Physics questions, chapter 2

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Force and Motion I

43 marks. Use Practice to check each answer as you go, or Exam to answer everything against the clock first. Take g = 9.81 m s⁻² where needed.

Section A: objective questions

16 marks

Choose the best answer for each question.

Question 1

Which of the following is a vector quantity?

Distance
Speed
Displacement
Time

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Displacement has both magnitude and direction. Distance, speed and time have magnitude only (textbook Table 2.1, p. 26).

For the teacher: Remembering, low difficulty, standard 2.1

Textbook p. 26

Question 2

A ticker timer is connected to an alternating current supply of frequency 50 Hz. What is the time interval of one tick?

0.01 s
0.02 s
0.05 s
0.20 s

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50 ticks are made in 1 s, so one tick = 1/50 = 0.02 s.

For the teacher: Remembering, low difficulty, standard 2.1

Textbook p. 30

Question 3

A runner moves 150 m due east and then 50 m due west. What is the displacement of the runner?

200 m east
100 m east
100 m west
50 m east

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Displacement = 150 − 50 = 100 m, in the east direction. (The distance is 200 m.)

For the teacher: Applying, low difficulty, standard 2.1

Textbook p. 28

Question 4

A car increases its velocity from rest to 24 m s⁻¹ in 8 s. What is its acceleration?

0.33 m s⁻²
2.0 m s⁻²
8.0 m s⁻²
3.0 m s⁻²

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a = (v − u) ÷ t = (24 − 0) ÷ 8 = 3.0 m s⁻².

For the teacher: Applying, low difficulty, standard 2.1

Textbook p. 28

Question 5

A car travelling at 20 m s⁻¹ brakes uniformly and stops after 40 m. What is its acceleration?

−5.0 m s⁻²
−2.5 m s⁻²
−10 m s⁻²
5.0 m s⁻²

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v² = u² + 2as gives 0 = 20² + 2a(40), so 80a = −400 and a = −5.0 m s⁻². The negative sign shows deceleration.

For the teacher: Applying, medium difficulty, standard 2.1

Textbook p. 36

Question 6

A strip of ticker tape with 5 ticks is 4.0 cm long. What is the average velocity of the object that pulled the tape?

8 cm s⁻¹
20 cm s⁻¹
4 cm s⁻¹
40 cm s⁻¹

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Time = 5 × 0.02 = 0.10 s. Velocity = 4.0 ÷ 0.10 = 40 cm s⁻¹.

For the teacher: Applying, medium difficulty, standard 2.1

Textbook p. 31

Question 7

Diagram 1 shows the velocity-time graph of a toy car. What is the total displacement of the toy car in 12 s?

Velocity-time graph rising from 0 to 8 m/s in 4 s, constant until 10 s, then falling to 0 at 12 s — Diagram 1

48 m
64 m
72 m
96 m

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Displacement = area under the graph = ½(6 + 12) × 8 = 72 m (a trapezium).

For the teacher: Applying, medium difficulty, standard 2.2

Textbook p. 42

Question 8

What does the gradient of a displacement-time graph represent?

Velocity
Acceleration
Distance
Momentum

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Gradient of a displacement-time graph = change in displacement ÷ time = velocity. The gradient of a velocity-time graph is acceleration.

For the teacher: Remembering, low difficulty, standard 2.2

Textbook p. 37

Question 9

What is meant by free fall?

Falling from a very great height
Falling at a constant velocity
Falling with air resistance only
Motion under the force of gravity only

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An object is in free fall when the only force acting on it is gravity, so there is no air resistance or friction.

For the teacher: Understanding, low difficulty, standard 2.3

Textbook p. 46

Question 10

A stone is released from rest and falls freely for 3.0 s. What is its velocity just before it hits the ground? [g = 9.81 m s⁻²]

9.8 m s⁻¹
19.6 m s⁻¹
29.4 m s⁻¹
44.1 m s⁻¹

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v = u + gt = 0 + 9.81 × 3.0 = 29.4 m s⁻¹.

