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2 Force and Motion
3
Chapter 3 Force and Motion (I)
Force and Motion (I)
Practice 3.2 (p.107)
Practice 3.1 (p.99)
1
C
1
C
2
D
2
A
3
C
3
(a)
Since the ferry moves at a constant velocity,
the net force acting on it is zero.
Take forwards as positive.
Tf =0
T = f = 2000 N
4
The bowling ball is harder to stop since it has
a larger inertia.
(b)
4
5
Yes
Take upwards as positive.
spaceship is negligible. When the rockets are
F + W = 12
shut down, no net force acts on the spaceship.
F = 12  W = 12  (10) = 22 N
5
In space, the gravitational force acts on the
(a)
By Newton’s first law, the spaceship keeps
moving at a uniform velocity.
normal force
pushing
force
6
When the cup rotates, the friction between the
tea and the cup is so small that the tea and the
tea leaf remain stationary due to inertia.
friction
7
weight
(b)
When the vehicle accelerates, we still remain
in the original state of motion. Therefore, we
may move (to the opposite direction of the
normal force
acceleration) relative to the vehicle. Holding
friction
the handrails helps us stand still.
8
The net force acting on it is zero.
Let T be the tension in the string.
weight
In method 1,
(c)
2T = W
W
T=
2
In method 2,
weight
6
tension in
upper string
X
weight
tension in
lower string
Y
tension in
lower string
weight
New Senior Secondary Physics at Work (Second Edition)
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T=W
The tension is larger in method 2 and the
string would break more easily.
Practice 3.3 (p.114)
1
2 Force and Motion
Chapter 3 Force and Motion (I)
1
C
(1) + (2),
2
A
7  5 = 2ma
F = ma = 40  0.5 = 20 N
3
ma = 1 N
C
Net force acting on P = ma = 1 N
a 1
=
F m
1
1
 Mass =
=
= 2 kg
slope 3  1
84
Alternative solution:
Slope of graph =
4
Consider the blocks as one single object.
By F = ma,
7  5 = 2ma
ma = 1 N
A
When F is 6 N, a is 2 m s2.
7
Take the moving direction of the car as
By F = ma,
positive.
6f=22
(a)
acceleration of the car
f =2N
5
2
 72 
0

2
2
v u
 3.6  = 5 m s2
=
=
2s
2  40
A
By F = ma,
1000  500 = 1500a
a = 0.333 m s2
1
1
s = ut + at 2 = 0 + (0.333)102 = 16.7 m
2
2
6
By v2 = u2 + 2as,
(b)
Braking force
= ma = 1000(5) = 5000 N
8
The object is at rest during 05 s.
A
It speeds up at 0.8 m s2 in the positive
Let m be the mass of each block and T be the
direction during 510 s.
tension in the string.
It speeds up at 0.4 m s2 in the positive
The figures below show the horizontal forces
direction during 1020 s.
acting on the blocks.
It still moves in the positive direction but
slows down at 0.4 m s2 during 2025 s.
5N
P
T
It moves at a constant velocity in the positive
direction during 2530 s.
T
Q
7N
Since the blocks are connected by a light
9
Take the direction to the right as positive.
(a)
By F = ma,
inextensible string, they move with the same
10  f = 4  2
acceleration a.
f =2N
Take the direction to the right as positive.
When the force is increased to 20 N,
F 20  2
acceleration = =
= 4.5 m s2
m
4
Apply F = ma.
For P,
T  5 = ma
(1)
For Q,
7  T = ma
2
(2)
Let f be the friction acting on the box.
(b)
Displacement
1
1
= ut + at 2 = 0 + (4.5)52 = 56.3 m
2
2
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2 Force and Motion
(c)
Chapter 3 Force and Motion (I)
Add lubrication oil to the road.
starts to fall, B is zero and A is the net
Practice 3.4 (p.126)
force acting on it. After a certain period
1
C
of time, A and B are equal in magnitude.
