2}A} = 4 \left(\frac{I}{A}\right)\)
Therefore, the drift velocity in the second wire is \(4v\).
Marking scheme
[1 mark] - Identifies the relationship \(I = nAve\) and uses the ratios of current and area to find the new drift velocity. Award for selecting option D.
Question 17 · multiple-choice
1 marks
A spacecraft of mass \(M\) is travelling in deep space at speed \(v\). It ejects a booster stage of mass \(0.20M\) in the opposite direction to its motion. Immediately after ejection, the speed of the booster stage relative to the remaining part of the spacecraft is \(2.0v\). What is the speed of the remaining part of the spacecraft?
A.\(1.2v\)
B.\(1.4v\)
C.\(1.5v\)
D.\(1.6v\)
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Worked solution
Let the velocity of the remaining part of the spacecraft of mass \(0.80M\) be \(V_f\) in the original direction of motion. The booster stage of mass \(0.20M\) is ejected in the opposite direction, so its velocity is \(V_e\). Since the relative speed between them is \(2.0v\), we have: \(V_f - V_e = 2.0v\) which gives \(V_e = V_f - 2.0v\). By the conservation of linear momentum: \(M v = 0.80M V_f + 0.20M V_e\). Substituting \(V_e\) into the equation: \(M v = 0.80M V_f + 0.20M (V_f - 2.0v)\). Dividing both sides by \(M\): \(v = 0.80 V_f + 0.20 V_f - 0.40v\), which simplifies to \(1.4v = V_f\). Therefore, the speed of the remaining part is \(1.4v\).
Marking scheme
1 mark for the correct answer B.
Question 18 · multiple-choice
1 marks
A uniform metal wire of resistance \(R\), length \(L\), and cross-sectional area \(A\) is stretched so that its length increases by \(10\%\). The volume of the wire remains constant during stretching, and its resistivity is unaffected. What is the percentage increase in the electrical resistance of the wire?
A.\(10\%\)
B.\(11\%\)
C.\(21\%\)
D.\(23\%\)
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Worked solution
The volume \(V = L A\) is constant. When length increases to \(L' = 1.10L\), the new cross-sectional area \(A'\) is: \(A' = V / L' = (L A) / (1.10 L) = A / 1.10\). Using the resistivity formula \(R = \rho L / A\), the new resistance \(R'\) is: \(R' = \rho (1.10 L) / (A / 1.10) = 1.21 (\rho L / A) = 1.21R\). The percentage increase is: \(((R' - R) / R) \times 100\% = (1.21 - 1) \times 100\% = 21\%\).
Marking scheme
1 mark for the correct answer C.
Question 19 · multiple-choice
1 marks
A metal wire of length \(L\) and diameter \(d\) is suspended vertically. When a load \(W\) is hung from the lower end, the extension of the wire is \(x\). Another wire made of the same metal has length \(2L\) and diameter \(2d\). What is the extension when a load of \(2W\) is hung from this second wire, assuming both wires obey Hooke's law?
A.\(\frac{1}{2}x\)
B.\(x\)
C.\(2x\)
D.\(4x\)
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Worked solution
The Young modulus \(E\) of the metal is given by: \(E = \text{Stress} / \text{Strain} = (F / A) / (x / L) = F L / (A x)\). Since \(A = \pi d^2 / 4\), the extension is: \(x = 4 F L / (\pi d^2 E)\). For the first wire, this is: \(x = 4 W L / (\pi d^2 E)\). For the second wire, the load is \(2W\), the length is \(2L\), and the diameter is \(2d\): \(x_2 = 4 (2W) (2L) / (\pi (2d)^2 E) = 16 W L / (4 \pi d^2 E) = 4 W L / (\pi d^2 E) = x\). Therefore, the extension of the second wire is also \(x\).
Marking scheme
1 mark for the correct answer B.
Question 20 · multiple-choice
1 marks
A ball is thrown vertically upwards from the edge of a cliff of height \(h\). The ball reaches a maximum height and then falls to the ground at the base of the cliff. The total time of flight is \(T\). If the ball is thrown with an initial speed \(u\), which equation correctly relates \(h\), \(u\), \(T\), and the acceleration of free fall \(g\)?
A.\(h = uT + \frac{1}{2}gT^2\)
B.\(h = uT - \frac{1}{2}gT^2\)
C.\(h = \frac{1}{2}gT^2 - uT\)
D.\(h = -uT - \frac{1}{2}gT^2\)
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Worked solution
Taking the upward direction as positive: Displacement \(s = -h\) (since the final position at the base of the cliff is a distance \(h\) below the starting point), initial velocity \(v_i = +u\), acceleration \(a = -g\), and time \(t = T\). Using the equation of motion \(s = v_i t + \frac{1}{2}a t^2\): \(-h = uT + \frac{1}{2}(-g)T^2\), which simplifies to \(-h = uT - \frac{1}{2}gT^2\). Multiplying both sides by \(-1\) gives: \(h = \frac{1}{2}gT^2 - uT\).
Marking scheme
1 mark for the correct answer C.
Question 21 · multiple-choice
1 marks
A uniform beam of length \(4.0\text{ m}\) and weight \(120\text{ N}\) is pivoted at a distance of \(1.0\text{ m}\) from its left end. A load \(W\) is placed at the extreme left end. To keep the beam horizontal, a vertical force of \(80\text{ N}\) is applied downwards at the extreme right end. What is the value of \(W\)?
A.\(160\text{ N}\)
B.\(200\text{ N}\)
C.\(360\text{ N}\)
D.\(440\text{ N}\)
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Worked solution
The beam is uniform, so its weight of \(120\text{ N}\) acts at its midpoint, which is \(2.0\text{ m}\) from either end. The pivot is \(1.0\text{ m}\) from the left end. The load \(W\) is at the extreme left, which is at a distance of \(1.0\text{ m}\) to the left of the pivot, producing an anticlockwise moment of \(W \times 1.0\text{ N m}\). The center of gravity is \(2.0\text{ m}\) from the left end, which is \(2.0 - 1.0 = 1.0\text{ m}\) to the right of the pivot, producing a clockwise moment of \(120 \times 1.0\text{ N m}\). The extreme right end is \(4.0\text{ m}\) from the left end, which is \(4.0 - 1.0 = 3.0\text{ m}\) to the right of the pivot. The downward force of \(80\text{ N}\) at this point produces a clockwise moment of \(80 \times 3.0\text{ N m}\). For rotational equilibrium, the sum of anticlockwise moments equals the sum of clockwise moments about the pivot: \(W \times 1.0 = (120 \times 1.0) + (80 \times 3.0)\), which gives \(W = 120 + 240 = 360\text{ N}\).
