PM - TB Solutions - C03 PDF

 

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Physics Matters for GCE ‘O’ Level (4  Edition): Full Solutions to Textbook Questions

Chapter 3

Chapter 3

Forces

Quick Check (page 52) 

The direction towards the right has been assigned as positive. Since the 3 N force points to the left, it is taken to be negative.

Test Yourself 3.1 & 3.2 (page 54)

1.

Magnetic force, gravitational force, electric force

2.

With a suitable suitable scale, such as 1 cm : 500 N, we can use the the paral parallelogram lelogram method or the tip-to-tail method to find the resultant force. tip-to-tail Parallelogram method

lorry

3000 N 20˚ 20˚

5600 N

lorry 20˚

3000 N

3.

Tip-to-tai Tip-to-taill method 5600 N 20˚ 3000 N

20˚

3000 N

With a suitable scale, such as 1 cm : 2 N, we we can use the parallelogram method or the tip-to-tail tip-to-tail method to find the value of W . Parallelogram method 17.3 N

Tip-to-tail Tip-to-tail method 60˚ 10 N

60˚  60˚ 10 N

10 N

17.3 N 60˚ 10 N

Since the object is in equilibrium, the weight W  is  is equal to 17.3 N but acts in the direction opposite to that of the resultant of the two tensions (i.e. downwards).

Quick Check (page 57)

No. The resultant force must be equivalent to the push on the book and the friction combined in terms of magnitude and direction.

Quick Check (page 59)

The weight of the egg acts downwards on the egg cup. The egg cup exerts an equal but opposite o pposite reaction force on the egg. These two forces form an action–reaction pair.

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Physics Matters for GCE ‘O’ Level (4  Edition): Full Solutions to Textbook Questions

Chapter 3

Test Yourself 3.3 (page 60)

1.

(a) (b)

When an object is moving moving at constant speed in a straight straight line, the result resultant ant fforce orce is zero. If the object object is is accelerating, accelerating, then ther there e must be a resultant force iin n the same direct direction ion as the acceleration. This resultant force is given by F = ma.

2.

Initially, Initially, the forces that act on the object are bal balanced. anced. B By y Newton’s Newton’s First Law of Motion, the object moves at a constant speed because it does not n ot experience a resultant force. Subsequently, the forces that act on the object become unbalanced, for example, because a larger pushing force isit applied to theaobject. Byforce. Newton’s Second Law of Motion, the object accelerates because experiences resultant

3.

4.

Let the mass of the unloaded van be M , and let the forward thrust be F for both vans. We use Newton’s Second Law of Motion, F = ma.  –2 For the unloaded van, F  =  = M  ! 5 m s For the loaded van, F  =  = (2M ) ! a where a = acceleration of the loaded van Since F  is  is the same for both vans,  –2 M  ! 5 m s  = (2M ) ! a  –2 5ms  –2   = 2.5 m s   a= 2  –2 Hence, the acceleration of the loaded van is 2.5 m s . We use Newton’s Second Law of Motion, F = ma. Given: mass m = 86.5 kg v–u  Acceleration  Accelera tion a =    "  –1  –1 700t m s  – 300 m s   =  6 s  –2 = 66.7 m s    –2  Average  Averag e resultant resultan t force F  exerted  exerted on missile = 86.5 kg ! 66.7 m s   = 5770 N

Test Yourself 3.4 (page 63)

1.

No. To lean lean against a wal walll without slipping, there must be friction friction between the s soles oles of our feet and the ground, and between our body and the wall.

2.

Use wheels and ball bearings, or apply lubricants.

Quick Check (page 64) 

Forces on apple being pushed by squirrel   Push force P  by  by squirrel   Weight W  of  of apple   Friction f  between  between apple and ground   Normal reaction force F  by  by ground •

•

F

• •

P

f

W

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Physics Matters for GCE ‘O’ Level (4  Edition): Full Solutions to Textbook Questions

Chapter 3

Test Yourself 3.5 (page 66)

1.

The trolley trolley is moving at a uniform uniform velocity, which means the forces acting on the trolley are balanced. Thus, a force of 50 5 0 N that keeps the trolley in motion must be balanced ba lanced by a frictional force of 50 N.

2.

