Question 1
Scientists are investigating the variation in air pressure at different locations on a mountain.
(a) The scientists make measurements of several physical quantities at each location.
Complete Table 1.1 by stating the SI base unit for each quantity and identifying with a tick (\(\checkmark\)) whether each quantity is a scalar or a vector. Use the space for any working.
Table 1.1
| quantity measured | SI base unit | scalar | vector |
|---|---|---|---|
| air temperature | |||
| air pressure |
(b) (i) At one location, the density of the air is \(1.1\,\mathrm{kg\,m^{-3}}\). A spherical weather balloon is filled with a gas and released from rest. The balloon has radius \(0.90\,\mathrm{m}\).
Calculate the upthrust acting on the balloon. (2 marks)
upthrust = __________________________________________ \( \mathrm{N} \)
(ii) Explain why an upthrust acts on the balloon. (2 marks)
________________________________
________________________________
________________________________
(iii) The balloon has weight \(19\,\mathrm{N}\).
Calculate the magnitude of the initial acceleration of the balloon. (3 marks)
acceleration = __________________________________________ \( \mathrm{m\,s^{-2}} \)
(c) A quantity \(c\) relating to the motion of the balloon is calculated from three measured quantities \(k\), \(F\) and \(v\) using the formula
\(c=\dfrac{2kF}{v^2}\)
The percentage uncertainties in the measured quantities are given in Table 1.2.
Table 1.2
| measured quantity | percentage uncertainty |
|---|---|
| \(k\) | 5% |
| \(F\) | 3% |
| \(v\) | 4% |
The calculated value of \(c\) is \(1.8\).
Determine the absolute uncertainty in \(c\). (2 marks)
absolute uncertainty = __________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 1.2: SI units — part (a)
• 1.3: Errors and uncertainties — part (c)
• 1.4: Scalars and vectors — part (a)
• 3.1: Momentum and Newton’s laws of motion — part (b)(iii)
• 4.3: Density and pressure — parts (b)(i) and (b)(ii)
▶️ Answer/Explanation
(a) SI base units and scalar/vector quantities [2 marks]
Air temperature has the SI base unit kelvin, \( \mathrm{K} \).
Air pressure has the SI base units \( \mathrm{kg\,m^{-1}\,s^{-2}} \).
Both air temperature and air pressure are scalars, because they have magnitude but no direction.
Answer:
| quantity | SI base unit | scalar | vector |
|---|---|---|---|
| air temperature | \(\mathrm{K}\) | \(\checkmark\) | |
| air pressure | \(\mathrm{kg\,m^{-1}\,s^{-2}}\) | \(\checkmark\) |
(b)(i) Upthrust on the balloon [2 marks]
The upthrust is equal to the weight of air displaced by the balloon:
\(F_{\mathrm{up}}=\rho Vg\)
For a sphere, \(V=\dfrac{4}{3}\pi r^3\).
Therefore,
\(F_{\mathrm{up}}=1.1\times9.81\times\dfrac{4}{3}\pi(0.90)^3\)
\(F_{\mathrm{up}}=33\,\mathrm{N}\)
Answer: \( \boxed{33\,\mathrm{N}} \)
(b)(ii) Explanation of upthrust [2 marks]
There is a difference in height or depth between the top and bottom of the balloon, so there is a difference in pressure between the top and bottom.
The pressure is greater at the bottom, so the upward force on the bottom of the balloon is greater than the downward force on the top.
Therefore, the resultant force is upwards and this is the upthrust.
(b)(iii) Initial acceleration of the balloon [3 marks]
The upthrust is \(33\,\mathrm{N}\) and the weight is \(19\,\mathrm{N}\), so the resultant force is
\(\Sigma F=33-19\)
\(\Sigma F=14\,\mathrm{N}\)
The mass of the balloon is obtained from \(W=mg\):
\(m=\dfrac{19}{9.81}\)
\(m=1.94\,\mathrm{kg}\)
Using \(F=ma\),
\(a=\dfrac{33-19}{19/9.81}\)
\(a=7.2\,\mathrm{m\,s^{-2}}\)
Answer: \( \boxed{7.2\,\mathrm{m\,s^{-2}}} \)
(c) Absolute uncertainty in \(c\) [2 marks]
The quantity is
\(c=\dfrac{2kF}{v^2}\)
For multiplication and division, percentage uncertainties are added. Since \(v\) is squared, its percentage uncertainty is doubled.
