A Level Physics 9702 Paper 22, Oct/Nov 2024

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Question paper

9702/22 · Oct/Nov 2024

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1
Figure 1 of 1 for question 1
a

State what is meant by a vector quantity.

[1]
b(i)

A sphere falls vertically through a liquid that has density 830 kg m−3. The sphere has radius r and constant velocity v, as shown in Fig. 1.1. The drag force D acting on the sphere is given by D=6πrηv where η is a property of the liquid. Determine the SI base units of η. SI base units .........................................................

[3]
b(ii)

State an equation showing the relationship between the magnitudes of the weight W, drag force D and upthrust U acting on the sphere.

[1]
b(iii)

The volume of the sphere is 4.6 cm3. The drag force D is 0.32 N. Calculate the weight of the sphere. weight = ...................................................... N

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

Define momentum.

[1]
b

A child stands on a scooter on horizontal ground. The combined mass of the child and the scooter is 16 kg. The child starts from rest and pushes once on the ground with her foot which causes her to accelerate. The push lasts for a time of 1.1 s. The speed of the child and the scooter after the push is 0.60 ms−1. Determine the average resultant force acting horizontally on the child and the scooter during the push. average force = ...................................................... N

[2]
c

Later, the child in (b) travels down a slope at a constant angle to the horizontal, as shown in Fig. 2.1. At point A her speed is 0.60 ms−1. She has a constant acceleration of 0.85 ms−2 parallel to the slope. After a time of 3.7 s, she reaches point B. Calculate the distance x travelled by the child along the slope from A to B. x= ...................................................... m

[2]
d(i)

At point B, the child in (c) applies the brake with a constant force to maintain a constant velocity. Point C is 18 m from point B, as shown in Fig. 2.2. The work done by the braking force between B and C is 250 J. Determine the magnitude of the braking force. force = ...................................................... N

[2]
d(ii)

On Fig. 2.3, sketch the variation of the kinetic energy of the child and scooter with distance travelled from point A to point C. Numerical values for kinetic energy are not required.

[3]
3
Figure 1 of 2 for question 3
Figure 2 of 2 for question 3
a(i)

The variation of stress with strain for a metal P is shown in Fig. 3.1. Point E is the elastic limit of the metal. Use Fig. 3.1 to determine the Young modulus for P. Young modulus = .................................................... Pa

[2]
a(ii)

On the line in Fig. 3.1, draw a cross (×) to show the limit of proportionality. Label this point Q.

[1]
b

State the conditions necessary for an object to be in equilibrium.

[2]
c(i)

A wire is used to hold a uniform shelf AB horizontally in equilibrium as shown in Fig. 3.2. The wire is connected to the midpoint of shelf AB at an angle of 50∘ to the horizontal. The shelf is attached to a wall by a hinge at A. The length of shelf AB is 0.65 m and its weight is 33 N. A cup of weight 1.5 N rests on the shelf with its centre of gravity at a horizontal distance of 0.12 m from B. By taking moments about A, determine the tension in the wire. tension = ...................................................... N

[3]
c(ii)

The stress in the wire is 1.5×107 Pa. Determine the radius of the wire. radius = ...................................................... m

[2]
c(iii)

More items are added to the shelf, doubling the stress in the wire. The wire is made of the metal P from (a). Use Fig. 3.1 to state and explain whether the wire will behave plastically or elastically as the stress doubles.

[2]
4
Figure 1 of 2 for question 4
Figure 2 of 2 for question 4
a

With reference to the direction of transfer of energy, compare the oscillations of transverse and longitudinal progressive waves.

[2]
b(i)

A pipe is open at one end and closed at the other with a piston. The piston can slide freely and is at a distance of 4.5×10−2 m from the open end of the pipe. A loudspeaker is positioned near the open end of the pipe and emits a sound wave of a single constant frequency. A stationary wave is formed in the pipe, as illustrated in Fig. 4.1. On Fig. 4.1, draw a letter A at the position of an antinode.

[1]
b(ii)

The speed of sound in air is 340 ms−1. Determine the frequency of the sound wave. frequency = .................................................... Hz

[3]
b(iii)

The piston is moved to the left. The frequency of the sound wave emitted by the loudspeaker is then changed so that a stationary wave is formed with same number of antinodes as in Fig. 4.1. State and explain the change that is made to the frequency of the sound wave.

