Question 1
(a) Table 1.1 lists some SI quantities. Complete the table by indicating with a tick (✓) which rows are SI base quantities. (1 mark)
(b) Use the definition of power to determine its SI base units. (2 marks)
SI base units = ________________________________________________
(c) A light meter is used to measure the intensity of light in a classroom. Daylight is incident normally on the sensor of the meter. The sensor has an area of \(2.2\,\mathrm{cm^2}\). The reading on the meter is \(950\,\mathrm{W\,m^{-2}}\).
Calculate the power of the daylight incident on the sensor. (3 marks)
power = ________________________________________________ \( \mathrm{W} \)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
▶️ Answer/Explanation
(a) SI base quantities [1 mark]
The SI base quantities listed in the table are current and mass.
Therefore, the table should be completed as follows:
| Quantity | Base quantity |
|---|---|
| Current | ✓ |
| Energy | |
| Force | |
| Mass | ✓ |
Answer: Current and mass
(b) SI base units of power [2 marks]
Power is defined as the rate of doing work:
\( P=\frac{W}{t} \)
The SI unit of work is the joule, so:
\( \mathrm{unit\ of\ power}=\mathrm{J\,s^{-1}} \)
Since:
\( \mathrm{J}=\mathrm{kg\,m^2\,s^{-2}} \)
Therefore:
\( \mathrm{J\,s^{-1}}=\mathrm{kg\,m^2\,s^{-2}\,s^{-1}} \)
\( \mathrm{J\,s^{-1}}=\mathrm{kg\,m^2\,s^{-3}} \)
Answer: \( \boxed{\mathrm{kg\,m^2\,s^{-3}}} \)
(c) Power of daylight incident on the sensor [3 marks]
The intensity is power per unit area:
\( I=\frac{P}{A} \)
Rearranging:
\( P=IA \)
The sensor area must be converted from \(\mathrm{cm^2}\) to \(\mathrm{m^2}\):
\( A=2.2\,\mathrm{cm^2}=2.2\times10^{-4}\,\mathrm{m^2} \)
Using \(I=950\,\mathrm{W\,m^{-2}}\):
\( P=(950)(2.2\times10^{-4}) \)
\( P=0.209\,\mathrm{W} \)
To an appropriate number of significant figures:
\( P=0.21\,\mathrm{W} \)
Answer: \( \boxed{0.21\,\mathrm{W}} \)
Question 2
(a) Define acceleration. (1 mark)
______________________
(b) An Olympic diver stands on a platform above a pool of water, as shown in Fig. 2.1.
When the diver is on the platform his centre of gravity is a vertical height of \(9.0\,\mathrm{m}\) above the surface of the water. The diver jumps from the platform with a velocity of \(5.9\,\mathrm{m\,s^{-1}}\) at an angle of \(60^\circ\) to the horizontal.
Air resistance is negligible.
When the diver hits the surface of the water, his centre of gravity is a vertical height of \(1.2\,\mathrm{m}\) above the surface of the water.
Calculate the speed of the diver at the instant he hits the surface of the water. (3 marks)
speed = ________________________________________________ \( \mathrm{m\,s^{-1}} \)
(c) The diver in (b) enters the water and decelerates.
(i) Describe and explain the variation of the viscous drag force acting on the diver in the water as he moves downwards. (2 marks)
____________________________________________________________
____________________________________________________________
____________________________________________________________
(ii) The diver has a volume of \(7.5\times10^{-2}\,\mathrm{m^3}\). The density of the water is \(1.0\times10^3\,\mathrm{kg\,m^{-3}}\).
Show that the upthrust acting on the diver when he is entirely underwater is \(740\,\mathrm{N}\). (1 mark)
____________________________________________________________
(iii) At a particular instant when the diver is entirely underwater his horizontal velocity is zero. The viscous drag force acting on him at this instant is \(950\,\mathrm{N}\) vertically upwards. The diver has mass \(78\,\mathrm{kg}\).
