Question 1
(a)
(i) State what is indicated by the direction of the gravitational field line at a point in a gravitational field. (1 mark)
______________________________________________
(ii) Explain, with reference to gravitational field lines, why the gravitational field near the surface of the Earth is approximately constant for small changes in height. (2 marks)
______________________________________________
______________________________________________
(b) A large isolated uniform sphere has mass \(M\) and radius \(R\).
Point P lies on a straight line passing through the centre of the sphere, at a variable displacement \(x\) from the centre, as shown in Fig. 1.1.
Fig. 1.2 shows the variation of \(x\) of the gravitational field \(g\) at P due to the sphere for the values of \(x\) for which P is inside the sphere.
The magnitude of the gravitational field at the surface of the sphere is \(Y\).
(i) Determine an expression for \(Y\) in terms of \(M\) and \(R\). Identify any other symbols that you use. (2 marks)
\(Y=\) ______________________________________________
(ii) Explain why, at the surface of the sphere, \(g\) always has the opposite sign to \(x\). (2 marks)
______________________________________________
______________________________________________
(iii) Complete Fig. 1.2 to show the variation of \(g\) with \(x\) for values of \(x\), up to \(\pm3R\), for which point P is outside the sphere. (3 marks)
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 13.2: Gravitational force between point masses — part (b)(i), using the gravitational field expression at the surface
• 13.3: Gravitational field of a point mass — parts (b)(i) and (b)(iii)
▶️ Answer/Explanation
(a)(i) Direction of a gravitational field line [1 mark]
The direction of a gravitational field line indicates the direction of the force acting on a test mass placed at that point.
Answer: \( \boxed{\text{direction of the force acting on a test mass}} \)
(a)(ii) Approximately constant gravitational field near Earth [2 marks]
For a small change in height near the Earth’s surface, the change in height is negligible compared with the radius of the Earth.
Therefore, the gravitational field lines are effectively parallel and their spacing changes negligibly.
Answer: \( \boxed{\text{The field lines are effectively parallel because the change in height is small compared with Earth’s radius.}} \)
(b)(i) Expression for \(Y\) [2 marks]
At the surface of the sphere, the distance from the centre is \(R\). The magnitude of the gravitational field is therefore
\( g=\frac{GM}{r^2} \)
At the surface, \(r=R\), so
\( Y=\frac{GM}{R^2} \)
where \(G\) is the gravitational constant.
Answer: \( \boxed{Y=\frac{GM}{R^2}} \)
(b)(ii) Sign of \(g\) relative to \(x\) [2 marks]
Gravitational force is always attractive, so the gravitational field always acts towards the centre of the sphere.
The displacement \(x\) is measured from the centre. Therefore, the gravitational field acts in the direction opposite to the displacement.
Hence, when \(x\) is positive, \(g\) is negative, and when \(x\) is negative, \(g\) is positive.
Answer: \( \boxed{g\text{ always has the opposite sign to }x} \)
(b)(iii) Variation of \(g\) outside the sphere [3 marks]
Outside the sphere, the gravitational field behaves as though the entire mass \(M\) were concentrated at the centre.
Thus, for \(x>R\),
\( g=-\frac{GM}{x^2} \)
and for \(x<-R\),
\( g=\frac{GM}{x^2} \)
The completed graph should therefore consist of two smooth curves outside the sphere. On the positive-\(x\) side, the curve starts at \((R,-Y)\) and approaches \(g=0\) as \(x\) increases. On the negative-\(x\) side, the curve starts at \((-R,Y)\) and approaches \(g=0\) as \(x\) becomes more negative.
Useful points are
\(x=\pm2R \Rightarrow g=\mp\frac{Y}{4} \)
\(x=\pm3R \Rightarrow g=\mp\frac{Y}{9}\approx\mp0.11Y \)
Therefore, the curve passes through \( (\pm2R,\mp0.25Y) \) and \( (\pm3R,\mp0.11Y) \), with the signs reversed on the negative-\(x\) side.
Answer: \( \boxed{\text{Smooth inverse-square curves outside }x=\pm R,\text{ approaching }g=0\text{ as }|x|\text{ increases.}} \)
Question 2
(a) Define specific heat capacity. (2 marks)
______________________________________________
______________________________________________
(b) An ideal gas of mass \(0.35\,\mathrm{kg}\) is heated at a constant pressure of \(2.0\times10^5\,\mathrm{Pa}\) so that its internal energy increases by \(7600\,\mathrm{J}\). During this process, the volume of the gas increases from \(0.038\,\mathrm{m^3}\) to \(0.063\,\mathrm{m^3}\) and the temperature increases by \(56^\circ\mathrm{C}\).
(i) Show that the magnitude of the work done on the gas is \(5000\,\mathrm{J}\). (1 mark)
______________________________________________
(ii) Explain whether the work done on the gas is positive or negative. (2 marks)
______________________________________________
______________________________________________
(iii) Determine the magnitude of the thermal energy \(q\) transferred to the gas. (2 marks)
\(q=\) __________________________________________ \( \mathrm{J} \)
(iv) Calculate the specific heat capacity of the gas for this process. Give a unit with your answer. (2 marks)
specific heat capacity = __________________________ \( \mathrm{unit} \)
(c) The gas in (b) is now heated at constant volume rather than at constant pressure. The increase in internal energy of the gas is the same as in (b).