Percentage uncertainty in \(c=5+3+(2\times4)\)
\(=16\%\)
The absolute uncertainty is
\(\text{absolute uncertainty}=1.8\times0.16\)
\(=0.288\)
Answer: \( \boxed{\pm0.3} \)
Question 2
A spacecraft in deep space uses jets of hot gas from its thrusters to change its velocity. Fig. 2.1 shows a side view of the spacecraft and some of its thrusters.
Thruster A is a distance of \(1.6\,\mathrm{m}\) leftwards from the centre of gravity of the spacecraft. Thruster C is a distance of \(0.40\,\mathrm{m}\) upwards from the centre of gravity of the spacecraft.
Thrusters A and B can produce forces on the spacecraft in the upwards direction only. Thruster C can produce a force on the spacecraft in the leftwards direction only.
All the thrusters produce forces entirely in the plane of the centre of gravity.
(a) (i) Thruster A is activated, producing a force of \(60\,\mathrm{N}\) upwards on the spacecraft. Thruster C is also activated, producing a force of \(220\,\mathrm{N}\) in the leftwards direction on the spacecraft.
Calculate the resultant moment due to these forces about the centre of gravity. (2 marks)
resultant moment = __________________________________________ \( \mathrm{N\,m} \)
(ii) State and explain whether the forces from A and C are a couple. (1 mark)
________________________________
________________________________
(b) Thrusters A and C are now switched off and the spacecraft is stationary. Thruster B is activated at time \(t_1\), producing a constant force on the spacecraft until the fuel runs out at time \(t_2\). As the fuel is used, the total mass of the spacecraft decreases.
On Fig. 2.2, sketch the variation of speed of the spacecraft with time from \(t_1\) to \(t_2\). (2 marks)
(c) The spacecraft now splits apart into a carrier and a payload as shown in Fig. 2.3.
During the split, an average force of \(5500\,\mathrm{N}\) acts on the payload for a time of \(0.36\,\mathrm{s}\). The velocity of the payload increases by \(8.5\,\mathrm{m\,s^{-1}}\) in the upwards direction.
The combined mass of the carrier and payload is \(2.5\times10^3\,\mathrm{kg}\).
(i) State the principle of conservation of momentum. (2 marks)
________________________________
________________________________
(ii) Show that the mass of the payload is \(230\,\mathrm{kg}\). (2 marks)
________________________________
(iii) Calculate the magnitude of the change in velocity of the carrier. (3 marks)
change in velocity = __________________________________________ \( \mathrm{m\,s^{-1}} \)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 3.2: Non-uniform motion — part (b)
• 3.3: Linear momentum and its conservation — parts (c)(i), (c)(ii) and (c)(iii)
▶️ Answer/Explanation
(a)(i) Resultant moment about the centre of gravity [2 marks]
The moment due to thruster A is
\(M_A=60\times1.6\)
\(M_A=96\,\mathrm{N\,m}\)
The moment due to thruster C is
\(M_C=220\times0.40\)
\(M_C=88\,\mathrm{N\,m}\)
These moments act in opposite directions, so
\(M_{\mathrm{resultant}}=96-88\)
\(M_{\mathrm{resultant}}=8.0\,\mathrm{N\,m}\)
Answer: \( \boxed{8.0\,\mathrm{N\,m}} \)
(a)(ii) Whether the forces form a couple [1 mark]
The forces do not form a couple because the resultant force is not zero. The forces are also not equal in magnitude and do not act along parallel lines in opposite directions.