Determine the magnitude and direction of the acceleration of the diver. (4 marks)
acceleration = ________________________________________________ \( \mathrm{m\,s^{-2}} \)
direction = ________________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 3.1: Momentum and Newton’s laws of motion — part (c)(iii)
• 3.2: Non-uniform motion — part (c)(i)
• 4.3: Density and pressure — part (c)(ii)
▶️ Answer/Explanation
(a) Definition of acceleration [1 mark]
Acceleration is the rate of change of velocity.
Answer: \( \boxed{\text{rate of change of velocity}} \)
(b) Speed of the diver when he reaches the water [3 marks]
The diver falls through a vertical distance of
\( \Delta h=9.0-1.2=7.8\,\mathrm{m} \)
Using the equation of motion for the vertical component, or equivalently conservation of mechanical energy:
\( v^2=u^2+2a\Delta h \)
The initial speed is \(5.9\,\mathrm{m\,s^{-1}}\), and the vertical acceleration is \(9.81\,\mathrm{m\,s^{-2}}\).
\( v^2=(5.9)^2+2(9.81)(7.8) \)
\( v^2=188 \)
\( v=\sqrt{188}=13.7\,\mathrm{m\,s^{-1}} \)
To two significant figures:
\( v=14\,\mathrm{m\,s^{-1}} \)
Answer: \( \boxed{14\,\mathrm{m\,s^{-1}}} \)
(c)(i) Variation of viscous drag force [2 marks]
As the diver moves downwards through the water, he decelerates, so his speed decreases.
The viscous drag force depends on the speed of the diver. Therefore, as the speed decreases, the viscous drag force also decreases.
Answer: The speed decreases, so the viscous drag force decreases.
(c)(ii) Upthrust on the diver [1 mark]
The upthrust is equal to the weight of water displaced:
\( F=\rho gV \)
Substituting \( \rho=1.0\times10^3\,\mathrm{kg\,m^{-3}} \), \(g=9.81\,\mathrm{m\,s^{-2}}\) and \(V=7.5\times10^{-2}\,\mathrm{m^3}\):
\( F=(1.0\times10^3)(9.81)(7.5\times10^{-2}) \)
\( F=735.75\,\mathrm{N} \)
Therefore, to an appropriate number of significant figures:
\( F=740\,\mathrm{N} \)
Answer: \( \boxed{740\,\mathrm{N}} \)
(c)(iii) Acceleration of the diver [4 marks]
The forces acting vertically on the diver are:
• Upthrust \(=740\,\mathrm{N}\) upwards
• Viscous drag \(=950\,\mathrm{N}\) upwards
• Weight \(=mg\) downwards
The weight of the diver is:
\( W=mg=(78)(9.81) \)
\( W=765.18\,\mathrm{N} \)
Taking upwards as positive, the resultant force is:
\( F=740+950-765.18 \)
\( F=924.82\,\mathrm{N} \)
Using Newton’s second law:
\( F=ma \)
\( a=\frac{F}{m} \)
\( a=\frac{924.82}{78} \)
\( a=11.9\,\mathrm{m\,s^{-2}} \)
To two significant figures:
\( a=12\,\mathrm{m\,s^{-2}} \)
Since the resultant force is upwards, the acceleration is vertically upwards.
Answer: \( \boxed{12\,\mathrm{m\,s^{-2}}} \), vertically upwards
Question 3
A thin metal wire X, of diameter \(1.2\times10^{-3}\,\mathrm{m}\), is used to suspend a model planet, as shown in Fig. 3.1.
The variation with strain of the stress for wire X is shown in Fig. 3.2.
(a) The strain in X is \(5.4\times10^{-3}\).
(i) Use Fig. 3.2 to calculate the force exerted on the wire by the model planet. (3 marks)
force = ________________________________________________ \( \mathrm{N} \)
(ii) The elastic potential energy of X is \(0.31\,\mathrm{J}\).
Calculate the original length of the wire before the model planet was attached. (3 marks)
original length = ________________________________________________ \( \mathrm{m} \)
(b) Wire X is replaced by a new wire, Y, with the same original length and diameter but double the Young modulus of X. Wire Y also obeys Hooke’s law.
