Question 1
(a) Explain what is meant by the accuracy of a measured value. (1 mark)
(b) Two solid cubes, A and B, are measured to determine the density of their materials.
Table 1.1 shows the measurements for cube A.
Table 1.1
| quantity | measurement |
| length of side | \( (1.53 \pm 0.01)\,\mathrm{cm} \) |
| mass | \( (31.3 \pm 0.5)\,\mathrm{g} \) |
(i) Show that the calculated density of the material of cube A is \(8.7 \times 10^3\,\mathrm{kg\,m^{-3}}\). (2 marks)
(ii) Calculate the percentage uncertainty in the density of the material of cube A. (2 marks)
percentage uncertainty = ____________________ %
(iii) The density of the material of cube B is determined to be \(9.2 \times 10^3\,\mathrm{kg\,m^{-3}} \pm 6\%\).
State and explain whether cube A and cube B could be made from the same material. (2 marks)
Syllabus Topic Codes (A Level Physics P2 syllabus):
• 4.3: Density and pressure — parts (b)(i) and (b)(iii)
▶️ Answer/Explanation
(a) Accuracy [1 mark]
Accuracy describes how close a measured value is to the true value of the quantity.
(b)(i) Density [2 marks]
For a cube,
\(V=l^3\)
\(V=(1.53\times10^{-2})^3\,\mathrm{m^3}\)
Density is given by
\(\rho=\dfrac{m}{V}\)
\(\rho=\dfrac{31.3\times10^{-3}}{(1.53\times10^{-2})^3}\)
\(\rho=8.7\times10^3\,\mathrm{kg\,m^{-3}}\)
Answer: \( \boxed{8.7\times10^3\,\mathrm{kg\,m^{-3}}} \)
(b)(ii) Percentage uncertainty [2 marks]
The percentage uncertainty in the mass is
\(\dfrac{0.5}{31.3}\times100=1.6\%\)
Since \(V=l^3\), the percentage uncertainty in the volume is
\(3\times\dfrac{0.01}{1.53}\times100=2.0\%\)
Therefore, the percentage uncertainty in density is
\(\%\text{ uncertainty}=1.6+2.0\)
\(=3.6\%\approx4\%\)
Answer: \( \boxed{4\%} \)
(b)(iii) Comparison of densities [2 marks]
For cube A, the density is approximately \(8.7\times10^3\,\mathrm{kg\,m^{-3}}\) with an uncertainty of \(4\%\).
The range for cube A is therefore approximately
\(8.7\times10^3\pm4\%\)
which gives a range of approximately \(8.35\times10^3\) to \(9.05\times10^3\,\mathrm{kg\,m^{-3}}\).
For cube B, the density range is
\(9.2\times10^3\pm6\%\)
which gives approximately \(8.65\times10^3\) to \(9.75\times10^3\,\mathrm{kg\,m^{-3}}\).
The two ranges overlap, so the difference between the measured densities is within the uncertainty.
Answer: \( \boxed{\text{Yes, they could be made from the same material.}} \)
The density ranges overlap, so the measurements are consistent with the cubes being made from the same material.
Question 2
(a) State the principle of moments. (2 marks)
(b) A solid plastic cylinder floats in water. It is used to support one end of a horizontal uniform beam AB as shown in Fig. 2.1.
Fig. 2.1 (not to scale)
The beam has length \(6.0\,\mathrm{m}\) and weight \(1700\,\mathrm{N}\). The beam is attached to solid ground with a hinge at end A.
The cylinder is floating vertically in the water. The top of the cylinder is attached at its centre to the beam at a horizontal distance of \(5.0\,\mathrm{m}\) from end A. The cylinder applies a vertical force of \(1300\,\mathrm{N}\) to the beam.
A person of weight \(660\,\mathrm{N}\) stands on the beam at point P.
The beam AB is in equilibrium.
