A Level Physics 9702 Paper 21, May/June 2024

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Question paper

9702/21 · May/June 2024

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1

The drag force FD acting on an object falling through air is given by FD=12CρAv2 where A is the cross-sectional area of the object, v is the velocity of the object in the air, ρ is the density of the air and C is a constant called the drag coefficient.

Figure 1 of 1 for question 1
a

Use SI base units to show that the drag coefficient has no units.

[3]
b

Fig. 1.1 shows a sphere falling at terminal velocity in air. Assume that the upthrust on the sphere is negligible. On Fig. 1.1, draw and label arrows to show the directions of the two forces acting on the sphere.

[2]
c

The mass of the sphere is 49 g. Calculate the drag force FD acting on the sphere. FD=...................................................... N

[2]
d

The sphere is falling in air at a terminal velocity of 25 in SI base units. The density of the air is 1.2 in SI base units. The diameter of the sphere is 0.060 in SI base units. Use your answer in (c) to calculate the drag coefficient C for the sphere. C=.........................................................

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

Define velocity.

[1]
b(i)

A student throws a ball over a vertical wall of height h, as shown in Fig. 2.1. The ball leaves the hand of the student at a height of 1.2 m above the horizontal ground. The ball has an initial velocity of 22 m s−1 at an angle of 40∘ to the horizontal. The wall is a horizontal distance of 36 m from where the student releases the ball. Air resistance is negligible. Determine the time taken for the ball to reach the wall. time taken=....................................................... s

[2]
b(ii)

Calculate the vertical component u of the initial velocity of the ball. u=................................................. ms−1

[1]
b(iii)

The ball just goes over the wall. Calculate the height h of the wall. h=...................................................... m

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

State the principle of conservation of momentum.

[2]
b

An object of mass 2m is travelling at a speed of 5.0 m s−1 in a straight line. It collides with an object of mass 3m which is initially stationary, as shown in Fig. 3.1. After the collision, the object of mass 2m moves with velocity v at an angle of 30∘ to its original direction of motion. The object of mass 3m moves with velocity w also at an angle of 30∘, as shown in Fig. 3.2. By considering the conservation of momentum in two dimensions, calculate the magnitudes of v and w. v=...................................................... ms−1 w=...................................................... ms−1

[4]
c

An object of mass 4.2 kg is travelling in a straight line at a speed of 6.0 m s−1. The object is brought to rest in a distance of 0.050 m by a constant force. Calculate the magnitude of this force. force=...................................................... N

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

Define strain.

[1]
b(i)

A copper wire of length 4.0 m has a uniform cross-sectional area of 4.5×10−7 m2. A tensile force of 18 N is applied to the wire. This causes the wire to extend by 1.4 mm up to its limit of proportionality. Calculate the Young modulus of the wire. Young modulus=..................................................... Pa

[3]
b(ii)

On Fig. 4.1, draw a line to show how the stress varies with the strain for the wire up to its limit of proportionality.

[2]
c

A second copper wire has the same length as the wire in (b) but a larger diameter. Both wires are subjected to a tensile force of 18 N. By placing a tick (✓) in each row, complete Table 4.1 to compare the stress and strain of the two wires.

[2]
5

A stretched string PQ has length 1.2 m. One end of the string is attached to a vibration generator and the other end is attached to a wall, as shown in Fig. 5.1. The vibration generator is switched on and a stationary wave is formed on the string. The string is shown at one instant of time in Fig. 5.2.

Figure 1 of 7 for question 5
Figure 2 of 7 for question 5
Figure 3 of 7 for question 5
Figure 4 of 7 for question 5
Figure 5 of 7 for question 5
Figure 6 of 7 for question 5
Figure 7 of 7 for question 5
a

Explain how a stationary wave is formed between the vibration generator and the wall.

[2]
b

Calculate the wavelength of the stationary wave shown in Fig. 5.2. wavelength=...................................................... m

[1]
c

Fig. 5.3 shows the stationary wave at time t=0 when all points on the wave are at their maximum displacements. The period of the wave is 0.16 s. On Fig. 5.3, sketch the shape of the stationary wave at time t=0.24 s.

[2]
d

Points R and T on the string are a horizontal distance of 0.30 m apart and in the positions shown in Fig. 5.4. State the phase difference between the oscillations of points R and T. phase difference=....................................................... ∘

[1]
e

Calculate the speed of the progressive waves on the stretched string. speed=................................................. ms−1

[2]
6
Figure 1 of 3 for question 6
Figure 2 of 3 for question 6
Figure 3 of 3 for question 6
a

State Kirchhoff’s first law.

[1]
b

A cell with internal resistance r is connected to two resistors of resistances R1 and R2 as shown in Fig. 6.1. The potential differences (p.d.s) across R1 and R2 are V1 and V2 respectively. The terminal p.d. across the cell is V. The current in the circuit is I. Use Kirchhoff’s laws to show that the total resistance RT of the external circuit is given by RT=R1+R2 .

[2]
c

The electromotive force (e.m.f.) of the cell in Fig. 6.1 is 1.50 V. The values of R1 and R2 are 10 Ω and 15 Ω respectively. The terminal p.d. of the cell is 1.35 V. Calculate the internal resistance r of the cell. r=...................................................... Ω

[3]
d(i)

A resistor of resistance R3 is added to the circuit in Fig. 6.1, so that the circuit is as shown in Fig. 6.2. State and explain the effect, if any, of this change on: the current in the cell

[2]
d(ii)

State and explain the effect, if any, of this change on: the terminal p.d. of the cell.

[2]
7

Nuclei of an isotope of copper (Cu) each have 29 protons and 37 neutrons. This isotope is a β− emitter.

Figure 1 of 2 for question 7
Figure 2 of 2 for question 7
a

State the nuclide notation in the form ZAX for this nucleus of copper.

