Physics 9702/21 May/June 2025: question paper and mark scheme
Cambridge AS & A Level Physics (9702), Paper 2 AS Level Structured Questions, May/June 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Define acceleration.
…… [1]
(b) In an experiment, two objects A and B are released from the side of a building, as shown in
Fig. 1.1.
building
A
14 m u
B
3.6 m
ground
Fig. 1.1 (not to scale)
Object A is released from rest at a height of 14 m above the horizontal ground.
Object B is released with an initial upwards vertical velocity u at a height of 3.6 m above the ground.
Both objects take the same time to reach the ground and they do not collide with each other.
Air resistance is negligible.
(i) Calculate the time taken for object A to reach the ground.
time = …… s [2]
(ii) Use your answer in (b)(i) to calculate u.
u = …… m s−1 [2]
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(c) In a second experiment, object B is released from the same height and given the same initial speed as in (b) but at a release angle θ to the vertical, as shown in Fig. 1.2.
building θ u
B
Fig. 1.2
(i) State and explain whether the time taken for object B to reach the ground is less than, the same as or greater than the time in (b)(i).
…… [2]
(ii) By considering energy, state and explain the effect of the change in release angle on the speed at which B reaches the ground.
…… [2]
[Total: 9]
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Mark scheme
1(a) rate of change of velocity B1
1(b)(i) s = ut + ½at 2 and u = 0 or s = ½at 2 C1
t = √(2 × 14 / 9.81)
t = 1.7 s A1
OR (C1)
v = √(02 + 2 × 9.81 × 14)
v = 17 (m s−1)
17 = (0 + 9.81) × t or 17 = 9.81 × t
or 14 = ½ × (0 + 17) × t or 14 = ½ × 17 × t
t = 1.7 s (A1)
1(b)(ii) s = ut + ½at 2 C1
u = (3.6 – ½ × 9.81 × 1.72) / 1.7
u = (–) 6.2 m s–1 A1
OR (C1)
v = (3.6 + ½ × 9.81 × 1.72) / 1.7
v = 10 (m s−1)
10 = u + 9.81 × 1.7
or 102 = u2 + (2 × 9.81 × 3.6)
or 3.6 = ½ × (u + 10) × 1.7
u = (−) 6.2 m s−1 (A1)
1(c)(i) Less time (as it reaches a lower height) B1
(because) the initial vertical (component of the) velocity is smaller (than in part (b)) B1
1(c)(ii) The (total) initial energy is the same (as in part (b)) B1
change in gravitational potential energy is same, so speed is the same B1
(a) State the principle of moments.
…… [2]
(b) Three objects A, B and C are placed on a horizontal beam. The beam is in equilibrium, as shown in Fig. 2.1.
A beam
B C
pivot x
6.0 m
Fig. 2.1 (not to scale)
The beam is uniform and has length 6.0 m.
The pivot is at the midpoint of the beam.
Object A has mass 60 kg and is at one end of the beam.
Object B has mass 45 kg and is at a distance x from the pivot.
Object C has mass 80 kg and is at the other end of the beam.
Calculate x.
x = …… m [3]
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(c) The beam is 0.80 m above horizontal ground.
Object A is removed and replaced by a spring connected to the ground and the beam, as shown in Fig. 2.2.
beam
B C
spring
0.80 m ground pivot
Fig. 2.2
After the change, the beam is again horizontal and in equilibrium. The positions of B and C are unchanged.
The spring has an unstretched length of 0.59 m and obeys Hooke’s law.
(i) Calculate the spring constant of the spring.
spring constant = …… N m–1 [3]
(ii) Calculate the elastic potential energy of the spring.
elastic potential energy = …… J [2]
[Total: 10]
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Mark scheme
2(a) in (rotational) equilibrium B1
sum / total of clockwise moments about a point = sum / total of anticlockwise moments about the (same) point. B1
2(b) 80 × 9.81 × 3 or 60 × 9.81 × 3 or 45 × 9.81 × x C1
80 × 9.81 × 3 = (60 × 9.81 × 3) + (45 × 9.81 × x) C1
x = 1.3 m A1
2(c)(i) k = F/x C1
x = 0.80 – 0.59 C1
= 0.21 m
k = (60 × 9.81) / 0.21 A1
= 2800 N m−1
2(c)(ii) E = ½ kx2 or E = ½ Fx or E = ½ F 2/k C1
= ½ × 2800 × 0.212 or
= ½ × 60 × 9.81 × 0.21 or
= ½ × (60 × 9.81)2 / 2800
= 62 J A1
(a) Define power.