(i) By taking moments about end A, determine the distance \(x\) from A to P. (3 marks)
distance = ____________________ \(\mathrm{m}\)
(ii) The bottom of the cylinder is submerged in the water to depth \(y\) as shown in Fig. 2.2. The beam is still attached to the cylinder but not shown.
The cylinder has mass \(11\,\mathrm{kg}\) and diameter \(0.78\,\mathrm{m}\). The beam exerts a vertical force of \(1300\,\mathrm{N}\) on the cylinder. The cylinder is in equilibrium.
Show that the upthrust acting on the cylinder is \(1400\,\mathrm{N}\). (1 mark)
(iii) The water has density \(990\,\mathrm{kg\,m^{-3}}\).
Calculate the depth \(y\). (2 marks)
\(y=\) ____________________ \(\mathrm{m}\)
(iv) The person can stand anywhere between A and B.
On Fig. 2.3, sketch the variation of the depth of the bottom of the cylinder with the distance of the person from A, for distances between \(0\) and \(6.0\,\mathrm{m}\). Numerical values are not required. (2 marks)
Syllabus Topic Codes (A Level Physics P2 syllabus):
• 4.3: Density and pressure — parts (b)(ii), (b)(iii) and (b)(iv)
▶️ Answer/Explanation
(a) Principle of moments [2 marks]
For an object in rotational equilibrium, the sum of the clockwise moments about a point equals the sum of the anticlockwise moments about the same point.
(b)(i) Distance \(x\) from A to P [3 marks]
The beam is in equilibrium, so take moments about A.
Clockwise moment = anticlockwise moment
\(1700\times3.0+660x=1300\times5.0\)
\(5100+660x=6500\)
\(660x=1400\)
\(x=2.12\,\mathrm{m}\)
Answer: \( \boxed{2.1\,\mathrm{m}} \)
(b)(ii) Upthrust [1 mark]
The cylinder is in equilibrium, so the upthrust equals the total downward force.
Weight of cylinder:
\(W=mg=11\times9.81=107.91\,\mathrm{N}\)
Therefore,
\(U=1300+107.91\)
\(U=1407.91\,\mathrm{N}\)
Hence, to the required precision,
Answer: \( \boxed{1400\,\mathrm{N}} \)
(b)(iii) Depth \(y\) [2 marks]
The upthrust is given by
\(U=\rho gV\)
The submerged volume of the cylinder is
\(V=\pi r^2y\)
The radius is
\(r=\dfrac{0.78}{2}=0.39\,\mathrm{m}\)
Therefore,
\(1400=990\times9.81\times\pi(0.39)^2y\)
\(y=0.302\,\mathrm{m}\)
Answer: \( \boxed{0.30\,\mathrm{m}} \)
(b)(iv) Variation of depth [2 marks]
Taking moments about A, if the person is a distance \(x\) from A,
\(1700\times3.0+660x=5.0F\)
As the distance of the person from A increases, the downward force exerted by the cylinder on the beam must increase.
The upthrust on the cylinder therefore increases. Since \(U=\rho g\pi r^2y\), the depth \(y\) also increases.
Therefore, the graph should be a straight line with a positive gradient and a non-zero depth when the distance from A is \(0\).
Answer: A straight line increasing from a non-zero value at \(0\,\mathrm{m}\) to a greater value at \(6.0\,\mathrm{m}\).
Question 3
(a) A truck R of mass \(9400\,\mathrm{kg}\) moves with constant acceleration in a straight line down a slope, as illustrated in Fig. 3.1.
At point A the speed of the truck is \(13\,\mathrm{m\,s^{-1}}\) and at point B the speed of the truck is \(22\,\mathrm{m\,s^{-1}}\). A and B are a distance of \(180\,\mathrm{m}\) apart.
