2 Force and Motion
3
Chapter 3 Force and Motion
Force and Motion
Practice 3.1 (p. 104)
6
(a) The MTR train is accelerating in the
1
C
forward direction. The man tends to
2
C
move at his original speed (smaller
3
(b), (e), (f)
speed), so he would move backwards
relative to the MTR train.
4
(b) The MTR train is slowing down. The
man tends to move at his original speed
(greater speed), so he would move
forwards relative to the MTR train.
(c)
The MTR train is moving forwards at
constant velocity. The man moves
forwards with the same constant velocity,
so he would remain at rest relative to the
MTR train.
(d) The MTR train is turning a corner. The
5
(a) Stretching a rubber band
man tends to move at his original
(b) Standing on the floor
direction, so he would move outwards
(c)
relative to the MTR train.
Walking
7
(d) Exists in every object on the earth at any
In space, the gravitational force acts on the
time
spaceship is negligible. When the rockets are
(e)
A compass
shut down, they do not exert a force on the
(f)
A rubbed plastic ruler attracts small bits
spaceship. Therefore, no net force acts on the
of paper
spaceship. By Newton’s first law, the
spaceship is in uniform motion and can travel
Practice 3.2 (p. 111)
far out in space.
8
Joan moves on the ice surface with a constant
1
C
2
C
3
D
4
C
Practice 3.3 (p. 122)
5
(a) No. Athletes would hit the wall of the
1
D
stadium if it is too close to the finishing
2
A
line.
3
B
(b) The mat is used to protect the athletes if
4
A
they hit the wall after passing the
5
D
velocity.
finishing line.
New Senior Secondary Physics at Work
1
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2 Force and Motion
6
Chapter 3 Force and Motion
7
(a)
(a) Horizontal component
= 40 + 30 cos 30 = 66.0 N
Vertical component
= 30 sin 30 = 15 N
Resultant = 66 2  15 2 = 67.7 N
Let  be the angle between the resultant
and the horizontal.
15
tan θ 
  = 12.8
66
Resultant’s magnitude is 67 N and the
angle between the resultant and the
horizontal is 13.
Resultant’s magnitude is 67.7 N and the
(b)
angle between the resultant and the
horizontal is 12.8.
(b) Horizontal component
= 40 + 30 cos 45 = 61.2 N
Vertical component
= 30 sin 45 = 21.2 N
Resultant’s magnitude is 65 N and the
Resultant = 61.2 2  21.2 2 = 64.8 N
angle between the resultant and the
Let  be the angle between the resultant
horizontal is 19.
and the horizontal.
21.2
tan θ 
  = 19.1
61.2
(c)
Resultant’s magnitude is 64.8 N and the
angle between the resultant and the
horizontal is 19.1.
(c)
Horizontal component
= 40 + 30 cos 60 = 55 N
Resultant’s magnitude is 60 N and the
Vertical component
angle between the resultant and the
= 30 sin 60 = 26.0 N
horizontal is 25.
Resultant = 55 2  26.0 2 = 60.8 N
(d)
Let  be the angle between the resultant
and the horizontal.
26.0
  = 25.3
tan θ 
55
Resultant’s magnitude is 60.8 N and the
angle between the resultant and the
Resultant’s magnitude is 50 N and the
horizontal is 25.3.
angle between the resultant and the
horizontal is 37.
New Senior Secondary Physics at Work
2
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2 Force and Motion
Chapter 3 Force and Motion
(d) Resultant = 40 2  30 2 = 50 N
8
Hence, the angle between the two 5-N forces
Let  be the angle between the resultant
is 120.
and the horizontal.
30
tan θ 
  = 36.9
40
Alternative method:
Resultant’s magnitude is 50 N and the
triangle. It is known that each angle of an
angle between the resultant and the
equilateral triangle is 60. Therefore, the
horizontal is 36.9.
angle between the two 5-N forces is 120.
By tip-to-tail method, the two 5-N forces and
the resultant 5-N force form an equilateral
(a)
10
Resultant force = 2  400 = 800 N
(b)
The resultant force provided by the cable is
800 N.
11
(c)
For the 2-kg mass:
R = weight  cos  = 20 cos 30
= 17.3 N
9
Suppose the two forces act in the direction as
T = 20 N
shown.
Therefore we have:
Vertical component Fx = 5 sin 
Horizontal component Fy
= 5  5 cos  = 5  (1  cos )
(magnitude of the resultant)2 = Fx2 + Fy 2
52 = (5 sin )2 + [5  (1  cos )]2
2T cos 45 = W
2  20  cos 45 = W
1 = sin2 + 1  2 cos  + cos2
cos  = 0.5
W = 28.3 N
 = 60
New Senior Secondary Physics at Work
3
 Oxford University Press 2009
2 Force and Motion
12
Chapter 3 Force and Motion
(a) 2T sin 10 = 500
T = 1440 N
The tension of the string is 1440 N.
3
B
4
C
5
A
Net force = ma = 40  0.5 = 20 N
(b) Component of force
= T cos 10
6
= 1440  cos 10
C
By v2 – u2 = 2as,
0 – u2 = 2a(20)
= 1420 N
u2 = 40a
The component of the force that pulls
the car is 1420 N.
13
a=
(a)
u2
40
 u2 
 = –0.03u2
Resistance = ma = 12   
 40 


7
‘A bag of sugar weighs 10 N.’ or ‘A bag of
sugar has a mass of 1 kg.’