Use the first law of thermodynamics to explain whether the specific heat capacity of the gas for this process is less than, the same as, or greater than the answer in (b)(iv). (3 marks)
______________________________________________
______________________________________________
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 16.1: Internal energy — parts (b) and (c)
• 16.2: The first law of thermodynamics — parts (b)(iii), (b)(iv) and (c)
▶️ Answer/Explanation
(a) Definition of specific heat capacity [2 marks]
Specific heat capacity is the thermal energy required per unit mass per unit change in temperature.
It can be expressed by
\( c=\frac{q}{m\Delta T} \)
Answer: \( \boxed{\text{thermal energy required per unit mass per unit temperature change}} \)
(b)(i) Work done on the gas [1 mark]
At constant pressure, the magnitude of the work done is
\( |W|=p\Delta V \)
\( |W|=(2.0\times10^5)(0.063-0.038) \)
\( |W|=(2.0\times10^5)(0.025) \)
\( |W|=5000\,\mathrm{J} \)
Answer: \( \boxed{5000\,\mathrm{J}} \)
(b)(ii) Sign of the work done on the gas [2 marks]
The volume of the gas increases, so the gas is expanding against the external pressure.
Therefore, the gas does work on its surroundings. Hence, the work done on the gas is negative.
Thus,
\( W=-5000\,\mathrm{J} \)
Answer: \( \boxed{W=-5000\,\mathrm{J}} \)
(b)(iii) Thermal energy transferred to the gas [2 marks]
Using the first law of thermodynamics with the convention that \(W\) is the work done on the gas,
\( \Delta U=q+W \)
The increase in internal energy is \(7600\,\mathrm{J}\), and \(W=-5000\,\mathrm{J}\).
\( 7600=q+(-5000) \)
\( q=7600+5000 \)
\( q=12600\,\mathrm{J} \)
Answer: \( \boxed{1.26\times10^4\,\mathrm{J}} \)
(b)(iv) Specific heat capacity [2 marks]
The specific heat capacity is
\( c=\frac{q}{m\Delta T} \)
\( c=\frac{12600}{(0.35)(56)} \)
\( c=642.9\,\mathrm{J\,kg^{-1}\,K^{-1}} \)
To an appropriate number of significant figures,
\( c=640\,\mathrm{J\,kg^{-1}\,K^{-1}} \)
Answer: \( \boxed{640\,\mathrm{J\,kg^{-1}\,K^{-1}}} \)
(c) Specific heat capacity at constant volume [3 marks]
The increase in internal energy is the same as in part (b), so the temperature rise is also the same for the same mass.
At constant volume, there is no change in volume, so no work is done by the gas.
Therefore,
\( W=0 \)
Using the first law,
\( \Delta U=q+W \)
so, when \(W=0\),
\( \Delta U=q \)
At constant pressure, some of the thermal energy supplied was used to do work during expansion. At constant volume, no energy is used for work, so less thermal energy is required for the same temperature change.
Since \( c=\frac{q}{m\Delta T} \), a smaller \(q\) for the same \(m\) and \(\Delta T\) gives a smaller specific heat capacity.
Answer: \( \boxed{\text{The specific heat capacity is less than the value in (b)(iv).}} \)
Question 3
(a) The product \(pV\) for an ideal gas is given by
\( pV=\frac{1}{3}Nm\langle c^2\rangle \)
where \(p\) is the pressure of the gas and \(V\) is the volume of the gas.
(i) State the meaning of the symbols \(N\), \(m\) and \(\langle c^2\rangle\) in this equation. (3 marks)
\(N:\) ______________________________________________
\(m:\) ______________________________________________
\(\langle c^2\rangle:\) ______________________________________________
(ii) Use the equation of state for an ideal gas to show that the average translational kinetic energy \(E_{\mathrm{K}}\) of a gas at thermodynamic temperature \(T\) is given by
\( E_{\mathrm{K}}=\frac{3}{2}kT \)
______________________________________________
(b) The surface of a star consists mainly of a gas that may be assumed to be ideal. The molecules of the gas have a root-mean-square (r.m.s.) speed of \(9300\,\mathrm{m\,s^{-1}}\).
The mass of a molecule of the gas is \(3.34\times10^{-27}\,\mathrm{kg}\).
Determine, to three significant figures, the temperature of the surface of the star.
temperature = __________________________________ \( \mathrm{K} \)
(c) The radiant flux intensity of the radiation from the star in (b) is \(2.52\times10^{-8}\,\mathrm{W\,m^{-2}}\) when observed at a distance of \(4.16\times10^{16}\,\mathrm{m}\) from the star.
(i) Calculate the luminosity of the star. Give a unit with your answer. (2 marks)
luminosity = ______________________________ unit __________
(ii) Determine the radius of the star. (2 marks)
radius = __________________________________ \( \mathrm{m} \)
(d) The gas at the surface of a star has a very high pressure.
