A Level Physics 9702 Paper 42, May/June 2023

Question paper and mark scheme100 marks10 questions

Solve it on ExamPal

Practise

Untimed. Write on the paper and check as you go.

Start

Exam

Timed, with a clock you set, like sitting the real paper.

Start

Opens full screen.

Papers open in

Opening the question paper

Read as textThe questions and the mark scheme as plain text. Easier to read on a phone.

Question paper

9702/42 · May/June 2023

Download PDF
1
a

State Newton’s law of gravitation.

[2]
b

A satellite is in a circular orbit around a planet. The radius of the orbit is R and the period of the orbit is T. The planet is a uniform sphere. Use Newton’s law of gravitation to show that R and T are related by 4π2R3=GMT2 where M is the mass of the planet and G is the gravitational constant.

[2]
c

The Earth may be considered to be a uniform sphere of mass 5.98×1024 kg and radius 6.37×106 m. A geostationary satellite is in orbit around the Earth. Use the expression in (b) to determine the height of the satellite above the Earth’s surface.

[3]
d(i)

Another satellite is in a circular orbit around the Earth with the same orbital radius and period as the satellite in (c). Calculate the angular speed of the satellite in this orbit. Give a unit with your answer.

[2]
d(ii)

Despite having the same orbital period, the orbit of this satellite is not geostationary. Suggest two ways in which the orbit of this satellite could be different from the orbit of the satellite in (c).

[2]
2
Figure 1 of 1 for question 2
a(i)

State what is meant by an ideal gas.

[2]
a(ii)

State the temperature, in degrees Celsius, of absolute zero.

[1]
b(i)

A sealed vessel contains a mass of 0.0424 kg of an ideal gas at 227 ∘C. The pressure of the gas is 1.37×105 Pa and the volume of the gas is 0.640 m3. Calculate: the number of molecules of the gas in the vessel

[3]
b(ii)

Calculate: the mass of one molecule of the gas

[1]
b(iii)

Calculate: the root-mean-square (r.m.s.) speed v of the molecules of the gas.

[3]
c

The gas in (b) is now cooled gradually to absolute zero. On Fig. 2.1, sketch the variation with thermodynamic temperature T of the r.m.s. speed of the molecules of the gas.

[2]
3
Figure 1 of 2 for question 3
Figure 2 of 2 for question 3
a

State the first law of thermodynamics. Identify the meaning of any symbols that you use.

[2]
b(i)

The state of an ideal gas is continuously changed according to the cycle ABCDA shown in Fig. 3.1. Complete Table 3.1 for the changes A to B and B to C by placing two ticks (✓) in each row.

[4]
b(ii)

Use the first law of thermodynamics to describe and explain the energy transfers associated with one complete cycle ABCDA.

[3]
4

A small steel sphere is oscillating vertically on the end of a spring, as shown in Fig. 4.1. The velocity v of the sphere varies with displacement x from its equilibrium position according to v=±9.711.6−x2 where v is in cm s−1 and x is in cm.

Figure 1 of 2 for question 4
Figure 2 of 2 for question 4
a(i)

Calculate the frequency of the oscillations.

[2]
a(ii)

Show that the amplitude of the oscillations is 3.4 cm.

[1]
a(iii)

Calculate the maximum acceleration a0 of the sphere.

[2]
b

On Fig. 4.2, sketch the variation with x of the acceleration a of the sphere.

[3]
c

Describe, without calculation, the interchange between the potential energy and the kinetic energy of the oscillations.

[3]
5

Two capacitors A and B are connected into the circuit shown in Fig. 5.1. Capacitor A has capacitance C and capacitor B has capacitance 3C. The electromotive force (e.m.f.) of the cell is V. The two-way switch S is initially at position X, and capacitor B is initially uncharged.

Figure 1 of 1 for question 5
a(i)

State, in terms of V and C, expressions for: the initial charge QA on the plates of capacitor A

[1]
a(ii)

State, in terms of V and C, expressions for: the initial energy EA stored in capacitor A.

[1]
b(i)

The two-way switch S is now moved to position Y. State and explain what happens to the charge that was initially on the plates of capacitor A.

[2]
b(ii)

Show that the final potential difference (p.d.) VB across capacitor B is given by VB=V4. Explain your reasoning.

[3]
b(iii)

Determine an expression, in terms of V and C, for the decrease ΔE in the total energy that is stored in the capacitors as a result of the change of the position of the switch.

[2]
6

A heavy aluminium disc has a radius of 0.36 m. The disc rotates with the wheels of a vehicle and forms part of an electromagnetic braking system on the vehicle. In order to activate the braking system, a uniform magnetic field of flux density 0.17 T is switched on. This magnetic field is perpendicular to the plane of rotation of the disc, as shown in Fig. 6.1.

Figure 1 of 1 for question 6
a(i)

Define magnetic flux.