8
(b) As the mass is stationary, the net force
(c)
By F = ma,
F 800 000
= 2 m s–2
a 
m 4  10 5
When it flies horizontally, its acceleration is
acting on it is zero.
2 m s–2.
(i)
100
(
)0
vu
(a) a =
= 3.6
= 4.63 m s–2
t
6
y-component of F1
9
= weight of mass
= 10 N
The acceleration of the car is
y-component of F1 = F1 sin 30 
4.63 m s–2.
F1 sin 30  = 10 N
(b) F = ma = 1500  4.63 = 6945 N
F1 = 20 N
The force provided by the car engine is
x-component of F1 = F1 cos 30 
6945 N.
= 20 cos 30 
10
(a)
= 17.3 N
(ii) y-component of F2 = 0
x-component of F2
= x-component of F1 = 17.3 N
(d) From (c)(i), F1 = 20 N.
F2 = x-component of F2 = 17.3 N
Practice 3.4 (p. 140)
(b) (i)
Downwards along the slide
1
D
(ii) No net force
2
B
(iii) No net force
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4
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2 Force and Motion
11
Chapter 3 Force and Motion
Take the upward direction as positive.
14
Weight = mg
Take the downward direction as positive.
Let R be the reading of the balance.
= 3  10  10
5
(a) By F = ma,
= 3  10 N
R  mg = 0
6
Net force = ma
R = 20 N
= 3  10  12
5
The reading of the balance is 20 N.
= 3.6  10 N
6
(b) By F = ma,
mg  R = ma
Net force = thrust – weight of the rocket
20  R = 2  1.5
Thrust = net force + weight of the rocket
= 3.6  10 + 3  10
6
6
R = 17 N
= 6.6  10 N
6
The reading of the balance is 17 N.
The thrust of the rocket is 6.6  10 N.
6
12
(c)
By F = ma,
R  mg = 0
(a)
R = 20 N
The reading of the balance is 20 N.
(d) By F = ma,
mg  R = ma
20  R = 2  (0.5)
R = 21 N
The reading of the balance is 21 N.
(b) The friction acting on the box is 3 N.
(c)
15
By F = ma,
F 53
m 
 1 kg
a
2
The frictional force acting on the trolley
is 3.47 N.
The mass of the box is 1 kg.
13
(a) (i)
(a) f = mg sin  = (2)(10)sin 10 = 3.47 N
(b) By F = ma,
mg sin   f  ma
Weight, air resistance
(2)(10) sin 30  3.47  2a
(ii) Weight
a = 3.27 m s–2
(iii) Weight, air resistance
In the above 3 cases, the net force acts
When the trolley moves down the
downwards.
runway, its acceleration is 3.27 m s–2.
(b)
16
(a) Take the direction of the car movement
as positive.
By F = ma,
F 6000
a= =
= –4 m s–2
m 1500
By v = u + at，
108
0  (
)
vu
3.6 = 7.5 s
t=
=
a
4
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5
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2 Force and Motion
Chapter 3 Force and Motion
It takes 7.5 s to stop the car.
1
(b) By s  ut  at 2 ,
2
1
s = (30)(7.5)  (4)( 7.5) 2 = 112.5 m
2
3
D
4
A
5
C
6
(a)
The braking distance is 112.5 m.
17
(a)
Acceleration
a / m s–2
Net force
F/N
AB
BC
CD
DE
1
2
0
–3
3
6
0
–9
(b) His comment is correct. From the graph,
the velocity of the object starts to
(b)
decrease from t = 30 s onwards and
becomes zero at t = 40 s. If the force
continues to act on the object, its
velocity will become negative. That
means it will change its moving
direction.
18
(a)
Time t / s
Acceleration
a / m s–2
0–5
5–10
10–20
20–30
0
4
1
0
(c)
(b) During 0–5 s, the object is moving at a
on B by A.
constant velocity as no net force acts on
7
it. During 5–10 s, the object is moving
(a) When the roller skater exerts a force on
the wall, the wall also exerts an equal
with an acceleration of 4 m s–2 as a net
but opposite force on the skater.
force of 20 N acts on it. During 10–20 s,
Therefore the skater moves backwards.
the object is moving with an acceleration
(b) When the diver pushes the platform, the
of 1 m s–2 as a net force of 5 N acts on it.
platform also exerts an equal but
During 20–30 s, the object is moving at
opposite force on the diver. Therefore
constant velocity as no net force acts on
the diver gains speed and dives.
it.
(c)
D
2
C
New Senior Secondary Physics at Work
When we push ourselves against the side
of the pool, the pool exerts an equal but
Practice 3.5 (p. 148)
1
Force acting on A by B and force acting
opposite force on us. Therefore we
accelerate forwards.
6
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
(d) When the runner exerts a force on the
(c)
but opposite force on the runner.
By F = ma,
20
F
a= =
= 2.5 m s–2
m 35
Therefore the runner moves forwards.
The accelerations of the blocks are
starting block, the block exerts an equal
8
(a) (i)
2.5 m s–2.