(i) Calculate the acceleration of the truck between A and B. (2 marks)
acceleration = ____________________ \(\mathrm{m\,s^{-2}}\)
(ii) Determine the gain in kinetic energy of the truck between A and B. (3 marks)
gain in kinetic energy = ____________________ \(\mathrm{J}\)
(b) A short time after passing point B truck R moves in a straight line on horizontal ground. The driver of the truck applies the brakes. Fig. 3.2 shows the variation with time of the momentum of the truck.
(i) Define force. (1 mark)
____________________________________________________________
(ii) Show that the average resultant force \(F\) acting on truck R between time \(t=0\) and \(t=15\,\mathrm{s}\) is \(-1.2\times10^4\,\mathrm{N}\). (1 mark)
____________________________________________________________
(iii) An identical truck S has the same initial momentum as truck R. Truck S experiences a constant force equal to the force \(F\) in (b)(ii).
State and explain whether truck S will take more, less or the same amount of time to come to rest as truck R. (3 marks)
____________________________________________________________
Syllabus Topic Codes (A Level Physics P2 syllabus):
• 5.2: Gravitational potential energy and kinetic energy — part (a)(ii)
• 3.1: Momentum and Newton’s laws of motion — parts (b)(i), (b)(ii) and (b)(iii)
▶️ Answer/Explanation
(a)(i) Acceleration [2 marks]
Use the equation
\(v^2=u^2+2as\)
Rearranging,
\(a=\dfrac{v^2-u^2}{2s}\)
\(a=\dfrac{22^2-13^2}{2\times180}\)
\(a=0.875\,\mathrm{m\,s^{-2}}\)
Answer: \( \boxed{0.88\,\mathrm{m\,s^{-2}}} \)
(a)(ii) Gain in kinetic energy [3 marks]
The change in kinetic energy is
\(\Delta E_{\mathrm{k}}=\dfrac{1}{2}m(v^2-u^2)\)
\(\Delta E_{\mathrm{k}}=\dfrac{1}{2}\times9400\times(22^2-13^2)\)
\(\Delta E_{\mathrm{k}}=1.487\times10^6\,\mathrm{J}\)
Answer: \( \boxed{1.5\times10^6\,\mathrm{J}} \)
(b)(i) Definition of force [1 mark]
Force is the rate of change of momentum.
(b)(ii) Average resultant force [1 mark]
The average resultant force is given by
\(F=\dfrac{\Delta p}{\Delta t}\)
From the graph, the momentum changes from approximately \(21\times10^4\,\mathrm{kg\,m\,s^{-1}}\) at \(t=0\) to \(2.5\times10^4\,\mathrm{kg\,m\,s^{-1}}\) at \(t=15\,\mathrm{s}\).
\(F=\dfrac{2.5\times10^4-21\times10^4}{15}\)
\(F=-1.23\times10^4\,\mathrm{N}\)
To the required precision,
Answer: \( \boxed{-1.2\times10^4\,\mathrm{N}} \)
(b)(iii) Comparing the stopping times [3 marks]
Both trucks have the same initial momentum and both must have the same change in momentum to come to rest.
Truck R has an average braking force of approximately
\(\dfrac{21\times10^4}{23}\approx0.91\times10^4\,\mathrm{N}\)
This is less than the constant force of \(1.2\times10^4\,\mathrm{N}\) acting on truck S.
Since the change in momentum is the same and truck S experiences the greater force, truck S takes less time to undergo the required change in momentum.
Answer: \( \boxed{\text{Truck S takes less time to come to rest.}} \)
Equivalently, for truck S, the stopping time is approximately
\(t=\dfrac{-21\times10^4}{-1.2\times10^4}\approx17.5\,\mathrm{s}\)
whereas truck R takes approximately \(23\,\mathrm{s}\).
Question 4
A device containing a microwave emitter and receiver is placed in front of a large metal sheet in a vacuum as shown in Fig. 4.1.
The line XY is perpendicular to the metal sheet. The device emits microwaves of frequency \(6.3\,\mathrm{GHz}\).
(a) When the device is at position P, a stationary wave is formed between the device and the sheet.