2
C
The net force on the raindrop becomes
By F = ma,
F 23 .3
m= =
= 72.6 kg
a 0.321
zero. By Newton’s first law, the raindrop
3
falls at a constant speed.
8
D
Take the direction to the right as positive.
(a)
normal force
Take upwards as positive.
thrust T
F
friction f
rocket
weight
weight mg
(b)
By F = ma,
f=F=3N
T  4.26  105  9.81 = 4.26  105  4.09
The friction is 3 N towards the left.
T = 5.95  106 N
4
By Newton’s first law,
(c)
D
Take upwards as positive.
By F = ma,
F 53
m= =
= 1 kg
a
2
The mass of the box is 1 kg.
air resistance f
9
(a)
The gravitational acceleration is same at
both ends of the balance. Therefore, the
skydiver
result is the same as that on the Earth.
(b)
weight mg
100  (190 )
3.6
Acceleration =
0.8
= 31.25 m s
The acceleration of the masses is the
same at both ends of the balance.
Therefore, the result is the same as that
on the Earth.
2
(c)
When the balance undergoes free fall,
By F = ma,
the force acting on the balance by a mass
f  70  9.81 = 7  31.25
becomes zero. Therefore, no
f = 2870 N
5
A
6
A bag of sugar has a mass of 1 kg. /
measurement can be taken.
10
(a)
A bag of sugar weighs 9.81 N.
7
(a)
A: weight of raindrop
B: air resistance
(b)
The magnitude of B increases with the
speed of the raindrop. When the raindrop
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3
2 Force and Motion
(b)
(i)
Chapter 3 Force and Motion (I)
Consider the trolley. Take the
12
(a)
air resistance f
direction to the right as positive.
Jackie
By F = ma,
F 4
a = = = 4 m s2
m 1
weight mg
(b)
Since the block and the trolley are
Her weight remains unchanged.
connected by an inextensible string,
The air resistance increases with time at
they accelerate at the same rate.
first and becomes constant after she has
 The acceleration of the block is
reached the terminal speed.
4 m s2 downwards.
(c)
At first, she accelerates from rest. Her
acceleration decrease in magnitude as
(ii) Consider the block. Take
downwards as positive. By F = ma,
she falls. When the air resistance has the
mg  T = ma
T
m=
g a
same magnitude as her weight, she falls
at a constant velocity.
(d)
v / m s1
4
=
9.81  4
= 0.688 kg
The mass of the block is 0.688 kg.
11
Take upwards as positive. Apply F = ma.
0
t/s
force by balance R
Area under graph
= distance that she falls
mass
= 6000  2000
weight mg
= 4000 m
R  mg = ma
(a)
R = ma + mg
Practice 3.5 (p.133)
Reading = 1.2(0) + 1.2(9.81)
1
D
2
C
3
C
4
A
5
When he pushes the platform, by Newton’s
= 11.8 N
(b)
Reading = 1.2(1.5) + 1.2(9.81)
= 9.97 N
(c)
Reading = 1.2(0) + 1.2(9.81)
third law, the platform also pushes him. He
= 11.8 N
(d)
accelerates because of this pushing force from
Reading = 1.2(0.5) + 1.2(9.81)
the platform.
= 12.4 N
6
The force acting on the ground by the tyres
points backwards. By Newton’s third law, the
4
New Senior Secondary Physics at Work (Second Edition)
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2 Force and Motion
7
Chapter 3 Force and Motion (I)
ground exerts a forward force on the tyres.
T  (m + M)g = (m + M)a
This force pushes the car forwards.
 T = (m + M)(a + g)
When he pushes the ground in order to jump,
= (65 + 200)(0.6 + 9.81)
the normal reaction acting on him by the
= 2760 N
ground is larger than his weight.
The tension is 2760 N.
This does not violate Newton’s third law since
8
(ii) Consider the man and the lift as
the two forces are not an action-and-reaction
one body.
pair.