Trolley A’s weight component
(ii) By F = ma,
down the plane
= mg sin 
20 – force acting on A by B
= (3)(10 ) sin 20 
= mAa = 3  2.5 = 7.5 N
= 10.3 N
Force acting on A by B
= 12.5 N (towards the left)
(ii) Net force acting on it
(b) (i)
= 10.3 N – T (down the plane)
By Newton’s third law,
Trolley B’s weight component
force acting on B by A
down the plane
= mg sin 
= force acting on A by B (opposite
= (2)(10 ) sin 30 
= 12.5 N (towards the right)
direction)
10
= 10 N
(a) F = ma = (1)(1) = 1 N
The net force acting on toy car B during
(ii) Net force acting on it
collision is 1 N towards the right.
= T – 10 N (up the plane)
(c)
(b) By Newton’s third law of motion, the
Trolley A moves down the plane while
force acting on B by A has the same
trolley B moves up the plane.
9
(i)
magnitude as that acting on A by B, but
(a)
their directions are opposite.
Therefore, the net force acting on toy car
A is 1 N towards the left.
(c)
Take the direction towards the right as
positive.
By F = ma,
F
a=
m
1
=
3
= 0.333 m s2
v = u + at
= 3 + (–0.333)(0.5)
= 2.83 m s–1
(b) Net force acting on A
= 20 N – force acting on A by B
The velocity of toy car A after the
Net force acting on B
collision is 2.83 m s–1 towards the right.
= force acting on B by A
New Senior Secondary Physics at Work
7
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
Practice 3.6 (p. 165)
1
(b)
A
Moment of force
2
= 30 sin 15  0.7 = 5.44 N m (clockwise)
(In the same direction, with equal
B
distance from O and on different sides of
Let W be the weight of the girl nearer to the
O.)
boy.
(c)
Take moment about the joint, in equilibrium,
clockwise moment = anticlockwise moment
600  3 = 400  (2+1) + W  2
(In opposite directions and acting at the
W = 300 N
same position.)
The weight of the girl nearer to the boy is
8
300 N.
3
Take moment about the elbow contact
D
point.
f  0.05 = 100  0.3
Clockwise moment
f = 600 N
4
= 5  10  0.3 + 1.5  10  0.15
(a) A door handle is placed well away from
= 17.25 N m
the hinge to give a large moment for
Anticlockwise moment
turning the door.
= F  0.05
(b) A mechanic uses a long spanner to give
In equilibrium,
a large moment for undoing the nut.
5
(a) Let F be the force exerted by the biceps.
clockwise moment = anticlockwise
The centre of gravity of the bat is outside the
moment
edge of the table. The weight of the bat seems
17.25 = F  0.05
to act on the centre of gravity and produces a
F = 345 N
torque which tips the bat over. The bat then
The force exerted by the biceps is
falls down.
345 N.
6
(b) Take moment about the shoulder joint,
the clockwise moment (= weight of
dumb-bell  length of the whole arm) is
greatly increased. In order to balance the
7
(Accept other reasonable answers.)
dumb-bell, the shoulder muscle has to
(a)
exert a great force to provide a sufficient
anticlockwise moment. Therefore the
man feels more tired.
9
(a) Torque = 5 N  0.5 m = 2.5 N m
(In opposite directions and acting at
different positions.)
New Senior Secondary Physics at Work
8
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
(b) Maximum force that can be applied
maximum torque
50
=
=
= 100 N
perpendicular distance 0.5
10
On the Moon:
1
By s  ut  at 2 ,
2
1  10  2
2  0   t
2 6 
(a) Let m be the maximum mass that the
system can withstand.
t = 2.4 = 6 tE = 2.45 tE
Take moment about A.
4
C
5
D
The maximum mass that the system can
6
B
withstand is 1.44 kg.
7
A
8
B
bench or add a heavy weight on the
9
D
platform of the stand.
10
D
11
B
Revision exercise 3
12
A
Multiple-choice (p. 170)
13
D
1
A
14
(HKCEE 2006 Paper II Q31)
Moment of force about O
15
(HKCEE 2007 Paper II Q6)
=Fd
16
(HKCEE 2007 Paper II Q30)
= 8 sin 45  0.4
17
(HKCEE 2007 Paper II Q27)
m  10  0.1 = 2.4  10  0.06
m = 1.44 kg
(b) Use a G-clamp to fix the stand on the
= 2.26 N m (clockwise)
2
A
Conventional (p. 173)
By F = ma,
1
1000 – 500 = 1500a
a = 0.333 m s–2
1
1
s  ut  at 2  0  (0.333 )(10) 2 = 16.7 m
2
2
3
(a) Gravitational acceleration of Mars
1
=  10
3
= 3.33 m s–2
(1A)
(b) The block dropped on Mars has a
smaller acceleration than that on Earth.
B
(1A)
On the Earth:
1
By s  ut  at 2 ,
2
1
2  0  (10 )t E 2
2
Thus, it takes more time for the block on
Mars to reach the ground.
2
(1A)
(a) (i)
normal
force
tE = 0.4
tension
T1 from
m1
M
tension
T2 from
m2
weight
New Senior Secondary Physics at Work
9
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
(2 tensions with T1 > T2)
(1A)
4
Take moment about the left trestle.
