Question 1
A well has a depth of \(36\,\mathrm{m}\) from ground level to the surface of the water in the well, as shown in Fig. 1.1.
A student wishes to find the depth of the well. The student plans to drop a stone down the well and record the time taken from releasing the stone to hearing the splash made by the stone as it enters the water.
(a) Assume that air resistance is negligible and that the stone is released from rest.
Calculate the time taken for the stone to fall from ground level to the surface of the water. (2 marks)
time = ……………………………………………… \(\mathrm{s}\)
(b) The time recorded by the student using a stop-watch is not equal to the time in (a).
Suggest three possible reasons, other than the effect of air resistance, for this difference. (3 marks)
1. ________________________________________________
2. ________________________________________________
3. ________________________________________________
(c) The student repeats the experiment three times and uses the results to calculate the depth of the well. The values are shown in Table 1.1.
The true depth of the well is \(36.0\,\mathrm{m}\). Explain why these results may be described as precise but not accurate. (2 marks)
________________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 1.3: Errors and uncertainties — parts (b) and (c)
▶️ Answer/Explanation
(a) Time taken for the stone to fall [2 marks]
The stone is released from rest, so \(u=0\). Using the equation of motion
\(s=ut+\frac{1}{2}at^2\)
Here, \(s=36\,\mathrm{m}\), \(u=0\) and \(a=g=9.81\,\mathrm{m\,s^{-2}}\).
Therefore,
\(36=\frac{1}{2}(9.81)t^2\)
\(t=\sqrt{\frac{2(36)}{9.81}}\)
\(t=2.71\,\mathrm{s}\)
Answer: \(\boxed{2.7\,\mathrm{s}}\)
(b) Possible reasons for the difference [3 marks]
Any three valid reasons can be given. For example:
• There is a reaction time between hearing the splash and stopping the stopwatch.
• The sound of the splash takes time to travel from the water to the student.
• The student may not release the stone exactly from ground level.
Other valid reasons include:
• The student may not release the stone and start the stopwatch at exactly the same time.
• The stopwatch may not be properly calibrated or may have a zero error.
• The local value of \(g\) may not be exactly \(9.81\,\mathrm{m\,s^{-2}}\).
• The stone may have an initial velocity, so it is not released exactly from rest.
• The stone may not fall exactly vertically.
Answer: Any three suitable reasons, \(\boxed{1\text{ mark each}}\)
(c) Precision and accuracy [2 marks]
The three calculated depths are \(54.4\,\mathrm{m}\), \(53.9\,\mathrm{m}\) and \(54.1\,\mathrm{m}\).
These values are close together and have little scatter, so the results are precise.
However, the true depth is \(36.0\,\mathrm{m}\), which is significantly different from the measured values. Therefore, the results are not accurate.
Answer: \(\boxed{\text{Precise because the results are close together, but not accurate because they are far from the true value.}}\)
Question 2
A sphere floats in equilibrium on the surface of sea water of density \(1050\,\mathrm{kg\,m^{-3}}\), as shown in Fig. 2.1.
(a) \(21\%\) of the volume of the sphere is below the surface of the water.
Calculate the density of the sphere. [2]
density = ………………………………………. \(\mathrm{kg\,m^{-3}}\)
(b) The sphere is now held so that its entire volume is below the surface of the water. The sphere is then released.
(i) Calculate the initial acceleration of the sphere. [3]
acceleration = ………………………………………… \(\mathrm{m\,s^{-2}}\)
(ii) The sphere accelerates upwards but remains entirely below the surface of the water.
State and explain what happens to the acceleration of the sphere as its velocity begins to increase. [3]
________________________________________________
________________________________________________
________________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
▶️ Answer/Explanation
(a) Density of the sphere [2 marks]
When the sphere is floating in equilibrium, the upthrust is equal to the weight of the sphere.
The volume of water displaced is \(0.21V\), where \(V\) is the total volume of the sphere.
