A Level Physics 9702 Paper 21, Oct/Nov 2024

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9702/21 · Oct/Nov 2024

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1
Figure 1 of 1 for question 1
a

Define density.

[1]
b(i)

Fig. 1.1 shows a cuboidal glass block. A student measures the mass m of the block and the side lengths x, y and z. The measurements are shown in Table 1.1. Table 1.1

quantitymeasurement
m(0.243±0.001) kg
x(5.41±0.01) cm
y(11.09±0.01) cm
z(1.62±0.01) cm

Determine the density of the glass.

[2]
b(ii)

Calculate the percentage uncertainty in the density.

[3]
c

The true value of the density of the glass is different from the answer in (b)(i) because of a systematic error in the measurements. Suggest one possible cause of this systematic error.

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

Define linear momentum.

[1]
b(i)

A car of mass 1800 kg is moving in a straight line. Fig. 2.1 shows the variation with time t of the momentum p of the car. Calculate the maximum speed reached by the car.

[1]
b(ii)

Calculate the maximum kinetic energy of the car.

[2]
b(iii)

Show that the acceleration of the car at time t=4.0 s is 5.0 m s−2.

[2]
b(iv)

Determine the distance travelled by the car between times t=0 and t=12.0 s.

[2]
c

On Fig. 2.2, sketch the variation with time t of the acceleration a of the car in (b) from t=0 to t=12.0 s.

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

State what is meant by the work done by a force.

[1]
b

A block of mass m is raised vertically at constant speed. The vertical height gained by the block is Δh, as shown in Fig. 3.1. Derive an expression, in terms of m and Δh, for the change in gravitational potential energy ΔEP of the block. State the meaning of any other symbols you use.

[2]
c(i)

An electric motor has an input power of 900 W. The motor takes 1.0 minute to lift a load of weight 240 N at constant speed through a vertical height of 150 m. Resistive forces are negligible. Show that the work done by the motor on the load in 1.0 minute is 36 kJ.

[1]
c(ii)

Determine the useful output power of the motor.

[2]
c(iii)

Use your answer in (c)(ii) to determine the efficiency of the motor.

[2]
c(iv)

Some of the power wasted in the motor is dissipated by the resistance of its coil. This dissipated power is 280 W. The coil of the motor is made from wire of total length 23 m. The wire has a cross-sectional area of 2.6×10−8 m2 and is made from metal of resistivity 1.7×10−8 Ω m. Calculate the current in the coil.

[3]
4
Figure 1 of 1 for question 4
a

Define the Young modulus of a material.

[1]
b(i)

A metal wire P that obeys Hooke’s law is stretched within its limit of proportionality. On Fig. 4.1, sketch the variation of tensile force F in the wire with its extension x.

[1]
b(ii)

State the name of the quantity represented by the gradient of the line in Fig. 4.1.

[1]
b(iii)

State the name of the quantity represented by the area under the line in Fig. 4.1.

[1]
c

Another wire Q is made from a metal that has twice the Young modulus of the metal of wire P in (b). Wire Q has the same volume as wire P but has double the cross-sectional area of wire P. The two wires are extended by equal tensile forces within their limits of proportionality. State and explain how the extension of wire Q compares with the extension of wire P.

[3]
5
a(i)

Potassium-40 (1940K) undergoes β− decay to form a nuclide of element X. Particle Z is emitted during the decay. The equation for the decay is shown. 1940K→QPX+SRβ−+Z State the values of P, Q, R and S. P=................................................. R=................................................. Q=................................................. S=.................................................

[2]
a(ii)

State the name of particle Z.

[1]
a(iii)

State the name of the class of fundamental particle to which both the β− particle and particle Z belong.

[1]
b

Determine the quark composition of an alpha-particle.

[3]
6

Two coherent sources X and Y of microwaves of frequency 2.5×1010 Hz are a distance of 0.18 m apart in a vacuum, as shown in Fig. 6.1. There is a phase difference of 90∘ between the waves emitted at the two sources. A microwave detector moves along the line PQ, which is parallel to the line joining the two sources and 2.3 m away from it. Point O is on the line PQ at a position that is equidistant from the two sources. Point A is the position on line PQ where the intensity of the microwaves is the greatest.