…… [1]
(b) An electric car is powered by a motor. The car is travelling at a constant speed of 35 m s−1 along a straight horizontal road, as shown in Fig. 3.1.
35 m s–1
car
1750 N
road
Fig. 3.1
There is a total resistive force of 1750 N acting on the car.
(i) Calculate the power transmitted to the wheels of the car by the motor.
power = ……W [2]
(ii) Calculate the useful work done by the motor when the car travels a distance of 17 km.
work done = …… J [2]
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(iii) The potential difference (p.d.) across the motor has a constant value of 600 V and the motor has an efficiency of 85%.
Calculate the current in the motor.
current = ……A [3]
(c) The car in (b) now reaches a slope, as shown in Fig. 3.2.
car 35 m s–1
road
Fig. 3.2
The car continues down the slope at the same speed as in (b).
State and explain the effect, if any, of the slope on:
(i) the air resistance acting on the car
…… [1]
(ii) the current in the motor.
…… [1]
[Total: 10]
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Mark scheme
3(a) work done per unit time B1
3(b)(i) P =F×v C1
= 1750 × 35
= 6.1 × 104 W A1
3(b)(ii) W = Fs C1
= 1750 × 17 000
= 3.0 × 107 J A1
or (C1)
W = Pt
= 6.1 × 104 × (17 000 / 35)
= 3.0 × 107 J (A1)
3(b)(iii) P=V×I C1
Power in = 600 × I
useful power output C1
Efficiency =
(total) power input
0.85 = 6.1 × 104 / (600 × I) A1
I = 120 A
3(c)(i) Air resistance is the same, as the speed is the same B1
3(c)(ii) The motor is producing less power (because of gravitational force / conversion of gravitational potential energy to kinetic B1
energy) so the current will be smaller.
(a) State the principle of superposition.
…… [2]
(b) Light of wavelength 7.2 × 10−7 m is incident normally on a double slit, as shown in Fig. 4.1.
screen
0.16 mm light, wavelength
7.2 × 10−7 m
D double slit
Fig. 4.1 (not to scale)
A screen is at a distance D from the double slit. The double slit and the screen are parallel.
The separation of the slits in the double-slit arrangement is 0.16 mm. The resulting interference pattern on the screen contains nine dark fringes, as shown in Fig. 4.2.
3.2 cm
dark fringes bright fringes
Fig. 4.2 (not to scale)
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The distance between the centres of the first and ninth dark fringes is 3.2 cm.
(i) Calculate D.
D = …… m [3]
(ii) The slit separation is now gradually decreased from 0.16 mm to 0.04 mm. The distance between the centres of adjacent dark fringes is x.
On Fig. 4.3, sketch the variation of x with slit separation.
1.6
1.2
x / cm
0.8
0.4
0
0 0.04 0.08 0.12 0.16 slit separation / mm
Fig. 4.3
[3]
[Total: 8]
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Mark scheme
4(a) (when two or more) waves meet / overlap (at a point) B1
(resultant) displacement is sum of the individual displacements B1
4(b)(i) Fringe width, x = 3.2 × 10−2 / 8 C1
= 4.0 × 10−3 (m)
D = ax / λ C1
= (4.0 × 10−3 × 0.16 × 10−3) / 7.2 × 10−7
= 0.89 m A1
4(b)(ii) Curved line with a negative gradient of decreasing magnitude throughout, from slit separation 0.04 mm to 0.16 mm B1
Line of negative gradient ending at (0.16, 0.4), from slit separation 0.04 mm B1
Line of negative gradient passing through (0.08, 0.8) and (0.04, 1.6) B1
(a) Use the definitions of speed v, frequency f and wavelength λ to derive the wave equation
v = f λ.
[2]
(b) A source of sound waves of frequency 236 Hz is travelling at a constant velocity of 20 m s−1.
A stationary observer has a microphone connected to a cathode-ray oscilloscope (CRO). The microphone detects the sound waves as the source moves directly towards the observer.
The resulting trace on the CRO is shown in Fig. 5.1.
1 div.
Fig. 5.1
The time-base on the CRO is set to 1.0 ms div−1.