Therefore,
\(1050\times9.81\times0.21V=\rho V\times9.81\)
Cancelling \(V\) and \(9.81\),
\(\rho=1050\times0.21\)
\(\rho=220.5\,\mathrm{kg\,m^{-3}}\)
Answer: \(\boxed{220\,\mathrm{kg\,m^{-3}}}\)
(b)(i) Initial acceleration of the sphere [3 marks]
When the entire sphere is below the surface, the full volume \(V\) displaces water.
The upthrust is
\(F_{\mathrm{U}}=1050\times9.81\times V\)
The weight of the sphere is
\(W=220\times9.81\times V\)
The resultant upward force is therefore
\(F=(1050\times9.81V)-(220\times9.81V)\)
Using \(F=ma\),
\((1050\times9.81V)-(220\times9.81V)=(220V)a\)
\(a=\frac{(1050-220)\times9.81}{220}\)
\(a=37.0\,\mathrm{m\,s^{-2}}\)
Answer: \(\boxed{37\,\mathrm{m\,s^{-2}}}\)
(b)(ii) Variation of acceleration [3 marks]
As the sphere moves upwards, its velocity increases.
The downward drag / viscous force therefore increases with speed.
The upthrust and weight remain constant because the sphere remains entirely below the surface, so the resultant upward force decreases.
Since \(F=ma\), the acceleration decreases as the velocity increases.
Answer: \(\boxed{\text{The acceleration decreases.}}\)
Key formulae:
\(\mathrm{upthrust}=\rho gV\)
\(F=ma\)
Question 3
(a) State the principle of conservation of momentum. (2 marks)
___________________________
(b) A firework is initially stationary. It explodes into three fragments A, B and C that move in a horizontal plane, as shown in the view from above in Fig. 3.1.
Fragment A has a mass of \(3m\) and moves away from the explosion at a speed of \(4.0\,\mathrm{m\,s^{-1}}\).
Fragment B has a mass of \(2m\) and moves away from the explosion at a speed of \(6.0\,\mathrm{m\,s^{-1}}\) at right angles to the direction of A.
Fragment C has a mass of \(m\) and moves away from the explosion at a speed \(v\) and at an angle \(\theta\) as shown in Fig. 3.1.
Calculate:
(i) the angle \(\theta\). (3 marks)
\(\theta=\) ………………………………………………°
(ii) the speed \(v\). (2 marks)
\(v=\) ………………………………………… \(\mathrm{m\,s^{-1}}\)
(c) The firework in (b) contains a chemical that has mass \(5.0\,\mathrm{g}\) and has chemical energy per unit mass \(700\,\mathrm{J\,kg^{-1}}\). When the firework explodes, all of the chemical energy is transferred to the kinetic energy of fragments A, B and C.
(i) Show that the total chemical energy in the firework is \(3.5\,\mathrm{J}\). (1 mark)
________________________________________________
(ii) Calculate the mass \(m\). (3 marks)
\(m=\) ……………………………………………. \(\mathrm{kg}\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 5.1: Energy conservation — part (c)(i)
• 5.2: Gravitational potential energy and kinetic energy — part (c)(ii)
▶️ Answer/Explanation
(a) Principle of conservation of momentum [2 marks]
For an isolated system, the total momentum before an interaction is equal to the total momentum after the interaction.
Therefore,
\(\text{total momentum before}=\text{total momentum after}\)
Answer: \(\boxed{\text{Total momentum is conserved when there is no resultant external force.}}\)
(b)(i) Angle \(\theta\) [3 marks]
The firework is initially stationary, so its total momentum before the explosion is zero. Hence, the vector sum of the momenta of A, B and C must be zero.