The air air resistance experienced by a falling object increases with its speed in air. However, the weight of the stone is larger than that of the feather. The air resistance experienced by the stone does not balance its weight as quickly as in the case of the feather. Thus, the stone accelerates to higher speed before the it attains terminalofvelocity. why the hits the ground before theafeather, even though acceleration free fall This is theissame for stone bo both th objects.

Get It Right (page 67)

(a) (b) (c)

(d) (e)

(f)

(g) (h)

False When an object does not move, this means that the forces acting on the object are balanced. True False To find the combined effect of two forces on an object, we find the resultant force equivalent to the individual forces in terms of magnitude and direction. False The product of mass and acceleration is the resultant of the forces acting on a body. False When there are forces acting on an object, the object o bject will accelerate only if the forces are unbalanced. If the forces are balanced, the resultant force is zero, and there is no acceleration. False Newton’s Third Law states that if body A exerts a force F  AB on body B, then body B will exert an equal and opposite force F BBAA on body A. True True

Let’s Review (page 68) Section A: Multiple Choice Questions

1.

B We take the rightward direction as positive and the leftward direction as negative. Resultant force = 4 N + 7 N + (–3 N) + (–2 N) = 6 N to the right

2.

A Because of the absence of friction between A and B, the 10 N force acts on B alone. B will be pulled away by the 10 N force, but A will drop vertically to the ground. Since no resultant horizontal force acts on A, A will experience no horizontal acceleration.

3.

C Since the parachutist falls with terminal velocity, the resultant force acting on him is zero, as his weight is balanced by the upward air resistance acting on him. The air resistance must be equal to his weight, which is given as 700 N.

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Physics Matters for GCE ‘O’ Level (4  Edition): Full Solutions to Textbook Questions

Chapter 3

Section B: Structured Questions

1.

(a)

(i) 

Tow truck

(ii) 

Car

1000 N tow truck

f t 

(i)

(b)

(ii)

(iii)

1000 N

F

car f c 

The car experiences a frictional force f c. By Newton’s Second Law of Motion, resultant force = 1000 N + (–f c)  –2 = 1000 kg ! 0.50 m s   ! f c = 500 N The tow truck experiences a forward tractive force F  and  and a frictional force f t of 750 N. By Newton’s Second Law of Motion, resultant force = F  +  + (–1000 N) + (– f t)  –2 = 1500 kg ! 0.50 m s  = 2500 N ! F  = We now consider the tow truck and the the car as one system. This system exper experiences iences a resultant force F R.  –2 F R = (1000 kg + 1500 kg) kg ) ! 0.5 m s   = 1250 N  Alternative ly,  Alternatively, F R = F  +  + (–f t) + (–f c) = 2500 N + (–750 N) + (–500 N) = 1250 N

2.

Use an appropriate scale, such as 1 cm : 0.5 km. N 5 km W

E

3 km 53˚

S 4 km

The resultant displacement of the car is 5 km.

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Physics Matters for GCE ‘O’ Level (4  Edition): Full Solutions to Textbook Questions

Chapter 3

Section C: Free-Response Questions

1.

Use an appropriate scale, such as 1 cm : 10 N. T 2 

45˚

T 1 

W  =  = 100 N

45˚

 Alternative ly, we can use the sine rule.  Alternatively, W  T 1 T 2  =  =    sin 45˚  sin 90˚  sin 45˚  = 100 N along the horizontal ! T 2 = W  = T 1 = 142 N at 45˚ to the horizontal ho rizontal 2.

We take the upward direction as positive and the downward direction as negative. Consider the forces that act on m1, and those on m2. By Newton’s Second Law of Motion, T  –  – m2g   = = m2a ——— (1) T  –  – m1g  =  = –m1a ——— (2)

From (2), T   = = m1g  –  – m1a ——— (3) Substituting (3) into (1), we have ( m1g  –  – m1a) – m2g  =  = m 2 a m1g – m2g = m1a + m2a (m1 – m2)g  =  = (m1 + m2)a m1 – m2  )g   ! a  = ( m1 + m2 We can find tension T   by by considering just one of the two masses. Here, we consider only the forces that act on m2. T  –  – m2g   = = m2a  T   = = m2a + m2g   = m2(a + g ) m1 – m2 = m2[(  )g + g ] m1 + m2  m1 – m2 = m2[(  ) + 1]g  m1 + m2  m1 – m2 + m1 + m2 = m2(  )g    m1 + m2 2m 1m 2 =(  )g   m1 + m2

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