For the teacher: Applying, medium difficulty, standard 2.3

Textbook p. 50

Question 11

A ball is thrown vertically upwards at 14 m s⁻¹. What is the maximum height it reaches? [Ignore air resistance, g = 9.81 m s⁻²]

7.0 m
10.0 m
14.3 m
20.0 m

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At the top v = 0. v² = u² + 2gs gives 0 = 14² − 2(9.81)s, so s = 196 ÷ 19.62 = 10.0 m.

For the teacher: Applying, high difficulty, standard 2.3

Textbook p. 50

Question 12

Passengers in a moving bus are thrown forward when the bus brakes suddenly. This happens because of

the inertia of the passengers
the gravitational force on the passengers
friction between the seat and the passengers
the reaction force of the bus

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Inertia is the tendency to keep the state of motion. The passengers continue to move forward while the bus slows down.

For the teacher: Understanding, low difficulty, standard 2.4

Textbook p. 56

Question 13

A lorry of mass 1 200 kg moves at 15 m s⁻¹. What is its momentum?

80 kg m s⁻¹
1 215 kg m s⁻¹
1 800 kg m s⁻¹
18 000 kg m s⁻¹

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p = mv = 1 200 × 15 = 18 000 kg m s⁻¹.

For the teacher: Applying, low difficulty, standard 2.5

Textbook p. 59

Question 14

A trolley of mass 2 kg moving at 3 m s⁻¹ collides with a stationary trolley of mass 1 kg. They stick together. What is their common velocity?

1.0 m s⁻¹
1.5 m s⁻¹
2.0 m s⁻¹
3.0 m s⁻¹

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2(3) + 1(0) = (2 + 1)v, so v = 6 ÷ 3 = 2.0 m s⁻¹ (conservation of momentum).

For the teacher: Applying, medium difficulty, standard 2.5

Textbook p. 62

Question 15

A resultant force of 600 N acts on a car of mass 1 500 kg. What is the acceleration of the car?

0.25 m s⁻²
0.40 m s⁻²
2.5 m s⁻²
9 000 m s⁻²

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a = F ÷ m = 600 ÷ 1 500 = 0.40 m s⁻².

For the teacher: Applying, medium difficulty, standard 2.6

Textbook p. 66

Question 16

A cricketer pulls his hands backwards while catching a fast ball. How does this reduce the pain in his hands?

The time of impact is longer, so the impulsive force is smaller
The change of momentum of the ball is smaller
The time of impact is shorter, so the impulsive force is smaller
The mass of the ball is reduced

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The change of momentum is the same, but F = (mv − mu) ÷ t. A longer impact time gives a smaller impulsive force.

For the teacher: Understanding, medium difficulty, standard 2.7

Textbook p. 68

Section B: structured questions

27 marks

Answer all questions. Show your working.

Question 17

Dina walks 60 m due east and then 20 m back due west. The journey takes 40 s.

[1 mark] (a)Calculate the distance travelled.

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60 + 20 = 80 m1

Tick each point you got.

[1 mark] (b)Calculate the displacement.

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60 − 20 = 40 m (due east)1

Tick each point you got.

[1 mark] (c)Calculate the average speed.

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80 ÷ 40 = 2 m s⁻¹1

Tick each point you got.

[1 mark] (d)Calculate the average velocity.

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40 ÷ 40 = 1 m s⁻¹ (due east)1

Tick each point you got.

[1 mark] (e)Why are the average speed and the magnitude of the average velocity different?

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Any 1 points

Speed uses distance (total path) but velocity uses displacement (shortest distance in a direction)1
She changed direction, so displacement is less than distance1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 2.1

Textbook p. 29

Question 18

A trolley pulls a ticker tape through a ticker timer of frequency 50 Hz. The tape is cut into strips of 5 ticks each. The first strip is 2.0 cm long and the fifth strip is 6.0 cm long.

[1 mark] (a)What is the time taken for 5 ticks?

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5 × 0.02 = 0.10 s1

Tick each point you got.

[1 mark] (b)Calculate the initial velocity, using the first strip.

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2.0 ÷ 0.10 = 20 cm s⁻¹1

Tick each point you got.