Use the basic assumptions of the kinetic theory to suggest why, in practice, a gas at the surface of a star is unlikely to behave as an ideal gas. (2 marks)
______________________________________________
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 15.3: Kinetic theory of gases — parts (a)(i), (a)(ii), (b) and (d)
• 25.2: Stellar radii — parts (c)(i) and (c)(ii)
▶️ Answer/Explanation
(a)(i) Meaning of the symbols [3 marks]
\(N\) is the number of molecules of the gas.
\(m\) is the mass of one molecule of the gas.
\(\langle c^2\rangle\) is the mean square speed of the molecules.
Answers:
\( \boxed{N=\text{number of molecules}} \)
\( \boxed{m=\text{mass of one molecule}} \)
\( \boxed{\langle c^2\rangle=\text{mean square speed}} \)
(a)(ii) Derivation of \(E_{\mathrm{K}}=\frac{3}{2}kT\) [2 marks]
The equation of state for an ideal gas is
\( pV=NkT \)
From the kinetic theory equation,
\( pV=\frac{1}{3}Nm\langle c^2\rangle \)
Equating the two expressions for \(pV\),
\( NkT=\frac{1}{3}Nm\langle c^2\rangle \)
The average translational kinetic energy of one molecule is
\( E_{\mathrm{K}}=\frac{1}{2}m\langle c^2\rangle \)
Therefore,
\( E_{\mathrm{K}}=\frac{3}{2}kT \)
Answer: \( \boxed{E_{\mathrm{K}}=\frac{3}{2}kT} \)
(b) Temperature of the star [2 marks]
For a molecule, the average translational kinetic energy is
\( E_{\mathrm{K}}=\frac{1}{2}mc^2=\frac{3}{2}kT \)
Hence,
\( \frac{1}{2}mc^2=\frac{3}{2}kT \)
Rearranging,
\( T=\frac{mc^2}{3k} \)
Using \(m=3.34\times10^{-27}\,\mathrm{kg}\), \(c=9300\,\mathrm{m\,s^{-1}}\) and \(k=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}\):
\( T=\frac{(3.34\times10^{-27})(9300)^2}{3(1.38\times10^{-23})} \)
\( T=6980\,\mathrm{K} \)
Answer: \( \boxed{6980\,\mathrm{K}} \)
(c)(i) Luminosity of the star [2 marks]
The radiant flux intensity \(F\) at distance \(d\) is related to the luminosity \(L\) by
\( F=\frac{L}{4\pi d^2} \)
Therefore,
\( L=4\pi d^2F \)
Substituting \(F=2.52\times10^{-8}\,\mathrm{W\,m^{-2}}\) and \(d=4.16\times10^{16}\,\mathrm{m}\):
\( L=4\pi(4.16\times10^{16})^2(2.52\times10^{-8}) \)
\( L=5.48\times10^{26}\,\mathrm{W} \)
Answer: \( \boxed{5.48\times10^{26}\,\mathrm{W}} \)
(c)(ii) Radius of the star [2 marks]
Using the Stefan-Boltzmann relation,
\( L=4\pi r^2\sigma T^4 \)
where \(\sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}\).
Substituting \(L=5.48\times10^{26}\,\mathrm{W}\) and \(T=6980\,\mathrm{K}\):
\( 5.48\times10^{26}=4\pi(5.67\times10^{-8})r^2(6980)^4 \)
Rearranging and solving for \(r\),
\( r=5.69\times10^8\,\mathrm{m} \)
Answer: \( \boxed{5.69\times10^8\,\mathrm{m}} \)
(d) Why the gas may not behave as an ideal gas [2 marks]
The pressure at the surface of the star is very high, so the molecules are very close together.
Consequently, the forces between molecules are no longer negligible. Also, the volume occupied by the molecules may become significant compared with the volume of the gas.
Answer: \( \boxed{\text{At very high pressure, intermolecular forces and molecular volume are no longer negligible.}} \)
Question 4
A heavy metal sphere of mass \(0.81\,\mathrm{kg}\) is suspended from a string. The sphere is undergoing small oscillations from side to side, as shown in Fig. 4.1.
The oscillations of the sphere may be considered to be simple harmonic with amplitude \(0.036\,\mathrm{m}\) and period \(3.0\,\mathrm{s}\).
(a) State what is meant by simple harmonic motion. (2 marks)
______________________________________________
______________________________________________
(b) Calculate:
(i) the angular frequency of the oscillations. (2 marks)
angular frequency = __________________________ \( \mathrm{rad\,s^{-1}} \)
(ii) the total energy of the oscillations. (2 marks)
total energy = ______________________________ \( \mathrm{J} \)
(c) The suspended sphere is now lowered into water. The sphere is given a sideways displacement of \(+0.036\,\mathrm{m}\) from its equilibrium position and is then released at time \(t=0\). The water causes the motion of the sphere to be critically damped.
On Fig. 4.2, sketch the variation of the displacement \(x\) of the sphere from its equilibrium position from \(t=0\) to \(t=6.0\,\mathrm{s}\).