Marking scheme
1 mark for the correct answer C.
Question 22 · multiple-choice
1 marks
A baryon has a charge of \(+e\) and a strangeness of \(-1\). Which combination of quarks could make up this baryon?
A.\(\text{uds}\)
B.\(\text{uus}\)
C.\(\text{uss}\)
D.\(\text{uud}\)
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Worked solution
A baryon consists of three quarks. Since the strangeness is \(-1\), it must contain exactly one strange (\(s\)) quark (which has a strangeness of \(-1\) and a charge of \(-1/3\,e\)). Let the other two quarks be \(q_1\) and \(q_2\). The total charge of the baryon is \(+1\,e\), so we have: \(Q(q_1) + Q(q_2) - 1/3 = +1\) which gives \(Q(q_1) + Q(q_2) = +4/3\). Since up (\(u\)) quarks have a charge of \(+2/3\,e\), two up quarks will sum to \(+4/3\,e\). Therefore, the quark composition is \(uus\).
Marking scheme
1 mark for the correct answer B.
Question 23 · multiple-choice
1 marks
A battery of electromotive force (e.m.f.) \(12\text{ V}\) and negligible internal resistance is connected to three resistors of resistances \(3.0\ \Omega\), \(6.0\ \Omega\), and \(R\) in series. The power dissipated in the \(6.0\ \Omega\) resistor is \(6.0\text{ W}\). What is the resistance \(R\)?
A.\(1.5\ \Omega\)
B.\(3.0\ \Omega\)
C.\(4.5\ \Omega\)
D.\(6.0\ \Omega\)
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Worked solution
The power \(P\) dissipated in a resistor of resistance \(R_0\) is given by: \(P = I^2 R_0\). For the \(6.0\ \Omega\) resistor, \(6.0 = I^2 \times 6.0\), which gives \(I = 1.0\text{ A}\). Since the resistors are in series, the same current of \(1.0\text{ A}\) flows through all of them. The total resistance of the circuit is \(R_{\text{total}} = 3.0 + 6.0 + R = 9.0 + R\). Using Ohm's law for the whole circuit: \(I = E / R_{\text{total}}\), which gives \(1.0 = 12 / (9.0 + R)\). This simplifies to \(9.0 + R = 12\), so \(R = 3.0\ \Omega\).
Marking scheme
1 mark for the correct answer B.
Question 24 · multiple-choice
1 marks
A ball of mass \(0.15\text{ kg}\) is moving horizontally to the right at a speed of \(20\text{ m s}^{-1}\). It is struck by a bat, which sends it back to the left at a speed of \(30\text{ m s}^{-1}\). The bat is in contact with the ball for a time of \(0.050\text{ s}\). What is the average force exerted by the bat on the ball?
A.\(30\text{ N}\)
B.\(75\text{ N}\)
C.\(150\text{ N}\)
D.\(300\text{ N}\)
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Worked solution
Let the direction to the right be positive. The initial velocity is \(u = +20\text{ m s}^{-1}\) and the final velocity is \(v = -30\text{ m s}^{-1}\). The change in momentum \(\Delta p\) of the ball is: \(\Delta p = m(v - u) = 0.15 \times (-30 - 20) = 0.15 \times (-50) = -7.5\text{ kg m s}^{-1}\). The magnitude of the change in momentum is \(7.5\text{ N s}\). Using Newton's second law, the average force \(F\) is: \(F = \Delta p / \Delta t = 7.5 / 0.050 = 150\text{ N}\).
Marking scheme
1 mark for the correct answer C.
Question 25 · multiple-choice
1 marks
A block of mass \(3m\) slides on a smooth horizontal surface at speed \(u\). It collides head-on with a stationary block of mass \(m\). After the collision, the first block continues in its original direction with speed \(0.5u\).
What is the speed of the second block and is the collision elastic or inelastic?
A.speed is \(1.5u\); collision is elastic
B.speed is \(1.5u\); collision is inelastic
C.speed is \(2.5u\); collision is elastic
D.speed is \(2.5u\); collision is inelastic
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Worked solution
By conservation of linear momentum in the direction of motion:
\(3m u + m(0) = 3m(0.5u) + m v_2\)
\(3m u = 1.5m u + m v_2 \implies v_2 = 1.5u\)
To determine if the collision is elastic, we compare total kinetic energy before and after the collision:
Since \(E_{ki} = E_{kf}\), kinetic energy is conserved, so the collision is elastic.
Marking scheme
Award 1 mark for the correct option A. - Reject other options because they either have the wrong final velocity or incorrect classification of the collision type.
Question 26 · multiple-choice
1 marks
A sand-blasting gun expels sand of density \(\rho\) through a nozzle of cross-sectional area \(A\) at a constant speed \(v\). The sand strikes a vertical wall normally and loses all its horizontal velocity, falling vertically down.
What is the average force exerted on the wall by the sand?
A.\(\rho A v\)
B.\(\rho A v^2\)
C.\(\frac{1}{2} \rho A v^2\)
D.\(\rho^2 A v\)
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Worked solution
The volume of sand striking the wall per unit time is \(\frac{\Delta V}{\Delta t} = A v\).
The mass of sand striking the wall per unit time is \(\frac{\Delta m}{\Delta t} = \rho \frac{\Delta V}{\Delta t} = \rho A v\).
Since the sand loses all of its horizontal velocity upon impact (final velocity = 0), the magnitude of the change in velocity of the sand is \(\Delta v = v\).
By Newton's second law, the force is equal to the rate of change of momentum:
\(F = \frac{\Delta p}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right) \Delta v = (\rho A v) v = \rho A v^2\).
Marking scheme
Award 1 mark for the correct option B. - 1 mark for using rate of mass flow multiplied by velocity change to get force.
Question 27 · multiple-choice
1 marks
A uniform metal wire of resistance \(R\) is stretched such that its length increases by \(10\%\) with no change in its volume or density.
What is the resistance of the stretched wire in terms of \(R\)?
A.\(1.10 R\)
B.\(1.21 R\)
C.\(1.33 R\)
D.\(0.91 R\)
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Worked solution
Resistance is given by \(R = \rho \frac{L}{A}\).
Since the volume \(V = A L\) is constant, if the length increases to \(L' = 1.10 L\), the cross-sectional area must decrease to \(A' = \frac{A}{1.10}\).
Award 1 mark for the correct option B. - Note that a common mistake is to assume area remains constant, which leads to \(1.10 R\) (option A).
Question 28 · multiple-choice
1 marks
Two cylindrical copper wires, X and Y, are connected in series. Wire X has length \(L\) and diameter \(d\). Wire Y has length \(2L\) and diameter \(2d\).
