JUN25 74083 BC01 - For Examiner’s Use Question Mark TOTAL Tuesda...

*JUN25 74083 BC01*

IB/M/Jun25/G4006/E7 7408/3BC

For Examiner’s Use Question Mark 1 2 3 4 TOTAL

Tuesday 17 June 2025 Morning

Materials

For this paper you must have:

• a pencil and a ruler • a scientific calculator • a Data and Formulae Booklet • a protractor.

Instructions • Use black ink or black ball-point pen.• Fill in the boxes at the top of this page.• Answer all questions.• You must answer the questions in the spaces provided. Do not write outside the box around each page or on blank pages.• If you need extra space for your answer(s), use the lined pages at the end of 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 marked.• Show all your working.

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.Please write clearly in block capitals.Centre number

Candidate number

Surname

Forename(s)

Candidate signature

I declare this is my own work.A-level

PHYSICS

Paper 3 Section B Engineering physics

Time allowed: The total time for

both sections of this paper is

• hours. You are advised to

spend approximately 50 minutes on this section. 1 / 3

2

*02* IB/M/Jun25/7408/3BC

Do not write outside the box Section B

Answer all questions in this section.

• 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 = t 1 .

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

. 1 Determine t 1 .[2 marks]

t 1 =

s

• / 3

3

*03* Turn over ►

IB/M/Jun25/7408/3 BC

Do not write outside the box

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 = t 2 , the angular velocity of the rotor is −120 rad s −1 .

• 1

. 2 Determine the number of anticlockwise revolutions made by the rotor between t 1 and t 2 .[2 marks]

number of revolutions =

• 1

. 3

The moment of inertia of the rotor about the axis of rotation is 9.60 × 10 −2 kg m 2 .

Calculate the angular impulse on the rotor between t = 0 and t 2 .[1 mark]

angular impulse =

N m s

Question 1 continues on the next page

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