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 17.2: Energy in simple harmonic motion — part (b)(ii)
• 17.3: Damped and forced oscillations, resonance — part (c)
▶️ Answer/Explanation
(a) Definition of simple harmonic motion [2 marks]
Simple harmonic motion is motion in which the acceleration is directly proportional to the displacement from a fixed equilibrium position.
The acceleration is always directed in the opposite direction to the displacement, towards the equilibrium position.
Answer: \( \boxed{\text{Acceleration is proportional to displacement and directed towards the equilibrium position.}} \)
(b)(i) Angular frequency [2 marks]
The angular frequency is related to the period by
\( \omega=\frac{2\pi}{T} \)
Using \(T=3.0\,\mathrm{s}\),
\( \omega=\frac{2\pi}{3.0} \)
\( \omega=2.09\,\mathrm{rad\,s^{-1}} \)
Answer: \( \boxed{2.1\,\mathrm{rad\,s^{-1}}} \)
(b)(ii) Total energy of the oscillations [2 marks]
The total energy of a simple harmonic oscillator is
\( E=\frac{1}{2}m\omega^2x_0^2 \)
where \(x_0\) is the amplitude.
Using \(m=0.81\,\mathrm{kg}\), \(\omega=2.1\,\mathrm{rad\,s^{-1}}\) and \(x_0=0.036\,\mathrm{m}\):
\( E=\frac{1}{2}(0.81)(2.1)^2(0.036)^2 \)
\( E=2.32\times10^{-3}\,\mathrm{J} \)
Answer: \( \boxed{2.3\times10^{-3}\,\mathrm{J}} \)
(c) Critically damped motion [3 marks]
The sphere is released from \(x=+0.036\,\mathrm{m}\), so the graph must start at
\( (t,x)=(0,+0.036\,\mathrm{m}) \).
Because the motion is critically damped, the displacement decreases continuously towards the equilibrium position without oscillating or crossing \(x=0\).
The curve should therefore be smooth, with no sudden changes in gradient. The magnitude of \(x\) continuously decreases, with the displacement approaching zero.
The final displacement reaches zero between approximately \(t=0.75\,\mathrm{s}\) and \(t=3.0\,\mathrm{s}\), as indicated by the marking guidance.
Answer: \( \boxed{\text{A smooth non-oscillating curve from }+0.036\,\mathrm{m}\text{ towards }x=0} \)
Question 5
(a) Define electric potential at a point. (2 marks)
______________________________________________
______________________________________________
(b) Two isolated charged metal spheres X and Y are situated near to each other in a vacuum with their centres a distance of \(24\,\mathrm{m}\) apart. Point P is at a variable distance \(x\) from the centre of sphere X on the line joining the centres of the spheres.
Fig. 5.1 shows the variation with \(x\) of the electric potential \(V\) due to the spheres at point P.
State three conclusions that can be drawn about the spheres from Fig. 5.1. The conclusions may be qualitative or quantitative. (3 marks)
1. ______________________________________________
2. ______________________________________________
3. ______________________________________________
(c) A positively charged particle is placed at point P in (b), such that \(x=12\,\mathrm{m}\). The particle is released.
Describe and explain the subsequent motion of the particle. (3 marks)
______________________________________________
______________________________________________
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 18.3: Electric force between point charges — part (c)
▶️ Answer/Explanation
(a) Definition of electric potential [2 marks]
Electric potential at a point is the work done per unit positive charge in bringing a small positive test charge from infinity to that point.
Hence,
\( V=\frac{W}{Q} \)
Answer: \( \boxed{\text{work done per unit positive charge in bringing a charge from infinity to the point}} \)
(b) Conclusions from the graph [3 marks]
1. Radius of sphere X: The graph becomes very steep near \(x=2\,\mathrm{m}\), indicating that the radius of sphere X is approximately \(2.0\,\mathrm{m}\).
2. Radius of sphere Y: The graph becomes very steep near \(x=20\,\mathrm{m}\). Since the centre of Y is \(24\,\mathrm{m}\) from the centre of X, the radius of Y is approximately \(24-20=4.0\,\mathrm{m}\).
3. Signs of the charges: The potential is negative near X and positive near Y. Therefore, sphere X has a negative charge and sphere Y has a positive charge.
The graph also indicates that the magnitudes of the charges on X and Y are equal because the potential changes from the negative contribution of X to the positive contribution of Y in a manner consistent with equal and opposite charges.
Possible answers: \( \boxed{r_X=2.0\,\mathrm{m},\quad r_Y=4.0\,\mathrm{m}} \), \( \boxed{X\text{ is negative and }Y\text{ is positive}} \), \( \boxed{|Q_X|=|Q_Y|} \)
(c) Subsequent motion of the positively charged particle [3 marks]
At \(x=12\,\mathrm{m}\), the positively charged particle is between the two spheres.
Sphere X is negatively charged, so it attracts the positive particle towards X. Sphere Y is positively charged, so it repels the particle away from Y.
Both forces therefore act towards the left, towards X.
The particle consequently accelerates towards X.
As the particle moves towards X, its distance from X decreases, so the attractive electric force due to X increases. Therefore, the magnitude of its acceleration increases.