On Fig. 3.2, draw a line representing the variation with strain of the stress for Y. (2 marks)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 6.2: Elastic and plastic behaviour — part (a)(ii)
▶️ Answer/Explanation
(a)(i) Force exerted on the wire [3 marks]
From Fig. 3.2, when the strain is \(5.4\times10^{-3}\), the stress is approximately:
\( \sigma=0.72\times10^9\,\mathrm{Pa} \)
Stress is defined by:
\( \sigma=\frac{F}{A} \)
Therefore:
\( F=\sigma A \)
The cross-sectional area of the wire is:
\( A=\pi r^2 \)
\( r=\frac{1.2\times10^{-3}}{2}=0.6\times10^{-3}\,\mathrm{m} \)
Hence:
\( F=(0.72\times10^9)\pi(0.6\times10^{-3})^2 \)
\( F=814\,\mathrm{N} \)
To an appropriate number of significant figures:
\( F=810\,\mathrm{N} \)
Answer: \( \boxed{810\,\mathrm{N}} \)
(a)(ii) Original length of the wire [3 marks]
The elastic potential energy stored in the wire is:
\( E_{\mathrm{P}}=\frac{1}{2}Fx \)
Rearranging for the extension \(x\):
\( x=\frac{2E_{\mathrm{P}}}{F} \)
Substituting \(E_{\mathrm{P}}=0.31\,\mathrm{J}\) and \(F=810\,\mathrm{N}\):
\( x=\frac{2(0.31)}{810} \)
\( x=7.65\times10^{-4}\,\mathrm{m} \)
Strain is given by:
\( \varepsilon=\frac{x}{L} \)
Therefore:
\( L=\frac{x}{\varepsilon} \)
\( L=\frac{7.65\times10^{-4}}{5.4\times10^{-3}} \)
\( L=0.142\,\mathrm{m} \)
\( L\approx0.14\,\mathrm{m} \)
Answer: \( \boxed{0.14\,\mathrm{m}} \)
(b) Stress-strain graph for wire Y [2 marks]
For a material obeying Hooke’s law, the Young modulus is the gradient of the stress-strain graph:
\( E=\frac{\sigma}{\varepsilon} \)
Wire Y has double the Young modulus of wire X. Therefore, the gradient of the line for Y must be twice the gradient of the line for X.
The required line should therefore:
• pass through the origin, and
• have a gradient twice as large as the line for wire X.
Answer: A straight line through the origin with twice the gradient of the original line.
Question 4
A nucleus P undergoes \(\alpha\)-decay to form nucleus Q.
(a) Complete the equation for this decay. (2 marks)
(b)
(i) State the principle of conservation of momentum. (2 marks)
____________________________________________________________
____________________________________________________________
(ii) Before the decay, nucleus P has a speed of \(3.2\times10^5\,\mathrm{m\,s^{-1}}\). After the decay, nucleus Q is stationary.
Calculate the speed of the alpha particle after the decay. (2 marks)
speed = ________________________________________________ \( \mathrm{m\,s^{-1}} \)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 3.3: Linear momentum and its conservation — parts (b)(i) and (b)(ii)
▶️ Answer/Explanation
(a) Completing the nuclear decay equation [2 marks]
In \(\alpha\)-decay, the nucleus emits an alpha particle, which is a helium nucleus with nucleon number \(4\) and proton number \(2\).
Conservation of nucleon number gives:
\( A_Q=215-4=211 \)
Conservation of proton number gives:
\( Z_Q=84-2=82 \)
Therefore, the completed equation contains \(^{211}_{82}\mathrm{Q}\) and \(^{4}_{2}\mathrm{\alpha}\).
Answer: \( \boxed{^{215}_{84}\mathrm{P}\rightarrow{}^{211}_{82}\mathrm{Q}+{}^{4}_{2}\mathrm{\alpha}} \)
(b)(i) Principle of conservation of momentum [2 marks]
The total momentum of an isolated system remains constant.
Equivalently:
\( \text{total momentum before}=\text{total momentum after} \)
Answer: The total momentum of an isolated system remains constant.
(b)(ii) Speed of the alpha particle [2 marks]
Using conservation of momentum:
\( p_{\mathrm{P}}=p_{\mathrm{Q}}+p_{\alpha} \)
Since nucleus Q is stationary, \(p_{\mathrm{Q}}=0\).
