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AQA GCSE Physics Smart practice A mixed round aimed at your weak spots
Unit 1
Energy
Energy changes in a system, and the ways energy is stored before and after such changes
Energy stores and systems
0%
Changes in energy
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Energy changes in systems
0%
Power
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Conservation and dissipation of energy
Energy transfers in a system
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Efficiency
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Unit 2
Electricity
Unit 3
Particle model of matter
Unit 4
Atomic structure
Unit 5
Forces
Unit 6
Waves
Unit 7
Magnetism and electromagnetism
Unit 8
Space physics (physics only)
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Velocity

Practise 22 exam-style questions on Velocity for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

  • 22 questions
  • 1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

1 mark

An object moves in a circle at a steady rate.

Which of these quantities stays constant: its speed or its velocity?

Model answer

Its speed.

Explanation

In circular motion at a steady rate the speed stays the same, but the direction of motion keeps changing. Because velocity includes direction, the velocity is always changing.

Mark scheme
  1. 1 mark

    speed

    do not accept both

Question 2

2 marks

An aircraft flies due north at 250 m/s relative to the air. A wind is blowing due south at 30 m/s.

Calculate the velocity of the aircraft relative to the ground.

Model answer

The wind blows in the opposite direction to the aircraft, so velocity relative to the ground = 250 − 30 = 220 m/s due north.

Explanation

Velocities are vectors, so velocities in opposite directions are subtracted. A headwind reduces the aircraft's velocity over the ground, while a tailwind would increase it to 280 m/s.

Mark scheme
  1. 1 mark

    220 m/s

    allow 250 − 30; correct answer with no working scores full marks

  2. 1 mark

    due north

    direction needed

Question 3

3 marks

The diagram shows a satellite moving in a circular orbit around the Earth at a constant speed. The arrow shows the direction of its velocity at one moment.

Explain why the velocity of the satellite is changing even though its speed is constant.

Orbit diagram: the Earth labelled "Earth" at the centre with a circular orbit around it. On the orbit, a satellite labelled "Satellite" is up and to the right of the centre. An arrow labelled "Velocity" points from it along the orbit (at a tangent, anticlockwise). Small arrowheads on the orbit show the motion is anticlockwise.

Model answer

Velocity is speed in a given direction, so it is a vector. As the satellite moves round its circular orbit, its direction of motion is continuously changing - the velocity arrow always points along the tangent to the circle. Since the direction changes, the velocity changes, even though the speed stays the same.

Explanation

For circular motion at constant speed, the size of the velocity stays the same but its direction keeps turning. A changing velocity means the satellite is accelerating towards the centre of the orbit.

Mark scheme
  1. 1 mark

    velocity is a vector / is speed in a given direction

  2. 1 mark

    the direction of motion of the satellite is continuously changing (it is always along the tangent to the orbit)

  3. 1 mark

    a change in direction means the velocity changes

    allow so the satellite is accelerating

Question 4

4 marks

The diagram shows a satellite moving in a circular orbit around the Earth. The radius of the orbit is 6800 km. The satellite moves at a constant speed of 7.7 km/s.

Calculate the time taken for the satellite to complete one orbit.
Give your answer in minutes.

Orbit diagram: the Earth labelled "Earth" at the centre with a circular orbit around it. On the orbit, a satellite labelled "Satellite" is up and to the left of the centre. An arrow labelled "Velocity" points from it along the orbit (at a tangent, anticlockwise). Small arrowheads on the orbit show the motion is anticlockwise.

Model answer

Distance for one orbit = 2πr = 2 × π × 6800 = 42 726 km
t = s ÷ v = 42 726 ÷ 7.7 = 5549 s
t = 5549 ÷ 60 = 92.5 minutes

Explanation

The satellite travels one circumference (2πr) in each orbit at constant speed, so t = s ÷ v. Keeping both distance and speed in kilometres gives the time in seconds directly. Although the speed is constant, the velocity is not, because the direction of motion keeps changing.

Mark scheme
  1. 1 mark

    distance = circumference = 2 × π × 6800 (= 42 700 km)

    an answer of 92 or 93 (minutes) scores 4 marks; using the diameter instead of the circumference scores 0 for this mark

  2. 1 mark

    t = 42 700 ÷ 7.7

    allow the calculation in m and m/s; allow ecf from their distance

  3. 1 mark

    t = 5550 (s)

    allow 5540-5560 s

  4. 1 mark

    92.5 (minutes)

    allow 92-93 minutes; allow ecf: their time in s ÷ 60

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