[1 mark] (c)Calculate the final velocity, using the fifth strip.

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6.0 ÷ 0.10 = 60 cm s⁻¹1

Tick each point you got.

[2 marks] (d)The time between the first and the fifth strip is 0.40 s. Calculate the acceleration of the trolley.

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a = (v − u) ÷ t = (60 − 20) ÷ 0.401
= 100 cm s⁻²1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 2.1

Textbook p. 31

Question 19

Diagram 2 shows the velocity-time graph of a lorry for 14 s.

Velocity-time graph: 4 m/s at 0 s rising to 16 m/s at 6 s, constant to 10 s, then falling to 0 at 14 s — Diagram 2

[1 mark] (a)Calculate the acceleration of the lorry from 0 s to 6 s.

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(16 − 4) ÷ 6 = 2 m s⁻²1

Tick each point you got.

[1 mark] (b)Describe the motion of the lorry from 6 s to 10 s.

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Moves with a uniform velocity of 16 m s⁻¹1

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[2 marks] (c)Calculate the total displacement of the lorry.

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Area = ½(4 + 16)(6) + (4)(16) + ½(4)(16) [any correct use of area]1
= 60 + 64 + 32 = 156 m1

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[1 mark] (d)Calculate the average velocity of the lorry.

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156 ÷ 14 = 11.1 m s⁻¹1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 2.2

Textbook p. 43

Question 20

Diagram 3 shows the displacement-time graph of a cyclist.

Displacement-time graph: 0 m at 0 s, 50 m at 10 s, constant until 20 s, then 20 m at 30 s — Diagram 3

[1 mark] (a)Calculate the velocity of the cyclist from 0 s to 10 s.

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50 ÷ 10 = 5 m s⁻¹1

Tick each point you got.

[1 mark] (b)What is the cyclist doing from 10 s to 20 s?

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At rest (stationary), because the displacement does not change1

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[1 mark] (c)Calculate the velocity from 20 s to 30 s and state what the sign shows.

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(20 − 50) ÷ 10 = −3 m s⁻¹, so he moves back towards the starting point (opposite direction)1

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[1 mark] (d)Calculate the total distance travelled in 30 s.

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50 + 30 = 80 m1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 2.2

Textbook p. 38

Question 21

Aiman drops a stone from rest from a bridge. The stone hits the water after 2.0 s. [Ignore air resistance, g = 9.81 m s⁻²]

[1 mark] (a)Calculate the velocity of the stone just before it hits the water.

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v = 0 + 9.81 × 2.0 = 19.6 m s⁻¹ (downwards)1

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[1 mark] (b)Calculate the height of the bridge above the water.

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s = ½(9.81)(2.0)² = 19.6 m1

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[2 marks] (c)Aiman says, "A heavy stone always falls faster than a light stone." Evaluate his statement.

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Not correct when air resistance is ignored (in a vacuum): all objects fall with the same gravitational acceleration1
Light objects with a large surface (e.g. a feather) fall slower in air because air resistance is large compared with their weight1

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For the teacher: Evaluating, high difficulty, standard 2.3

Textbook p. 47

Question 22

In a crash test, a car of mass 1 000 kg moving at 20 m s⁻¹ hits a wall and stops in 0.50 s.

[1 mark] (a)Calculate the momentum of the car before the crash.

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1 000 × 20 = 20 000 kg m s⁻¹1

Tick each point you got.

[1 mark] (b)Calculate the impulse acting on the car.

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Impulse = change of momentum = 0 − 20 000 = −20 000 N s (magnitude 20 000 N s)1

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[1 mark] (c)Calculate the average impulsive force.

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F = 20 000 ÷ 0.50 = 40 000 N (opposite to the motion)1

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[1 mark] (d)Suggest one safety feature that you would design for the car to reduce the impulsive force on the passengers, and explain how it works.

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Crumple zone / airbag: increases the impact time, so the impulsive force is smaller for the same change of momentum1

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For the teacher: Creating, high difficulty, standard 2.7

Textbook p. 68

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