Explain how the stationary wave, including the nodes and the antinodes, is formed. (4 marks)
(b)(i) Calculate the wavelength of the microwaves. (2 marks)
wavelength = ____________________ \(\mathrm{m}\)
(ii) At point P the receiver detects a maximum amplitude of the stationary wave.
The device is moved slowly from point P along the line XY and the receiver detects a series of minimum and maximum amplitudes. The first time a minimum amplitude is detected by the receiver is when the device is at point Q.
Determine the distance between P and Q. (1 mark)
distance = ____________________ \(\mathrm{m}\)
(iii) The intensity of the microwaves emitted by the device is increased. The frequency of the microwaves is unchanged. The device is moved slowly along the line XY from point Q until the next maximum amplitude is detected at point R.
State and explain whether the distance QR is greater than, less than or the same as distance PQ. (1 mark)
____________________________________________________________
Syllabus Topic Codes (A Level Physics P2 syllabus):
• 7.1: Progressive waves — part (b)(i)
▶️ Answer/Explanation
(a) Formation of the stationary wave [4 marks]
The microwaves emitted by the device travel towards the metal sheet and are reflected from the sheet.
The incident and reflected waves travel in opposite directions and superpose, producing a stationary wave.
At an antinode, constructive interference occurs and the resultant amplitude is maximum.
At a node, destructive interference occurs and the resultant amplitude is minimum or zero.
Answer: Reflection of the microwaves at the metal sheet produces an incident and reflected wave. Their superposition forms a stationary wave with maximum amplitude at antinodes and minimum or zero amplitude at nodes.
(b)(i) Wavelength [2 marks]
Use the wave equation
\(c=f\lambda\)
Therefore,
\(\lambda=\dfrac{c}{f}\)
\(\lambda=\dfrac{3.0\times10^8}{6.3\times10^9}\)
\(\lambda=4.76\times10^{-2}\,\mathrm{m}\)
Answer: \( \boxed{0.048\,\mathrm{m}} \)
(b)(ii) Distance \(PQ\) [1 mark]
Point P is at a maximum amplitude, so it corresponds to an antinode.
The first minimum amplitude occurs at the nearest node. The distance from an antinode to the adjacent node is \(\dfrac{\lambda}{4}\).
\(PQ=\dfrac{\lambda}{4}\)
\(PQ=\dfrac{0.048}{4}=0.012\,\mathrm{m}\)
Answer: \( \boxed{0.012\,\mathrm{m}} \)
(b)(iii) Distance \(QR\) [1 mark]
The intensity of the microwaves affects the amplitude of the wave but does not change its frequency.
Since the frequency is unchanged, the wavelength is unchanged.
The distance between successive nodes and antinodes therefore remains unchanged.
Answer: \( \boxed{\text{QR is the same as PQ}} \)
Both distances are determined by the wavelength, and \(QR=PQ=\dfrac{\lambda}{4}\).
Question 5
A stationary loudspeaker emits sound of constant frequency. A microphone is placed near to the loudspeaker and connected to a cathode-ray oscilloscope (CRO). The trace on the screen of the CRO is shown in Fig. 5.1.
The time-base of the CRO is set to \(5.0\times10^{-4}\,\mathrm{s\,cm^{-1}}\).
(a) The speed of the sound emitted by the loudspeaker is \(330\,\mathrm{m\,s^{-1}}\). Determine the wavelength of the sound. (3 marks)
wavelength = ____________________ \(\mathrm{m}\)
(b) The loudspeaker now moves in a straight line while emitting the same sound of constant frequency. The period of the trace on the CRO increases continuously.
Describe the motion of the loudspeaker. (2 marks)
____________________________________________________________
Syllabus Topic Codes (A Level Physics P2 syllabus):
• 7.3: Doppler effect for sound waves — part (b)
▶️ Answer/Explanation
(a) Wavelength [3 marks]
From the CRO trace, one complete cycle occupies \(5.8\,\mathrm{cm}\).