Answer: \( \boxed{\text{The particle accelerates towards X, with increasing magnitude of acceleration.}} \)
Question 6
(a) Define magnetic flux density. (2 marks)
______________________________________________
______________________________________________
(b) Electrons are moving in a vacuum with speed \(1.7\times10^7\,\mathrm{m\,s^{-1}}\). The electrons enter a uniform magnetic field of flux density \(4.8\,\mathrm{mT}\). Fig. 6.1 shows the path of the electrons.
The path of the electrons remains in the plane of the page.
(i) State the direction of the magnetic field. (1 mark)
______________________________________________
(ii) Show that the magnitude of the force exerted on each electron by the magnetic field is \(1.3\times10^{-14}\,\mathrm{N}\). (2 marks)
______________________________________________
(iii) On Fig. 6.1, draw an arrow to indicate the direction of the centripetal acceleration of the electron when it enters the magnetic field at point X. (1 mark)
______________________________________________
(iv) Use the information in (b)(ii) to calculate the distance \(d\) between the path of the electrons entering the magnetic field and the path of the electrons leaving it. (3 marks)
\(d=\) __________________________________ \( \mathrm{m} \)
(c) The electrons in (b) are replaced with positrons that are moving with speed \(3.4\times10^7\,\mathrm{m\,s^{-1}}\) along the same initial path as the electrons.
The positrons enter the magnetic field at point X on Fig. 6.1.
On Fig. 6.1, draw a line to show the path of the positrons through the magnetic field. (3 marks)
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 20.3: Force on a moving charge — parts (b)(ii), (b)(iii), (b)(iv) and (c)
▶️ Answer/Explanation
(a) Definition of magnetic flux density [2 marks]
Magnetic flux density is the force per unit length per unit current on a current-carrying conductor placed perpendicular to the magnetic field.
Thus,
\( B=\frac{F}{IL} \)
Answer: \( \boxed{\text{force per unit length per unit current, with the conductor perpendicular to the field}} \)
(b)(i) Direction of the magnetic field [1 mark]
The electrons initially move to the right and curve downwards. Since an electron has negative charge, its magnetic force is opposite to the direction given by the conventional positive-charge rule.
Using Fleming’s left-hand rule or \( \mathbf{F}=q(\mathbf{v}\times\mathbf{B}) \), the magnetic field must be directed into the page.
Answer: \( \boxed{\text{into the page}} \)
(b)(ii) Magnetic force on each electron [2 marks]
The electrons enter the magnetic field perpendicular to the field, so
\( F=Bqv \)
Using \(B=4.8\times10^{-3}\,\mathrm{T}\), \(q=1.6\times10^{-19}\,\mathrm{C}\) and \(v=1.7\times10^7\,\mathrm{m\,s^{-1}}\):
\( F=(4.8\times10^{-3})(1.6\times10^{-19})(1.7\times10^7) \)
\( F=1.31\times10^{-14}\,\mathrm{N} \)
Answer: \( \boxed{1.3\times10^{-14}\,\mathrm{N}} \)
(b)(iii) Direction of centripetal acceleration [1 mark]
The electron follows a circular path, so its centripetal acceleration is directed towards the centre of the circular path.
At point X, the centre of the circular path is below X.
Answer: \( \boxed{\text{downwards}} \)
(b)(iv) Distance \(d\) [3 marks]
The magnetic force provides the centripetal force:
\( F=\frac{mv^2}{r} \)
Using \(F=1.3\times10^{-14}\,\mathrm{N}\), \(m=9.11\times10^{-31}\,\mathrm{kg}\) and \(v=1.7\times10^7\,\mathrm{m\,s^{-1}}\):
\( 1.3\times10^{-14}=\frac{(9.11\times10^{-31})(1.7\times10^7)^2}{r} \)
\( r=0.020\,\mathrm{m} \)
The electron follows a semicircular path, so the separation between the entering and leaving paths is the diameter:
\( d=2r \)
\( d=2(0.020) \)
\( d=0.040\,\mathrm{m} \)
Answer: \( \boxed{0.040\,\mathrm{m}} \)
(c) Path of the positrons [3 marks]
A positron has positive charge, whereas an electron has negative charge. Therefore, for the same initial velocity and magnetic field direction, the magnetic force on the positron is in the opposite direction to the force on the electron.
The positrons therefore curve upwards, so the curvature is anticlockwise within the magnetic field.
The speed of the positrons is twice the speed of the electrons. Since
\( r=\frac{mv}{Bq} \)
the radius is twice as large.
Thus, the path is a circular arc with a radius \(2r\), entering the field at X and leaving the field at a distance \(2d\) vertically from X.
Answer: \( \boxed{\text{An anticlockwise circular path with radius }2r\text{, curving upwards from X.}} \)
Question 7
A varying current \(I\) passes through a resistor of resistance \(R\) in the circuit shown in Fig. 7.1.
Fig. 7.2 shows the variation with time \(t\) of \(I\).
The current has magnitude \(2I_0\) when it is in the positive direction and \(I_0\) when it is in the negative direction. The period of the variation of the current is \(T\).