Resolving the momentum components horizontally and vertically:
\(3m(4.0)=mv\sin\theta\)
\(v\sin\theta=12\)
For the perpendicular direction,
\(2m(6.0)=mv\cos\theta\)
\(v\cos\theta=12\)
Therefore,
\(v\sin\theta=v\cos\theta\)
\(\tan\theta=1\)
\(\theta=45^\circ\)
Answer: \(\boxed{45^\circ}\)
(b)(ii) Speed \(v\) [2 marks]
Using \(v\cos45^\circ=12\),
\(v=\frac{12}{\cos45^\circ}\)
\(v=16.97\,\mathrm{m\,s^{-1}}\)
Answer: \(\boxed{17\,\mathrm{m\,s^{-1}}}\)
(c)(i) Total chemical energy [1 mark]
Convert the mass into kilograms:
\(5.0\,\mathrm{g}=0.0050\,\mathrm{kg}\)
Chemical energy \(=\text{mass}\times\text{energy per unit mass}\)
\(E=0.0050\times700\)
\(E=3.5\,\mathrm{J}\)
Answer: \(\boxed{3.5\,\mathrm{J}}\)
(c)(ii) Mass \(m\) [3 marks]
All the chemical energy is transferred into the kinetic energy of the three fragments.
Therefore,
\(3.5=\frac{1}{2}(3m)(4.0)^2+\frac{1}{2}(2m)(6.0)^2+\frac{1}{2}(m)(17)^2\)
\(3.5=24m+36m+144.5m\)
\(3.5=204.5m\)
\(m=0.0171\,\mathrm{kg}\)
Answer: \(\boxed{m=0.017\,\mathrm{kg}}\)
Question 4
(a) For a progressive wave, state what is meant by the frequency. (1 mark)
________________________________________________
(b) A loudspeaker, microphone and cathode-ray oscilloscope (CRO) are arranged as shown in Fig. 4.1.
The loudspeaker is emitting a sound wave which is detected by the microphone and displayed on the screen of the CRO as shown in Fig. 4.2.
The time-base on the CRO is set to \(0.50\,\mathrm{ms\,cm^{-1}}\) and the y-gain is set to \(0.20\,\mathrm{V\,cm^{-1}}\).
Calculate:
(i) the frequency of the sound wave. (2 marks)
frequency = ……………………………………………. \(\mathrm{Hz}\)
(ii) the amplitude of the signal received by the CRO. (1 mark)
amplitude = ……………………………………………… \(\mathrm{V}\)
(c) The intensity of the sound wave in (b) is reduced to a quarter of its original intensity without a change in frequency. Assume that the amplitude of the signal received by the CRO is proportional to the amplitude of the sound wave. On Fig. 4.2, sketch the trace that is now seen on the screen of the CRO. (3 marks)
[Sketch on Fig. 4.2]
(d) A metal sheet is now placed in front of the loudspeaker in (b), as shown in Fig. 4.3.
A stationary wave is formed between the loudspeaker and the metal sheet.
(i) State the principle of superposition. (2 marks)
________________________________________________
(ii) The initial position of the microphone is such that the trace on the CRO has an amplitude minimum. It is now moved a distance of \(1.05\,\mathrm{m}\) away from the loudspeaker along the line joining the loudspeaker and metal sheet.
As the microphone moves, it passes through three positions where the trace has an amplitude maximum before ending at a position where the trace has an amplitude minimum.
Determine the wavelength of the sound wave. (2 marks)
wavelength = …………………………………………….. \(\mathrm{m}\)
(iii) Use your answers in (b)(i) and (d)(ii) to determine the speed of the sound in the air. (2 marks)
speed = ………………………………………… \(\mathrm{m\,s^{-1}}\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 8.1: Stationary waves — parts (d)(i), (d)(ii) and (d)(iii)
▶️ Answer/Explanation
(a) Frequency [1 mark]
Frequency is the number of oscillations per unit time of a source, point on the wave or particle of the medium.
Answer: \(\boxed{\text{number of oscillations per unit time}}\)
(b)(i) Frequency [2 marks]
From Fig. 4.2, one complete cycle occupies \(4.0\,\mathrm{cm}\).