Answer: \( \boxed{\text{They are not a couple.}} \)
(b) Variation of speed with time [2 marks]
The thrust force is constant, but the mass of the spacecraft decreases as fuel is used.
Using \(F=ma\),
\(a=\dfrac{F}{m}\)
As \(m\) decreases while \(F\) remains constant, the acceleration increases.
Therefore, the speed-time graph starts at \( (t_1,0) \) and has an increasing positive gradient from \(t_1\) to \(t_2\).
Answer: \( \boxed{\text{A curve with an increasing positive gradient from }t_1\text{ to }t_2} \)
(c)(i) Principle of conservation of momentum [2 marks]
For an isolated system with no resultant external force, the total momentum remains constant.
Therefore,
\(\text{total momentum before}=\text{total momentum after}\)
Answer: \( \boxed{\text{The total momentum of an isolated system remains constant.}} \)
(c)(ii) Mass of the payload [2 marks]
The impulse acting on the payload is
\(\Delta p=F\Delta t\)
The change in momentum is also
\(\Delta p=m\Delta v\)
Therefore,
\(m=\dfrac{F\Delta t}{\Delta v}\)
\(m=\dfrac{5500\times0.36}{8.5}\)
\(m=230\,\mathrm{kg}\)
Answer: \( \boxed{230\,\mathrm{kg}} \)
(c)(iii) Change in velocity of the carrier [3 marks]
The change in momentum of the payload is
\(\Delta p_{\mathrm{payload}}=5500\times0.36\)
\(\Delta p_{\mathrm{payload}}=1980\,\mathrm{N\,s}\)
By conservation of momentum, the carrier has an equal change in momentum in the opposite direction.
The mass of the carrier is
\(m_{\mathrm{carrier}}=2.5\times10^3-230\)
\(m_{\mathrm{carrier}}=2270\,\mathrm{kg}\)
Hence,
\(1980=2270\Delta v\)
\(\Delta v=0.87\,\mathrm{m\,s^{-1}}\)
Answer: \( \boxed{0.87\,\mathrm{m\,s^{-1}}} \)
Question 3
A spring is fixed at one end and attached to the frame of a pulley at the other end. A cable is passed around the wheel of the pulley. The spring is stretched to a fixed length using the cable and pulley.
Fig. 3.1 shows the view from above of the spring, cable and pulley.
The spring obeys Hooke’s law and has a spring constant \(k\) of \(250\,\mathrm{N\,m^{-1}}\). A force \(F\) acts on the spring. The tension in the cable is \(T\). The pulley is in equilibrium.
(a) On Fig. 3.2, draw labelled arrows to show the directions of the forces acting on the pulley. (2 marks)
________________________________
________________________________
(b) The force \(F\) is \(110\,\mathrm{N}\).
(i) Determine \(T\). (1 mark)
\(T=\) __________________________ \( \mathrm{N} \)
(ii) Calculate the extension of the spring. (2 marks)
extension = __________________________ \( \mathrm{m} \)
(c) A second identical spring with the same spring constant of \(250\,\mathrm{N\,m^{-1}}\) is now also connected to the pulley, as shown in Fig. 3.3.
The tension in the cable is kept the same. The pulley is again in equilibrium.
(i) Determine the extension of the springs. (2 marks)
extension = __________________________ \( \mathrm{m} \)
(ii) The elastic potential energy stored in the spring in Fig. 3.1 is \(E_1\). The total elastic potential energy stored in the two springs in Fig. 3.3 is \(E_2\).
Calculate the ratio \(\dfrac{E_1}{E_2}\). (2 marks)
ratio = __________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 6.2: Elastic and plastic behaviour — parts (b)(ii), (c)(i) and (c)(ii)
▶️ Answer/Explanation
(a) Forces acting on the pulley [2 marks]
The spring exerts a force \(F\) on the pulley horizontally to the left.
The cable passes around the pulley, so it exerts two tension forces \(T\), both acting horizontally to the right.
Answer: \( \boxed{\text{one force }F\text{ to the left and two forces }T\text{ to the right}} \)
(b)(i) Tension in the cable [1 mark]
The pulley is in equilibrium, so the resultant force is zero.