Figure 1 of 2 for question 6
Figure 2 of 2 for question 6
a(i)

Explain why the position of greatest intensity is not at point O.

[2]
a(ii)

On Fig. 6.1, draw a cross (×) to show the position of the point on line PQ where the intensity minimum that is the closest to point O occurs. Label this point B.

[2]
b(i)

Show that the wavelength of the microwaves is 0.012 m.

[2]
b(ii)

For point A on line PQ, determine the difference in the distances Δx travelled by the microwaves from X and the microwaves from Y.

[1]
b(iii)

Use the formula for the double-slit interference of light to calculate the distance between adjacent intensity maxima on line PQ.

[2]
7
Figure 1 of 2 for question 7
Figure 2 of 2 for question 7
a(i)

Fig. 7.1 shows two resistors connected in series with a cell of electromotive force (e.m.f.) 1.50 V and internal resistance 0.28 Ω. One of the resistors has resistance 1.0 Ω. The other resistor has resistance R. The terminal potential difference (p.d.) across the cell is 1.36 V. Show that the current I in the circuit is 0.50 A.

[2]
a(ii)

Calculate the combined resistance of the two resistors.

[2]
a(iii)

Use your answer in (a)(ii) to determine resistance R.

[1]
b(i)

The circuit in Fig. 7.1 is disconnected and the two resistors are reconnected to the cell, now in parallel with each other. On Fig. 7.2, complete the circuit diagram to show this arrangement.

[1]
b(ii)

Explain, without calculation, whether the terminal p.d. across the cell is now less than, equal to or greater than 1.36 V.

[2]

Mark scheme

9702/21 · Oct/Nov 2024

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1a · 1 mark

  • B1

    mass per unit volume

1b(i) · 2 marks

  • C1

    density = 0.243 / (0.0541 × 0.1109 × 0.0162)

  • A1

    = 2500 kg m–3

1b(ii) · 3 marks

  • C1

    (0.001 / 0.243) or (0.01 / 5.41) or (0.01 / 11.09) or (0.01 / 1.62)

  • C1

    (Δρ / ρ) = (Δm / m) + (Δx / x) + (Δy / y) + (Δz / z) = (0.001 / 0.243) + (0.01 / 5.41) + (0.01 / 11.09) + (0.01 / 1.62) ( = 0.013)

  • A1

    percentage uncertainty in ρ = 0.013 × 100 = 1.3% (allow 1 s.f. answer of 1%)

1c · 1 mark

  • B1

    zero error (on calipers / balance) or incorrect calibration (of calipers / balance)

2a · 1 mark

  • B1

    product of mass and velocity

2b(i) · 1 mark

  • A1

    maximum speed = 7.2 × 10⁴ / 1800 = 40 m s–1

2b(ii) · 2 marks

  • C1

    kinetic energy = ½ mv²

  • A1

    = ½ × 1800 × 40² = 1.4 × 10⁶ J

2b(iii) · 2 marks

  • C1

    a = gradient of line / mass or a = (v – u) / t or a = Δv / t

  • A1

    a = e.g. (3.6 × 10⁴) / (4.0 × 1800) = 5.0 m s–2 or a = e.g. 20 (– 0) / 4 = 5.0 m s–2 or F = e.g. 3.6 × 10⁴ / 4.0 = 9.0 × 10³ and a = 9.0 × 10³ / 1800 = 5.0 m s–2

2b(iv) · 2 marks

  • C1

    distance = ½ × 40 × (8 + 4) or distance = ½ × 72 000 × (8 + 4) / 1800 or distance = [40² / (2 × 5)] + [40² / (2 × 10)] or distance = ½ × 5 × 8² + (40 × 4 – ½ × 10 × 4²)

  • A1

    distance = 240 m

2c · 3 marks

  • B1

    stepped shape, showing one constant value up to 8.0 s then stepping to a different constant value from 8.0 s with no time delay