[2]
5
Figure 1 of 1 for question 5
a

Define electric potential difference (p.d.).

[1]
b(i)

A power supply, three resistors and a component X are connected in the circuit shown in Fig. 5.1. The power supply has an electromotive force (e.m.f.) of 230 V and negligible internal resistance. The current in the power supply is 7.0 A. Identify component X.

[1]
b(ii)

Show that the p.d. across the resistor of resistance 0.86 Ω is 6.0 V.

[1]
b(iii)

Determine the current I1. I1= ....................................................... A

[2]
b(iv)

Calculate the p.d. across component X. p.d. = ...................................................... V

[2]
b(v)

Calculate the power dissipated in component X. power = ..................................................... W

[2]
b(vi)

The purpose of the circuit is to provide power to component X. Determine the percentage efficiency of the circuit. efficiency = ...................................................... %

[2]
b(vii)

The resistor of resistance 170 Ω is removed, leaving an open circuit in the lower branch of the circuit. There is no change to the resistance of component X. State whether the current in the power supply increases, decreases or remains the same.

[1]
6
Figure 1 of 2 for question 6
Figure 2 of 2 for question 6
a

Compare an α‑particle with a β+ particle in terms of their masses and charges.

[3]
b(i)

Nucleus P undergoes α‑decay to form nucleus Q. Nucleus Q then undergoes a further decay to form nucleus R. The proton and nucleon numbers of P and R are shown in Fig. 6.1. On Fig. 6.1, draw a cross (×) to show the proton number and nucleon number of Q. Label your cross Q.

[1]
b(ii)

State the names of the particles emitted as Q decays to form R.

[2]
c(i)

Before the α‑decay, P is travelling at a constant velocity. After the decay, Q has a velocity of 1.3×105 ms−1 at an angle of 68∘ to the original path of P. The α‑particle has a velocity of 150×105 ms−1 at an angle of θ to the original path of P, as shown in Fig. 6.2. Use the principle of conservation of momentum to determine θ. θ= ........................................................ ∘

[3]
c(ii)

Calculate the kinetic energy of the α‑particle. kinetic energy = ....................................................... J

[2]

Mark scheme

9702/22 · Oct/Nov 2024

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1a · 1 mark

  • B1

    a quantity with magnitude and direction

1b(i) · 3 marks

  • C1

    SI base units of D: kg m s–2

  • C1

    SI base units of r: m and v: m s–1

  • A1

    base units of η: kg m s–2 / (m × m s–1) = kg m–1 s–1

1b(ii) · 1 mark

  • A1

    W = U + D

1b(iii) · 2 marks

  • C1

    U = 830 × 9.81 × 4.6 × (10–2)3 ( = 0.037 N)

  • A1

    W = 0.037 + 0.32 = 0.36 N

2a · 1 mark

  • B1

    product of mass and velocity

2b · 2 marks

  • C1

    F = mΔv / Δt = 16 × 0.60 / 1.1

  • A1

    = 8.7 N

2c · 2 marks

  • C1

    x = ut + ½at2 x = 0.60 × 3.7 + ½ × 0.85 × 3.72 or v = 0.60 + 0.85 × 3.7 (= 3.75 m s–1) x = 3.75 × 3.7 – 0.5 × 0.85 × 3.72 or x = ½ × (0.60 + 3.75) × 3.7 or x = (3.752 – 0.602) / (2 × 0.85)

  • A1

    x = 8.0 m

2d(i) · 2 marks

  • C1

    F = W / s = 250 / 18

  • A1

    = 14 N

2d(ii) · 3 marks

  • B1

    any line starting at distance = 0 and a positive non-zero value of kinetic energy

  • B1

    a straight line from distance = 0 to distance = x with positive gradient

  • B1

    a straight horizontal line at a non-zero value of kinetic energy starting at distance = x and ending at distance = x + 18 m that is continuous with the previous line