Time-base \(=0.50\,\mathrm{ms\,cm^{-1}}\)
Therefore, the period is
\(T=4.0\times0.50\times10^{-3}\)
\(T=2.0\times10^{-3}\,\mathrm{s}\)
Using \(f=\frac{1}{T}\),
\(f=\frac{1}{2.0\times10^{-3}}\)
\(f=500\,\mathrm{Hz}\)
Answer: \(\boxed{500\,\mathrm{Hz}}\)
(b)(ii) Amplitude [1 mark]
From Fig. 4.2, the amplitude is \(2.8\,\mathrm{cm}\).
Using the y-gain of \(0.20\,\mathrm{V\,cm^{-1}}\),
\(\text{amplitude}=2.8\times0.20\)
\(\text{amplitude}=0.56\,\mathrm{V}\)
Answer: \(\boxed{0.56\,\mathrm{V}}\)
(c) Effect of reducing intensity [3 marks]
For a progressive wave,
\(I\propto A^2\)
The intensity is reduced to one quarter:
\(\frac{I_2}{I_1}=\frac{1}{4}\)
Therefore,
\(\frac{A_2}{A_1}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
Original amplitude \(=2.8\,\mathrm{cm}\), so the new amplitude is
\(A_2=1.4\,\mathrm{cm}\)
The frequency is unchanged, so the period and horizontal spacing of the trace remain unchanged. The new sinusoidal trace therefore has the same period but half the amplitude.
Sketch: a sinusoidal wave with the same period and amplitude \(1.4\,\mathrm{cm}\).
(d)(i) Principle of superposition [2 marks]
When two or more waves meet at a point, the resultant displacement is equal to the sum of the individual displacements.
Answer: \(\boxed{\text{Resultant displacement}=\text{sum of individual displacements}}\)
(d)(ii) Wavelength [2 marks]
In a stationary wave, the separation between adjacent nodes is
\(\frac{\lambda}{2}\)
The microphone starts at an amplitude minimum, which corresponds to a node. As it moves \(1.05\,\mathrm{m}\), it passes through three amplitude maxima and finishes at the next amplitude minimum.
Thus, it moves through three node-to-node separations:
\(1.05=3\left(\frac{\lambda}{2}\right)\)
\(1.05=1.5\lambda\)
\(\lambda=\frac{1.05}{1.5}\)
\(\lambda=0.70\,\mathrm{m}\)
Answer: \(\boxed{0.70\,\mathrm{m}}\)
(d)(iii) Speed of sound [2 marks]
Using the wave equation,
\(v=f\lambda\)
\(v=500\times0.70\)
\(v=350\,\mathrm{m\,s^{-1}}\)
Answer: \(\boxed{350\,\mathrm{m\,s^{-1}}}\)
Question 5
A student sets up a circuit with a battery, an ammeter, a heater and a light-dependent resistor (LDR) all in series.
The battery has negligible internal resistance.
A voltmeter is connected across (in parallel with) the heater.
(a) On Fig. 5.1, complete the circuit diagram of this arrangement. (3 marks)
Fig. 5.1
(b) The heater is a wire made of metal of resistivity \(1.1\times10^{-6}\,\Omega\,\mathrm{m}\). The wire has length \(2.0\,\mathrm{m}\) and cross-sectional area \(3.8\times10^{-7}\,\mathrm{m^2}\).
The reading on the voltmeter is \(4.8\,\mathrm{V}\).
Calculate:
(i) the resistance of the heater. (2 marks)
resistance = ………………………………………… \(\Omega\)
(ii) the reading on the ammeter. (1 mark)
reading on ammeter = ………………………….. \(\mathrm{A}\)
(c) The heater is replaced by a new wire. The new wire is made of the same metal as the wire in (b) and has the same length but a larger diameter.
The resistance of the LDR remains constant.