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(i) Calculate the frequency of the sound waves detected by the microphone.
frequency = …… Hz [2]
(ii) Determine the speed of the sound in air.
speed of sound = …… m s−1 [2]
[Total: 6]
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14
Mark scheme
5(a) wavelength: B1
• wavelength = distance between successive / adjacent in phase points / wavefronts / crests / troughs
• λ = d / (number of) oscillations
frequency:
• frequency = (number of) oscillations / cycles crests / troughs / wavefronts (passing a point) per unit time
• f = (number of) oscillations / t
One correct point from either list
One correct point from both lists and B1
speed = distance / time and
one of:
• wavelength × frequency (= distance per unit time) = speed
• [(number of) oscillations / t ] × [d / (number of) oscillations] = fλ
• v (= d / t) = λ / (1/f) = fλ or
v (= d / t) = λ / T = fλ
5(b)(i) T = 4 × 10−3 C1
f = 1 / T = 1 / 0.004
f = 250 Hz A1
5(b)(ii) fo = fs v / (v – vs) C1
250 = 236 × v / (v – 20)
v = (250 × 20) / (250 – 236)
= 360 m s−1 A1
A nichrome wire X of length 45 cm and cross-sectional area 4.7 × 10−7 m2 is connected into the circuit shown in Fig. 6.1.
A
X
Fig. 6.1
The resistance of X is 1.1 Ω. The cell has electromotive force (e.m.f.) 1.3 V and negligible internal resistance.
(a) (i) Calculate the current in X.
current = ……A [1]
(ii) The number density of charge carriers (electrons) in nichrome is 8.5 × 1028 m−3.
Calculate the average drift speed of the charge carriers in X.
average drift speed = …… m s−1 [2]
resistivity = …… Ω m [3]
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(b) Wire Y is identical to wire X. Wire Y is added to the circuit in parallel with wire X, as shown in
Fig. 6.2.
A
X
Y
Fig. 6.2
State and explain the effect, if any, of this change on:
(i) the reading on the ammeter
…… [2]
(ii) the average drift speed of the charge carriers in X.
…… [1]
[Total: 9]
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16
12
Mark scheme
6(a)(i) I = 1.3 / 1.1 A1
= 1.2 A
6(a)(ii) v = I / nqA C1
= 1.2 / (8.5 × 1028 × 1.60 × 10−19 × 4.7 × 10−7)
= 1.9 × 10−4 m s−1 A1
6(a)(iii) ρ = RA / L C1
= (1.1 × 4.7 × 10−7) / 0.45 C1
= 1.1 × 10−6 Ω m A1
6(b)(i) (Total) resistance decreases (and the potential difference stays the same) M1
(so the reading on the ammeter) increases A1
6(b)(ii) (The average drift speed will be) the same because the current is the same (in X). B1
(a) Nitrogen-12 ( 7N) is an unstable isotope of nitrogen that decays by the emission of radiation
12 to carbon-12 ( 6C).
Complete the full nuclear equation for the decay, including all the particles involved.
12
7N
[3]
1
(b) Hydrogen-1 ( 1H) is an isotope of hydrogen. An atom of hydrogen-1 comprises a proton with an orbiting electron.
An antiparticle equivalent of hydrogen-1 comprises an antiproton with an orbiting positron.
The antiquarks in the antiproton are the antiparticles of the quarks in a proton.
(i) State the charge on the positron in terms of the elementary charge e.
charge = …… e [1]
(ii) State the group (class) of fundamental particle to which the positron belongs.
…… [1]
(iii) In Table 7.1, state the flavour and charge of the three antiquarks that comprise the antiproton.
Table 7.1
flavour charge / e
[3]
[Total: 8]
Mark scheme
7(a) 12 12 0 + ( 0) B1
7 N → 6 C + 1β + ( 0) n
beta-plus shown
neutrino shown B1
symbols, nucleon numbers and proton numbers all correct B1
7(b)(i) +1 B1
7(b)(ii) Lepton(s) B1
7(b)(iii)
flavour charge / e
up / u 2
–
3
up / u 2
–
3
down / d 1
(+)
3
3 correct quark flavours B1
Charge on anti-up quark –⅔(e) B1
Charge on anti-down quark (+)⅓(e) B1
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Physics 9702/21 for May/June 2025 is out of 60 marks, with 1 hour 15 minutes to answer it, across 7 questions.
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