Using the time-base, the period is
\(T=5.8\times5.0\times10^{-4}\)
\(T=2.9\times10^{-3}\,\mathrm{s}\)
The wave equation is
\(v=f\lambda\)
Since \(f=\dfrac{1}{T}\),
\(\lambda=vT\)
\(\lambda=330\times2.9\times10^{-3}\)
\(\lambda=0.957\,\mathrm{m}\)
Answer: \( \boxed{0.96\,\mathrm{m}} \)
(b) Motion of the loudspeaker [2 marks]
The period of the trace increases, so the frequency detected by the microphone decreases.
This is due to the Doppler effect. A decreasing detected frequency means that the loudspeaker is moving away from the microphone.
Because the period increases continuously, the detected frequency is continuously decreasing. Therefore, the loudspeaker is moving away from the microphone at an increasing speed, so it is accelerating away from the microphone.
Answer: \( \boxed{\text{The loudspeaker moves away from the microphone at an increasing speed.}} \)
Total: \(5\) marks
Question 6
A cylindrical copper wire P of length \(0.24\,\mathrm{m}\) is shown in Fig. 6.1.
The current in the wire is \(0.85\,\mathrm{A}\).
The resistance of the wire is \(3.3\,\mathrm{m\Omega}\).
The total number of charge carriers \(N\) in the wire is \(2.6\times10^{22}\).
The resistivity of copper is \(1.8\times10^{-8}\,\Omega\,\mathrm{m}\).
(a) Calculate the potential difference between the two ends of the wire. (2 marks)
potential difference = ____________________ \(\mathrm{V}\)
(b)(i) Show that the cross-sectional area of the wire is \(1.3\times10^{-6}\,\mathrm{m^2}\). (2 marks)
____________________________________________________________
(ii) Show that the number density of charge carriers in the wire is \(8.3\times10^{28}\,\mathrm{m^{-3}}\). (1 mark)
____________________________________________________________
(iii) Calculate the average drift speed of the charge carriers (electrons) in the wire. (2 marks)
average drift speed = ____________________ \(\mathrm{m\,s^{-1}}\)
(c) A different copper wire Q has the same volume as wire P, but non-uniform radius, as shown in Fig. 6.2.
The radius \(r_1\) at end X of wire Q is the same as the radius of wire P. Radius \(r_2\) is less than \(r_1\).
(i) State and explain how the resistance of wire Q compares with the resistance of wire P. (4 marks)
____________________________________________________________
(ii) On Fig. 6.3, sketch a graph of the variation of the average drift speed of the charge carriers with distance from end X of wire Q. (2 marks)
Syllabus Topic Codes (A Level Physics P2 syllabus):
• 9.3: Resistance and resistivity — parts (a), (b)(i) and (c)(i)
▶️ Answer/Explanation
(a) Potential difference [2 marks]
Use Ohm’s law:
\(V=IR\)
\(V=0.85\times3.3\times10^{-3}\)
\(V=2.805\times10^{-3}\,\mathrm{V}\)
Answer: \( \boxed{2.8\times10^{-3}\,\mathrm{V}} \)
(b)(i) Cross-sectional area [2 marks]
For a uniform wire,
\(R=\dfrac{\rho L}{A}\)
Rearranging,
\(A=\dfrac{\rho L}{R}\)
\(A=\dfrac{1.8\times10^{-8}\times0.24}{3.3\times10^{-3}}\)
\(A=1.31\times10^{-6}\,\mathrm{m^2}\)
Answer: \( \boxed{1.3\times10^{-6}\,\mathrm{m^2}} \)
(b)(ii) Number density of charge carriers [1 mark]
The volume of the wire is
\(V=AL\)
The number density is
\(n=\dfrac{N}{AL}\)
\(n=\dfrac{2.6\times10^{22}}{1.3\times10^{-6}\times0.24}\)
Answer: \( \boxed{8.3\times10^{28}\,\mathrm{m^{-3}}} \)
(b)(iii) Average drift speed [2 marks]
The current is related to drift speed by
\(I=nAqv\)
Therefore,
\(v=\dfrac{I}{nAq}\)
Using \(q=1.6\times10^{-19}\,\mathrm{C}\),
\(v=\dfrac{0.85}{(8.3\times10^{28})(1.3\times10^{-6})(1.6\times10^{-19})}\)
\(v=4.9\times10^{-5}\,\mathrm{m\,s^{-1}}\)
Answer: \( \boxed{4.9\times10^{-5}\,\mathrm{m\,s^{-1}}} \)
(c)(i) Comparison of resistance [4 marks]
Wire Q has the same volume as wire P, but its length is greater because its average cross-sectional area is smaller.
