Active Physics Full Solutions to Textbook Exercises 06 Force | p.

06 Force Checkpoint 2 (p.84)

Checkpoint
1. B
Since Fiona is lifting her son at a steady speed, by
Checkpoint 1 (p.78)
Newton’s first law, no net force is acting on him. So
1. (a) F her applied force is equal to his weight.
Force is a vector.
2.
(b) F
An object can exert a force on another object
without touching it. This kind of force is
called non-contact force. Weight and
magnetic force are non-contact forces. 3. (a) F
(c) F An object in uniform motion requires no
Consider an apple lying on your hand. The force acting on it.
apple exerts a downward force on your hand (b) F
even though it is stationary. Even though an object is acted on by several
(d) F forces, it may remain at rest (i.e. have no
Both forces have effect on the object, because motion) if the forces balance each other out.
the motion of the object is determined by the (c) T
net force (i.e. the vector sum of all the forces)
(d) F
acting on it.
An object moving upwards at a constant
2. (a) No. speed (i.e. in uniform motion) has no net
The force no longer acts on the ball after it force acting on it.
leaves the pitcher’s hand.
(e) T
(b) The weight of the ball (pointing downwards)
(f) T
3. (a) 1 N.
The weight of the apple (i.e. 1 N) is balanced
by the tension in the string attached to it. Checkpoint 3 (p.87)
The balance measures the tension and 1. (a) F
therefore reads 1 N. Inertia is the tendency of an object to resist
(b) Both are 1 N. any change in its state of motion.
As in (a), the weight of the apple is balanced (b) F
by the tension in the string attached to it. Two stationary objects of different mass
The lower balance measures the tension and have different amount of inertia.
therefore reads 1 N.
(c) T
The upper balance measures the tension in
the spring of the lower balance, and so it also (d) F
reads 1 N. The passengers tend to lean backwards in
the car because the inertia of their bodies
(c) Both are 0.5 N.
tends to resist the change in their motion (i.e.
In this case, the weight of the apple is
speeding up from rest).
balanced by the resultant force of the
tensions in the strings attached to it. Hence, 2. An elephant has a larger mass, and so it is more
the tension in each string is only 0.5 N, which difficult to change its motion.
is shown on each balance. 3. (a) When the magician pulls out the tablecloth,
4. the dishes are moved by the friction from the
tablecloth. But on the other hand, the dishes
tend to stay at rest due to their inertia. So, if
the pulling speed is high enough to shorten
the time at which the friction acts on the
dishes, the dishes would only move slightly.
(b) Heavier.
The heavier the dishes, the larger amount
Active Physics Full Solutions to Textbook Exercises 06 Force | p.2

their inertia, and hence the larger the The acceleration of the lift is 1.19 m s−2
tendency they would remain at rest. (upwards).
(c) This is to reduce the time at which the
friction acts on the dishes. Checkpoint 6 (p.108)
1. Table:
Checkpoint 4 (p.94)
pulling force friction motion of the
1. (a) F (magnitude) block
An object experiencing a constant net force 10 N 10 N at rest
must be undergoing acceleration, and hence 20 N 20 N at rest
its velocity must be changing. 40 N 30 N accelerating
(b) F 80 N 30 N accelerating
The fact that an object experiences a larger
net force only implies it has a larger 2. B
acceleration (i.e. a larger change in velocity), The limiting friction between the block and the
but not necessarily a larger velocity. For surface is 2 N. As the block keeps moving forward
example, a car moving at a very high velocity after the pulling force is removed, it still
experiences a zero net force when its experiences a friction of 2 N.
velocity remains unchanged.
3. A
(c) T At the instant an object is released from rest, its
2. A acceleration is equal to the acceleration due to
By Newton’s second law, we have gravity. As the object speeds up, the air resistance
Fnet = ma ⇒ 1 = m(2) ⇒ m = 0.5 kg acting on it increases, reducing its acceleration
gradually. Eventually, the object falls at a constant
3. By Newton’s second law,
speed, and its acceleration becomes zero.
Fnet = ma
0.5 − f =(0.2)(2) 4. B
f = 0.1 N The net force acting on the feather is zero, and
hence the air resistance is equal to its weight.
Hence, the resistive force f has a magnitude of
0.1 N.
Checkpoint 7 (p.115)

