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AQA A-level PHYSICS 7408/3BC ACTUAL 2025 PAPER MERGED WITH MARK SCHEME

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A-level PHYSICS 7408/3BC ACTUAL 2025 PAPER MERGED WITH MARK SCHEME

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A-level PHYSICS
7408/3BC
ACTUAL 2025
PAPER MERGED
WITH MARK
SCHEME
Paper 3
Section B Engineering physics
Please write clearly in block capitals.

Candidate number
Centre number



Surname


Forename(s)


Candidate signature
I declare this is my own work.


A-level PHYSICS
Paper 3
Section B Engineering physics

Tuesday 17 June 2025 Morning Time allowed: The total time for
both sections of this paper is 2
Materials hours. You are advised to spend
For this paper you must have:
• a pencil and a ruler approximately
• a scientific calculator 50 minutes on this section.
• a Data and Formulae Booklet • a protractor.


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Instructions For Examiner’s Use
• Use black ink or black ball-point pen.
• Fill in the boxes at the top of this page. Question Mark
• Answer all questions.
1
• You must answer the questions in the spaces provided. Do not write outside
the box around each page or on blank pages. 2
• If you need extra space for your answer(s), use the lined pages at the end of 3
this book. Write the question number against your answer(s).
• Do all rough work in this book. Cross through any work you do not want to be 4

• Show all your working. TOTAL


Information
• The marks for questions are shown in brackets.
• The maximum mark for this paper is 35.
• You are expected to use a scientific calculator where appropriate.
• A Data and Formulae Booklet is provided as a loose insert.



*JUN2574083BC01*
IB/M/Jun25/G4006/E7
7408/3BC
marked.




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IB/M/Jun25/7408/3BC

, 3

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Section B box
0 1
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.



.1 Determine t1.
[2 marks]




t1 = s




Turn over ►


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IB/M/Jun25/7408/3BC

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*02*

The electrical supply remains connected and the rotor now accelerates uniformly
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.
.2
0 1 Determine the number of anticlockwise revolutions made by the rotor between t1
t
and 2.
[2 marks]




number
of revolutions =




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IB/M/Jun25/7408/3BC

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