Therefore:
\( p_{\alpha}=p_{\mathrm{P}} \)
Using \(p=mv\):
\( (4u)v=(215u)(3.2\times10^5) \)
where \(u\) is the atomic mass unit.
Hence:
\( v=\frac{215(3.2\times10^5)}{4} \)
\( v=1.72\times10^7\,\mathrm{m\,s^{-1}} \)
Answer: \( \boxed{1.7\times10^7\,\mathrm{m\,s^{-1}}} \)
Question 5
(a) By reference to the direction of propagation of energy, state what is meant by a transverse wave. (1 mark)
____________________________________________________________
(b) A space telescope is designed to detect electromagnetic radiation with wavelengths in the range \(12\,\mu\mathrm{m}\) to \(28\,\mu\mathrm{m}\).
State the region of the electromagnetic spectrum for this radiation. (1 mark)
____________________________________________________________
(c) A detector on another space telescope detects an electromagnetic wave. The signal from the detector is transmitted to Earth and displayed on an oscilloscope as shown in Fig. 5.1. The frequency of the signal displayed on the oscilloscope is equal to the frequency of the detected electromagnetic wave.
The time-base setting on the oscilloscope is \(5.0\times10^{-15}\,\mathrm{s\,cm^{-1}}\).
Calculate the wavelength of the detected electromagnetic wave. (3 marks)
wavelength = ________________________________________________ \( \mathrm{m} \)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 7.4: Electromagnetic spectrum — part (b)
• 7.1: Progressive waves — part (c)
▶️ Answer/Explanation
(a) Definition of a transverse wave [1 mark]
In a transverse wave, the vibrations or oscillations are perpendicular to the direction of propagation of energy.
Answer: \( \boxed{\text{vibrations are perpendicular to the direction of propagation of energy}} \)
(b) Region of the electromagnetic spectrum [1 mark]
Wavelengths from \(12\,\mu\mathrm{m}\) to \(28\,\mu\mathrm{m}\) lie in the infrared region of the electromagnetic spectrum.
Answer: \( \boxed{\text{infrared}} \)
(c) Wavelength of the detected electromagnetic wave [3 marks]
From Fig. 5.1, one complete cycle occupies \(6.0\,\mathrm{cm}\) horizontally.
The time-base setting is \(5.0\times10^{-15}\,\mathrm{s\,cm^{-1}}\).
Therefore, the period is:
\( T=6.0\times5.0\times10^{-15} \)
\( T=3.0\times10^{-14}\,\mathrm{s} \)
For an electromagnetic wave:
\( \lambda=cT \)
Using \(c=3.0\times10^8\,\mathrm{m\,s^{-1}}\):
\( \lambda=(3.0\times10^8)(3.0\times10^{-14}) \)
\( \lambda=9.0\times10^{-6}\,\mathrm{m} \)
Answer: \( \boxed{9.0\times10^{-6}\,\mathrm{m}} \)
Question 6
(a) Coherent visible light of a single frequency is incident normally on a double slit. This produces a pattern of bright and dark interference fringes on a screen, as illustrated in Fig. 6.1.
There are seven bright fringes.
(i) Explain how the pattern of bright and dark interference fringes is formed. (3 marks)
____________________________________________________________
____________________________________________________________
____________________________________________________________
(ii) The distance between the centres of bright fringe X and bright fringe Y in the pattern is \(10.2\,\mathrm{mm}\). The slit spacing is \(1.2\,\mathrm{mm}\). The distance from the slits to the screen is \(3.1\,\mathrm{m}\).
Calculate the wavelength of the light incident on the slits. (3 marks)
wavelength = ________________________________________________ \( \mathrm{m} \)
(iii) The light is replaced by different visible light with a shorter wavelength.
State how the new fringe separation will compare to the original fringe separation. (1 mark)
____________________________________________________________
(b) A stationary wave is formed on a stretched string AB, as shown in Fig. 6.2.
P, Q and R are points on the string.