Checkpoint 5 (p.101) 1. C
By Newton’s third law, the force acting on Slimy
1. (a) The weight of the object W' on planet B is
Amy by Fat Bob is equal (in magnitude) to the
given by
force acting on Fat Bob by Slimy Amy, regardless of
W'/W = g'/g
the difference in their mass.
∴ W' = 1/2 W = 0.5 × 100 = 50 N
2. B
(b) The mass of the object is
By Newton’s third law, the force acting on the
m = W/g = 100/10 = 10 kg
expert’s hand by the brick is equal to the force
(c) The acceleration of the object is acting on the brick by the expert’s hand.
a' = 1/2 × 10 = 5 m s−2 (downwards)
3. A
2. Table: The gorilla does not experience any external net
motion of the direction of the how Bob feels force.
lift acceleration his weight
4. (a) No.
speeding up upwards gain in weight Although the forces are of equal magnitude
moving at a zero his actual weight and opposite directions, they act on the same
constant speed
object (i.e. the carrot).
slowing down downwards loss in weight
(b) Yes.
3. Take the upward direction as positive.
Note that the weight of the carrot is the
By Newton’s second law, we get
gravitational force exerted on the carrot by
R − mg = ma
the Earth.
550 − (50)(9.81) = (50)a
a =1.19 m s−2 (c) Yes.
Active Physics Full Solutions to Textbook Exercises 06 Force | p.3

5. (a) Free body diagram of the beam:

Exercise

Exercise 6.1 (p.78)

1. (a) Net force = 4 N
(b) Net force = −3 N
(c) Net force = 13 N
2. (a) Free body diagram of the person: (b) Given that the net force acting on the beam is
zero. Hence,
Fnet = W − (90 + 90) = 0 ⇒ W = 180 N
The weight of the beam is 180 N
downwards.
6. Free body diagrams of bottles A and B:

(b) Free body diagram of the book:

7. (a) Free body diagram of the wooden block:

(c) Free body diagram of the T-shirt:

The net force acting on the block is zero.

Hence, R = W = 20 N.
The force acting on the block by the tank is
(d) Free body diagram of the box:
20 N upwards.
(b) Free body diagram of the floating wooden
block:

3. A
At this moment, the ball has no contact with any
object (e.g. the player’s foot or the ground). Hence, Take the upward direction as positive.
the only force acting on it is its weight. Fnet = F − W ⇒ 10 = F − 20
∴ F =30 N
4. (a) Free body diagram of the box:
Hence, the force exerted on the block by the
water is 30 N upwards.

Exercise 6.2 (p.88)

1. C
Net force = 0 ⇒ acceleration = 0.
(b) Take the direction to the right as positive. 2. A
The net force acting on the box is Since the box moves at a constant speed, no net
Fnet = 250 + 200 − 400 = 50 N to the right. force acts on it. Hence, the pushing force applied
by the man balances the friction from the ground.
Since the applied force acts horizontally, we need
not consider the weight, which acts vertically.
Active Physics Full Solutions to Textbook Exercises 06 Force | p.4

3. (a) The spacecraft will continue to move at a 8. (a) No.