T  (m + M)g = (m + M)a
Take the direction to the right as positive.
 T = (m + M)(a + g)
(a)
(b)
Average force acting on B
= (65 + 200)(0.6 + 9.81)
= ma = 1  3 = 3 N
= 2760 N
The force acting on B by A and the force
The tension is 2760 N.
acting on A by B forms a pair of action
(c)
and reaction.
Revision exercise 3
 Average force acting on A = 3 N
F 3
Acceleration of A = =
= 1 m s2
m 3
Concept traps (p.137)
F
A moving object remains in a state of uniform
The time of acceleration of A is the same
motion unless acted on by an unbalanced force.
as that of B.
An unbalanced force makes an object
Velocity of A after collision
accelerate.
= u + at = 1.2 + (1)0.5 = 0.7 m s1
9
1
normal reaction
from lift N
(a)
2
tension in
cable T
F
The net force acting on each object is its own
weight. The statement is true only if the two
man
weight mg
weight Mg
(b)
objects have the same mass.
lift
force acting
on lift by
man N
3
T
4
T
Take upwards as positive.
Multiple-choice questions (p.137)
Apply F = ma.
5
B
(i)
6
B
7
C
By Newton’s third law,
8
A
N = N
9
B
Consider the lift.
10
C
Consider the man.
N  mg = ma
(1)
T  Mg  N = Ma
T  Mg  N = Ma
Take the moving direction of the ball as
(2)
(1) + (2),
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positive.
By v2 = u2 + 2as,
5
2 Force and Motion
a=
Chapter 3 Force and Motion (I)
v2  u 2 0  u 2
u2
=
=
2s
2  20
40
tension T
 u2 
Resistive force = ma = 1.2     = 0.03u2
 40 
2-kg mass
Magnitude of resistive force is 0.03u2.
11
tension T
weight mg
weight Mg
B
normal reaction
from lift N
3-kg mass
Take upwards as positive. Apply F = ma.
For the 2-kg mass,
T  mg = ma
woman
T  2(9.81) = 2a
weight mg
T  19.62 = 2a
Take upwards as positive.
For the 3-kg mass,
T  3(9.81) = 3(a)
During 02 s, by F = ma,
N  mg = ma
T  29.43 = 3a
= 55(1) + 55(9.81)
= 594.6 N
12
T = 23.5 N
A
15
A
Take the moving direction of the car as
16
C
positive. By F = ma,
F 6000
a= =
= 4 m s2
m 1500
17
B
2
 108 
0

v2  u 2
 3.6  = 113 m
s=
=
2(4)
2a
13
3T + 58.86 = 2T  58.86
594 .6
= 60.6 kg
9.81
By v2 = u2 + 2as,
1 2
1
at = 0 + at 2
2
2
2s
a= 2
t
s = ut +
By F = ma,
F  f = ma
=m
C
Consider the 3-kg mass.
 T = mg = 3  9.81 = 29.4 N
14
C
After the 2-kg mass is released, it moves
upwards and the 3-kg mass moves downwards.
The magnitudes of their accelerations are a.
2s
t2
 t2 
 t2 
F  

s = 

 2m  f
 2m 


Since it is stationary, the net force acting on it
is zero.
(2)
(1)  (2),
T  19 .62
2
=
T  29 .43  3
N = ma + mg
Reading of scale R =
(1)
When F  fmax, f = F

s=0
When F > fmax, f = constant
 t2 
 F  constant
 s = 

 2m 
 s varies linearly with F.
6
New Senior Secondary Physics at Work (Second Edition)
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2 Force and Motion
Chapter 3 Force and Motion (I)
18
(HKCEE 2006 Paper 2 Q31)
The gravitational acceleration on Mars is
19
(HKCEE 2009 Paper 2 Q3)
3.27 m s2.
20
(HKCEE 2010 Paper 2 Q30)
21
(HKDSE 2012 Paper 1A Q8)
lower rate
22
(HKDSE 2013 Paper 1A Q7)
and takes a longer time to reach the
(b)
On Mars, the object accelerates at a
1A
ground.