[2]
a(ii)

Calculate the magnetic flux through the disc. Give a unit with your answer.

[2]
b

The disc is rotating at a rate of 25 revolutions per second. Calculate the magnitude of the electromotive force (e.m.f.) induced between the axle and the rim of the disc.

[3]
c

The axle and the rim are connected into an external circuit that enables the energy of the rotation of the disc to be stored for future use. The direction of rotation is shown in Fig. 6.1. Use Lenz’s law of electromagnetic induction to determine whether the current in the disc is from the rim to the axle or from the axle to the rim. Explain your reasoning.

[3]
7

Four diodes are used in a bridge rectifier circuit to produce rectification of a sinusoidal a.c. input voltage VIN. Fig. 7.1 shows part of the circuit, but three of the diodes are missing. The p.d. across the load resistor R is the output p.d. VOUT of the bridge rectifier.

Figure 1 of 4 for question 7
Figure 2 of 4 for question 7
Figure 3 of 4 for question 7
Figure 4 of 4 for question 7
a(i)

State the name of the type of rectification produced by a bridge rectifier.

[1]
a(ii)

Complete Fig. 7.1 by drawing the three missing diodes, correctly connected.

[2]
a(iii)

On Fig. 7.1, draw an arrow to indicate the direction of the current in resistor R.

[1]
b(i)

VIN has amplitude V0 and period T. Fig. 7.2 shows the variation with time t of VIN. On Fig. 7.3, sketch the variation of VOUT with t between t=0 and t=2.0T.

[3]
b(ii)

The power dissipated in the resistor is P. On Fig. 7.4, sketch the variation of P with t between t=0 and t=2.0T.

[2]
b(iii)

Suggest, with a reason, how the root-mean-square (r.m.s.) value of VOUT compares with the r.m.s. value of VIN.

[1]
8

Fig. 8.1 shows the lowest four energy levels of an electron in an isolated atom. Fig. 8.2 shows the lines in the emission spectrum of the atom that correspond to the transitions of the electron from n=3 to n=1 and from n=4 to n=1.

Figure 1 of 2 for question 8
Figure 2 of 2 for question 8
a

Explain, with reference to photons, why there is a single frequency of electromagnetic radiation that corresponds to each of these transitions.

[2]
b(i)

On Fig. 8.2, draw a line that corresponds to the transition of the electron from n=2 to n=1. Label this line A.

[2]
b(ii)

On Fig. 8.2, draw a line that corresponds to the transition of the electron from n=3 to n=2. Label this line B.

[2]
c

The frequency of radiation represented by line A is fA. The frequency of radiation represented by line B is fB. The energy of the ground state (n=1) is E1. Determine an expression, in terms of fA, fB, E1 and the Planck constant h, for the energy E3 of the energy level n=3.

[2]
9
a

Define mass defect.

[2]
b(i)

Table 9.1 shows the mass defects of three nuclei. Table 9.1

nucleusmass defect/u
12H0.002388
13H0.009105
24He0.030377

The nuclear fusion process in a particular star is described by 12H+13H→24He+X where X is a particle that has no mass defect. State the name of particle X.

[1]
b(ii)

Show that the energy released when one nucleus of 24He is formed in this fusion reaction is 2.8×10−12 J.

[3]
c(i)

The star in (b) has a radius of 2.3×109 m and a luminosity of 1.4×1028 W. All the energy released from the formation of 24He is radiated away from the star. All the energy that is radiated from the star has been released in the formation of 24He. Determine: the mass of 24He produced per unit time by the fusion process

[3]
c(ii)

Determine: the surface temperature of the star.

[2]
10

X-rays for use in medical diagnosis are produced in an X-ray tube. In the X-ray tube, charged particles are accelerated towards a metal target by an applied potential difference (p.d.).

a(i)

State the name of the charged particles that are accelerated by the applied p.d.

[1]
a(ii)

Explain how X-rays are produced at the metal target.

[2]
a(iii)

Calculate the minimum wavelength of X-rays produced when the applied p.d. is 5.80 kV.

[3]
b

X-rays pass through a medium that has an attenuation coefficient of 1.4 cm−1. Calculate the percentage of the X-ray energy that is absorbed by a 2.8 cm thickness of this medium.