constant velocity because it tends to The string on the left tends keep the block in
maintain its state of motion due to inertia. its place. To do so, the tension in the string is
(b) The spacecraft will move at a constant 10 N. This is within the limit of the block, and
velocity. After the fuel runs out, the net force therefore it would not be torn apart.
on the spacecraft is zero. Because of inertia, (b) Yes.
the spacecraft tends to maintain its state of In this case, the tension in the string on the
motion just before the fuel runs out. left has to be 20 N in order to keep the block
4. Both the cotton and the iron nails have the same in its place. This exceeds the limit of the clay
mass and hence the same amount of inertia. and it would be torn apart.
Therefore, they are equally difficult to move.
5. (a) Since the trolley moves at a constant velocity, Exercise 6.3 (p.94)
the friction f from the ground must be 1. (a) Under negligible resistive force, the
balanced by the pushing force F exerted by acceleration of the car is given by
Amy. Thus, the pushing force is F = f = 20 N
a=
forwards.
Therefore, we have
(b) If the pushing force is larger than friction,
there will be a net forward force acting on driving force F / N acceleration a / m s−2
the trolley that makes it accelerate forwards. 5000 5
6. (a) (i) No net force is acting on the brothers if 10000 10
they are sliding on a frictionless 15000 15
surface. According to Newton’s first law, (b) Now the car experiences a constant resistive
they are moving at constant velocities. force of 1000 N. The acceleration is given by
(ii) A backward net force is acting on the a= =
brothers because of the friction exerted Therefore, we have
on them by the ice surface. They are
slowing down. driving force F / N acceleration a / m s−2
5000 4
(b) Anthony will find it more difficult to slow
10000 9
down. Because he has greater mass, i.e.
15000 14
larger inertia, it is more difficult for him to
change the motion of his body.
2. C
7. (a) Free body diagram of the magnet A:
Note that the net force acting on the block and the
acceleration of the block always point to the same
direction.
3. B
By Newton’s second law (i.e. Fnet = ma), if both the
net force Fnet and the mass m are doubled, then the
acceleration a remains unchanged.
Take the downward direction as positive.
Since magnet A remains stationary, the net 4. B
force acting on it is zero. Hence, we have By Newton’s second law, a ∝ 1/m for a constant
W + F − T = 0 ⇒ 10 + F − 15 = 0 net force on the trolley.
∴ F =5 N 5. (a) (i) Horizontal forces on the skier:
The magnetic force acting on magnet A is 5 N
downwards.
(b) Since magnet A remains stationary, the net
force acting on it is zero. Hence, we have
10 + F − 5 = 0 ⇒ F = −5 N. Both the boat and the skier move with
The magnetic force acting on magnet A is 5 N a constant velocity, so the net force
upwards. acting on the skier is zero. The tension
in the cord balances the resistive force.
Hence, the tension is 40 N forward.
Active Physics Full Solutions to Textbook Exercises 06 Force | p.5

(ii) Take the direction to the right as positive. • By definition, W = mg, where g is the
By Newton’s second law, we have acceleration due to the gravity of the specified
T − f = ma ⇒ T − 40 = (60)(0.8) planet.
∴ T =88 N
2. B
So the tension in the cord is 88 N forward.
Since the Kenny’s actual weight W is greater than
(b) No. the scale reading R, the acceleration of the lift
Once the cord breaks, the skier would points downwards.
experience a backward net force while still
3. D
moving forwards. He would slow down
Since the box moves with a constant speed, we can
uniformly and come to a stop eventually.
deduce that the pulling force balances the friction
After that, the resistive force would no
from the surface, i.e. the force arrows have the
longer act on him, and so he would not move
same length.
backwards.
4. D
6. Take the direction to the right as positive. By
At the instant just after the skydiver opens his
Newton’s second law, we have
parachute, the air resistance acting on him
F − 900 − 1100 = (1500)(0.2) ⇒ F = 2300 N
increases greatly and becomes much greater than
So the driving force acting on the drive is 2300 N
his weight (which is a fixed value).
to the right.
5. Free body diagram:
7. Take the forward direction as positive. By
Newton’s second law, we have
a= = = -4 m s-2

By v2 − u2 = 2as, we have
02 − 202 = 2(−4)s ⇒ s = 50 m
It has moved for 50 m after the brake is applied. Take the upward direction as positive. By
Newton’s second law, we have
8. (a) The acceleration is the slope of the v–t graph,
F − mg = ma
and hence we have
(8 × 106) − (5.5 × 105)(9.81) = (5.5 ×105 )a
a = (0.8 − 0) / (4 − 0) = 0.2 m s−2
∴ a ≈ 4.74 m s−2
(b) By Newton’s second law, we have So the initial acceleration of the rocket is
F = ma = (0.6)(0.2) = 0.12 N 4.74 m s−2 upwards.
(c) Since a ∝ 1/m, if m is doubled, then a is 6. (a)
halved. Hence, the v–t graph becomes:

When the balance remains stationary, the net

force acting on the book is zero (i.e. the
forces acting on the book are balanced).
Hence, we have
R = mg
Exercise 6.4 (p.108)
⇒ 10 = m(9.81) ⇒ m = 1.019 ≈ 1.02 kg
1. (a) 0.6; 6; 1 The mass of the book is 1.02 kg.
(b) 180; 1800; 300 (b) (i) Take the upward direction as positive.
(c) 1200; 12000; 2000 By Newton’s second law, we have
R − mg = ma
 Note:
Rearranging the equation gives
• The mass of an object is independent of where it R = ma + mg = (1.019)(2) + 10 ≈ 12.0 N
is placed. So the reading is 12.0 N.
Active Physics Full Solutions to Textbook Exercises 06 Force | p.6

(ii) Given that the acceleration a =−5 m s−2. deduce that the acceleration of the lift points
So the reading is R = ma + mg = upwards.
(1.019)(−5) + 10 ≈ 4.90 N.
During t = 2 to 8 s, Peter’s apparent weight
(iii) Since the set-up falls at a constant equals his actual weight (i.e. R = W), so we
velocity, the net force acting on the deduce that the acceleration of the lift is 0.
book is zero. The reading is the same as
(b) By Newton’s second law, we have
in (a), i.e. 10.0 N.
R − W = ma ⇒ R − 981=(100)(−0.5)
(iv) When the set-up falls freely, both the ∴ R =931 N
balance and the book have the same Hence, the R–t graph of Peter becomes:
acceleration, i.e. a = −9.81 m s−2. So the
reading is
R = ma + mg
= (1.019)(−9.81 + 9.81) = 0
7. (a) Since the block remains at rest, the friction
acting on the block balances the horizontal
force F. Hence, the friction is 10 N to the left.
(b) No. 10. (a) The maximum friction is 1 N.
The block would remain at rest as long as the (b) (i) No, the block will not move. In this case,
applied force F is lower than the limiting the friction on the block is 0.5 N, which
friction. Hence, we can only deduce that the balances the pulling force completely.
limiting friction is larger than 10 N.
(ii) Yes, the block will accelerate and move.
8. (a) From the graph, we know that the terminal In this case, the friction on the block is
speed of the feather is 0.2 m s−1. 1 N, which is the maximum friction
(b) between the block and the surface.
(c) Add some lubricants between the block and
the ground.
11. (a) From the graph, we know that the fluid
resistance is 0.06 N.
By Newton’s second law, we have (b) When the bead reaches terminal speed, the
W − f = ma fluid resistance acting on it is balanced by its
where W is the weight and f is the air weight. Hence, we have W = 0.06 N.
resistance. The acceleration a is the rate of Free body diagram of the bead:
change of the velocity, which is represented
by the slope of the v–t graph.
At the beginning, the air resistance f is very
small, so the feather accelerates at a rate
close to g. As its speed increases, the air
resistance increases. As a result, the Take the downward direction as positive.
acceleration drops gradually and so does the The net force acting on the bead is
rate of increase of the velocity. Fnet = W − F = 0.06 − F
F–t graph of the bead:
9. (a) Take the upward direction as positive. By
W = mg, the mass of Peter is
m= = = 100 kg

During t = 0 to 2 s, by Newton’s second law,

we have
R − W = ma ⇒ 1071 − 981 = 100a
∴ a = 0.9 m s−2
The acceleration of the lift is 0.9 m s−2
upwards.
 When Peter’s apparent weight is greater
than his actual weight (i.e. R > W ), we can
Active Physics Full Solutions to Textbook Exercises 06 Force | p.7