(a) Determine expressions, in terms of \(I_0\) and \(R\), for the power \(P\) dissipated in the resistor for the times when:
(i) the current is in the negative direction. (1 mark)
\(P=\) __________________________________________
(ii) the current is in the positive direction. (1 mark)
\(P=\) __________________________________________
(b) On Fig. 7.3, sketch the variation of \(P\) with \(t\) between \(t=0\) and \(t=2.0T\). Label the power axis with an appropriate scale. (3 marks)
(c) Use your answer in (b) to determine an expression, in terms of \(I_0\) and \(R\), for:
(i) the mean power \(\langle P\rangle\) in the resistor. (1 mark)
\(\langle P\rangle=\) __________________________________________
(ii) the root-mean-square (r.m.s.) current \(I_{\mathrm{r.m.s.}}\) in the resistor. (2 marks)
\(I_{\mathrm{r.m.s.}}=\) __________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
▶️ Answer/Explanation
(a)(i) Power when the current is in the negative direction [1 mark]
The power dissipated in a resistor is
\( P=I^2R \)
When the current is in the negative direction, its magnitude is \(I_0\). Therefore,
\( P=I_0^2R \)
Answer: \( \boxed{I_0^2R} \)
(a)(ii) Power when the current is in the positive direction [1 mark]
When the current is in the positive direction, its magnitude is \(2I_0\).
\( P=(2I_0)^2R \)
\( P=4I_0^2R \)
Answer: \( \boxed{4I_0^2R} \)
(b) Variation of power with time [3 marks]
Since \(P=I^2R\), the sign of the current does not affect the power. The power is therefore always positive.
For the half-period when the current has magnitude \(I_0\),
\( P=I_0^2R \)
For the half-period when the current has magnitude \(2I_0\),
\( P=4I_0^2R \)
Therefore, the power-time graph is a square wave with period \(T\).
From \(0\) to \(0.5T\) and from \(T\) to \(1.5T\),
\( P=I_0^2R \)
From \(0.5T\) to \(T\) and from \(1.5T\) to \(2.0T\),
\( P=4I_0^2R \)
Answer: \( \boxed{\text{Square wave between }I_0^2R\text{ and }4I_0^2R} \)
(c)(i) Mean power [1 mark]
Each power level occurs for half of one period. Hence,
\( \langle P\rangle=\frac{1}{2}(I_0^2R)+\frac{1}{2}(4I_0^2R) \)
\( \langle P\rangle=\frac{5}{2}I_0^2R \)
Answer: \( \boxed{\langle P\rangle=\frac{5}{2}I_0^2R} \)
(c)(ii) r.m.s. current [2 marks]
The mean power can also be written as
\( \langle P\rangle=I_{\mathrm{r.m.s.}}^2R \)
Therefore,
\( I_{\mathrm{r.m.s.}}^2R=\frac{5}{2}I_0^2R \)
Cancelling \(R\),
\( I_{\mathrm{r.m.s.}}^2=\frac{5}{2}I_0^2 \)
Hence,
\( I_{\mathrm{r.m.s.}}=\sqrt{\frac{5}{2}}I_0 \)
Answer: \( \boxed{I_{\mathrm{r.m.s.}}=\sqrt{\frac{5}{2}}I_0} \)
Question 8
(a)
(i) Show that the momentum \(p\) of a photon of electromagnetic radiation with wavelength \(\lambda\) is given by
\( p=\frac{h}{\lambda} \)
where \(h\) is the Planck constant. (2 marks)
______________________________________________
(ii) Use the expression in (a)(i) to show that a photon in free space that has a momentum of \(9.5\times10^{-28}\,\mathrm{N\,s}\) is a photon of red light. (1 mark)
______________________________________________
(b) A beam of red light of intensity \(160\,\mathrm{W\,m^{-2}}\) is incident normally on a plane mirror, as shown in Fig. 8.1. The momentum of each photon in the beam is \(9.5\times10^{-28}\,\mathrm{N\,s}\).
All of the light is reflected by the mirror in the opposite direction to its original path. The cross-sectional area of the beam is \(2.5\times10^{-6}\,\mathrm{m^2}\).
(i) Show that the number of photons incident on the mirror per unit time is \(1.4\times10^{15}\,\mathrm{s^{-1}}\). (2 marks)
______________________________________________
(ii) Use the information in (b)(i) to determine the pressure exerted by the light beam on the mirror. (3 marks)
pressure = __________________________________ \( \mathrm{Pa} \)
(c) The beam of red light in (b) is now replaced with a beam of blue light of the same intensity.