Conventional questions (p.140)
23
(a)
(c)
1A
The motion is the same on Mars and on
the Earth.
normal reaction
26
(a)
1A
upward force by hand U
friction
weight
X
(1 correct force with correct name) 1A
(All correct)
(b)
weight Mg
1A
tension in
string T
(1 correct force with correct name) 1A
normal reaction
(All correct)
friction
1A
tension in string T
weight
Y
(1 correct force with correct name) 1A
(All correct)
24
(a)
weight mg
1A
magnetic force
(1 correct force with correct name) 1A
(All correct)
(b)
1A
Take upwards as positive.
The masses move at the same
acceleration a as they are connected by
weight
(1 correct force with correct name) 1A
(b)
(c)
(All correct)
1A
The magnetic force is 1 N
1A
upwards.
1A
The magnetic force acting on the globe
by the holder and the magnetic force
25
acting on the holder by the globe
mg E
(a) mgM =
3
9.81
g
gM = E =
= 3.27 m s2
3
3
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1A
1M
1A
an inextensible string.
Apply F = ma.
Consider X.
U  Mg  T = Ma
(1)
1M
(2)
1M
Consider Y.
T  mg = ma
(1)  (2),
U  Mg  T
M
=
m
T  mg
mU  mMg  mT = MT  mMg
7
2 Force and Motion
Chapter 3 Force and Motion (I)
mU
M m
0.8  25
=
1 .2  0 . 8
(b)
T=
The free-body diagram of Joan is as
shown. Take upwards as positive.
normal reaction
from balance N
= 10 N
Joan
1A
Tension in the string is 10 N.
27
weight mg
(c)
Zero
1A
(a)
His idea is incorrect.
1A
(i)
By F = ma,
N  mg = ma
The ball moves horizontally at the same
N = ma + mg
velocity as the train before it is thrown.
= 50(1.5) + 50(9.81)
1A
= 566 N
By Newton’s first law, its horizontal
(ii) Reading of balance
thrown since no horizontal force acts on
(b)
= 50(0) + 50(9.81) = 491 N
1A
forwards as positive.
By F = ma,
F 2
a= =
= 10 m s2
m 0 .2
1A
(iii) Reading of balance
Consider the horizontal motion. Take
(i)
1A
Reading of balance is 566 N.
motion remains unchanged after it is
it.
1M
(c)
1M
= 50(1.2) + 50(9.81)
1M
= 431 N
1A
No,
1A
this is because the normal reaction and
The ball is stationary relative to
Joan’s weight are not a pair of action and
Bobby before it is thrown.
reaction.
v = u + at
= 0 + (10)0.5 = 5 m s1
1M
1A
29
1A
Take the direction to the right as positive.
(a)
The velocity of the ball is 5 m s1
Consider the boxes as one single object.
By F = ma,
50
F
a= =
= 1.25 m s2
m 30  10
backwards relative to Bobby.
(ii) The ball is moving at 20 m s1
1M
1A
The acceleration of Y is 1.25 m s2
forwards relative to a person on the
towards the right.
ground before it is thrown.
(b)
v = u + at
normal reaction
from ground
= 20 + (10)0.5
= 15 m s1
1A
X
tension in
string T
The velocity of the ball is 15 m s1
forwards relative to a person on the
ground.
28
(a)
weight
(1 correct force with correct name) 1A
One newton of force is defined as the
(All correct)
1A
force that produces an acceleration of
1 m s2 on a mass of 1 kg.
8
1A
New Senior Secondary Physics at Work (Second Edition)
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2 Force and Motion
Chapter 3 Force and Motion (I)
normal reaction
from ground
M will accelerate towards the left, m1
will accelerate downwards and m2 will
Y
tension in
string T
50 N
weight
(c)
(c)
accelerate upwards.
1A
If fmax  T1  T2,
1A
The masses will remain at rest.