What is the ratio \(\frac{\text{drift speed of free electrons in X}}{\text{drift speed of free electrons in Y}}\)?
A.\(0.5\)
B.\(1\)
C.\(2\)
D.\(4\)
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Worked solution
The relationship between electric current \(I\) and drift speed \(v\) is given by:
\(I = n A v e\)
Since wires X and Y are in series, they carry the same current \(I\). Both wires are made of copper, so the number density of charge carriers \(n\) is the same.
Thus, drift speed \(v\) is inversely proportional to the cross-sectional area \(A\):
Award 1 mark for the correct option D. - Reject option C if diameter ratio was not squared.
Question 29 · multiple-choice
1 marks
A metal wire of original length \(L\) and cross-sectional area \(A\) is suspended vertically from a rigid support. When a load \(F\) is applied to its lower end, the extension is \(x\). The stress in the wire does not exceed the limit of proportionality.
A second wire made of the same metal has original length \(2L\) and cross-sectional area \(2A\). What is the extension of this second wire when a load of \(2F\) is applied?
A.\(0.5x\)
B.\(x\)
C.\(2x\)
D.\(4x\)
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Award 1 mark for the correct option C. - 1 mark for setting up stress and strain ratios correctly or using Young modulus equation.
Question 30 · multiple-choice
1 marks
A ball is thrown vertically upwards from the edge of a cliff with an initial speed of \(15\text{ m s}^{-1}\). It rises to its maximum height and then falls to the ground at the base of the cliff. The total time of flight from the moment it is thrown to the moment it hits the ground is \(5.0\text{ s}\).
Assume air resistance is negligible. What is the height of the cliff?
A.\(11\text{ m}\)
B.\(48\text{ m}\)
C.\(75\text{ m}\)
D.\(120\text{ m}\)
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Worked solution
Define the upward direction as positive.
Initial velocity \(u = +15\text{ m s}^{-1}\)
Acceleration \(a = -9.81\text{ m s}^{-2}\)
Time of flight \(t = 5.0\text{ s}\)
Using the equation of motion for displacement \(s\):
Since displacement \(s\) is negative, the landing point is \(47.6\text{ m}\) below the starting point.
Thus, the height of the cliff is approximately \(48\text{ m}\) (to 2 significant figures).
Marking scheme
Award 1 mark for the correct option B. - Correct use of kinematic equations with consistent signs.
Question 31 · multiple-choice
1 marks
A uniform beam of length \(3.0\text{ m}\) and weight \(120\text{ N}\) is supported horizontally by a pivot at \(1.0\text{ m}\) from its left end, and a vertical cable attached to its right-hand end.
What is the tension in the cable?
A.\(20\text{ N}\)
B.\(30\text{ N}\)
C.\(60\text{ N}\)
D.\(90\text{ N}\)
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Worked solution
Since the beam is uniform, its center of gravity is at its geometric center, \(1.5\text{ m}\) from the left end.
Let's choose the pivot as the axis of rotation (located \(1.0\text{ m}\) from the left end): - The perpendicular distance from the pivot to the line of action of the weight is \(1.5\text{ m} - 1.0\text{ m} = 0.5\text{ m}\) (acting downwards, creating a clockwise moment). - The perpendicular distance from the pivot to the cable at the right end is \(3.0\text{ m} - 1.0\text{ m} = 2.0\text{ m}\) (acting upwards, creating an anticlockwise moment).
Taking moments about the pivot for rotational equilibrium:
\(\frac{2n}{3} - 1 + \frac{n}{3} = 1 \implies n - 1 = 1 \implies n = 2\)
Therefore, the baryon must have 2 up quarks and 1 strange quark, giving the combination \(uus\).
Marking scheme
Award 1 mark for the correct option B. - 1 mark for correctly matching quark charges to get a total of +1e.
Question 33 · multiple-choice
1 marks
A ball of mass \(0.15\text{ kg}\) travels horizontally at a speed of \(12\text{ m s}^{-1}\) towards a vertical wall. It hits the wall and rebounds horizontally with a speed of \(8.0\text{ m s}^{-1}\). The collision lasts for a total time of \(0.080\text{ s}\). The force \(F\) exerted by the wall on the ball varies with time \(t\) as a symmetric triangular pulse, rising from zero to a maximum value of \(F_{\text{max}}\) and then falling back to zero.
What is the value of \(F_{\text{max}}\)?
A.\(37.5\text{ N}\)
B.\(75\text{ N}\)
C.\(150\text{ N}\)
D.\(300\text{ N}\)
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Worked solution
First, calculate the change in momentum (impulse) of the ball. Let the direction towards the wall be positive. Initial velocity \(u = 12\text{ m s}^{-1}\) Final velocity \(v = -8.0\text{ m s}^{-1}\)
Change in momentum: \(\Delta p = m(v - u) = 0.15 \times (-8.0 - 12) = -3.0\text{ N s}\)
The magnitude of the impulse is \(3.0\text{ N s}\).
The impulse is represented by the area under the force-time graph. For a triangular pulse: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\) \(3.0 = \frac{1}{2} \times 0.080 \times F_{\text{max}}\) \(3.0 = 0.040 \times F_{\text{max}}\) \(F_{\text{max}} = \frac{3.0}{0.040} = 75\text{ N}\)
Marking scheme
Award 1 mark for the correct calculation of impulse (3.0 N s), equating it to the area of a triangle (0.5 * 0.080 * F_max), and correctly resolving to 75 N.
Question 34 · multiple-choice
1 marks
Two wires, X and Y, are made of the same material. Wire X has length \(L\) and diameter \(d\). Wire Y has length \(2L\) and diameter \(2d\).
The wires are connected in parallel across a power supply of constant potential difference.
What is the ratio \(\frac{\text{power dissipated in wire X}}{\text{power dissipated in wire Y}}\)?
A.\(0.25\)
B.\(0.50\)
C.\(1.0\)
D.\(2.0\)
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Worked solution
The resistance of a wire is given by \(R = \rho \frac{L}{A} = \rho \frac{4L}{\pi d^2}\).
Since the wires are connected in parallel, the potential difference \(V\) across both wires is the same. Power dissipated is \(P = \frac{V^2}{R}\).
Therefore, the ratio of power dissipated is: \(\frac{P_X}{P_Y} = \frac{V^2 / R_X}{V^2 / R_Y} = \frac{R_Y}{R_X} = \frac{0.5 R_X}{R_X} = 0.50\)
Marking scheme
Award 1 mark for calculating the correct resistance ratio R_Y/R_X = 0.50, identifying that the voltage is constant in parallel, and determining the correct ratio of power as 0.50.