There are two equal tension forces acting to the right, therefore
\(F=2T\)
\(T=\dfrac{110}{2}\)
\(T=55\,\mathrm{N}\)
Answer: \( \boxed{55\,\mathrm{N}} \)
(b)(ii) Extension of the spring [2 marks]
Using Hooke’s law,
\(F=kx\)
Therefore,
\(x=\dfrac{F}{k}\)
\(x=\dfrac{110}{250}\)
\(x=0.44\,\mathrm{m}\)
Answer: \( \boxed{0.44\,\mathrm{m}} \)
(c)(i) Extension of the two springs [2 marks]
The tension in the cable remains \(55\,\mathrm{N}\). The two identical springs share the force equally.
Therefore, the force on each spring is
\(F_{\mathrm{spring}}=55\,\mathrm{N}\)
Using Hooke’s law,
\(x=\dfrac{F}{k}\)
\(x=\dfrac{55}{250}\)
\(x=0.22\,\mathrm{m}\)
Answer: \( \boxed{0.22\,\mathrm{m}} \)
(c)(ii) Ratio of elastic potential energies [2 marks]
The elastic potential energy stored in a spring is
\(E=\dfrac{1}{2}kx^2\)
For Fig. 3.1, there is one spring with extension \(0.44\,\mathrm{m}\):
\(E_1=\dfrac{1}{2}k(0.44)^2\)
For Fig. 3.3, there are two springs, each with extension \(0.22\,\mathrm{m}\):
\(E_2=2\left(\dfrac{1}{2}k(0.22)^2\right)\)
Therefore,
\(\dfrac{E_1}{E_2}=\dfrac{\frac{1}{2}k(0.44)^2}{2\left(\frac{1}{2}k(0.22)^2\right)}\)
Since \(0.44=2(0.22)\),
\(\dfrac{E_1}{E_2}=2.0\)
Answer: \( \boxed{\dfrac{E_1}{E_2}=2.0} \)
Question 4
A laser emits visible light of a single frequency in a vacuum. The light is incident normally on a double slit and then forms a pattern of bright and dark fringes on a screen, as shown in Fig. 4.1.
The separation of the slits is \(1.0\times10^{-3}\,\mathrm{m}\). The distance from the slits to the screen is \(4.8\,\mathrm{m}\). The distance between the centres of adjacent bright fringes on the screen is \(3.3\,\mathrm{mm}\).
(a) Explain how the pattern of bright and dark fringes is formed. (3 marks)
________________________________
________________________________
________________________________
(b) Calculate the frequency of the light emitted by the laser. (4 marks)
frequency = __________________________________________ \( \mathrm{Hz} \)
(c) The double slit is removed. A second laser is placed beside the first laser. The second laser produces visible light of a different frequency from that of the first laser. The beams of light from the two lasers overlap on the screen.
Explain why a steady pattern of bright and dark fringes is not formed on the screen. (1 mark)
________________________________
________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
▶️ Answer/Explanation
(a) Formation of bright and dark fringes [3 marks]
Light diffracts or spreads out as it passes through each slit.
The light waves from the two slits superpose and interfere at the screen.
When the path difference is \(n\lambda\), where \(n\) is an integer, constructive interference occurs and a bright fringe is formed. When the path difference is \((n+\frac{1}{2})\lambda\), destructive interference occurs and a dark fringe is formed.
Answer: \( \boxed{\text{The fringes are produced by interference of the waves from the two slits.}} \)
(b) Frequency of the laser light [4 marks]
For double-slit interference, the fringe spacing is
\(x=\dfrac{\lambda D}{a}\)
Therefore,
\(\lambda=\dfrac{ax}{D}\)
\(a=1.0\times10^{-3}\,\mathrm{m}\), \(x=3.3\times10^{-3}\,\mathrm{m}\) and \(D=4.8\,\mathrm{m}\).
