Force and Motion II: Physics Tingkatan 5 questions

Section A: objective questions

10 marks

Choose the best answer for each question.

What is the resultant force on an object?

The largest of the forces acting on the object
The single force that represents the vector sum of the forces acting on the object
The weight of the object
The force that the object exerts on its surroundings

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The resultant force is the single force equal to the vector sum of all forces on the object (textbook p. 3).

For the teacher: Remembering, low difficulty, standard 1.1.1

Textbook p. 3

A force of 9 N to the right and a force of 5 N to the left act on the same point. What is the resultant force?

4 N to the right
14 N to the right
4 N to the left
14 N to the left

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The forces are in opposite directions, so subtract: 9 − 5 = 4 N, in the direction of the larger force (right).

For the teacher: Applying, low difficulty, standard 1.1.2

Textbook p. 4

Two forces of 12 N and 5 N act on an object at right angles to each other. What is the magnitude of the resultant force?

7 N
17 N
60 N
13 N

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F = √(12² + 5²) = √169 = 13 N.

For the teacher: Applying, low difficulty, standard 1.1.2

Textbook p. 5

A car moves along a straight road with a constant velocity of 20 m s⁻¹. Which statement about the forces on the car is correct?

The engine thrust is larger than the friction
The engine thrust is smaller than the friction
The engine thrust is equal to the friction
There is no friction on the car

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Constant velocity means a = 0, so the resultant force is zero. The engine thrust balances the friction and air resistance.

For the teacher: Understanding, low difficulty, standard 1.1.3

Textbook p. 9

A passenger of mass 50 kg is in a lift that accelerates upwards at 2.0 m s⁻². What is the normal reaction on the passenger from the floor of the lift?

490.5 N
100 N
590.5 N
390.5 N

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R − W = ma, so R = mg + ma = 50 × 9.81 + 50 × 2.0 = 490.5 + 100 = 590.5 N.

For the teacher: Applying, medium difficulty, standard 1.1.4

Textbook p. 11

A force of 50 N acts at 30° above the horizontal. What is the horizontal component of the force?

25.0 N
43.3 N
50.0 N
28.9 N

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Fx = F cos θ = 50 cos 30° = 43.3 N. (25.0 N is the vertical component, 50 sin 30°.)

For the teacher: Applying, low difficulty, standard 1.2.1

Textbook p. 14

A block of weight 20 N rests on a smooth slope inclined at 30° to the horizontal. What is the component of the weight parallel to the slope?

10 N
17.3 N
20 N
40 N

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Parallel component = W sin θ = 20 sin 30° = 10 N. The perpendicular component is 20 cos 30° = 17.3 N.

For the teacher: Applying, low difficulty, standard 1.2.2

Textbook p. 16

Which statement describes an object in equilibrium of forces?

The object must be at rest
Only one force acts on the object
The object moves with increasing velocity
The resultant force is zero, so the object is at rest or moves with constant velocity

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In equilibrium the resultant force is zero, so a = 0. The object may be stationary or moving with constant velocity.

For the teacher: Understanding, low difficulty, standard 1.3.1

Textbook p. 18

A spring extends by 4.0 cm when a force of 10 N is applied. Assuming the elastic limit is not exceeded, what is the extension when a force of 25 N is applied?

6.0 cm
10.0 cm
12.5 cm
14.0 cm

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Extension is directly proportional to force: 4.0 × (25 ÷ 10) = 10.0 cm.

For the teacher: Applying, medium difficulty, standard 1.4.2

Textbook p. 27

A spring has a spring constant of 200 N m⁻¹. What is the elastic potential energy stored when it is stretched by 0.10 m?

0.10 J
2.0 J
1.0 J
10 J

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Ep = ½kx² = ½ × 200 × 0.10² = 1.0 J.

For the teacher: Applying, medium difficulty, standard 1.4.2

Textbook p. 28

Section B: structured questions

32 marks

Answer all questions. Show your working.

A passenger of mass 60 kg stands in a lift.

[2 marks] (a)Calculate the normal reaction on the passenger when the lift moves upwards with a constant velocity of 8.0 m s⁻¹.

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Resultant force is zero, so R = W = mg = 60 × 9.81 [1 mark]1
= 588.6 N [1 mark]1

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[3 marks] (b)The lift then moves downwards with an acceleration of 1.5 m s⁻². Calculate the normal reaction on the passenger.

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W − R = ma [1 mark]1
588.6 − R = 60 × 1.5 [1 mark]1
R = 498.6 N [1 mark]1

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For the teacher: Applying, low difficulty, standard 1.1.4

Textbook p. 11

Diagram 1 shows a block of mass 5.0 kg sliding down a smooth slope inclined at 30° to the horizontal.