(i) State and explain whether the new wire has a resistance that is greater than, less than or the same as that of the wire in (b). (2 marks)
________________________________________________
________________________________________________
(ii) State and explain whether the new reading on the voltmeter is greater than, less than or equal to \(4.8\,\mathrm{V}\). (2 marks)
________________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 9.3: Resistance and resistivity — parts (b)(i) and (c)(i)
• 9.1: Electric current — part (b)(ii)
• 10.3: Potential dividers — part (c)(ii)
▶️ Answer/Explanation
(a) Circuit diagram [3 marks]
The battery, ammeter, heater and LDR are connected in series.
The voltmeter is connected in parallel across the heater.
Answer: \(\boxed{\text{Ammeter in series, heater and LDR in series, voltmeter in parallel across the heater.}}\)
(b)(i) Resistance of the heater [2 marks]
Using the resistivity equation,
\(R=\frac{\rho L}{A}\)
\(R=\frac{(1.1\times10^{-6})(2.0)}{3.8\times10^{-7}}\)
\(R=5.79\,\Omega\)
Answer: \(\boxed{5.8\,\Omega}\)
(b)(ii) Ammeter reading [1 mark]
The heater and LDR are in series, so the same current flows through the heater and the LDR.
Using \(V=IR\),
\(I=\frac{V}{R}\)
\(I=\frac{4.8}{5.79}\)
\(I=0.829\,\mathrm{A}\)
Answer: \(\boxed{0.83\,\mathrm{A}}\)
(c)(i) Resistance of the new wire [2 marks]
The new wire is made of the same metal, so its resistivity \(\rho\) is unchanged.
It also has the same length, but its diameter is larger, so its cross-sectional area \(A\) is larger.
Since
\(R=\frac{\rho L}{A}\)
a larger cross-sectional area gives a smaller resistance.
Answer: \(\boxed{\text{The resistance is less.}}\)
(c)(ii) New voltmeter reading [2 marks]
The resistance of the heater decreases while the resistance of the LDR remains constant.
Therefore, the total resistance of the series circuit decreases, so the current in the circuit increases.
The heater has a smaller resistance and therefore takes a smaller share of the total potential difference.
Hence, the potential difference across the heater decreases.
Answer: \(\boxed{\text{The voltmeter reading is less than }4.8\,\mathrm{V}.}\)
Key formulae:
\(R=\frac{\rho L}{A}\)
\(V=IR\)
\(A\propto d^2\)
Question 6
(a) Define the Young modulus. [1]
____________________________________________________________
(b) A uniform wire is suspended from a fixed support. Masses are added to the other end of the wire, as shown in Fig. 6.1.
Fig. 6.1
The variation of the length \(l\) of the wire with the force \(F\) applied to the wire by the masses is shown in Fig. 6.2.
Fig. 6.2
The cross-sectional area of the wire is \(0.95\,\mathrm{mm^2}\).