1A
Consider m1. Take downwards as
(1 correct force with correct name) 1A
positive.
(All correct)
Apply F = ma.
1A
Consider X.
m1g  T1 = m1a
Tension = ma = 30(1.25) = 37.5 N
1A
T1 = m1g  m1a  m1g
(d)
Net force = ma = 10(1.25) = 12.5 N 1A
Consider m2. Take upwards as positive.
(e)
Her statement is incorrect.
T2  m2g = m2a
1A
Since the surface is frictionless, the net
T2 = m2g + m2a  m2g
force acting on X becomes zero after the
string breaks.
30
 T1  T2  m1g  m2g
1A
= 0.8(9.81)  0.5(9.81)
By Newton’s first law, X will continue to
= 2.94 N
move at a constant velocity.
1A
< fmax
(a)
T
1A
(b)
Consider Q.
 The masses remain at rest.
Q
(a)
1M
Tension in string hanging m2
= m2g
TP
31
1M
TR
= 0.5(9.81)
By F = ma,
1A
TR  TP = ma = 0
1A
 TR = TP
1A
32
(a)
1A
No, I do not agree with her.
1A
At the beginning, the air resistance is
normal reaction
from ground
tension T2
due to m2
tension T1
due to m1
= 4.91 N
smaller than the weight of the pot.
1A
The net force acting on the pot points
friction f
(b)
weight
downwards,
1A
so it speeds up as it falls.
1A
F/N
(1 correct force with correct name) 1A
(3 correct forces with correct names)
1A
(b)
(All correct)
1A
If fmax < T1  T2,
1A
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0
t/s
(Correct axes)
1A
(F decreases gradually)
1A
9
2 Force and Motion
Chapter 3 Force and Motion (I)
(F remains zero after a certain period of
(c)
33
time)
1A
At the beginning.
1A
(a)
=
8000  7000
7000
9.81
= 1.4014  1.40 m s2 1A
normal reaction
from ground
(b)
A
The weight of the passengers remains
unchanged
F
1A
because their masses and the
force on A
by B, FAB
gravitational acceleration remain
unchanged.
weight
(c)
(1 correct force with correct name) 1A
(All correct)
1A
Consider the balloon.
By v2 = u2 + 2as,
20 2  0
v2  u 2
s=
=
= 142.7 m
2(1.4014 )
2a
1A
normal reaction
from ground
1M
Consider the sandbag.
B
force on B
by A, FBA
By Newton’s first law, its initial velocity
is 20 m s1 upwards when it leaves the
weight
balloon.
Take the direction to the right as
positive.
 4.905t 2  20t  142.7 = 0
(1 correct force with correct name) 1A
(All correct)
(b)
(i)
1A
Consider the blocks as one object.
F = ma
 t = 7.80 s or
= 6.4 N
The sandbag needs 7.80 s to reach the
towards the right.
1A
the force acting on A by B is 4 N
towards the left.
1A
By F = ma,
(a)
By F = ma,
F
a=
m
1 .5
=
1 .2  1 . 2
1M
= 0.625 m s2
1A
2
The acceleration of X is 0.625 m s
towards the right.
Take upwards as positive.
(a)
ground.
35
1A
The force acting on B by A is 4 N
By Newton’s third law,
3.73 s (rejected)
1A
1A
(ii) Consider B.
FBA = mBa = 5(0.8) = 4 N
1M
1M
= (3 + 5)0.8
34
1M
1
By s = ut + at 2,
2
1
142.7 = 20t + (9.81)t 2
2
1M
F
acceleration =
m
10
New Senior Secondary Physics at Work (Second Edition)
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2 Force and Motion
(b)
Chapter 3 Force and Motion (I)
By F = ma,
F 10  5
a= =
= 1 m s2
m 23
F/N
0.75
force on Y
0
2
0.75
= mX a = 2  1 = 2 N
force on X
on Y by X is 2 N towards the right.
opposite in direction)
1A
(Correct sign)
1A
(Correct value)
1A
1A
(c)
Push Y with a force larger than 42.5 N.