Question 35 · multiple-choice
1 marks
A metal wire of length \(2.5\text{ m}\) and cross-sectional area \(1.2 \times 10^{-7}\text{ m}^2\) is suspended vertically. A load of \(60\text{ N}\) is suspended from the lower end of the wire. The Young modulus of the metal is \(2.0 \times 10^{11}\text{ Pa}\).
Assume that the wire obeys Hooke's law.
What is the elastic potential energy stored in the wire?
A.\(0.094\text{ J}\)
B.\(0.19\text{ J}\)
C.\(0.38\text{ J}\)
D.\(1.5\text{ J}\)
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Worked solution
First, calculate the extension \(x\) of the wire using the definition of Young modulus \(E\): \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\)
Next, calculate the elastic potential energy stored in the wire: \(E_p = \frac{1}{2} F x = \frac{1}{2} \times 60 \times 6.25 \times 10^{-3} = 0.1875\text{ J} \approx 0.19\text{ J}\)
Marking scheme
Award 1 mark for calculating the extension of 6.25 mm and applying the elastic strain energy formula to find 0.19 J.
Question 36 · multiple-choice
1 marks
An object is projected vertically upwards from the edge of a high cliff with an initial speed of \(15\text{ m s}^{-1}\). It rises to its maximum height and then falls past the cliff edge to the sea below. The total time of flight from projection to hitting the sea is \(6.0\text{ s}\).
What is the height of the cliff above the sea? (Ignore air resistance and use \(g = 9.81\text{ m s}^{-2}\).)
A.\(87\text{ m}\)
B.\(177\text{ m}\)
C.\(267\text{ m}\)
D.\(353\text{ m}\)
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Worked solution
Use the equation of motion with constant acceleration: \(s = u t + \frac{1}{2} a t^2\)
Define the upward direction as positive: - Initial velocity, \(u = +15\text{ m s}^{-1}\) - Acceleration, \(a = -g = -9.81\text{ m s}^{-2}\) - Time, \(t = 6.0\text{ s}\)
Substitute the values into the equation: \(s = (15 \times 6.0) + \frac{1}{2}(-9.81)(6.0)^2\) \(s = 90 - 4.905 \times 36\) \(s = 90 - 176.58 = -86.58\text{ m}\)
The displacement is \(-86.6\text{ m}\), which means the cliff edge is \(87\text{ m}\) (to 2 s.f.) above the sea level.
Marking scheme
Award 1 mark for the correct formulation using the displacement equation of motion and solving for the height of the cliff to be approximately 87 m.
Question 37 · multiple-choice
1 marks
A uniform plank of length \(4.0\text{ m}\) and weight \(120\text{ N}\) is supported horizontally by two vertical ropes, A and B. Rope A is attached at a distance of \(0.50\text{ m}\) from the left end. Rope B is attached at a distance of \(1.0\text{ m}\) from the right end.
A child of weight \(300\text{ N}\) stands on the plank at a distance of \(1.5\text{ m}\) from the left end.
What is the tension in rope B?
A.\(120\text{ N}\)
B.\(192\text{ N}\)
C.\(228\text{ N}\)
D.\(420\text{ N}\)
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Worked solution
Since the plank is uniform, its center of gravity is at its geometric center, which is \(2.0\text{ m}\) from the left end.
Let's define the positions of all forces measured from the left end: - Rope A: \(x_A = 0.50\text{ m}\) - Rope B: \(x_B = 4.0 - 1.0 = 3.0\text{ m}\) - Weight of plank (\(120\text{ N}\)): \(x_W = 2.0\text{ m}\) - Weight of child (\(300\text{ N}\)): \(x_C = 1.5\text{ m}\)
Take moments about the point of attachment of rope A (at \(x = 0.50\text{ m}\)) to eliminate the tension in rope A: - Distance to child: \(1.5 - 0.50 = 1.0\text{ m}\) - Distance to plank's weight: \(2.0 - 0.50 = 1.5\text{ m}\) - Distance to rope B: \(3.0 - 0.50 = 2.5\text{ m}\)
Applying the principle of moments for rotational equilibrium: \(\text{Sum of clockwise moments} = \text{Sum of anticlockwise moments}\) \((300 \times 1.0) + (120 \times 1.5) = T_B \times 2.5\) \(300 + 180 = 2.5 T_B\) \(480 = 2.5 T_B\) \(T_B = \frac{480}{2.5} = 192\text{ N}\)
Marking scheme
Award 1 mark for setting up the correct moments equation about Rope A (or another valid pivot point) and solving to find the tension in Rope B as 192 N.
Question 38 · multiple-choice
1 marks
The decay of a neutron is represented by the equation:
Which row in the table correctly describes the change in quark composition and the class of particles to which the electron and the antineutrino belong?
| | Change in quark composition | Class of \(\text{e}^-\) and \(\bar{\nu}_e\) | |---|---|---| | **A** | down quark to up quark | leptons | | **B** | down quark to up quark | hadrons | | **C** | up quark to down quark | leptons | | **D** | up quark to down quark | hadrons |
A.A
B.B
C.C
D.D
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Worked solution
A neutron has a quark composition of \(\text{udd}\). A proton has a quark composition of \(\text{uud}\). Therefore, during beta-minus decay, one down quark (\(\text{d}\)) changes into an up quark (\(\text{u}\)).
The electron (\(\text{e}^-\)) and the electron antineutrino (\(\bar{\nu}_e\)) are elementary particles that do not feel the strong force, meaning they belong to the lepton class of particles.
Thus, row A is the correct option.
Marking scheme
Award 1 mark for identifying the correct quark transformation (d to u) and the correct classification of leptons for the products.
Question 39 · multiple-choice
1 marks
A block of mass \(2.0\text{ kg}\) is moving along a frictionless horizontal track at a speed of \(6.0\text{ m s}^{-1}\). It collides with a stationary block of mass \(4.0\text{ kg}\). After the collision, the \(2.0\text{ kg}\) block rebounds in the opposite direction with a speed of \(1.0\text{ m s}^{-1}\).
What is the loss in total kinetic energy during this collision?