(i) Determine the unstretched length of the wire. [1]
unstretched length = …………………………………………….. \(\mathrm{m}\)
(ii) For an applied force \(F\) of \(30\,\mathrm{N}\), determine:
• the stress in the wire [3]
stress = ………………………………………………… \(\mathrm{Pa}\)
• the strain of the wire
strain = …………………………………………………
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 6.1 Stress and strain
• 6.2 Elastic and plastic behaviour
▶️ Answer/Explanation
(a) [1 mark]
Young modulus is the ratio of tensile stress to tensile strain:
\(E=\frac{\mathrm{stress}}{\mathrm{strain}}\)
Answer: \(\boxed{E=\frac{\mathrm{stress}}{\mathrm{strain}}}\)
(b)(i) Unstretched length [1 mark]
From the graph, when \(F=0\),
\(l=1.9980\,\mathrm{m}\)
Answer: \(\boxed{1.9980\,\mathrm{m}}\)
(b)(ii) Stress [2 marks]
Convert the cross-sectional area:
\(A=0.95\,\mathrm{mm^2}=0.95\times10^{-6}\,\mathrm{m^2}\)
Stress is given by
\(\mathrm{stress}=\frac{F}{A}\)
\(\mathrm{stress}=\frac{30}{0.95\times10^{-6}}\)
\(\mathrm{stress}=3.16\times10^{7}\,\mathrm{Pa}\)
Answer: \(\boxed{3.2\times10^{7}\,\mathrm{Pa}}\)
Strain [1 mark]
From the graph, at \(F=30\,\mathrm{N}\),
\(l=2.0030\,\mathrm{m}\)
Extension:
\(\Delta l=2.0030-1.9980=0.0050\,\mathrm{m}\)
Strain is
\(\mathrm{strain}=\frac{\Delta l}{l_0}\)
\(\mathrm{strain}=\frac{0.0050}{1.9980}\)
\(\mathrm{strain}=2.5\times10^{-3}\)
Answer: \(\boxed{2.5\times10^{-3}}\)
Key formulae:
\(\mathrm{stress}=\frac{F}{A}\)
\(\mathrm{strain}=\frac{\Delta l}{l_0}\)
\(E=\frac{\mathrm{stress}}{\mathrm{strain}}\)
Question 7
(a) Table 7.1 shows incomplete data for three flavours (types) of quark. The elementary charge is \(e\).
Complete Table 7.1 by inserting the missing charges. (2 marks)
| flavour | quark | antiquark | ||
|---|---|---|---|---|
| symbol | charge/\(e\) | symbol | charge/\(e\) | |
| up | \(u\) | \(+\frac{2}{3}\) | \(\bar{u}\) | ……………. |
| down | \(d\) | ……………. | \(\bar{d}\) | ……………. |
| charm | \(c\) | ……………. | \(\bar{c}\) | ……………. |
(b) Using the symbols given in Table 7.1, state a possible quark combination for the following hadrons:
(i) a neutral baryon (1 mark)
________________________________________________
(ii) a meson with a charge of \(+e\). (1 mark)
________________________________________________
(c) Quarks are fundamental particles.
Electrons are in another group (class) of fundamental particle.
(i) State the name of this group. (1 mark)
________________________________________________
(ii) State the name of another particle in this group. (1 mark)
________________________________________________
Syllabus Topic Code (Cambridge International AS & A Level Physics 9702):
▶️ Answer/Explanation
(a) Missing charges [2 marks]
The down quark has charge \(-\frac{1}{3}e\), while the charm quark has charge \(+\frac{2}{3}e\).
Antiquarks have the opposite sign and the same magnitude of charge as the corresponding quarks.
| flavour | quark charge/\(e\) | antiquark charge/\(e\) |
|---|---|---|
| up | \(+\frac{2}{3}\) | \(-\frac{2}{3}\) |
| down | \(-\frac{1}{3}\) | \(+\frac{1}{3}\) |
| charm | \(+\frac{2}{3}\) | \(-\frac{2}{3}\) |
(b)(i) Neutral baryon [1 mark]
A possible combination is \(udd\).
Charge \(=+\frac{2}{3}e-\frac{1}{3}e-\frac{1}{3}e=0\)
Answer: \(\boxed{udd}\)
(b)(ii) Meson with charge \(+e\) [1 mark]
A possible combination is \(u\bar{d}\).
Charge \(=+\frac{2}{3}e+\frac{1}{3}e=+e\)
Answer: \(\boxed{u\bar{d}}\)
(c)(i) [1 mark]
Electrons belong to the group of leptons.
Answer: \(\boxed{\text{lepton}}\)
(c)(ii) [1 mark]
Another particle in this group may be a positron, neutrino or antineutrino.
Answer: \(\boxed{\text{positron / neutrino / antineutrino}}\)
Key facts:
\(u=+\frac{2}{3}e\)
\(d=-\frac{1}{3}e\)
\(c=+\frac{2}{3}e\)
Antiquarks have the opposite charge to their corresponding quarks.