(Weight and normal force, both of
In equilibrium,
the same magnitude)
clockwise moment = anticlockwise moment
(1A)
(1M)
(ii)
normal
force
tension
T1 from
m1
500  3 = 700  1 + Y  4
Y = 200 N
(1A)
Besides,
tension
T2 from
m2
M
(1M)
net force = 0
(1M)
X + Y = 700 + 500
friction
X + 200 = 1200
X = 1000 N
weight
5
(2 tensions with T1 > T2)
(1A)
(Weight and normal force, both of
(b) (i)
the same magnitude)
(1A)
(Friction)
(1A)
(a) By F = ma,
F 30  10
a= =
= 5 m s–2
m
4
m2 accelerates upwards and M
accelerates to the left.
(1A)
(ii) Let f be the friction acting on M.
(1A)
The displacement of the box is
82.5 m.
If T1 > T2 + f, the masses will move
(c)
in a way similar to that in (b)(i) but
Any one of the following:
(1A)
Add a layer of oil / polystyrene beads
the magnitude of the acceleration
along the path of the block.
of the system will be smaller. (1A)
Use air cushion.
If T1 = T2 + f, the masses will
3
(1M)
The acceleration of the box is 5 m s–2.
1
(b) s = ut  at 2
(1M)
2
1
= 4  5 + 552 = 82.5 m
(1A)
2
Mass m1 accelerates downwards,
remain at rest.
(1A)
6
(1A)
(a)
(a) Moment about P
= Fd
(0.5A)
= 10  3
= 30 N m (clockwise)
(1A + 0.5A)
(b) Moment about Q
= 10  1
= 10 N m (clockwise)
(c)
(1A + 0.5A)
Moment about R = 10  0 = 0
(1A)
(d) Moment about S
(Weight of Joan)
= 10  1
(Reaction from the balance on Joan)
= 10 N m (anticlockwise) (1A + 0.5A)
New Senior Secondary Physics at Work
(1A)
(1A)
10
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
7
(b)
Reading of
Weight that Joan
the scale
(＜, = or
feels (heavier,
lighter or normal
＞500 N)
weight)
(a) A force of 50 N is used to pull blocks of
total mass 40 kg.
By F = ma,
(1M)
50 = (10 + 30)  a
a = 1.25 m s2
(1A)
The acceleration of the boxes is
Lift
accelerates
upwards
1.25 m s2.
> 500 N
heavier
(1A)
(1A)
= 500 N
normal weight
(b)
Lift moves
up at
constant
(Correct force)
(1A)
(Correct label)
(1A)
Joan by the balance (the reading of the
(Correct force)
(1A)
scale) and W be the weight of Joan.
(Correct label)
(1A)
speed
Lift slows
down and
stops
(c)
< 500 N
lighter
(1A)
(1A)
Let R be the normal reaction acting on
Take the upward direction as positive.
(i)
By F = ma
(c)
(1M)
For the 30-kg box,
R  W = ma
R  500 = 50  3
R = 650 N
Let T be the tension in the string.
(1A)
By F = ma,
(1M)
T = 30  1.25 = 37.5 N
(1A)
The tension in the string is 37.5 N.
The reading of the scale is 650 N.
(d) Net force = 50  T
(ii) Since acceleration is 0 and, by
= 50  37.5
F = ma, the reading of the scale
R = W = 500 N.
(iii) By F = ma,
= 12.5 N
The net force acting on the 10-kg box is
(1M)
12.5 N.
R  W = ma
(e)
R  500 = 50  (2)
R = 400 N
(1A)
(1A)
Her statement is not correct.
(1A)
When the string breaks, the net force
(1A)
acting on the 30-kg box is zero.
The reading of the scale is 400 N.
(1A)
By Newton’s first law of motion, the
box will continue to move and its
velocity will be constant.
New Senior Secondary Physics at Work
11
(1A)
 Oxford University Press 2009
2 Force and Motion
8
(a)
v u
t
2 1
=
= 20 m s–2
0.05
a
F = ma
Chapter 3 Force and Motion
gently momentarily.
(1M)
(c)
The box will slide down the plane by
either reducing the friction acting on the
box or increasing the weight component
(1M)
of the box down the plane.
= 0.5(20)
= 10 N
Any two of the following:
(1A)
(2  1A)
Add rollers on the plane.
The force acting on the stone during the
Add a layer of wax/oil on the plane.
collision is 10 N.
Tilt the plane more such that the weight
(b) Force acting on the can
component of the box along the plane is
= force acting on the stone
(c)
(1A)
= 10 N
(1A)
By F = ma,
F 10
a 
 25 m s–2
m 0.4
(1M)
v = u + at
(1M)
greater than the friction acting on it.
(Or other reasonable answers)
10
(a)
= 0 + (25)(0.05)
= 1.25 m s–1
(1A)
The velocity of the can after collision is
1.25 m s–1.
9
(a) When the box tends to move along the
plane, friction acts on it to oppose its
motion.
(Correct forces)
(1A)
(Correct labels)
(1A)
(Correct forces)
(1A)
(Correct labels)
(1A)
(1A)
Unless the net force acting on the box
down the plane is greater than zero (i.e.
when the weight component of the box
along the plane is larger than the friction
acting on it), the box will not slide down
the plane.
(1A)
(b) Samuel assumes that the plane is
friction-compensated, such that the
The reaction of m1 (R) and the force
weight component of the box along the
acting on pan A by m1 (R) form an
plane balances the friction acting on the
action-and-reaction pair.
box.
(1A)
(b) The pans and masses would move
(1A)
Therefore, the net force acting on the
up/down at constant speed
(1A)
box along the plane is zero and the box
or remain at rest.