[3]

Mark scheme

9702/42 · May/June 2023

Download PDF

1a · 2 marks

  • B1

    (gravitational) force is (directly) proportional to product of masses

  • B1

    force (between point masses) is inversely proportional to the square of their separation

1b · 2 marks

  • M1

    GMm / R^2 = mRω^2 or GMm / R^2 = mv^2 / R

  • A1

    ω = 2π / T and algebra leading to 4π^2R^3 = GMT^2 or v = 2πR / T and algebra leading to 4π^2R^3 = GMT^2

1c · 3 marks

  • C1

    4π^2 × R^3 = 6.67 × 10^–11 × 5.98 × 10^24 × (24 × 60 × 60)^2 (R = 4.22 × 10^7 m)

  • C1

    h = R – (6.37 × 10^6)

  • A1

    h = (4.22 × 10^7) – (6.37 × 10^6) = 3.6 × 10^7 m

1d(i) · 2 marks

  • C1

    ω = 2π / T

  • A1

    = 2π / (24 × 60 × 60) = 7.3 × 10^–5 rad s^–1

1d(ii) · 2 marks

  • B1

    orbit is from east to west

  • B1

    orbit is not equatorial / orbit is polar

2a(i) · 2 marks

  • M1

    (gas that obeys) pV ∝ T (for all values of p, V and T)

  • A1

    where T is thermodynamic temperature

2a(ii) · 1 mark

  • A1

    temperature = –273.15 °C

2b(i) · 3 marks

  • C1

    pV = NkT

  • C1

    N = (1.37 × 10^5 × 0.640) / (1.38 × 10^–23 × (227 + 273))

  • A1

    = 1.27 × 10^25

2b(ii) · 1 mark

  • A1

    mass = 0.0424 / (1.27 × 10^25) = 3.34 × 10^–27 kg

2b(iii) · 3 marks

  • C1

    ½m<c^2> = (3 / 2)kT or pV = ⅓(Nm)<c^2> and Nm = mass of gas

  • C1

    3.34 × 10^–27 × v^2 = 3 × 1.38 × 10^–23 × 500 or 0.0424 × v^2 = 3 × 1.37 × 10^5 × 0.640

  • A1

    v = 2490 m s^–1

2c · 2 marks

  • B1

    sketch: line from (0, 0) to (500, v)

  • B1

    line with decreasing positive gradient throughout

3a · 2 marks

  • C1

    change in internal energy = work done + energy transfer by heating

  • A1

    increase in internal energy = work done on system + energy transferred to the system by heating

3b(i) · 4 marks

  • B1

    AB change in internal energy: decrease

  • B1

    AB work done on gas: positive

  • B1

    BC change in internal energy: increase

  • B1

    BC work done on gas: zero

3b(ii) · 3 marks

  • B1

    more work done by gas in CD than is done on gas in AB or (no work done on gas in BC and DA so) (overall) gas does work

  • B1

    (overall) change in internal energy is zero

  • B1

    (must be an overall) input of thermal energy

4a(i) · 2 marks

  • C1

    ω = 2πf

  • A1

    f = 9.7 / 2π = 1.5 Hz

4a(ii) · 1 mark

  • A1

    amplitude = √(11.6) = 3.4 cm

4a(iii) · 2 marks

  • C1

    a_0 = ω^2 x_0

  • A1

    = 9.7^2 × 3.4 × 10^–2 = 3.2 m s^–2

4b · 3 marks

  • B1

    sketch: straight line through the origin with negative gradient

  • B1

    line with negative gradient passing through (+3.4, –a_0) and (–3.4, +a_0)

  • B1

    line with ends at x = ± 3.4 cm and a = ± a_0

4c · 3 marks

  • B1

    sum of potential energy and kinetic energy is constant

  • B1

    at maximum displacement, kinetic energy is zero or at maximum displacement, potential energy is maximum

  • B1

    at zero displacement, kinetic energy is maximum or at zero displacement, potential energy is minimum

5a(i) · 1 mark

  • A1

    Q_A = CV

5a(ii) · 1 mark

  • A1

    E_A = ½CV^2

5b(i) · 2 marks

  • B1

    some of the charge transfers to (the plates of) capacitor B

  • B1

    transfer is because the p.d.s across the capacitors are not equal or transfer stops when the p.d.s across the capacitors become equal

5b(ii) · 3 marks

  • M1

    V_A = V_B or C_T = 4C

  • M1

    charge on A + charge on B = CV or Q_T = CV

  • A1

    CV_B + 3CV_B = CV leading to V_B = V / 4 or V_B = CV / 4C = V / 4

5b(iii) · 2 marks

  • C1

    ΔE = ½CV^2 – nCV^2, where n is a multiple that is less than ½ or total final energy = ½ × 4C × (V / 4)^2 = ⅛CV^2

  • A1

    ΔE = ½CV^2 – ⅛CV^2 = ⅜CV^2

6a(i) · 2 marks

  • M1

    product of (magnetic) flux density and area

  • A1

    area perpendicular to the (magnetic) field

6a(ii) · 2 marks

  • C1

    flux = B × πr^2 = 0.17 × π × 0.36^2

  • A1

    = 6.9 × 10^–2 Wb

6b · 3 marks

  • C1

    time for one revolution = 1 / 25 s

  • C1

    e.m.f. = rate of cutting flux or ΔΦ / Δt

  • A1

    = 0.069 × 25 = 1.7 V

6c · 3 marks

  • B1

    current (in disc) is perpendicular to magnetic field or current causes force to act on disc