(c) The weight of the bead is 0.06 N. By W = mg, 2. D

the mass of the bead is The force acting on A by B and that on B by A have
the same magnitude but are in opposite directions.
m= ≈ 0.00612 kg
3. A
At time t = 1 s, the magnitude of the fluid The force acting on Katie by Tom is equal to the
resistance is 0.04 N. By Newton’s second law, force acting on Tom by Katie. So the acceleration of
we have Tom is given by
W − F = ma ⇒ 0.06 − 0.04 = 0.00612a m1a1 = m2a2 ⇒ (60)a1 = (50)(6)
∴ a ≈3.27 m s−2 ⇒ a1 = 5 m s−2
The acceleration of the bead is 3.27 m s−2
downward.
4. C
12. (a) Since the block initially moves with a
constant velocity, it implies that a 8 N friction 5. (a) When a runner exerts a backward pushing
(pointing towards the left) acts on the block. force against a starting block with his foot,
We can also deduce that the block moves to the reaction force exerted by the block acts
the right. on him to push him forwards. In this way, the
Just after the 10 N force is removed, the starting block can help a runner accelerate.
friction continues to oppose the rightward (b) When the boxer’s fist hits the punch bag, the
motion of the block, and the net force on the fist exerts a force on the punch bag. At the
block is 2+8 = 10 N to the left. Thus the same time, the punch bag exerts a reaction
block’s acceleration is 10 m s−2 to the left. force on the fist which gives the boxer the
(b) No. feel of pain.
As the 2 N force is smaller than the limiting 6. (a) These two forces form an action–reaction
friction between the block and the ground pair, and so they have the same magnitudes
(8 N), the block will remain at rest as the 2 N but are in opposite directions.
will be balanced by the friction from the
(b) The mass of the bowling ball is much larger
ground.
than that of the pin. Although the pin and the
bowling ball experience forces of the same
Exercise 6.5 (p.116) magnitude, the pin has a much larger
acceleration. Hence, the pin bounces off but
1. (a) the table; the book
the motion of the ball is almost unaffected.
7.

(b) the swimming platform; the girl Take the direction to the right as positive.
Considering the forces acting on B only, by
Newton’s second law, we have
6 − FB by A = (2)(2) ⇒ FB by A = 2 N
The force acting on B by A is 2 N to the left.
Since the force acting on B by A and the force
acting on A by B form an action–reaction pair, the
(c) the box; the string force acting on A by B is 2 N to the right.

Chapter Exercise

Multiple-choice Questions (p.120)

(d) the wall; the basketball 1. C
Only (1) and (3) could be the motion of the train.
The football in the train moves due to inertia.
Since the ball accelerates to the right in the train, it
implies that the train accelerates to the left.
Active Physics Full Solutions to Textbook Exercises 06 Force | p.8

2. B Peter is at rest, and so we have W = R + F'. Since F

The displacement of the hammer is and F' form an action–reaction pair and F = F', we
have W = R + F.
9. A
If the acceleration a decreases, then the time t
Since object A pushes object B forwards as it
increases. By W = mg, when g decreases, the force
moves, both objects move at the same velocity and
of gravity (i.e. W ) experienced by the hammer
hence the same acceleration.
decreases.
10. D
3. C
Before the string breaks, the ball moves up with a
Only (1) and (3) will cause the string to break.
uniform speed. So the corresponding v–t graph
Take the upward direction as positive. Considering
shows a horizontal line above the t -axis.
the forces acting on the object, we have
After the string breaks, the ball continues to move
T − W = ma ⇒ T = W + ma
upwards due to inertia. But since its acceleration
If the acceleration a is positive (i.e. pointing
points downwards (i.e. the acceleration due to
upwards), the tension T will increase and break
gravity), its velocity decreases until it reverses
the string. Note that the acceleration points
direction of motion (i.e. v is momentarily zero). So
upwards when the lift is moving downwards while
the answer is option D.
slowing down.
11. When the ball has zero velocity (i.e. momentarily
4. C
at rest) and is in contact on the pan, it experiences
As the object falls and accelerates, the weight of
its downward weight W and the upward normal
the object remains unchanged but the air
reaction R exerted by the pan, which is also the
resistance acting on it increases with its speed.
scale reading.
When the air resistance is large enough to balance
At the next instant, the ball is bounced upwards.
the weight, the speed of the object no longer
We therefore know that it experiences an upward
increases, and so does the air resistance.
net force when it is still in contact with the pan.
5. D Hence, we have R >W.
Note that the weight of the skydiver is the
12. D
gravitational force acting on him by the Earth.
Statement (1) is correct. When the object is
6. D moving up, the air resistance f is opposing its
Statement (1) is correct but statement (2) is motion and hence is acting downwards (i.e. in the
incorrect. Since the cat is at rest, the weight of the same direction as the weight W ). So F = W + f >W.
cat (i.e. the gravitational force acting on the cat by
Statement (2) is correct. When the object reaches
the Earth) and the normal reaction acting on the
the highest point, it is momentarily at rest and no
cat by the pillow balance each other and therefore
air resistance is acting on it. So F = W.
have the same magnitude.
Statement (3) is correct. When the object is
Statement (3) is incorrect. Both the forces act on
moving down, the air resistance f is opposing its
the same object, i.e. the cat.
motion and hence is acting upwards (i.e. in the
7. A opposite direction as the weight W ). So F =W − f <
Both statements (1) and (2) are correct. It is the W.
friction acting on the hook by the wall that
13. B
balances the weight of the hook and prevents the
Free body diagram of the man on the balance:
hook from falling.
Statement (3) is incorrect. Both the forces act on
the same object, i.e. the hook.
8. B
The free body diagram of Peter when he pushes
the table is as follows.
The weight of the man is
W = mg =(50)(10) = 500 N
Take the downward direction as positive.
By Newton’s second law, we have
W − R = ma ⇒ 500 − R = 50a
⇒ R = 500 − 50a
Active Physics Full Solutions to Textbook Exercises 06 Force | p.9