(1A)
will move along the plane with a
uniform speed after pushing the box
New Senior Secondary Physics at Work
12
 Oxford University Press 2009
2 Force and Motion
11
Chapter 3 Force and Motion
v / m s1
(a)
t/s
(Axes with correct labels)
(Correct forces)
(1A)
(Correct labels)
(1A)
(b) The weight of Jackie is constant.
(1A)
(The speed of the pot increases at a
decreasing rate.)
(1A)
13
The air resistance acting on her increases
(1A)
(a) (i)
normal
reaction
gradually from zero as her velocity
increases. When the air resistance is
friction
equal to her weight, the net force acting
on her becomes zero.
(c)
(1A)
Jackie will fall at a constant speed. (1A)
weight
When the air resistance is equal to her
weight, the net force acting on her is
(Weight)
(1A)
zero.
(Normal reaction)
(1A)
(Friction)
(1A)
(1A)
By Newton’s second law, she will fall at
a constant speed.
12
(a) I do not agree with Gloria.
(1A)
(ii) Take the direction down the plane
(1A)
as positive.
The air resistance acting on the flower
1 2
at ,
2
1
2 = 0 +  a  42
2
a = 0.25 m s–2
By s = ut +
pot increases from zero as the pot falls in
air.
(1A)
Since the maximum magnitude of air
resistance acting on the pot is equal to
the weight of the flower pot,
(1M)
(1A)
The acceleration of the trolley is
(1A)
0.25 m s–2.
the downward net force acting on the pot
is always greater than or equal to zero.
(iii) F = ma
By F = ma, the pot will not slow down.
(1M)
= 1  0.25 = 0.25 N
(1A)
The resultant force acting on the
(b) Take the downward direction as
trolley is 0.25 N (down the plane).
positive.
(1A)
New Senior Secondary Physics at Work
13
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
(b) In order to allow the trolley to move
(b)
down the runway at uniform velocity,
we should make the runway
friction-compensated, i.e. reduce the size
(c)
of the angle .
(1A)
The student is wrong.
(1A)
normal reaction
(Axes with correct labels)
(1A)
(The graph decreases linearly from A to
B.)
friction
weight
(The graph is horizontal between B and
When the trolley moves up along the
C.)
runway, friction on the trolley acts
force acting on the trolley is not zero.
15
(1A)
Instead of moving at a uniform speed,
and C.)
vu
(a) a =
t
80  0
=
= 2 m s–2
40
By v 2  u 2  2as ,
the trolley decelerates as it moves up
14
(1A)
(The graph is on the x-axis between B
downwards along the runway and the net
along the runway.
(1A)
(1M)
(1M)
802  02 = 2  2  s
(1A)
s = 1600 m
(a) When the food parcel is thrown from the
(1A)
The minimum length of the runway is
plane, it accelerates at first. As it gains
1600 m.
speed, the air resistance acting on it
(b) Net force acting on the aeroplane
increases. The net force acting on the
= ma
(1M)
food parcel and thus the acceleration
= 2.5  105  2
decreases (from point A to point B).
= 5  105 N
(1A)
(c)
air resistance and the weight of the
the weight of the food parcel. The net
aeroplane.
16
force acting on the food parcel and thus
acceleration of 6 m s–2.
point B to point C).
4–8 s: The object moves with zero
As a result, the food parcel moves with a
acceleration.
constant speed called terminal speed
8–12 s: The object moves with an
(50 m s–1).
acceleration of –6 m s–2.
(1A)
14
(1A)
(a) 0–4 s: The object moves with an
the acceleration becomes zero (from
(1A)
(1A)
I would adjust the thrust to balance the
Eventually, the air resistance balances
New Senior Secondary Physics at Work
(1A)
(1A)
(1A)
(1A)
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
(b) During 0–4 s:
(b) Net force acting on the block
F = ma = 2  6 = 12 N
= 12  5 2 (Pythagoras’ theorem)
(1A)
The force acting on the object is 12 N.
= 5.10 N
5
tan  =
1
During 4–8 s:
F = ma = 2  0 = 0
(1A)
 = 78.7
The force acting on the object is 0.
(1A)
The net force is 5.10 N (S 78.7 E).
During 8–12 s:
F = ma = 2  (–6) = –12 N
(c)
(1A)
The force acting on the object is –12 N.
17
(1A)
(a)
By F = ma,
5.10
a=
= 2.04 m s–2
2 .5
(1M)
(1A)
The acceleration of the block is
2.04 m s–2.
19
(a)
F
F
weight
(Axes with correct labels)
(1A)
(Correct shape)
(1A)
(Forces F normal to the wings)
(1A)
(Weight)
(1A)
(b)
F 
(Correct slopes : during 09 s,
slope = 3 m s–2; then slope = 0; final part
F 
steeper than the first part with negative
(3  1A)
slope)

(v = 0 at the starting point and the end
point)
weight
(1A)
–1
(v = 27 m s at t = 9 s)
Consider the forces in the vertical
(1A)
direction.
(b) The magnitude of the maximum
2F  cos  = mg
acceleration of the train is 4 m s–2. (1A)
18
(a) (i)
The aeroplane does not fly with uniform
Net force along vertical direction
velocity.