  • B1

    horizontal straight line from (0, 5.0) to (8.0, 5.0)

  • B1

    horizontal straight line from (8.0, –10.0) to (12.0, –10.0)

3a · 1 mark

  • B1

    product of force and displacement in direction of force

3b · 2 marks

  • B1

    weight/force = mg, and g identified as acceleration of free fall

  • B1

    ΔEP identified as work (done by lifting force/weight), and so = mg × Δh

3c(i) · 1 mark

  • A1

    240 × 150 = 36 kJ

3c(ii) · 2 marks

  • C1

    P = W / t

  • A1

    = 36 000 / 60 = 600 W

3c(iii) · 2 marks

  • C1

    efficiency = useful output power / total input power

  • A1

    = 600 / 900 = 0.67

3c(iv) · 3 marks

  • C1

    P = I²R and R = ρL / A

  • C1

    280 = I² × (1.7 × 10–8 × 23) / (2.6 × 10–8)

  • A1

    I = 4.3 A

4a · 1 mark

  • B1

    stress per unit strain

4b(i) · 1 mark

  • B1

    straight line with positive gradient passing through origin

4b(ii) · 1 mark

  • B1

    spring constant

4b(iii) · 1 mark

  • B1

    elastic potential energy (stored in wire)

4c · 3 marks

  • B1

    length (of Q) is half (the length of P)

  • B1

    extension is proportional to length, and inversely proportional to area and Young modulus

  • B1

    extension of Q is = ½ / (2 × 2) = ⅛ times the extension of P

5a(i) · 2 marks

  • A1

    P = 40 and R = 0

  • A1

    Q = 20 and S = –1

5a(ii) · 1 mark

  • B1

    (electron) antineutrino

5a(iii) · 1 mark

  • B1

    leptons

5b · 3 marks

  • C1

    (alpha particle consists of) 2 protons and 2 neutrons

  • C1

    quark composition of proton = up up down or quark composition of neutron = up down down

  • A1

    quark composition (of alpha particle) = 6 up, 6 down

6a(i) · 2 marks

  • B1

    for maximum intensity, waves must be in phase (at detection)

  • B1

    (phase difference at source means) waves are not in phase / have a 90° phase difference at O

6a(ii) · 2 marks

  • M1

    point B labelled on the line PQ below O

  • A1

    distance OB = distance OA

6b(i) · 2 marks

  • C1

    v = fλ

  • A1

    λ = (3.00 × 10⁸) / (2.5 × 10¹⁰) = 0.012 m

6b(ii) · 1 mark

  • A1

    Δx = 0.012 / 4 = 0.0030 m

6b(iii) · 2 marks

  • C1

    λ = ax / D

  • A1

    distance = λD / a = 0.012 × 2.3 / 0.18 = 0.15 m

7a(i) · 2 marks

  • B1

    (V =) 1.50 – 1.36 (= 0.14 V)

  • A1

    current = V / R = 0.14 / 0.28 = 0.50 A

7a(ii) · 2 marks

  • C1

    resistance = V / I

  • A1

    = 1.36 / 0.50 = 2.7 Ω

7a(iii) · 1 mark

  • A1

    R = 2.7 – 1.0 = 1.7 Ω

7b(i) · 1 mark

  • B1

    two resistors correctly shown in parallel with cell and no other components

7b(ii) · 2 marks

  • B1

    (external) resistance is now smaller and (so) current (in cell) is greater / (external) resistance smaller fraction of total resistance / internal resistance larger fraction of total resistance

  • B1

    (greater p.d. across internal resistance so) terminal p.d. is less

Questions

  1. 1
    Question 17 marks
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  2. 2
    Question 211 marks
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  3. 3
    Question 311 marks
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  4. 4
    Question 47 marks
    Mark scheme ›
  5. 5
    Question 57 marks
    Mark scheme ›
  6. 6
    Question 69 marks
    Mark scheme ›
  7. 7
    Question 78 marks
    Mark scheme ›

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