Statement (1) is incorrect. Between t = 0 and 3 s, The driving force at this stage is 1500 N
we should have Fnet =W − R = 500 – 400 = 100 N forward. (1A)
Statement (2) is incorrect. If the lift starts to move (c) The velocity of the car before the brakes are
upwards (i.e. a < 0), we should have R >500 N. applied is
Statement (3) is correct. If the lift is moving at a v = u + at = 0 + (4)(4) = 16 m s−1 (1M)
constant velocity (i.e. a = 0), we should have By Newton’s second law, the acceleration of
R = 500 N . the car is

14. A a= = = -1 m s−2 (1M)

Let f be the friction acting on the block. By v2 −u2 = 2as, we have

0 − 162 = 2(−1)s ⇒ s = 128 m
Therefore, the distance travelled by the car is
128 m . (1A)
When the force F is less than the limiting friction
17. (a) Take the forward direction as positive.
between the block and the surface, it is always
By v =u + at, we have
balanced by the friction f. The block does not move,
2.4 = 0 + a(2) ⇒ a = 1.2 m s−2 (1M)
and so a =0.
By Newton’s second law, we have
When the force F is larger than the limiting friction,
F − f = ma
by Newton’s second law, we have Fnet = F − f = ma.
30 − f = (15)(1.2) (1M)
Since m and f are constant, a increases linearly
f = 12 N
with increasing F.
So the friction exerted on the trolley is 12 N
15. A (backward). (1A)
After t = 2 s, the only force acting on the
trolley is the friction. By Newton’s second
law, we have
a= = = 0.8 m −2 (1M)

By v = u + at, we obtain
At the instant shown, there is no net force acting
0 = 2.4 + (−0.8)(T − 2) ⇒ T=5s (1A)
on the object, because it is moving at a constant
(b) New result obtained:
speed. So, the friction acting on the block is f =10 N
(to the right).
If the 12 N force is removed, the friction f
continues to oppose the motion of the block.
Hence, the net force acting on the block is
Fnet =2 + 10 = 12 N

• Correct graph for t = 0 to 2 s: 1A

Structured Questions (p.122) • Correct graph for t = 2 to 5 s: 1A
16. (a) 18. (a) Free body diagram of the stone:

Take the forward direction as positive.