=3N–2N
= 1 N (downwards)
on the aeroplane towards the left. By
F = ma, the aeroplane has an
direction
acceleration.
= 10 N – 5 N
New Senior Secondary Physics at Work
(1A)
This is because a net force, 2F sin , acts
(1A)
(ii) Net force along horizontal
= 5 N (towards the right)
(1A)
(1A)
(1A)
15
 Oxford University Press 2009
2 Force and Motion
20
Chapter 3 Force and Motion
(a)
21
(a)
(Weight of passenger)
(1A)
(2 tensions)
(1A)
(Reaction from the platform to the
(Weight)
(1A)
passenger)
(1A)
(b) Take the upward direction as positive.
(b) The net force acting on the picture is
(i)
zero.
Total distance during initial rise
= 50 – 2 = 48 m
Consider the vertical components.
2T cos  = mg
Total time = 24 s
(1M)
Average speed =
 80  
2T  cos 
 = 1  10
 2 
T = 6.53 N
48
24
= 2 m s–1
(1M)
(1A)
The average speed of the platform
(1A)
when it rises from the ground to
The tension in the string is 6.53 N.
the top of the tower is 2 m s–1.
(c)
(ii)
Total distance during the first
downward thrust = 50 – 9 = 41 m
Total time = 43 – 39 = 4 s
41
Average speed =
(1M)
4
= 10.25 m s–1(1A)
The average speed of the platform
during the first downward thrust
is 10.25 m s–1.
(iii)
If a longer string is used,  will be
smaller.
Since T =
(1A)
mg
, T decreases with .
2 cos 
(1A)
Therefore, the tension in a longer string
is smaller and it is harder for the string
to break.
New Senior Secondary Physics at Work
(1A)
16
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
Let P be the pulling force.
23
(a) The net force acting on the case is 0.
By F = ma,
(1A)
P + mg = ma
(1M)
(b) Let T be the tension.
4T cos 20 = 225  10
P = ma – mg
= m(1.5g) – mg
T = 599 N
= 70 (–15) – 70(–10)
= –350 N
22
(1A)
The tension in each string is 599 N.
(1A)
(c)
It is safer to hang the case with a longer
During the first downward thrust,
string,
the pulling force acting on the
because the angle between the string and
passenger by the chain is 350 N.
the vertical will be smaller.
(a) By F = ma,
(1M)
a=6ms
(1A)
(1A)
Therefore, the tension in the string is
8000 – 5000 = 500a
smaller and it is harder for the strings to
–2
(1A)
break.
The acceleration of the balloon is
24
–2
6ms .
v u
By a 
,
t
v  u 20  0
t

 3.33 s
a
6
(1A)
(a) Take moment about the left trestle.
In equilibrium,
clockwise = anticlockw ise (1M)
moment
moment
(1M)
600  1 + 200  2 = Y  4
(1A)
Y = 250 N
The balloon reaches a velocity of
20 m s in 3.33 s.
net force = 0
(b) He feels his weight heavier than
expected.
(1M)
(1A)
Besides,
–1
(1M)
X + Y = 600 + 200
(1A)
X + 250 = 800
The upward net force acting on him is
X = 550 N
R – W = ma > 0, where R is the normal
(b) (i)
reaction and W is his weight. He feels
(ii)
By Newton’s first law, the sandbag
(1A)
When the plank begins to tip, the
reaction Y is zero.
heavier because R is greater than W.(1A)
(c)
(1M)
(1A)
Let d be the distance the painter is
away from the left trestle when
–1
moves up at 20 m s when it leaves the
the plank begins to tip.
balloon.
From (i), we have Y = 0.
(1A)
Then it slows down due to gravity. (1A)
When the plank just begins to tip,
After reaching the maximum height, it
the conditions of equilibrium still
changes its moving direction and
apply.
accelerates downwards.
(1A)
Take moment about the left
As its velocity increases, the air
trestle.
clockwise = anticlockw ise (1M)
moment
moment
resistance increases. As a result, the
acceleration of the sandbag decreases.
200  2 = 600  d
(1A)
New Senior Secondary Physics at Work
d = 0.667 m
17
(1A)
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
When the plank begins to tip, the
25
27
(a) The ball bearing accelerates at first. As it
painter is 0.667 m away from the
gains speed, the fluid friction acting on it
left trestle.
increases. The net force acting on the
Note that the centre of gravity of the can is
ball bearing and thus the acceleration
approximately at its centre. The can will
decreases.
topple when its centre of gravity is outside the
Eventually, the fluid friction increases to
edge of the runway.
a value that balances the weight of the
Maximum distance travelled by the can
ball bearing. The net force acting on the
(1A)
= 3  0.035 = 2.965 m
(1M)
ball bearing and thus the acceleration
By F = ma,
(1M)
becomes zero.
0.2 = 0.5  a
a = 0.4 m s
Then the ball-bearing moves with a
2
constant speed called terminal speed.
By v  u = 2as,
2
2
(1M)
(1A)
0  u = 2  (0.4)  2.965
2
(1A)
2
1
u = 1.54 m s
(b)
(1A)
The maximum velocity of the can just after
the impact is 1.54 m s1.
26
(a) The trolley remains at rest until
t = 0.8 s.
(1A)
Then it moves with a uniform
acceleration.
vu
(b) a =
t
1.15  0
=
= 0.575 m s–2
2 .8  0 . 8
(1A)
(1M)
(1A)
(Axes with correct labels.)