A block of mass 5.0 kg on a slope that makes an angle of 30 degrees with the horizontal — Diagram 1

[1 mark] (a)Calculate the weight of the block.

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W = 5.0 × 9.81 = 49.05 N1

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[2 marks] (b)Calculate the component of the weight parallel to the slope.

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W sin 30° [1 mark]1
= 49.05 × 0.5 = 24.5 N [1 mark]1

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[2 marks] (c)Calculate the acceleration of the block down the slope.

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a = F ÷ m = 24.5 ÷ 5.0 [1 mark]1
= 4.9 m s⁻² [1 mark]1

On a smooth slope a = g sin θ, independent of mass.

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For the teacher: Analysing, low difficulty, standard 1.2.2

Textbook p. 16

Diagram 2 shows a lamp of weight W hanging at rest from two identical strings. Each string has tension T and makes an angle of 40° with the horizontal. The weight of the lamp is 10 N.

A lamp hanging from two strings; each string makes 40 degrees with the horizontal ceiling — Diagram 2

[1 mark] (a)State the resultant force on the lamp.

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Zero1

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[3 marks] (b)Calculate the tension T in each string.

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Vertical forces balance: 2T sin 40° = W [1 mark]1
T = 10 ÷ (2 sin 40°) [1 mark]1
= 7.78 N [1 mark]1

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[1 mark] (c)What happens to T if the strings are pulled so that they become closer to the horizontal? Give a reason.

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T increases, because the vertical component of each tension (T sin θ) decreases, so T must be larger to support the same weight1

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For the teacher: Applying, medium difficulty, standard 1.3.3

Textbook p. 21

Diagram 3 shows the graph of force F against extension x for a spring.

Graph of F against x: a straight line through the origin reaching 12 N at an extension of 0.06 m — Diagram 3

[2 marks] (a)Does the spring obey Hooke's law? Give a reason.

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Yes1
The graph is a straight line through the origin / F is directly proportional to x1

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[2 marks] (b)Calculate the spring constant.

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k = gradient = 12 ÷ 0.06 [1 mark]1
= 200 N m⁻¹ [1 mark]1

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[2 marks] (c)Calculate the elastic potential energy stored when the extension is 0.06 m.

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Ep = ½ × 12 × 0.06 [1 mark] (or ½ × 200 × 0.06²)1
= 0.36 J [1 mark]1

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For the teacher: Analysing, low difficulty, standard 1.4.2

Textbook p. 28

A spring with a spring constant of 4 N cm⁻¹ is used to build two arrangements from identical springs. Each arrangement carries a load of 8 N.

[2 marks] (a)Two springs are arranged in series. Calculate the total extension.

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Each spring carries the full 8 N: x = 8 ÷ 4 = 2 cm [1 mark]1
Total extension = 2 × 2 = 4 cm [1 mark]1

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[2 marks] (b)Two springs are arranged in parallel. Calculate the total extension.

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Each spring carries 8 ÷ 2 = 4 N [1 mark]1
x = 4 ÷ 4 = 1 cm [1 mark]1

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[1 mark] (c)Which arrangement is stiffer?

SeriesParallel

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Parallel

The same load gives a smaller extension (1 cm compared with 4 cm).

For the teacher: Applying, medium difficulty, standard 1.4.4

Textbook p. 31

Two forces of 17 N and 13 N act on a point. Aini says that the resultant force cannot be smaller than 4 N or larger than 30 N.

[2 marks] (a)Justify Aini's statement.

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Largest resultant when the forces act in the same direction: 17 + 13 = 30 N [1 mark]1
Smallest resultant when the forces act in opposite directions: 17 − 13 = 4 N [1 mark]1

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[1 mark] (b)Is a resultant of 22 N possible? Explain.

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Yes, 22 N lies between 4 N and 30 N; it occurs when the forces are at a suitable angle to each other1

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For the teacher: Evaluating, high difficulty, standard 1.1.2

Textbook p. 4

A workshop platform must support a load of 2 400 N and the springs may compress by at most 3.0 cm. Spring X has k = 400 N cm⁻¹. Suggest a design using spring X.

[1 mark] (a)Calculate the minimum total spring constant needed.

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k = F ÷ x = 2 400 ÷ 3.0 = 800 N cm⁻¹1

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[2 marks] (b)State the number of springs and how they should be arranged. Justify your design.

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At least 2 springs (3 for a safety margin) arranged in parallel1
In parallel the spring constants add (2 × 400 = 800 N cm⁻¹); the load is shared so each spring is not overloaded and the compression stays within 3.0 cm1

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For the teacher: Creating, high difficulty, standard 1.4.4

Textbook p. 37

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Section B: structured questions

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17