By Fnet = ma, we have
• Correct arrows: 1A
F – 1500 = (1500)(4) (1M)
F = 7500 N • Correct labels: 1A
So the driving force is 7500 N forward. (1A) (b) From the graph, when the stone falls at the
(b) The net force acting on the car must be zero terminal speed of 40 m s−1, the air resistance
for it to move at a constant velocity. acting on the stone is 16 N . (1M)
Therefore, a forward driving force is needed When the stone falls at the terminal speed,
to balance the backward friction acting on the weight of the stone is balanced by the air
the car. (1A) resistance. So we have
Mathematically, mg = f
F = f = 1500 N m(9.81) = 16 (1M)
Active Physics Full Solutions to Textbook Exercises 06 Force | p.10

m = 1.631 ≈ 1.63 kg the thrust, the car travels at a constant

The mass of the stone is 1.63 kg . (1A) speed. (1A)
(c) From the graph, when the stone falls at a 21. (a) At t =20 s, the parachutist opens his
speed of 30 m s−1, the air resistance acting on parachute.
the stone is 9 N . (1M) At t = 55 s, the parachutist reaches the
By Newton’s second law, we have ground and comes to a stop. (1A)
mg − f = ma (b) The speed of the parachutist increases. (1A)
(1.631)(9.81) – 9 = 1.631a But the rate of increase decreases with time.
a ≈ 4.29 m s−2 (1A)

The acceleration of the stone is 4.29 m s−2 (c) At both instants, the air resistance acting on
downwards. (1A) the parachutist balances his weight. (1A)
19. Suspend the spring with the mass holder attached The net force acting on the parachutist
to its end, and measure the length of the spring ℓ . becomes zero, and so does his acceleration.
(1A) Hence, his speed is constant. (1A)
Add one weight onto the mass holder at a time.
(d) Displacement of the parachutist:
Measure the length of the extended spring ℓx .
displacement = area under the graph (1A)
Record five sets of data. (1A)
= (30)(5) = 150 m
Calculate the tension T ( = mg ) in the spring,
The distance travelled by the parachutist
where m is the total mass of the mass holder. (1A)
during t =25 to 55 s is 150 m . (1A)
Plot a graph of the tension T against the extended
length x ( = ℓx − ℓ ). (1A) 22. (a) (i) Converting the speed into SI units, we
have
20. (a) By Newton’s second law, we have
F = ma 380 km h-1 = = 105.6 m s-1
3 = 0.8a (1M) Take the forward direction as positive.
a = 3.75 m s−2 By v2 − u2 = 2as, we have
The initial acceleration of the car is 105.62 – 0 = 2a( 1.7×103) (1M)
3.75 m s−2 forward. (1A) ∴ a = 3.280 ≈ 3 m s−2
(b) (i) When the speed of the car increases, The acceleration of a Concorde was
the air resistance acting on the car also about 3 m s−2 . (1A)
increases. (1A) (ii) By Newton’s second law, we have
The net force acting on the car and F = ma = (85)(3.280) ≈ 279 N (1M)
hence its acceleration therefore So the average forward push on the
decrease. (1A) passenger is 279 N . (1A)
(ii) The air resistance acting on the car no (iii) When a Concorde lands, it has already
longer increases when it balances the consumed a lot of fuel, and so its mass
thrust of the car. (1A) is much less than that when it takes off.
At this stage, the net force acting on the (1A)

car becomes zero, and so does its By Newton’s second law, when the
acceleration. Hence, the car reaches a mass of Concorde decreases, its
constant speed. (1A) acceleration increases (in magnitude).
(1A)
(c) (i) The velocity of the car decreases at a
(b) (i)
decreasing rate. (1A)
It eventually reaches a constant
velocity. (1A)
(ii) At the moment when the parachute • Correct arrows: 1A
opens, the air resistance is much • Forward thrust = 210 kN: 1A
greater than the forward thrust. Hence,
(ii) The engine of Concorde pushes the air
the velocity of the car decreases rapidly.
(1A)
backwards. (1A)
As the velocity of the car is greatly By Newton’s third law, the air exerts a
reduced, the air resistance also reaction force that pushes Concorde
decreases. When the air resistance forwards. (1A)
decreases to the extent that balances Clarity of the answer: 1A
Active Physics Full Solutions to Textbook Exercises 06 Force | p.11

(c) Assume that air is an ideal gas. (1M) 25. (a) Frictional force acting on the parcel:
By the kinetic theory equation, we have

(1M)

Also, as vrms2 ∝ T, we have

Correct arrow: 1A
(1M)