(The velocity firstly increases with time
The acceleration of the trolley is
linearly.)
0.575 m s–2.
(c)
F = ma
= (1)(0.575) = 0.575 N
(1A)
(1A)
(1M)
(Then the slope of the curve decreases
(1A)
continuously.)
(1A)
The net force acting on the trolley is
(Finally, the velocity becomes constant
0.575 N.
and the slope of curve becomes zero.)
(d) He is incorrect.
(1A)
(1A)
This is because he ignores the friction of
(c)
the runway. The spring balance reading
speed in air, it experiences a great air
is equal to the pulling force only. The
resistance which opposes its motion of it.
net force is equal to the pulling force
minus the friction.
New Senior Secondary Physics at Work
When an aeroplane travels at a high
(1A)
(1A)
18
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
Air resistance increases with the speed
The discrepancy may be due to the
of the moving object. Therefore, it is
friction acting on the trolley; the friction
difficult for an aeroplane to travel at the
acting on the trolley may not be
speed of sound in air.
negligible.
(1A)
(1A)
The student must use a
In order to improve the speed of an
friction-compensated runway to carry
aeroplane, the body of the aeroplane
out this experiment.
should be streamlined so that air can
29
(1A)
(a) (i)
flow smoothly over its surface and air
resistance can be reduced.
28
(1A)
(a) In Figure v, from t = 0.1 s to t = 0.4 s,
the average value of tension is 2.44 N.
Tension T equals to the weight of the
weights.
Let m be the mass of the weights.
T = mg
(1M)
2.44 = m  10
m = 0.244 kg
(1A)
(b) (i)
As fans B and C blow air
backwards, an action force acts on
equal to the slope of the graph in
the air by the fans.
Figure w.
(1A)
Thus, an equal and opposite
The acceleration is 1.24 m s2.(1A)
reaction force acts on the fans by
According to Newton’s second law
the air.
(1A)
(1A)
Therefore, the boat moves forwards.
the tension pulling the force sensor and
(1A)
the trolley T = ma.
(ii) By F = ma,
T = (0.333 + 0.718)  1.24 = 1.30 N(1A)
0.2 + 0.2 = 1a
This theoretical result is not close to the
a = 0.4 m s–2
result in (b)(i). The results are not in
(1M)
v = u + at
accordance with Newton’s second law.
(1M)
–1
= 0 + (0.4)(5) = 2 m s
(1A)
New Senior Secondary Physics at Work
(1A)
(1A)
(1A)
(ii) The acceleration of the trolley is
(F = ma),
(1A)
towards the direction it is pushed.
of the string is 1.93 N
(c)
(Force on boat by air)
velocity
By the data in Figure v, the tension
(t = 1.1 s to 1.5 s).
(1A)
(ii) The boat moves with a constant
The mass of the weights is 0.244 kg.
(b) (i)
(Weight)
(1A)
–1
Its speed is 2 m s .
19
 Oxford University Press 2009
2 Force and Motion
(iii) Any one of the following:
(c)
(i)
Chapter 3 Force and Motion
(1A)
(ii) The reading on the balance is not
Switch off fan C.
zero.
Control fan B to blow air
The fan blows the air downwards
backwards and control fan C to
and hits the balance.
blow air forwards.
The air therefore exerts a force on
The boat still moves forwards (1A)
the balance.
with a constant velocity.
(ii) Any one of the following:
(1A)
31
(1A)
(1A)
(a) Take the moving direction of the trucks
(1A)
as positive.
Switch off fan A, then the boat will
stopped by the friction.
By F = ma,
(1M)
6000
F
a= =
= 0.706 m s–2
m 5.5  10 3  3000
Control both fans B and C to blow
By v2 – u2 = 2as,
land on the ground and will be
(1M)
0 – u = 2(–0.706)(30)
2
air forwards.
30
(1A)
u = 6.51 m s–1
(a)
(1A)
The speed of the trucks after the
collision is 6.51 m s–1.
vu
(b) a =
t
6.51  0
=
= 130.2 m s–2
0.05
(1M)
Net force = ma = (5.5  103)(130.2)
= 716 000 N
(1M)
Force acting on truck P by truck Q
(Weight)
(1A)
= net force  friction
(Force on toy by air)
(1A)
= 716 000  (3000)
= 719 000 N
(b) The powerful fan of the toy blows air
(c)
downwards. Therefore, an action force
acts on the air by fan.
= force acting on truck P by truck Q (but
(1A)
in opposite direction)
force acts on the fan of the toy by the air.
= –719 000 N
= force acting on truck Q by truck P +
Such upward reaction force is larger
friction
than the weight of the toy. Therefore, the
(i)
(1A)
(d) Net force acting on truck Q
(1A)
(c)
Force acting on truck Q by truck P
Then, an equal and opposite reaction
toy can go up in mid-air.
(1A)
= –719 000 + (3000)
(1A)
Minimum upward force
= –722 000 N
(1M)
= weight of the toy
50
=
 10
1000
By F = ma,
(1M)
= 0.5 N
New Senior Secondary Physics at Work
–722 000 = 3000a
a = –240.7 m s–2
(1A)
20
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
By v = u + at,
(b) (i)
u = v  at = 6.51  (240.7)  0.05
= 18.5 m s–1
(1A)
The speed of truck Q before the collision
is 18.5 m s–1.