Suggest and explain whether the pressure exerted on the mirror by the beam of blue light is less than, the same as, or greater than the pressure exerted by the beam of red light. (2 marks)
______________________________________________
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
▶️ Answer/Explanation
(a)(i) Momentum of a photon [2 marks]
For a photon, the relationship between its energy and momentum is
\( p=\frac{E}{c} \)
The energy of a photon is
\( E=hf \)
Since \(c=f\lambda\),
\( f=\frac{c}{\lambda} \)
Therefore,
\( E=\frac{hc}{\lambda} \)
Substituting into \(p=\frac{E}{c}\),
\( p=\frac{1}{c}\left(\frac{hc}{\lambda}\right) \)
Hence,
\( p=\frac{h}{\lambda} \)
Answer: \( \boxed{p=\frac{h}{\lambda}} \)
(a)(ii) Wavelength of the photon [1 mark]
From
\( p=\frac{h}{\lambda} \)
rearrange to give
\( \lambda=\frac{h}{p} \)
\( \lambda=\frac{6.63\times10^{-34}}{9.5\times10^{-28}} \)
\( \lambda=7.0\times10^{-7}\,\mathrm{m}=700\,\mathrm{nm} \)
Answer: \( \boxed{\lambda=700\,\mathrm{nm}} \), so the photon is red light.
(b)(i) Number of photons incident per unit time [2 marks]
The power of the incident beam is
\( P=I A \)
\( P=(160)(2.5\times10^{-6}) \)
\( P=4.0\times10^{-4}\,\mathrm{W} \)
The energy of each photon is
\( E=pc \)
\( E=(9.5\times10^{-28})(3.00\times10^8) \)
\( E=2.85\times10^{-19}\,\mathrm{J} \)
Therefore, the number of photons incident per unit time is
\( N=\frac{P}{E} \)
\( N=\frac{160(2.5\times10^{-6})}{(9.5\times10^{-28})(3.00\times10^8)} \)
\( N=1.4\times10^{15}\,\mathrm{s^{-1}} \)
Answer: \( \boxed{1.4\times10^{15}\,\mathrm{s^{-1}}} \)
(b)(ii) Pressure exerted by the light beam [3 marks]
Pressure is force per unit area:
\( p=\frac{F}{A} \)
The force is the rate of change of momentum.
Because the photons are completely reflected, their momentum changes from \(+p\) to \(-p\). Thus, the magnitude of the change in momentum per photon is
\( \Delta p=2p \)
Therefore,
\( F=2pN \)
\( F=2(9.5\times10^{-28})(1.4\times10^{15}) \)
\( F=2.66\times10^{-12}\,\mathrm{N} \)
Hence,
\( p=\frac{2(9.5\times10^{-28})(1.4\times10^{15})}{2.5\times10^{-6}} \)
\( p=1.1\times10^{-6}\,\mathrm{Pa} \)
Answer: \( \boxed{1.1\times10^{-6}\,\mathrm{Pa}} \)
(c) Effect of replacing red light with blue light [2 marks]
Blue light has a shorter wavelength than red light. From
\( p=\frac{h}{\lambda} \)
the photons of blue light therefore have greater momentum.
However, the blue light has the same intensity, so the same total energy arrives per unit time. Since each blue photon has greater energy, there are fewer photons arriving per unit time.
The greater momentum per photon is therefore balanced by the smaller number of photons per unit time.
Answer: \( \boxed{\text{The pressure is the same.}} \)
Question 9
(a) State what is meant by nuclear fusion. (2 marks)
______________________________________________
______________________________________________
(b) On Fig. 9.1, sketch the variation of binding energy per nucleon with nucleon number \(A\) for values of \(A\) between 1 and 250. (2 marks)
(c) On your line in Fig. 9.1, label:
(i) a point X that could represent a nucleus that undergoes alpha-decay. (1 mark)
(ii) a point Y that could represent a nucleus that undergoes nuclear fusion. (1 mark)
(d) A nucleus Z undergoes nuclear fission to form strontium-93 \( \left(^{93}_{38}\mathrm{Sr}\right) \) and xenon-139 \( \left(^{139}_{54}\mathrm{Xe}\right) \) according to
\( ^1_0\mathrm{n}+Z\rightarrow{}^{93}_{38}\mathrm{Sr}+{}^{139}_{54}\mathrm{Xe}+2\,{}^1_0\mathrm{n} \)
Table 9.1 shows the binding energies of the strontium-93 and xenon-139 nuclei.
Table 9.1
| nucleus | binding energy / J |
|---|---|
| \(^{93}_{38}\mathrm{Sr}\) | \(1.25\times10^{-10}\) |
| \(^{139}_{54}\mathrm{Xe}\) | \(1.81\times10^{-10}\) |
The fission of \(1.00\,\mathrm{mol}\) of Z releases \(1.77\times10^{13}\,\mathrm{J}\) of energy.
Determine the binding energy per nucleon, in MeV, of Z. (4 marks)
binding energy per nucleon = __________________________ \( \mathrm{MeV} \)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 23.2: Radioactive decay — part (c)(i)
▶️ Answer/Explanation
(a) Nuclear fusion [2 marks]
Nuclear fusion is the process in which two small nuclei join together to form one larger nucleus.
Answer: \( \boxed{\text{two small nuclei join together to form one larger nucleus}} \)
(b) Binding energy per nucleon graph [2 marks]
The curve should rise steeply from small \(A\), reaching a maximum at approximately \(A=56\).
After the maximum, the curve should decrease with a shallower negative gradient and should not return to zero for values of \(A\) up to 250.