\(\lambda=\dfrac{(1.0\times10^{-3})(3.3\times10^{-3})}{4.8}\)
\(\lambda=6.875\times10^{-7}\,\mathrm{m}\)
Using the wave equation,
\(f=\dfrac{v}{\lambda}\)
For light in a vacuum, \(v=3.0\times10^8\,\mathrm{m\,s^{-1}}\).
\(f=\dfrac{3.0\times10^8}{6.875\times10^{-7}}\)
\(f=4.4\times10^{14}\,\mathrm{Hz}\)
Answer: \( \boxed{4.4\times10^{14}\,\mathrm{Hz}} \)
(c) Absence of a steady interference pattern [1 mark]
The two lasers have different frequencies, so the light waves are not coherent and do not have a constant phase difference.
Answer: \( \boxed{\text{The sources are not coherent, so no steady interference pattern is formed.}} \)
Question 5
Fig. 5.1 shows a circuit containing a battery, two fixed resistors X and Y, and a light-dependent resistor (LDR) Z.
The battery has electromotive force (e.m.f.) \(5.0\,\mathrm{V}\) and internal resistance \(4.7\,\Omega\). The current in X is \(I_1\) and the current in Y is \(I_2\).
The resistance of X is \(100\,\Omega\). The resistance of Z varies with the intensity of light incident on it as shown in Fig. 5.2.
(a) State Kirchhoff’s first law. (1 mark)
________________________________
(b) The intensity of light incident on Z is \(130\,\mathrm{W\,m^{-2}}\). The current in the battery is \(38\,\mathrm{mA}\).
(i) Show that the terminal potential difference of the battery is \(4.8\,\mathrm{V}\). (2 marks)
________________________________
(ii) Calculate the current \(I_2\) in Y. (3 marks)
\(I_2=\) __________________________ \( \mathrm{A} \)
(iii) Calculate the power dissipated in Y. (2 marks)
power = __________________________ \( \mathrm{W} \)
(iv) The intensity of the light incident on Z decreases.
State and explain the effect on the terminal potential difference of the battery. (3 marks)
________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 10.1: Practical circuits — part (b)(i) and (b)(iv)
• 9.3: Resistance and resistivity — parts (b)(ii) and (b)(iv)
• 9.2: Potential difference and power — part (b)(iii)
▶️ Answer/Explanation
(a) Kirchhoff’s first law [1 mark]
The sum of current(s) into a junction is equal to the sum of current(s) out of the junction.
Alternatively, the algebraic sum of current(s) at a junction is zero.
Answer: \( \boxed{\text{sum of currents into a junction = sum of currents out}} \)
(b)(i) Terminal potential difference [2 marks]
For a cell with internal resistance \(r\),
\(V=E-Ir\)
\(V=5.0-(38\times10^{-3}\times4.7)\)
\(V=4.8\,\mathrm{V}\)
Answer: \( \boxed{4.8\,\mathrm{V}} \)
(b)(ii) Current \(I_2\) in Y [3 marks]
From Fig. 5.2, at a light intensity of \(130\,\mathrm{W\,m^{-2}}\),
\(R_Z=120\,\Omega\)
The current \(I_1\) through the series combination of X and Z is
\(I_1=\dfrac{4.8}{100+120}\)
Using Kirchhoff’s first law,
\(I_2=38\times10^{-3}-\dfrac{4.8}{100+120}\)
\(I_2=0.016\,\mathrm{A}\)
Answer: \( \boxed{0.016\,\mathrm{A}} \)
Alternative method:
The external resistance is
\(R_{\mathrm{EXT}}=\dfrac{4.8}{38\times10^{-3}}=126\,\Omega\)
For the parallel combination,
\(\dfrac{1}{R_Y}=\dfrac{1}{126}-\dfrac{1}{120+100}\)
\(R_Y=297\,\Omega\)
\(I_2=\dfrac{4.8}{297}\)
\(I_2=0.016\,\mathrm{A}\)
(b)(iii) Power dissipated in Y [2 marks]
Using
\(P=IV\)
\(P=0.016\times4.8\)
\(P=0.077\,\mathrm{W}\)
Answer: \( \boxed{0.077\,\mathrm{W}} \)
(b)(iv) Effect of decreasing light intensity [3 marks]
The resistance of the LDR \(Z\) increases as the light intensity decreases.