1A
Hook the spring balances to each other.
Push X with a force larger than 15 N.
1A
Pull one of the trolleys so that the
trolleys move together.
1A
37
(a)
1A
(a)
During 05 s, the tension in the string is
larger than the friction between the 2-kg
The balances should show the same
36
1A
(iii) By Newton’s third law, the friction
(Two forces equal in magnitude but
(c)
1M
Friction on X by Y
t/s
4
1M
mass and the table,
1A
reading.
1A
so the mass accelerates to the right. 1A
Zero
1A
The mass m reaches the ground at t = 5 s
and the tension becomes zero. Friction
(b)
normal reaction
from ground
(i)
becomes the net force acting on the 2-kg
X
friction
from Y
(b)
weight
mass,
1A
so the mass slows down.
1A
Consider the motion of the 2-kg mass
during 57 s.
(1 correct force with correct name)
a = slope of vt graph
04
=
= 2 m s2
75
1A
(All correct)
normal reaction
from ground
1A
Friction = ma
= 2(2) = 4 N
friction from X
Y
(c)
F
(1 correct force with correct name)
(All correct)
T  4 = 2(0.8)
1A
T = 5.6 N
1M
1M
Consider the motion of the mass m
Take the direction to the left as
during 05 s. Take downwards as
positive.
positive.
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
1A
By F = ma,
1A
(ii) Consider the blocks as one object.
1M
Consider the motion of the 2-kg mass
during 05 s.
40
a=
= 0.8 m s2
50
weight
1M
11
2 Force and Motion
Chapter 3 Force and Motion (I)
By F = ma,
(c)
(i)
m(9.81)  5.6 = m(0.8)
a decreasing rate.
m = 0.622 kg
1A
(a)
1A
The car eventually travels at a
The mass of m is 0.622 kg.
38
The velocity of the car decreases at
constant velocity.
1A
(ii) When the parachute is opened, the
upward force U
produced by fan
air resistance becomes greater than
toy
the thrust in magnitude. As a result,
weight mg
the velocity decreases.
1A
Air resistance decreases with the
(1 correct force with correct name) 1A
(All correct)
(b)
speed, and is finally equal to the
1A
thrust in magnitude. Therefore the
When the fan pushes air downwards,
car travels at a constant velocity.
1A
by Newton’s third law, the air exerts an
1A
41
(HKCEE 2010 Paper 1 Q1)
equal and opposite force on the fan. 1A
The toy can hover in mid-air when this
force has the same magnitude as the
(c)
Experiment questions (p.144)
42
toy’s weight.
1A
No,
1A
(a)
T = weight of weights = mg
T
Mass of weight =
g
when the air pushed by the fan hits the
39
40
balance.
1A
it exerts a force on the balance.
1A
(HKCEE 2007 Paper 1 Q1)
F
(a) a =
m
3 .0
=
0.80
= 3.75 m s2
(b)
(i)
=
(b)
1M
= slope of Fig ai
1A
= 1.24 m s2
(c)
air resistance) decreases, and the
1A
the net force and hence the
ma = (0.333 + 0.718)1.24
= 1.30 N
1A
< tension in string
1A
friction acting on the trolley.
1A
Physics in article (p.145)
43
acceleration is zero, so the car
1A
The discrepancy may be due to the
1A
(ii) When the air resistance is equal to
12
1A
(ii) Acceleration of trolley
so the net force (= forward thrust 
reaches a constant speed.
1A
From Fig ah,
tension in string = 1.93 N
1A
the thrust in magnitude,
(i)
1M
2.437
9.81
= 0.248 kg
Air resistance increases with speed,
acceleration decreases.
Before the block is removed,
(HKCEE 2005 Paper 1 Q13)
1A
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
2 Force and Motion
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
Chapter 3 Force and Motion (I)
13