(i) On Fig. 6.2, draw a cross (×) to show the position of a node. (1 mark)
(ii) State the phase difference between P and Q.
phase difference = ________________________________________________ \( ^\circ \) (1 mark)
(iii) State the phase difference between P and R.
phase difference = ________________________________________________ \( ^\circ \) (1 mark)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 8.1: Stationary waves — parts (b)(i), (b)(ii) and (b)(iii)
▶️ Answer/Explanation
(a)(i) Formation of interference fringes [3 marks]
Light diffracts at each of the two slits.
The light waves from the two slits meet and superpose at the screen.
When the waves arrive in phase, the phase difference is \(0^\circ\), producing a bright fringe or intensity maximum.
When the waves arrive in antiphase, the phase difference is \(180^\circ\), producing a dark fringe or intensity minimum.
Answer: The two diffracted waves superpose. In-phase waves produce bright fringes, while waves with a phase difference of \(180^\circ\) produce dark fringes.
(a)(ii) Wavelength of the light [3 marks]
There are seven bright fringes, so the distance from bright fringe X to bright fringe Y corresponds to six fringe spacings.
Therefore, the fringe separation is:
\( x=\frac{10.2\times10^{-3}}{6} \)
\( x=1.70\times10^{-3}\,\mathrm{m} \)
For double-slit interference:
\( x=\frac{\lambda D}{a} \)
Therefore:
\( \lambda=\frac{ax}{D} \)
Substituting \(a=1.2\times10^{-3}\,\mathrm{m}\), \(x=1.70\times10^{-3}\,\mathrm{m}\) and \(D=3.1\,\mathrm{m}\):
\( \lambda=\frac{(1.2\times10^{-3})(1.70\times10^{-3})}{3.1} \)
\( \lambda=6.58\times10^{-7}\,\mathrm{m} \)
\( \lambda=6.6\times10^{-7}\,\mathrm{m} \)
Answer: \( \boxed{6.6\times10^{-7}\,\mathrm{m}} \)
(a)(iii) Effect of shorter wavelength [1 mark]
The fringe separation is proportional to wavelength:
\( x\propto\lambda \)
Therefore, a shorter wavelength produces a smaller fringe separation.
Answer: \( \boxed{\text{smaller}} \)
(b)(i) Position of a node [1 mark]
A node is a point on a stationary wave that has zero displacement at all times.
On Fig. 6.2, the node is at the point where the string crosses the mean position between two adjacent antinodes.
Answer: Place the cross at the intersection of the string and the mean-position line between two adjacent loops.
(b)(ii) Phase difference between P and Q [1 mark]
P and Q lie within the same loop of the stationary wave, so they oscillate in phase.
Therefore:
Answer: \( \boxed{0^\circ} \)
(b)(iii) Phase difference between P and R [1 mark]
P and R lie in adjacent loops of the stationary wave. Points in adjacent loops oscillate in antiphase.
Therefore:
Answer: \( \boxed{180^\circ} \)
Question 7
(a) Define electric potential difference. (1 mark)
____________________________________________________________
(b) A cell of electromotive force (e.m.f.) \(1.8\,\mathrm{V}\) and internal resistance \(r\) is connected in parallel with a resistor of resistance \(6.0\,\Omega\) and a filament lamp, as shown in Fig. 7.1.
The switch S is open. The ammeter reading is \(0.25\,\mathrm{A}\).
Determine the internal resistance of the cell. (3 marks)
\(r=\) ________________________________________________ \( \Omega \)
(c) At time \(t_1\) switch S in Fig. 7.1 is closed. Fig. 7.2 shows the variation with time \(t\) of the ammeter reading \(I\).
(i) State whether the e.m.f. of the cell after \(t_1\) is greater than, less than or the same as it was before \(t_1\). (1 mark)
____________________________________________________________
(ii) By considering the effect of the lamp on the total resistance of the circuit, explain the variation of the ammeter reading shown in Fig. 7.2. (3 marks)
____________________________________________________________
____________________________________________________________
____________________________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 10.1: Practical circuits — parts (b), (c)(i) and (c)(ii)
▶️ Answer/Explanation
(a) Definition of electric potential difference [1 mark]
Electric potential difference is the energy transferred per unit charge.