32
(a) (i)
(Weight)
(1A)
(Normal reaction)
(1A)
(ii)
(Force by the square object and the
masses)
(Weight acting at A)
(1A)
(Normal force and weight of the
(Friction)
(1A)
plank)
(Normal reaction)
(1A)
(1A)
(iii) The force acting on the plank by
the square object and the normal
(ii) The weight and the normal reaction
provide an anticlockwise moment
force acting on the square object by
about X.
the plank.
(1A)
moment is zero.
(1A)
(iv) (1) F should be the force acting on
On the other hand, the clockwise
the plank by the square object.
(1A)
(1A)
The anticlockwise moment is
larger than the clockwise moment.
From (iii), we know that the
(1A)
force acting on the plank by
Therefore, the box would topple
the square object and the
about X.
normal force acting on the
(1A)
square object by the plank are
(b) In this case, the weight of the box acts at
B,
an action-and-reaction pair, so
(1A)
they are equal in magnitude.
and provides a clockwise moment about
X,
33
(1A)
(1A)
(1A)
which balances the anticlockwise
Since the square object is in
moment provided by the normal reaction
equilibrium, by Newton’s first
and keeps the box in equilibrium. (1A)
law, the net force acting on it
is zero. Therefore, the normal
(a) He needs to make sure that x is the
horizontal distance between O and the
reaction acting on the square
centre of gravity of the object,
object is equal to the weight of
(1A)
the square object in magnitude.
because the weight acts at the centre of
gravity of the object.
New Senior Secondary Physics at Work
(1A)
(1A)
21
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
Therefore, F is equal to the
40  x = 60  (1  x)
(1M)
x = 0.6 m
(1A)
weight of the square object in
magnitude, which is 10 N.(1A)
(iii) (1) The centre of gravity is the
(2) Torque
point on which the (entire)
=Fd
weight of the body seems to
= 10  1
act.
(1A)
(Correct arrow)
(1A)
= 10 N m
(v)
(1A)
(2)
The result on the torque is not the
same.
(1A)
The square object is accelerating,
so the normal reaction acting on
the square object is not equal to
the weight of the square object
(10 N) in magnitude.
(1A)
36
(a)
Therefore, F is different and so is
the torque ( = F  d).
34
(HKCEE 2006 Paper I Q4)
35
(a)
(1A)
F
a=
m
3 .0
=
0.80
= 3.75 m s2
(Moment =) F  perpendicular distance
from pivot
(b) (i)
(b) (i)
(1A+1A)
1.5 m
(1M)
(1A)
Air resistance increases with speed,
(1A)
(1A)
so the net force (= forward thrust 
Assumption: uniform/regular beam.
air resistance) decreases, and the
(1A)
acceleration decreases.
(ii) (1)
(1A)
(ii) When the air resistance is equal to
the forward thrust in magnitude,
(1A)
the net force and hence the
acceleration is zero, so the car
reaches a constant speed.
(Pivot between the cube of
(c)
weight and the centre of the
beam)
(i)
The velocity of the car decreases at
a decreasing rate.
(1A)
(1A)
(1A)
The car eventually travels at a
(2) Let x be the distance between
constant velocity.
the new position of the pivot
(1A)
(ii) When the parachute is opened, the
and the original position.
Take moment about the joint.
air resistance becomes greater than
In equilibrium, clockwise
the thrust in magnitude. As a result,
moment is equal to the
the velocity decreases.
(1A)
anticlockwise moment. (1M)
New Senior Secondary Physics at Work
22
 Oxford University Press 2009
2 Force and Motion
Chapter 3 Force and Motion
Air resistance decreases with speed,
(b) (i)
and is finally equal to the thrust in
magnitude. Therefore the car
travels at constant velocity.
37
(a)
No net force.
(b) (i)
w1x1 = wx + w2x2
(ii) R = w + w1 + w2
(c)
(1A)
1 2
at ,
2
1
4 = 0  10  t2
2
By s = ut 
t = 0.894 s
(1A)
(1M)
(1A)
The man takes 0.894 s to fall from the
(1A)
first floor to the ground without the air
(1A)
bag.
(i)
(ii) 0.5 s + 0.1 s = 0.6 s
(1A)
The time interval between the man
jumps and the bag is fully inflated is
0.6 s.
(iii) From (i) and (ii), the time for air
resistance acting on the man is very
short and the velocity of the man cannot
be reduced to a small value.
(1A)
In this case, the man is mainly protected
(Downward force.)
(1A)
by the thick special cushion of the air
(Upward force)
(1A)
bag when he reaches the ground.
(1A)
(Correct distance of all downward
forces)
(1A)
(ii) Take moment about the pivot.
In equilibrium,
w  0.2 = 80  1 + 20  2
w = 600 N
(1M)
(1A)
(Correct perpendicular distances of
the forces from the pivot.)
(iii) R = 600 + 80 + 20 = 700 N
(1M)
(1A)
Physics in articles (p. 183)
(a) When the air bag inflates, the air resistance
acting on the air bag and the person increases.
(1A)
When the air resistance is greater than the
weight of the person, the person will slow
down (F = ma).
(1A)
It is easier for a person to land with a lower
falling speed.
New Senior Secondary Physics at Work
(1A)
23
 Oxford University Press 2009