Therefore, the average cross-sectional area of Q is less than that of P.
Since
\(R=\dfrac{\rho L}{A}\)
resistance is proportional to \(\dfrac{L}{A}\).
Wire Q has a greater length and a smaller average cross-sectional area, so its resistance is greater.
Answer: \( \boxed{R_Q>R_P} \)
(c)(ii) Variation of drift speed [2 marks]
The current through the wire is constant, so
\(I=nAqv\)
For copper, \(n\) and \(q\) remain constant. Therefore,
\(v\propto\dfrac{1}{A}\)
As the distance from X increases, the cross-sectional area decreases. Hence, the average drift speed increases.
The graph should start from a non-zero value at distance \(0\) and have an increasing positive gradient.
Answer: An increasing curve starting at a non-zero drift speed when the distance from X is \(0\), with an increasingly positive gradient.
Question 7
An isolated stationary nucleus Q decays into nucleus R and an \(\alpha\)-particle. The \(\alpha\)-particle has speed \(1.5\times10^7\,\mathrm{m\,s^{-1}}\).
(a) Complete the equation for the decay. (1 mark)
\(\,^{226}_{88}\mathrm{Q}\rightarrow\,^{222}_{86}\mathrm{R}+\,^{4}_{2}\alpha\)
(b) By considering momentum, calculate the speed of nucleus R after the decay. (3 marks)
speed = ____________________ \(\mathrm{m\,s^{-1}}\)
(c) State three quantities that are conserved during the decay. (3 marks)
1. ____________________________________________________________
2. ____________________________________________________________
3. ____________________________________________________________
Syllabus Topic Codes (A Level Physics P2 syllabus):
• 3.3: Linear momentum and its conservation — part (b)
▶️ Answer/Explanation
(a) Completing the nuclear equation [1 mark]
In \(\alpha\)-decay, the nucleus emits an \(\alpha\)-particle with nucleon number \(4\) and proton number \(2\).
Nucleon number of Q:
\(A_R=226-4=222\)
Proton number of R:
\(Z_R=88-2=86\)
Answer: \( \boxed{\,^{226}_{88}\mathrm{Q}\rightarrow\,^{222}_{86}\mathrm{R}+\,^{4}_{2}\alpha\,} \)
(b) Speed of nucleus R [3 marks]
The original nucleus is stationary, so its initial momentum is zero. Therefore, the momenta of the \(\alpha\)-particle and nucleus R are equal in magnitude and opposite in direction.
Using \(p=mv\),
\(4u\times1.5\times10^7=222u\times v\)
where \(u\) is the atomic mass unit.
Therefore,
\(v=\dfrac{4u\times1.5\times10^7}{222u}\)
\(v=2.70\times10^5\,\mathrm{m\,s^{-1}}\)
Answer: \( \boxed{2.7\times10^5\,\mathrm{m\,s^{-1}}} \)
(c) Conserved quantities [3 marks]
Any three of the following are acceptable:
- Momentum
- Charge
- Nucleon number
- Neutron number
- Proton number
Example answer: momentum, charge and nucleon number.
Total: \(7\) marks