3a(i) · 2 marks

  • C1

    E = σ / ε or E = gradient

  • A1

    E = e.g. 12 × 107 / 0.0050 = 2.4 × 1010 Pa

3a(ii) · 1 mark

  • B1

    cross drawn at (1.0%, 24 × 107 Pa), labelled Q

3b · 2 marks

  • B1

    resultant force (in any direction) is zero

  • B1

    resultant moment / torque (about any point) is zero

3c(i) · 3 marks

  • C1

    (moment =) 33 × 0.65 / 2 or 1.5 × (0.65 – 0.12) or T sin 50° × (0.65 / 2)

  • C1

    sum of clockwise moments = sum of anticlockwise moments 33 × (0.65 / 2) + 1.5 × (0.65 – 0.12) = T sin 50° × (0.65 / 2)

  • A1

    tension = 46 N

3c(ii) · 2 marks

  • C1

    σ = F / A

  • A1

    πr2 = 46 / (1.5 × 107) r = 9.9 × 10–4 m

3c(iii) · 2 marks

  • M1

    elastic limit is not reached or (new) stress is less than (stress at) elastic limit or (new) strain is less than (strain at) elastic limit

  • A1

    (so the wire behaves) elastically

4a · 2 marks

  • B1

    longitudinal waves have oscillations parallel to the (direction of) transfer of energy

  • B1

    transverse waves have oscillations perpendicular to the (direction of) transfer of energy

4b(i) · 1 mark

  • B1

    A marked at the open end of the pipe

4b(ii) · 3 marks

  • C1

    f = v / λ

  • C1

    λ = 4 × 4.5 × 10–2

  • A1

    f = 340 / (4 × 4.5 × 10–2) = 1900 Hz

4b(iii) · 2 marks

  • M1

    the node–antinode distance is longer or the wavelength (of the wave) is longer

  • A1

    (the speed of sound is constant so) the frequency (of the wave) is lower

5a · 1 mark

  • B1

    energy transferred per unit charge (from electrical to other forms)

5b(i) · 1 mark

  • A1

    heater

5b(ii) · 1 mark

  • A1

    (V =) 7.0 × 0.86 = 6.0 (V)

5b(iii) · 2 marks

  • C1

    I = (230 – 6.0) / 170 ( = 1.3 A)

  • A1

    I1 = 7.0 – 1.3 = 5.7 A

5b(iv) · 2 marks

  • C1

    V = 230 – 6.0 – (5.7 × 2.4) or R = ((230 – 6.0) / 5.7) – 2.4 ( = 36.9 Ω)

  • A1

    = 210 V or V = 5.7 × 36.9 = 210 V

5b(v) · 2 marks

  • C1

    P = IV or P = I2R or P = V2 / R

  • A1

    P = 5.7 × 210 or P = 5.72 × (210 / 5.7) or P = 2102 / (210 / 5.7) P = 1200 W

5b(vi) · 2 marks

  • C1

    % efficiency = useful power out / (total) power in (× 100)

  • A1

    = 1200 / (230 × 7.0) (×100) = 0.75 (× 100) = 75% or 74% (using 3 s.f. value from (v) gives 74%)

5b(vii) · 1 mark

  • A1

    (current) decreases

6a · 3 marks

  • B1

    mass of α (particle) is much greater (than β+ particle)

  • B1

    both particles are positively charged

  • B1

    (magnitude of) charge on α (particle) is twice the charge (on β+ particle)

6b(i) · 1 mark

  • B1

    cross labelled Q at (82, 212)

6b(ii) · 2 marks

  • B1

    either particle named

  • B1

    both particles named and no incorrect particles named

6c(i) · 3 marks

  • C1

    212(u) × 1.3 (× 105) sin 68° or 4(u) × 150 (× 105) sin θ

  • C1

    4(u) × 150 (×105) × sin θ = 212(u) × 1.3 (× 105) × sin 68°

  • A1

    sin θ = 0.426 θ = 25°

6c(ii) · 2 marks

  • C1

    E = ½mv2

  • A1

    = ½ × 4 u × (150 × 105)2 = ½ × 4 × 1.66 × 10–27 × (150 × 105)2 = 7.5 × 10–13 J

Questions

  1. 1
    Question 17 marks
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  2. 2
    Question 210 marks
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  3. 3
    Question 312 marks
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  4. 4
    Question 48 marks
    Mark scheme ›
  5. 5
    Question 512 marks
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  6. 6
    Question 611 marks
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