[1]
b(i)

The energy spectrum of the β− radiation emitted by a sample of this isotope is shown in Fig. 7.1. Use Fig. 7.1 to explain why other particles apart from the β− particles must be emitted during this decay.

[3]
b(ii)

State the name of the other particle emitted during the decay of this isotope.

[1]
b(iii)

The copper isotope decays to an isotope of zinc (Zn). Give the radioactive decay equation for this decay. Include the nucleon and proton numbers of all the particles involved.

[3]

Mark scheme

9702/21 · May/June 2024

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1a · 3 marks

  • M1

    units of FD: kg m s−2

  • M1

    units of ρ: kg m−3 and units of A: m2 and units of v: m s−1 or units of v2: m2 s−2

  • A1

    kg m s−2=C kg m s−2 and comment '(so) C has no units' / unit terms cancelled or C=kg m s−2/(kg m−3 m2 m2 s−2) and comment '(so) C has no units' / unit terms cancelled

1b · 2 marks

  • B1

    one arrow vertically downward labelled weight to within 10° of the vertical

  • B1

    one arrow vertically upwards labelled drag / drag force / FD / air resistance / viscous force to within 10° of the vertical

1c · 2 marks

  • C1

    (at terminal velocity) FD=mg

  • A1

    FD=0.049×9.81 =0.48 N

1d · 3 marks

  • C1

    area=π×(0.060/2)2

  • C1

    0.48=12×C×1.2×π×(0.060/2)2×252

  • A1

    C=0.45

2a · 1 mark

  • B1

    change in displacement / time (taken)

2b(i) · 2 marks

  • C1

    horizontal velocity = 22×cos⁡40∘

  • A1

    time taken = 36/(22×cos⁡40∘) =2.1 s

2b(ii) · 1 mark

  • A1

    u=22×sin⁡40∘ =14 m s−1

2b(iii) · 3 marks

  • C1

    s=ut+12at2 =(14×2.1)+(12×−9.81×2.12)

  • C1

    =7.8 (m)

  • A1

    (therefore) height of wall =7.8+1.2 =9.0 m

3a · 2 marks

  • M1

    sum / total momentum before (a collision) = sum / total momentum after (a collision) or sum / total momentum (of a system) is constant

  • A1

    if no (resultant) external force (acts) / for an isolated system

3b · 4 marks

  • C1

    along direction of motion: 10m=2mvcos⁡30∘+3mwcos⁡30∘

  • C1

    perpendicular to direction of motion: 2mvcos⁡60∘=3mwcos⁡60∘ (v=3w/2)

  • A1

    v=2.9 m s−1

  • A1

    w=1.9 m s−1

3c · 3 marks

  • C1

    EK=12×m×v2 (=12×4.2×6.02) (=76 J)

  • C1

    force = work done / distance

  • A1

    force =76/0.050 =1500 N

4a · 1 mark

  • B1

    extension / original length

4b(i) · 3 marks

  • C1

    Young modulus = stress / strain

  • C1

    =(18/4.5×10−7)/(1.4×10−3/4.0)

  • A1

    =(4.0×107)/(3.5×10−4) =1.1×1011 Pa

4b(ii) · 2 marks

  • B1

    straight line through the origin

  • B1

    ending at the point (3.5, 4.0)

4c · 2 marks

  • B2
    greater in second wireless in second wirethe same in both wires
    stress✓
    strain✓

5a · 2 marks

  • B1

    wave(s) (travel along string and) reflect at fixed point / wall / Q / end / vibration generator / P

  • B1

    incident and reflected waves superpose

5b · 1 mark

  • A1

    0.80 m

5c · 2 marks

  • B1

    same wavelength as original throughout and passing through intersection of solid and dashed lines

  • B1

    reflected in dashed line and of same amplitude

5d · 1 mark

  • A1

    180°

5e · 2 marks

  • C1

    v=fλ and f=1/T or v=λ/T

  • A1

    v=6.25×0.80 or 0.80/0.16 =5.0 m s−1

6a · 1 mark

  • B1

    sum of current(s) entering a junction = sum of current(s) leaving (the same junction) or (algebraic) sum of current (s) at a junction is zero

6b · 2 marks

  • B1

    (by Kirchhoff’s second law) V=V1+V2

  • B1

    so IRT=IR1+IR2 (and cancelling I gives) RT=R1+R2 or V/I=V1/I+V2/I (and substituting R gives) RT=R1+R2

6c · 3 marks

  • C1

    current in circuit =1.35/(10+15) (= 0.054 A)

  • C1

    r=(E−V)/I

  • A1

    =(1.5−1.35)/0.054 =2.8 Ω

6d(i) · 2 marks

  • M1

    the (total) resistance (of the circuit) has decreased (and e.m.f. is unchanged)

  • A1

    (the current (in the cell) will) increase

6d(ii) · 2 marks

  • M1

    (as the current is greater and so there is a) larger p.d. across the internal resistance

  • A1

    (terminal p.d. will) decrease

7a · 1 mark

  • B1

    2966Cu

7b(i) · 3 marks

  • B1

    the energy of the decay is fixed / constant

  • B1

    the energies of the beta particles have a (continuous) range of values / varies / not constant

  • B1

    another particle / an (anti)neutrino must possess the extra / remaining energy (difference between energy of the decay and the β kinetic energy)

7b(ii) · 1 mark

  • B1

    (electron) antineutrino

7b(iii) · 3 marks

  • B1

    values for Cu and Zn correct with no other extra particles on either side of the equation

  • B1

    second term correct (−10β)

  • B1

    third term correct (00νˉe)

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  2. 2
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  3. 3
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  4. 4
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  5. 5
    Question 58 marks
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  6. 6
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