Answer: \( \boxed{\text{A curve with a peak at approximately }A=56} \)
(c)(i) Position of X for alpha-decay [1 mark]
Alpha-decay is characteristic of heavy nuclei. Therefore, X should be placed at a value of \(A\) to the right of the peak.
Answer: \( \boxed{X\text{ to the right of the peak}} \)
(c)(ii) Position of Y for nuclear fusion [1 mark]
Nuclear fusion involves small nuclei combining to form a nucleus with greater binding energy per nucleon. Therefore, Y should be placed at a value of \(A\) close to 1, on the rising part of the curve.
Answer: \( \boxed{Y\text{ close to }A=1} \)
(d) Binding energy per nucleon of Z [4 marks]
The energy released per nucleus is found from the energy released by \(1.00\,\mathrm{mol}\) of Z.
\( E_{\mathrm{released\ per\ nucleus}}=\frac{1.77\times10^{13}}{6.02\times10^{23}} \)
\( E_{\mathrm{released\ per\ nucleus}}=2.94\times10^{-11}\,\mathrm{J} \)
The binding energy of the products is
\( E_{\mathrm{products}}=(1.25+1.81)\times10^{-10} \)
\( E_{\mathrm{products}}=3.06\times10^{-10}\,\mathrm{J} \)
The binding energy of Z is therefore
\( E_{\mathrm{B,Z}}=3.06\times10^{-10}-2.94\times10^{-11} \)
\( E_{\mathrm{B,Z}}=2.77\times10^{-10}\,\mathrm{J} \)
From conservation of nucleon number,
\( A_Z+1=93+139+2 \)
\( A_Z=233 \)
Therefore, the binding energy per nucleon is
\( \frac{E_{\mathrm{B,Z}}}{A_Z}=\frac{2.77\times10^{-10}}{233} \)
\( =1.19\times10^{-12}\,\mathrm{J} \)
Using \(1\,\mathrm{MeV}=1.60\times10^{-13}\,\mathrm{J}\),
\( E_{\mathrm{B\ per\ nucleon}}=\frac{1.19\times10^{-12}}{1.60\times10^{-13}} \)
\( E_{\mathrm{B\ per\ nucleon}}=7.43\,\mathrm{MeV} \)
Answer: \( \boxed{7.43\,\mathrm{MeV}} \)
Question 10
Ultrasound and X-rays are both types of wave that are used in medical diagnosis to form images of internal body structures.
(a) Complete Table 10.1 to state, for each type of wave:
- the method of production of the wave
- whether the wave that is detected and used to form the image is the wave that has been absorbed, reflected or transmitted by the internal body structure.
Table 10.1
| ultrasound | X-rays | |
|---|---|---|
| method of production | ________________________________ | ________________________________ |
| detected wave (absorbed, reflected or transmitted) | ________________________________ | ________________________________ |
(4 marks)
(b)
(i) For one type of wave passing through tissue, the wave has \(72\%\) of its initial intensity after it has passed through \(6.2\,\mathrm{cm}\) of the tissue.
Calculate the linear attenuation coefficient \(\mu\) of the tissue for this wave. (2 marks)
\(\mu=\) __________________________________ \( \mathrm{cm^{-1}} \)
(ii) Another wave of the same type as in (b)(i) passes through \(9.3\,\mathrm{cm}\) of the same tissue.
Calculate the percentage of the initial intensity of the wave that is attenuated by the tissue. (2 marks)
percentage attenuated = __________________________ \( \% \)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 24.2: Production and use of X-rays — part (a), X-ray production and detection, and part (b) where the wave is X-rays
▶️ Answer/Explanation
(a) Medical imaging methods [4 marks]
| Question | Answer |
|---|---|
| Ultrasound production | A vibrating piezoelectric crystal produces the ultrasound waves. |
| X-ray production | Electrons bombard a metal target. |
| Ultrasound detected wave | Reflected |
| X-ray detected wave | Transmitted |
Answers: Ultrasound → vibrating piezoelectric crystal; reflected. X-rays → electrons hitting a metal target; transmitted.
(b)(i) Linear attenuation coefficient [2 marks]
The attenuation equation is
\( I=I_0\exp(-\mu x) \)
Since \(I=0.72I_0\) and \(x=6.2\,\mathrm{cm}\),
\( 0.72=\exp(-6.2\mu) \)
Taking natural logarithms,
\( \ln(0.72)=-6.2\mu \)
\( \mu=\frac{-\ln(0.72)}{6.2} \)
\( \mu=0.053\,\mathrm{cm^{-1}} \)
Answer: \( \boxed{0.053\,\mathrm{cm^{-1}}} \)
(b)(ii) Percentage of intensity attenuated [2 marks]
For \(x=9.3\,\mathrm{cm}\),
\( \frac{I}{I_0}=\exp(-\mu x) \)
\( \frac{I}{I_0}=\exp[-(0.053)(9.3)] \)
\( \frac{I}{I_0}=0.61 \)
Therefore, the fraction attenuated is
\( 1.00-0.61=0.39 \)
Percentage attenuated is
\( 100(0.39)=39\% \)
Answer: \( \boxed{39\%} \)