2025 AQA A-LEVEL PHYSICS Paper 3 Section B Engineering Physics Question Paper & Mark
Scheme (Merged) Tuesday 17 June 2025 [VERIFIED]
, 2
Do not write
outside the
Section B
box
Answer all questions in this section.
0 1 The rotating part of an electric motor is called the rotor.
Figure 1 shows an end view of a rotor turning clockwise due to a driving torque from
the motor. In this question, the clockwise direction is treated as positive.
Figure 1 Figure 2
The rotor can be brought to rest rapidly by reversing the electrical supply connections
to the motor. Figure 2 shows the rotor at time t = 0 when the supply connections are
reversed.
The rotor then slows down due to a constant anticlockwise retarding torque so that it
stops at time t = t1.
The angular velocity of the rotor at t = 0 is 98.0 rad s−1 clockwise.
The applied torque on the rotor at t = 0 is anticlockwise.
The applied torque produces a constant angular acceleration of −303 rad s−2. Friction
torque is negligible.
0 1 . 1 Determine t1.
[2 marks]
t1 = s
for more: tyrionpapers.com
IB/M/Jun25/7408/3BC
, 3
Do not write
outside the
The electrical supply remains connected and the rotor now accelerates uniformly box
anticlockwise with an acceleration of magnitude 303 rad s−2.
At a later time t = t2, the angular velocity of the rotor is −120 rad s−1.
0 1 . 2 Determine the number of anticlockwise revolutions made by the rotor between
t1 and t2.
[2 marks]
number of revolutions =
0 1 . 3 The moment of inertia of the rotor about the axis of rotation is 9.60 × 10−2 kg m2.
Calculate the angular impulse on the rotor between t = 0 and t2.
[1 mark]
angular impulse = Nms
Question 1 continues on the next page
for more: tyrionpapers.com
IB/M/Jun25/7408/3BC
, 4
Do not write
outside the
Turn over ►
for more: tyrionpapers.com
IB/M/Jun25/7408/3BC
Scheme (Merged) Tuesday 17 June 2025 [VERIFIED]
, 2
Do not write
outside the
Section B
box
Answer all questions in this section.
0 1 The rotating part of an electric motor is called the rotor.
Figure 1 shows an end view of a rotor turning clockwise due to a driving torque from
the motor. In this question, the clockwise direction is treated as positive.
Figure 1 Figure 2
The rotor can be brought to rest rapidly by reversing the electrical supply connections
to the motor. Figure 2 shows the rotor at time t = 0 when the supply connections are
reversed.
The rotor then slows down due to a constant anticlockwise retarding torque so that it
stops at time t = t1.
The angular velocity of the rotor at t = 0 is 98.0 rad s−1 clockwise.
The applied torque on the rotor at t = 0 is anticlockwise.
The applied torque produces a constant angular acceleration of −303 rad s−2. Friction
torque is negligible.
0 1 . 1 Determine t1.
[2 marks]
t1 = s
for more: tyrionpapers.com
IB/M/Jun25/7408/3BC
, 3
Do not write
outside the
The electrical supply remains connected and the rotor now accelerates uniformly box
anticlockwise with an acceleration of magnitude 303 rad s−2.
At a later time t = t2, the angular velocity of the rotor is −120 rad s−1.
0 1 . 2 Determine the number of anticlockwise revolutions made by the rotor between
t1 and t2.
[2 marks]
number of revolutions =
0 1 . 3 The moment of inertia of the rotor about the axis of rotation is 9.60 × 10−2 kg m2.
Calculate the angular impulse on the rotor between t = 0 and t2.
[1 mark]
angular impulse = Nms
Question 1 continues on the next page
for more: tyrionpapers.com
IB/M/Jun25/7408/3BC
, 4
Do not write
outside the
Turn over ►
for more: tyrionpapers.com
IB/M/Jun25/7408/3BC