A.\(10.5\text{ J}\)
B.\(11.5\text{ J}\)
C.\(22.5\text{ J}\)
D.\(25.5\text{ J}\)
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Worked solution
First, calculate the initial kinetic energy of the system: \(E_{\text{ki}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 2.0 \times 6.0^2 = 36.0\text{ J}\)
Next, use conservation of linear momentum to find the velocity of the second block \(v_2\) after the collision. Let the initial direction of the \(2.0\text{ kg}\) block be positive: \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\) \(2.0 \times 6.0 + 0 = 2.0 \times (-1.0) + 4.0 \times v_2\) \(12.0 = -2.0 + 4.0 v_2\) \(14.0 = 4.0 v_2\) \(v_2 = 3.5\text{ m s}^{-1}\)
Calculate the total final kinetic energy of the system: \(E_{\text{kf}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2\) \(E_{\text{kf}} = \left(\frac{1}{2} \times 2.0 \times (-1.0)^2\right) + \left(\frac{1}{2} \times 4.0 \times 3.5^2\right) = 1.0 + 2.0 \times 12.25 = 25.5\text{ J}\)
The loss in total kinetic energy is: \(\Delta E_{\text{k}} = E_{\text{ki}} - E_{\text{kf}} = 36.0 - 25.5 = 10.5\text{ J}\)
Marking scheme
Award 1 mark for finding the correct rebound velocity of the 4.0 kg block (3.5 m/s) and using it to determine the kinetic energy loss of 10.5 J.
Question 40 · multiple-choice
1 marks
In an experiment to determine the resistivity \(\rho\) of a metal wire, the following measurements are made:
- resistance of the wire \(R = (4.8 \pm 0.1)\ \Omega\) - length of the wire \(L = (1.50 \pm 0.02)\text{ m}\) - diameter of the wire \(d = (0.32 \pm 0.01)\text{ mm}\)
What is the percentage uncertainty in the calculated value of the resistivity \(\rho\)?
A.\(4.5\%\)
B.\(6.5\%\)
C.\(9.7\%\)
D.\(12.8\%\)
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Worked solution
The formula for resistivity is: \(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\)
To find the percentage uncertainty, sum the relative uncertainties of the measured quantities (multiplying the diameter's relative uncertainty by 2 because it is squared): \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\)
Award 1 mark for using the correct percentage uncertainty formula, doubling the diameter uncertainty, and arriving at 9.7%.
Paper 23 (AS Level Structured)
Answer all structured questions in the spaces provided. Show clear working steps and state final answers with correct units and significant figures.
7 Question · 59.99 marks
Question 1 · structured
8.57 marks
A small stone is projected vertically upwards from the top of a cliff of height \(24.0\text{ m}\) with an initial speed of \(15.0\text{ m s}^{-1}\). Air resistance is negligible.
(a) Determine the maximum height reached by the stone above the top of the cliff.
(b) Calculate the total time elapsed from the moment the stone is projected to the instant it hits the ground at the base of the cliff.
(c) State and explain how the magnitude of the acceleration of the stone at its maximum height compares to its acceleration just before hitting the ground.
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Worked solution
(a) Use the equation of motion: \(v^2 = u^2 + 2as\) At maximum height, \(v = 0\), and taking upwards as positive: \(0 = (15.0)^2 + 2(-9.81)s\) \(19.62s = 225\) \(s = 11.47\text{ m} \approx 11.5\text{ m}\)
(b) Taking upwards as positive, the displacement \(s = -24.0\text{ m}\): \(s = ut + \frac{1}{2}at^2\) \(-24.0 = 15.0t - 4.905t^2\) Rearranging into a quadratic equation: \(4.905t^2 - 15.0t - 24.0 = 0\) Using the quadratic formula: \(t = \frac{-(-15.0) \pm \sqrt{(-15.0)^2 - 4(4.905)(-24.0)}}{2 \times 4.905}\) \(t = \frac{15.0 \pm \sqrt{225 + 470.88}}{9.81}\) \(t = \frac{15.0 \pm 26.38}{9.81}\) Since time must be positive: \(t = \frac{41.38}{9.81} = 4.22\text{ s}\)
(c) The acceleration is constant and equal to \(9.81\text{ m s}^{-2}\) downwards throughout the entire flight. Therefore, the acceleration at the maximum height is equal to the acceleration just before impact.
Marking scheme
(a) - Use of \(v^2 = u^2 + 2as\) with correct substitution [1 mark] - Correct calculation to find height = \(11.5\text{ m}\) (or \(11\text{ m}\) to 2 s.f.) [1 mark]
(b) - Clear statement of displacement \(s = -24.0\text{ m}\) or split-interval method [1 mark] - Correct setup of quadratic or equivalent kinematic equations [1 mark] - Correct final answer of \(4.22\text{ s}\) (accept \(4.2\text{ s}\)) [1 mark]
(c) - States that they are equal / both are \(9.81\text{ m s}^{-2}\) [1 mark] - Explains that gravity is the only force acting / air resistance is negligible so acceleration remains constant [1 mark]
Question 2 · structured
8.57 marks
A block of mass \(1.2\text{ kg}\) is held at rest on a frictionless ramp inclined at \(30^\circ\) to the horizontal. It is connected by a light, inextensible string passing over a frictionless pulley to a suspended block of mass \(0.80\text{ kg}\).
(a) Show that the block on the ramp will accelerate up the ramp when the system is released, and calculate the magnitude of this acceleration.
(b) Determine the tension in the string during this motion.
(c) State the Newton's third law pair force to the gravitational force acting on the suspended block.
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Worked solution
(a) Let \(m_1 = 1.2\text{ kg}\) (on ramp) and \(m_2 = 0.80\text{ kg}\) (suspended). Force acting down the ramp on \(m_1\) due to gravity: \(F_{g1} = m_1 g \sin 30^\circ = 1.2 \times 9.81 \times 0.50 = 5.89\text{ N}\) Force acting downwards on \(m_2\) due to gravity: \(F_{g2} = m_2 g = 0.80 \times 9.81 = 7.85\text{ N}\) Since \(F_{g2} > F_{g1}\), the system accelerates such that \(m_2\) moves down and \(m_1\) moves up the ramp.
Using \(F = ma\) for each mass: For \(m_2\): \(m_2 g - T = m_2 a \Rightarrow 7.85 - T = 0.80 a\) For \(m_1\): \(T - m_1 g \sin 30^\circ = m_1 a \Rightarrow T - 5.89 = 1.2 a\)
Add the two equations: \(7.85 - 5.89 = (0.80 + 1.2) a\) \(1.96 = 2.0 a\) \(a = 0.98\text{ m s}^{-2}\) (or \(0.981\text{ m s}^{-2}\))
(b) Substitute \(a\) back into one of the equations: \(T = 1.2(0.981) + 5.89 = 1.18 + 5.89 = 7.07\text{ N} \approx 7.1\text{ N}\)
(c) According to Newton's third law, if the Earth exerts a gravitational force on the suspended block downwards, the block exerts an equal and opposite gravitational force on the Earth upwards.