  • B1

    force opposes rotation of disc

  • B1

    left-hand rule indicates current is from rim to axle

7a(i) · 1 mark

  • B1

    full-wave (rectification)

7a(ii) · 2 marks

  • B1

    lower left diode shown pointing left

  • B1

    lower right and upper left diodes shown pointing left

7a(iii) · 1 mark

  • B1

    arrow indicating current direction in resistor to the right

7b(i) · 3 marks

  • B1

    sketch: periodic line showing minimum V_OUT = 0 and maximum V_OUT = +V_0

  • B1

    line showing peak V_OUT at t = 0, 0.5T, 1.0T, 1.5T and 2.0T, with V_OUT going to zero half-way in between each peak

  • B1

    line showing correct modulated sine shape

7b(ii) · 2 marks

  • B1

    sketch: sinusoidal curve with troughs sitting on the time axis

  • B1

    peak power at t = 0, 0.5T, 1.0T, 1.5T and 2.0T and zero power half-way in between each peak

7b(iii) · 1 mark

  • B1

    same power-time graph with or without rectification, so same V_rms or V^2-time graph is same for both V_OUT and V_IN, so same V_rms or power does not depend on sign of V, so same V_rms

8a · 2 marks

  • B1

    transition (emits) (one) photon with energy equal to the difference in energy between the two levels

  • B1

    frequency of radiation corresponds to energy of photon

8b(i) · 2 marks

  • B1

    line to the left of the pair in Fig. 8.2, labelled A

  • B1

    larger gap between line A and the nearest of the pair in Fig. 8.2 than between the lines in the pair

8b(ii) · 2 marks

  • B1

    line to the left of both the pair in Fig. 8.2 and line A, labelled B

  • B1

    larger gap between line B and line A than between line A and the nearest one of the pair in Fig. 8.2

8c · 2 marks

  • C1

    E = hf

  • A1

    E_3 = E_1 + h(f_A + f_B)

9a · 2 marks

  • M1

    difference between mass of nucleus and (total) mass of nucleons

  • A1

    when infinitely separated

9b(i) · 1 mark

  • B1

    neutron

9b(ii) · 3 marks

  • C1

    E = Δm c^2

  • C1

    Δm = (0.030377 – 0.002388 – 0.009105)u ( = 0.018884u)

  • A1

    energy release = (0.030377 – 0.002388 – 0.009105) × 1.66 × 10^–27 × (3.00 × 10^8)^2 = 2.8 × 10^–12 J

9c(i) · 3 marks

  • C1

    number of atoms per unit time = (1.4 × 10^28) / (2.8 × 10^–12) ( = 5.0 × 10^39 s^–1)

  • C1

    mass of one atom = 4 × 1.66 × 10^–27 or (4 × 10^–3) / (6.02 × 10^23) ( = 6.64 × 10^–27 kg)

  • A1

    mass per unit time = 6.64 × 10^–27 × 5.0 × 10^39 = 3.3 × 10^13 kg s^–1

9c(ii) · 2 marks

  • C1

    L = 4πσr^2 T^4 1.4 × 10^28 = 4π × 5.67 × 10^–8 × (2.3 × 10^9)^2 × T^4

  • A1

    T = 7800 K

10a(i) · 1 mark

  • B1

    electrons

10a(ii) · 2 marks

  • B1

    electrons are decelerated / stopped on impact with the target

  • B1

    (kinetic) energy lost by electrons emitted as (X-ray) photons

10a(iii) · 3 marks

  • C1

    eV = hc / λ

  • C1

    λ = (6.63 × 10^–34 × 3.00 × 10^8) / (1.60 × 10^–19 × 5800)

  • A1

    = 2.14 × 10^–10 m

10b · 3 marks

  • C1

    I = I_0 exp (–μx)

  • C1

    I_T / I_0 = exp (–(1.4 × 2.8)) = 0.020

  • A1

    % absorbed = (1.000 – 0.0198) × 100 = 98%

Questions

  1. 1
    Question 111 marks
    Mark scheme ›
  2. 2
    Question 212 marks
    Mark scheme ›
  3. 3
    Question 39 marks
    Mark scheme ›
  4. 4
    Question 411 marks
    Mark scheme ›
  5. 5
    Question 59 marks
    Mark scheme ›
  6. 6
    Question 610 marks
    Mark scheme ›
  7. 7
    Question 710 marks
    Mark scheme ›
  8. 8
    Question 88 marks
    Mark scheme ›
  9. 9
    Question 911 marks
    Mark scheme ›
  10. 10
    Question 109 marks
    Mark scheme ›

More 9702 papers

Other papers from May/June 2023

Solve it on ExamPalStart