Therefore, the total resistance of the circuit increases, so the current in the battery decreases.
The potential difference across the internal resistance decreases because \(V_{\mathrm{internal}}=Ir\).
Since \(V_{\mathrm{terminal}}=E-Ir\), the terminal potential difference increases.
Answer: \( \boxed{\text{terminal potential difference increases}} \)
Question 6
(a) State what is meant by a fundamental particle. (1 mark)
________________________________
(b) (i) Particle Q is a meson with a charge of \(0\).
Determine a possible quark composition for Q. (2 marks)
________________________________
(ii) Particle Q has a mass of \(0.67\,\mathrm{u}\) and a kinetic energy of \(2.1\times10^{-16}\,\mathrm{J}\).
Calculate the speed of particle Q. (3 marks)
speed = __________________________ \( \mathrm{m\,s^{-1}} \)
(c) Radium-228 \(\left(^{228}_{88}\mathrm{Ra}\right)\) is a radioactive nuclide.
(i) State the number of electrons in a neutral atom of radium-228. (1 mark)
number of electrons = __________________________
(ii) A nucleus of radium-228 undergoes a series of decays to form nucleus X. During the process, 5 \(\alpha\)-particles and 4 \(\beta^-\)-particles are emitted.
Determine the number of protons and the number of neutrons in nucleus X. (2 marks)
number of protons = __________________________
number of neutrons = __________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 5.2: Gravitational potential energy and kinetic energy — part (b)(ii)
• 11.1: Atoms, nuclei and radiation — parts (c)(i) and (c)(ii)
▶️ Answer/Explanation
(a) Fundamental particle [1 mark]
A fundamental particle is a particle that cannot be divided or subdivided into smaller particles.
Answer: \( \boxed{\text{a particle that cannot be divided into smaller particles}} \)
(b)(i) Quark composition of Q [2 marks]
A meson consists of one quark and one antiquark.
For a neutral meson, one possible combination is an up quark and an anti-up quark:
\(Q=u\bar{u}\)
The charges are \(+\dfrac{2}{3}e\) and \(-\dfrac{2}{3}e\), giving a total charge of zero.
Answer: \( \boxed{u\bar{u}} \)
(b)(ii) Speed of particle Q [3 marks]
The kinetic energy of a non-relativistic particle is
\(E_{\mathrm{K}}=\dfrac{1}{2}mv^2\)
The mass is \(0.67\,\mathrm{u}\). Using \(1\,\mathrm{u}=1.66\times10^{-27}\,\mathrm{kg}\),
\(m=0.67\times1.66\times10^{-27}\,\mathrm{kg}\)
\(2.1\times10^{-16}=\dfrac{1}{2}\times0.67\times1.66\times10^{-27}\times v^2\)
\(v=6.1\times10^5\,\mathrm{m\,s^{-1}}\)
Answer: \( \boxed{6.1\times10^5\,\mathrm{m\,s^{-1}}} \)
(c)(i) Number of electrons [1 mark]
The proton number of radium is \(88\). A neutral atom has equal numbers of protons and electrons.
Therefore,
\(\text{number of electrons}=88\)
Answer: \( \boxed{88} \)
(c)(ii) Protons and neutrons in nucleus X [2 marks]
Each \(\alpha\)-particle reduces the nucleon number by \(4\), so
\(A=228-(5\times4)=208\)
Each \(\alpha\)-particle also reduces the proton number by \(2\), while each \(\beta^-\)-decay increases the proton number by \(1\).
Therefore,
\(Z=88-(5\times2)+4\)
\(Z=82\)
The number of neutrons is
\(N=A-Z\)
\(N=208-82\)
\(N=126\)
Answer: \( \boxed{82\text{ protons},\ 126\text{ neutrons}} \)