\( V=\frac{W}{Q} \)
Answer: \( \boxed{\text{energy transferred per unit charge}} \)
(b) Internal resistance of the cell [3 marks]
Since the switch S is open, only the \(6.0\,\Omega\) resistor is connected across the cell.
The potential difference across the resistor is:
\( V=IR \)
\( V=(0.25)(6.0) \)
\( V=1.5\,\mathrm{V} \)
The lost volts across the internal resistance are:
\( Ir=\mathcal{E}-V \)
\( (0.25)r=1.8-1.5 \)
\( (0.25)r=0.3 \)
Therefore:
\( r=\frac{0.3}{0.25} \)
\( r=1.2\,\Omega \)
Answer: \( \boxed{1.2\,\Omega} \)
(c)(i) E.m.f. after \(t_1\) [1 mark]
The e.m.f. of the cell does not change simply because the switch is closed.
Answer: \( \boxed{\text{the same}} \)
(c)(ii) Variation of the ammeter reading [3 marks]
Before \(t_1\), the current is constant, so the total resistance of the circuit is constant.
At \(t_1\), the switch is closed and the lamp is connected in parallel with the \(6.0\,\Omega\) resistor. This causes the external resistance, and hence the total resistance of the circuit, to decrease.
The decrease in total resistance causes the current to increase suddenly, producing the sharp rise in the ammeter reading.
The increased current causes the temperature of the filament lamp to rise. The resistance of the lamp therefore increases, causing the total resistance of the circuit to increase.
As the total resistance increases, the current decreases gradually until it reaches a new steady value.
Answer: The lamp initially decreases the total resistance, so the current increases. As the lamp heats up, its resistance increases, causing the total resistance to increase and the current to decrease gradually.
Question 8
(a) State the name of the class (group) of fundamental particles that contains a neutrino. (1 mark)
____________________________________________________________
(b) A hadron P has a charge of \(+1e\), where \(e\) is the elementary charge. The hadron P is composed of a down antiquark and only one other quark.
(i) Identify a possible flavour for this other quark. (1 mark)
____________________________________________________________
(ii) State what type of hadron is P. (1 mark)
____________________________________________________________
(c) Nucleus Q undergoes radioactive decay to form nucleus R, emitting an antineutrino and another particle X, as shown in the decay equation.
(i) State what particle is represented by X. (1 mark)
____________________________________________________________
(ii) Compare the nucleon numbers of Q and R. (1 mark)
____________________________________________________________
(iii) Compare the charges of Q and R. (1 mark)
____________________________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
▶️ Answer/Explanation
(a) Class of fundamental particles containing a neutrino [1 mark]
A neutrino is a member of the lepton class of fundamental particles.
Answer: \( \boxed{\text{lepton}} \)
(b)(i) Possible flavour of the other quark [1 mark]
A down antiquark has charge \(+\frac{1}{3}e\).
The hadron has total charge \(+1e\), so the other quark must have charge:
\( +1e-\frac{1}{3}e=+\frac{2}{3}e \)
Quarks with charge \(+\frac{2}{3}e\) include up, charm and top quarks.
Answer: \( \boxed{\text{up, charm or top}} \)
(b)(ii) Type of hadron [1 mark]
A hadron composed of a quark and an antiquark is a meson.
Answer: \( \boxed{\text{meson}} \)
(c)(i) Particle represented by X [1 mark]
The emission of an antineutrino indicates \(\beta^-\) decay. The particle emitted in \(\beta^-\) decay is an electron.
Answer: \( \boxed{\beta^- \text{ particle (electron)}} \)
(c)(ii) Comparison of nucleon numbers [1 mark]
In \(\beta^-\) decay, a neutron changes into a proton, but the total number of nucleons remains unchanged.
Answer: The nucleon numbers of Q and R are \( \boxed{\text{equal}} \).
(c)(iii) Comparison of charges [1 mark]
In \(\beta^-\) decay, a neutron changes into a proton. Therefore, the nuclear charge increases by one elementary charge.
Answer: The charge of R is \( \boxed{\text{greater than that of Q}} \).