Denoting the situation at take-off with the (b) Sketch of the force–time graph:
subscript ’t’ and that at cruising with the
subscript ’c’, we have

=
Correct graphs: 2A

= (1M)
(c) The acceleration of the parcel is
acceleration = slope of the graph
≈ 9.45
The ratio of the density of air at take-off to = (1M)

that at crushing height is 9.45:1 . (1A) = 1.2 m s-2

23. (a) The volume of the disc is By Newton’s second law, we have
V = πr2h Fnet = ma = (15)(1.2) = 18 N
= π (15 × 10−3)2 ( 5×10−3) The magnitude of the net force is 18 N . (1A)

= 3.534 × 10−6 m3 (1M) (d) The distance travelled by the train is

By , the mass of the disc is distance = area under the curve
m = ρV = (8900) ( 3.534×10−6)
= (1M)
= 3.145 × 10−2 kg (1M)
= 5940 m
The weight of the disc is
W = mg = ( 3.145×10−2) (9.81) 26. (a) When the car decelerates suddenly, the
=0.309 ≈ 0.3 N (1M) pendulum has the tendency to stay in motion
due to inertia, and so it continues to swing
(b) (i) An object remains at rest or moves at a
forwards. (1A)
constant velocity, (1A)
if there is no net force acting on it. (1A) (b) When the car decelerates suddenly,
passengers tend to keep their original
(ii) P = Q = 0.3 N (1A)
motion and move forwards due to their
X = Y = 0.6 N (1A)
inertia. (1A)
(iii) magnitude of F = 0.3 N (1A) Passengers wearing seat belts may be
direction of F = upwards (1A) secured on their seats, but those who do not
type of force = gravitational (1A) may hit the dashboard or windscreen and
object upon which F acts = Earth (1A)
suffer serious injuries. (1A)
24. (a) Free body diagram of the bottle: (c) When the car is hit from behind, the car
experiences a forward force and accelerates
suddenly. Since the passengers have the
tendency to remain at rest due to inertia,
they may hit hard on the seats. (1A)
The headrest protects passengers from
• Weight: 1A suffering serious neck injuries. (1A)
• Normal reaction: 1A
(b) The force of gravity pulls the bottle down. Shoot-the-stars Questions (p.126)
The bottle pushes down on the table, so by
1. B
Newton’s third law, the table pushes up with
Take the downward direction as positive. Let h be
an equal and opposite force. (1A)
the height at which the ball falls when the lift is
According to Newton’s first law, if the forces
stationary. By , we have
are balanced, the bottle is either at rest or
in uniform motion. (2A) h= =
Active Physics Full Solutions to Textbook Exercises 06 Force | p.12

Let t' be the time of travel of the ball in the air Refer to the following graph.
when the lift is moving downwards at a constant
speed u . The ball travels with the speed u before it
is released.
During a time interval t', the lift moves downwards
for a distance of ut', and the ball travels for a
distance of h + ut'. So we have

2. A
Due to inertia, both the bottle and the water inside
the bottle will slide backward. The water level will
remain horizontal.
3. B
When the box moves at its terminal speed, the
spring balance and the object experience no net
force. So R = mg.
4. D
Free body diagram of the ping-pong ball:

Take the upward direction as positive. The

acceleration of the ball is

As the ball decelerates, the air resistance f acting

on the ball decreases, and so does the acceleration
(i.e. change in velocity v ). The slope of the v–t
graph decreases with time. So option D is correct.

5. C
Given that the air drag f is proportional to the
speed v of the car, we have f = kv, where k is a
constant.
By Newton’s second law, we have
F − f = ma ⇒ F = ma + f ⇒ F = ma + kv
When t = 0 – T, the acceleration of the car is a
constant of a = a' and its velocity is v = a't,
therefore we have
F = ma' + ka't
This part of the graph shows a straight line with a
slope of ka' and a y intercept ma'.
When t > T, the acceleration of the car is zero (i.e. a
=0 ), and its velocity is a constant of v = a'T. So,
F = 0 + kv = ka'T
This part of the graph shows a horizontal straight
line with F = ka'T.