Marking scheme
(a) - Calculates weight component down the ramp \(m_1 g \sin 30^\circ = 5.89\text{ N}\) [1 mark] - Formulates equations of motion for both blocks [1 mark] - Solves for acceleration to get \(0.98\text{ m s}^{-2}\) (accept \(0.981\text{ m s}^{-2}\)) [1 mark]
(b) - Substitutes \(a\) into either force equation [1 mark] - Obtains tension = \(7.1\text{ N}\) (or \(7.07\text{ N}\)) with correct unit [1 mark]
(c) - Identifies the force as gravitational [1 mark] - Identifies the force is acting on the Earth (and exerted by the block) [1 mark]
Question 3 · structured
8.57 marks
A uniform cylindrical wire made of a resistive alloy has a resistance of \(4.5\ \Omega\). The wire has a length of \(1.6\text{ m}\) and a diameter of \(0.48\text{ mm}\).
(a) Calculate the resistivity of the alloy.
(b) The wire is now stretched uniformly to twice its original length without changing its mass or density.
(i) Explain why the cross-sectional area of the wire is halved.
(ii) Determine the new resistance of the stretched wire.
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Worked solution
(a) The cross-sectional area \(A\) is: \(A = \frac{\pi d^2}{4} = \frac{\pi (0.48 \times 10^{-3})^2}{4} = 1.81 \times 10^{-7}\text{ m}^2\) Resistivity \(\rho\) is given by: \(\rho = \frac{R A}{L} = \frac{4.5 \times 1.81 \times 10^{-7}}{1.6} = 5.09 \times 10^{-7}\ \Omega\text{ m} \approx 5.1 \times 10^{-7}\ \Omega\text{ m}\)
(b) (i) Mass \(m\) and density \(\rho_d\) are constant, so the volume \(V = A \times L\) is constant. Since \(V = A \times L\), if the new length is \(2L\), then the new area must be \(A/2\) to maintain a constant product.
(ii) The new resistance \(R_{new}\) is: \(R_{new} = \rho \frac{L_{new}}{A_{new}} = \rho \frac{2L}{A/2} = 4 \left( \rho \frac{L}{A} \right) = 4 R\) \(R_{new} = 4 \times 4.5 = 18.0\ \Omega\)
Marking scheme
(a) - Calculates cross-sectional area correctly [1 mark] - Recalls and rearranges resistivity formula [1 mark] - Obtains correct resistivity of \(5.1 \times 10^{-7}\ \Omega\text{ m}\) (accept \(5.09 \times 10^{-7}\)) with correct unit [1 mark]
(b)(i) - Explains that volume remains constant because mass and density are constant [1 mark] - Connects constant volume to the inverse relationship between area and length [1 mark]
(b)(ii) - Recognizes that resistance is proportional to \(L/A\) so factor of change is 4 [1 mark] - Calculates new resistance of \(18\ \Omega\) (or \(18.0\ \Omega\)) [1 mark]
Question 4 · structured
8.57 marks
A steel wire of length \(2.2\text{ m}\) and diameter \(0.80\text{ mm}\) is suspended vertically from a rigid support. A mass of \(6.5\text{ kg}\) is attached to the lower end. The Young modulus of steel is \(2.0 \times 10^{11}\text{ Pa}\).
(a) Show that the stress in the wire is approximately \(1.3 \times 10^8\text{ Pa}\).
(b) Calculate the extension produced in the wire by this mass.
(c) Determine the elastic potential energy stored in the stretched wire.
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(b) Young Modulus \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{\sigma}{x / L}\), where \(x\) is the extension and \(L\) is the original length. \(x = \frac{\sigma L}{E} = \frac{1.2685 \times 10^8 \times 2.2}{2.0 \times 10^{11}} = 1.395 \times 10^{-3}\text{ m} \approx 1.4\text{ mm}\).
(c) Elastic potential energy \(E_p = \frac{1}{2} F x\) (assuming Hooke's law is obeyed): \(E_p = 0.5 \times 63.77 \times 1.395 \times 10^{-3} = 0.0445\text{ J} \approx 0.044\text{ J}\) (or \(0.045\text{ J}\) if using rounded intermediate values).
Marking scheme
(a) - Calculates tension \(F = 63.8\text{ N}\) and cross-sectional area \(A = 5.03 \times 10^{-7}\text{ m}^2\) [1 mark] - Shows final division leading to \(1.27 \times 10^8\text{ Pa}\) [1 mark]
(b) - Recalls formula for Young Modulus or relates it directly to extension [1 mark] - Substitutes values correctly [1 mark] - Obtains extension = \(1.4\text{ mm}\) (or \(1.40 \times 10^{-3}\text{ m}\)) [1 mark]
(c) - Recalls \(E_p = \frac{1}{2} F x\) or \(E_p = \frac{1}{2} k x^2\) [1 mark] - Calculates energy = \(0.044\text{ J}\) or \(0.045\text{ J}\) [1 mark]
Question 5 · structured
8.57 marks
A uniform horizontal shelf \(AB\) of length \(0.90\text{ m}\) and weight \(18\text{ N}\) is hinged to a vertical wall at end \(A\). The shelf is held horizontal at end \(B\) by a light cable attached to the wall at a point vertically above \(A\). The cable makes an angle of \(40^\circ\) with the shelf.
(a) State the two conditions required for a rigid body to be in static equilibrium.
(b) By taking moments about the hinge \(A\), calculate the tension in the cable.
(c) Determine the magnitude and direction of the horizontal force exerted by the hinge on the shelf.
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Worked solution
(a) The two conditions are: 1. The resultant force acting on the body in any direction must be zero. 2. The resultant moment about any point must be zero.
(b) The weight of the uniform shelf acts at its midpoint, \(0.45\text{ m}\) from \(A\). Clockwise moment about \(A = \text{Weight} \times 0.45\text{ m} = 18 \times 0.45 = 8.1\text{ N m}\).
The vertical component of the tension \(T\) acts upwards at \(B\), \(0.90\text{ m}\) from \(A\). Anticlockwise moment about \(A = T \sin 40^\circ \times 0.90 = 0.5785 T\).
Equating clockwise and anticlockwise moments: \(0.5785 T = 8.1\) \(T = \frac{8.1}{0.5785} = 14.0\text{ N} \approx 14\text{ N}\).
(c) For horizontal translational equilibrium, the horizontal forces must balance. The horizontal component of the tension acts to the left on the shelf: \(T_H = T \cos 40^\circ = 14.00 \times \cos 40^\circ = 10.7\text{ N}\).
Therefore, the hinge must exert an equal and opposite horizontal force of \(10.7\text{ N} \approx 11\text{ N}\) pointing to the right.
Marking scheme
(a) - Statement of zero resultant force [1 mark] - Statement of zero resultant moment [1 mark]
(b) - Identifies that the weight acts at the midpoint (\(0.45\text{ m}\)) and sets up moment equation [1 mark] - Correctly resolves the vertical component of tension (\(T \sin 40^\circ\)) [1 mark] - Solves to obtain tension = \(14\text{ N}\) (accept \(14.0\text{ N}\)) [1 mark]
(c) - Recognizes that the horizontal hinge force balances the horizontal component of tension [1 mark] - Calculates \(T \cos 40^\circ = 11\text{ N}\) (accept \(10.7\text{ N}\)) [1 mark] - States direction is to the right [1 mark]
Question 6 · structured
8.57 marks
A free neutron decays into a proton, an electron, and an electron antineutrino.
(a) Write a nuclear equation for this decay, including the nucleon number and proton number for each particle.
(b) Describe the change in the quark composition of the nucleon during this decay.
(c) State the fundamental interaction responsible for this decay, and name the corresponding exchange particle.
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Worked solution
(a) The equation is: \({^1_0\text{n}} \rightarrow {^1_1\text{p}} + {^0_{-1}\text{e}} + \bar{\nu}_e\) where \({^1_0\text{n}}\) is the neutron, \({^1_1\text{p}}\) is the proton, \({^0_{-1}\text{e}}\) is the electron, and \(\bar{\nu}_e\) is the electron antineutrino.
(b) A neutron has a quark composition of up-down-down (\(\text{udd}\)). A proton has a quark composition of up-up-down (\(\text{uud}\)). During the decay, one down (\(\text{d}\)) quark in the neutron is transformed into an up (\(\text{u}\)) quark.
(c) The force responsible for beta decay is the weak nuclear force (or weak interaction). The exchange particle carrying this force is the \(\text{W}^-\)\ boson.
Marking scheme
(a) - Correct symbols for neutron, proton, and electron with correct nucleon and proton numbers [1 mark] - Includes electron antineutrino symbol [1 mark]
(b) - States the quark composition of a neutron (\(\text{udd}\)) and a proton (\(\text{uud}\)) [1 mark] - Clearly states that a down quark changes into an up quark (\(\text{d} \rightarrow \text{u}\)) [1 mark]
(c) - Identifies the interaction as the weak interaction / weak force [1 mark] - Identifies the exchange particle as the \(\text{W}^-\)\ boson [1 mark]
Question 7 · structured
8.57 marks
Two gliders, \(P\) and \(Q\), are on a horizontal, frictionless linear air track. Glider \(P\) has a mass of \(0.35\text{ kg}\) and is moving to the right with a speed of \(2.4\text{ m s}^{-1}\). Glider \(Q\) has a mass of \(0.15\text{ kg}\) and is moving to the left with a speed of \(1.2\text{ m s}^{-1}\). The gliders collide and stick together.
(a) Calculate the common velocity of the gliders after the collision.
(b) Determine the loss of kinetic energy during the collision.
(c) State whether the collision is elastic or inelastic, and justify your answer.
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Worked solution
(a) Taking the direction to the right as positive: Initial momentum of \(P\), \(p_P = m_P v_P = 0.35 \times 2.4 = 0.84\text{ kg m s}^{-1}\) Initial momentum of \(Q\), \(p_Q = m_Q v_Q = 0.15 \times (-1.2) = -0.18\text{ kg m s}^{-1}\) Total initial momentum \(p_i = 0.84 - 0.18 = 0.66\text{ kg m s}^{-1}\)
Since the gliders stick together, the combined mass \(M = 0.35 + 0.15 = 0.50\text{ kg}\). Let \(V\) be the final common velocity: \(p_f = M V \Rightarrow 0.66 = 0.50 V\) \(V = 1.32\text{ m s}^{-1} \approx 1.3\text{ m s}^{-1}\) to the right.
Final kinetic energy: \(E_{kf} = \frac{1}{2} M V^2 = 0.5 \times 0.50 \times (1.32)^2 = 0.4356\text{ J}\)
Loss of kinetic energy: \(\Delta E_k = E_{ki} - E_{kf} = 1.116 - 0.4356 = 0.6804\text{ J} \approx 0.68\text{ J}\).
(c) The collision is inelastic. This is justified because there is a loss of kinetic energy (kinetic energy is not conserved during the collision).
Marking scheme
(a) - Calculates correct initial momentum taking direction into account [1 mark] - Equates initial and final momentum [1 mark] - Obtains final velocity = \(1.3\text{ m s}^{-1}\) (or \(1.32\text{ m s}^{-1}\)) to the right [1 mark]
(b) - Calculates initial kinetic energy = \(1.12\text{ J}\) [1 mark] - Calculates final kinetic energy = \(0.44\text{ J}\) [1 mark] - Obtains loss of kinetic energy = \(0.68\text{ J}\) (accept \(0.680\text{ J}\)) [1 mark]
(c) - States the collision is inelastic [1 mark] - Justifies by stating that kinetic energy is not conserved (or is converted to other forms of energy/thermal energy) [1 mark]
Paper 33 (Advanced Practical Skills)
Complete both practical investigations using the provided materials. Record all data, plot required linear trends, determine values of physical variables, and analyze limitations.
2 Question · 40 marks
Question 1 · Practical Investigation
20 marks
In this experiment, you will investigate the equilibrium of a pivoted wooden rule supported by a vertical spring.
**Apparatus:** - Wooden meter rule with a small hole drilled at the 5.0 cm mark - Stand, boss, and clamp - Steel pin to act as a pivot - Vertical spring - Second stand, boss, and clamp to support the spring - Mass hanger and slotted masses to total \(M = 200\text{ g}\) - Thread to attach the mass to the rule - Half-meter rule
**Procedure:** 1. Set up the apparatus with the steel pin passing through the hole at the 5.0 cm mark of the meter rule, clamping the pin horizontally so the rule can pivot freely. 2. Attach the vertical spring to the rule at the 85.0 cm mark (giving a distance \(d = 80.0\text{ cm}\) from the pivot). Clamping the upper end of the spring, adjust its height until the rule is horizontal. 3. Measure the unstretched length \(l_0\) of the spring. 4. Suspend the mass \(M = 200\text{ g}\) from the rule at a distance \(x = 10.0\text{ cm}\) from the pivot. 5. The rule will tilt. Raise the upper clamp of the spring until the rule is horizontal again. Measure the new length \(l\) of the spring and calculate the extension \(e = l - l_0\). 6. Repeat this process for several values of \(x\) in the range \(10.0\text{ cm} \le x \le 70.0\text{ cm}\) and record your results in a single table, including columns for \(x\), \(l\), and \(e\). 7. Plot a graph of \(e\) on the y-axis against \(x\) on the x-axis, draw the line of best fit, and find its gradient \(G\) and y-intercept \(I\). 8. The variables \(e\) and \(x\) are related by the equation: \(e = \left(\frac{M g}{k d}\right) x + \frac{m g d_G}{k d}\) where \(g = 9.81\text{ m s}^{-2}\), \(d = 80.0\text{ cm}\), \(d_G = 45.0\text{ cm}\) (the distance from the pivot to the center of gravity of the rule), \(k\) is the spring constant, and \(m\) is the mass of the rule.
Determine the values of \(k\) (in \(\text{N m}^{-1}\)) and \(m\) (in \(\text{g}\)).
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**Collection of Data (6 marks):** - [1] Setup: measuring \(l_0\) to nearest mm with appropriate units. - [1] Table: at least 6 sets of readings of \(x\) and \(l\) with no blank entries. - [1] Range: values of \(x\) cover at least 50.0 cm. - [1] Column headings: proper names/symbols with units, e.g., \(x / \text{cm}\), \(l / \text{cm}\), \(e / \text{cm}\). - [1] Consistency: all raw \(x\) and \(l\) values recorded to the nearest 0.1 cm (1 mm). - [1] Calculations: calculated values of \(e\) are correct based on \(e = l - l_0\).
**Graph (4 marks):** - [1] Axes: linear scales where the plotted points occupy more than half of the grid in both directions. - [1] Plotting: points plotted correctly to within half a small square. - [1] Best-fit line: single straight line showing a balanced distribution of points. - [1] Quality: points lie close to the best-fit line.
**Gradient and Intercept (2 marks):** - [1] Gradient: calculated using a triangle with hypotenuse longer than half the length of the drawn line. - [1] Intercept: read directly from the y-axis (at \(x=0\)) or calculated using a point on the line.
**Constants and Analysis (8 marks):** - [1] Value of \(k\) calculated correctly using \(k = \frac{M g}{G d}\). - [1] Value of \(m\) calculated correctly using \(m = M \frac{I}{G d_G}\). - [1] Appropriate units included for \(k\) (e.g., \(\text{N m}^{-1}\)) and \(m\) (e.g., \(\text{g}\)). - [1] Significant figures: both constants given to 2 or 3 SF. - [1] Correct SI unit conversions shown in calculations. - [1] Value of \(k\) in range \(5.0\text{ N m}^{-1}\) to \(30.0\text{ N m}^{-1}\). - [1] Value of \(m\) in range \(80\text{ g}\) to \(180\text{ g}\). - [1] Absolute uncertainty in \(e\) estimated (typically \(\pm 0.2\text{ cm}\)) and justified based on alignment precision.
Question 2 · Practical Investigation
20 marks
In this experiment, you will investigate how the period of vertical oscillation of a clamped horizontal cantilever rule depends on its overhang length.
**Apparatus:** - Wooden meter rule - G-clamp and a small wooden block - Mass of 200 g - Adhesive tape - Stopwatch reading to 0.01 s - Half-meter rule
**Procedure:** 1. Clamp the wooden meter rule near the edge of a bench so that an overhang length \(L_1 = 60.0\text{ cm}\) projects horizontally beyond the edge of the bench. Securely tape the 200 g mass at the very end of the rule (centered at the 98.0 cm mark). 2. Displace the free end of the rule downwards by a small distance (approx. 2 to 3 cm) and release it so that it performs vertical oscillations. 3. Measure and record the time \(t_1\) for 10 complete oscillations. Repeat this measurement and determine the average period \(T_1\). 4. Estimate the percentage uncertainty in your value of \(T_1\). Show your working. 5. Change the position of the rule so that the overhang length is \(L_2 = 45.0\text{ cm}\). Repeat step 3 to find the time for 10 oscillations and determine the new average period \(T_2\). 6. It is suggested that the period \(T\) and the overhang length \(L\) are related by: \(T^2 = C L^3\) where \(C\) is a constant.
Calculate the values of the constant \(C\) for both overhang lengths. State whether the experimental results support the suggested relationship, justifying your conclusion by comparing the percentage difference between your two values of \(C\) with a specified criterion.
7. Identify four limitations of this experiment and suggest four matching improvements to reduce experimental error.
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Worked solution
1. First measurements (\(L_1 = 60.0\text{ cm}\)): - Raw times: \(t_{1,a} = 8.12\text{ s}\), \(t_{1,b} = 8.18\text{ s}\) -> Average \(t_1 = 8.15\text{ s}\). - Period \(T_1 = 0.815\text{ s}\).
5. Comparison: - Percentage difference \(= \frac{|3.10 - 3.07|}{3.07} \times 100\% \approx 1.0\%\). - Since \(1.0\% < 10\%\) (or less than the experimental uncertainty), the relationship is supported.
Marking scheme
**Measurements and Uncertainty (6 marks):** - [1] Value of \(L_1\) recorded to the nearest mm with correct unit (m or cm). - [1] Measurement of raw times for 10 oscillations, repeated, with unit (s). - [1] Calculation of period \(T_1\) with correct unit. - [1] Absolute uncertainty in raw time of 10 oscillations set to \(0.1\text{ s}\) to \(0.2\text{ s}\) with justification. - [1] Correct calculation of percentage uncertainty in \(T_1\). - [1] Second measurements for \(L_2\) and \(T_2\) showing correct trend (shorter length gives shorter period).
**Analysis and Evaluation (6 marks):** - [1] Correct calculation of \(C_1\) with units (e.g., \(\text{s}^2 \text{ m}^{-3}\)). - [1] Correct calculation of \(C_2\). - [1] SF: Constants \(C_1\) and \(C_2\) given to 2 or 3 SF, consistent with raw measurements. - [1] Percentage difference between \(C_1\) and \(C_2\) calculated correctly. - [1] Criterion stated clearly (e.g., \'I will use 10% as the limit for agreement\'). - [1] Conclusion clearly stated and justified by comparing the percentage difference to the criterion.
**Limitations and Improvements (8 marks - 1 mark for each of 4 pairs):** - [1] Limitation 1: Two sets of readings are not enough to draw a firm conclusion. - [1] Improvement 1: Take more sets of readings and plot a graph of \(T^2\) against \(L^3\). - [1] Limitation 2: Rule slips or moves at the clamp during oscillation. - [1] Improvement 2: Use rubber pads or sandpaper between the clamp and the rule to increase friction. - [1] Limitation 3: Large amplitude causes non-linear oscillations or rule hits the edge of the bench. - [1] Improvement 3: Use a smaller initial displacement or increase bench clearance. - [1] Limitation 4: Difficulty in counting oscillations or determining exactly when an oscillation is complete. - [1] Improvement 4: Use a fiducial marker at the equilibrium position, or use a light gate connected to a data logger.
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