1
Motion I
Practice 1.1 (p. 6)
7
(a) From 1 January 2009 to 10 January 2009,
1
C
the watch runs slower than the actual
2
D
time by 9 minutes.
3
(a) Possible percentage error
10 −6
=
× 100%
24 × 3600
Therefore, when the actual time is
2:00 pm on 10 January 2009, the time
shown on the watch should be 1:51 pm
–9
= 1.16 × 10 %
1
(b)
= 1 000 000 days
10 −6
on 10 January 2009.
(b) Percentage error
9
=
× 100%
9 × 24 × 60
It would take 1 000 000 days to be in
= 6.94 × 10–2%
error by 1 s.
4
(a) One day
= 24 × 60 × 60
Practice 1.2 (p. 15)
= 86 400 s
1
C
2
B
= 365 × 86 400
3
D
= 31 500 000 s
4
D
5
(a) Total distance she travels
2 × × 10 2 × × 20 2 × × 15
+
+
=
2
2
2
(b) One year
5
Let t be the period of time recorded by a
stop-watch.
Percentage error =
0.4
× 100% ≤ 1%
t
= 141 m
(b) Magnitude of total displacement
t ≥ 40 s
6
The minimum period of time is 40 s.
= 10 × 2 + 20 × 2 + 15 × 2
(a) Percentage error
error due to reaction time
=
× 100%
time measured
= 90 m
0.3
=
× 100%
10
= 3%
Direction: east
Her total displacement is 90 m east.
6
His total displacement is 0.
7
With the notation in the figure below.
(b) From (a), the percentage error of a short
time interval (e.g. 10 s) measured by a
stop-watch is very large. Since the time
intervals of 110-m hurdles are very short
Since ZX = ZY = 1 m, α = β = 60°.
in the Olympic Games, stop-watches are
Therefore, XY = ZX = ZY = 1 m
not used to avoid large percentage
The magnitude of the displacement of the ball
errors.
is 1 m.

8
(a) The distance travelled by the ball will be
7
(a) Length of the path
= 0.8 × 120 = 96 m
longer if it takes a curved path.
(b) No matter which path the ball takes, its
(b) Length of AB along the dotted line
96
= 30.6 m
=
displacement remains the same.
Practice 1.3 (p. 23)
1
B
Total time
5000 5000
=
+
= 9821 s
1.4
0.8
5000 + 5000
= 1.02 m s–1
Average speed =
9821
2
3
(c)
Practice 1.4 (p. 31)
1
C
2
B
C
Final speed
Total time = 9821 + 10 × 60 =10 421 s
5000 + 5000
Average speed =
= 0.96 m s–1
10 421
= 1.5 × 1 – 0.2 × 1 = 1.3 m s–1
3
A
By a =
D
revolution, the spacecraft returned to its
starting point. Therefore, its displacement was
= 7 m s–1
zero and its average velocity was also zero.
4
D
5
(a) Average speed
100
=
= 10.3 m s–1
9.69
v −u
,
t
v = u + at
36
=
+ ( −1.5) × 2
3.6
When the spacecraft had just finished 1
= 7 × 3.6 km h–1
= 25.2 km h–1
Its speed after 2 s is 25.2 km h–1.
4
(b) Yes. This is because the magnitude of
6
Magnitude of Jack’s average velocity
30.6 × 2
=
= 0.51 m s–1
120
B
Take the direction of the original path as
the displacement is equal to the distance
positive.
in this case.
Average acceleration of the ball
−10 − 17
=
0.8
(a) Two cars move with the same speed, e.g.
50 km h–1, but in opposite directions.
= –33.8 m s–2
(b) A man runs around a 400-m playground.
The magnitude of the average acceleration of
When we calculate his average speed,
we can take 400 m as the distance and
his average speed is non-zero. But since
his displacement is zero (he returns to
his starting point), his average velocity
is zero.
5
the ball is 33.8 m s–2.
v −u
By a =
,
t
100
−0
v − u 3.6
t=
=
= 4.27 s
a
6.5
The shortest time it takes is 4.27 s.

4
6
Time / s
0
2
–1
4
Speed / m s
2
7
12
v − u 22 − 2
a=
= 2.5 m s–2
=
t
8
6
8
17
22
Average speed
80 + 60
=
5
= 28 km h–1
Average velocity
The acceleration of the car is 2.5 m s–2.
7
D
(a) I will choose ‘towards the left’ as the
=
positive direction.
80 2 + 60 2
5
= 20 km h–1
(b)
5
C
Total time
10 10
= +
2
3
= 8.33 s
(c)
Average speed
20
=
8.33
v −u
,
t
u = v − at = 9 − (−2) × 3 = 15 m s–1
By a =
= 2.4 m s–1
Her average speed for the whole trip is
The initial velocity of the skater is
–1
15 m s .
8
(a) The object initially moves towards the
left and accelerates towards the left. It
will speed up.
(b) The object initially moves towards the
right and accelerates towards the left. It
will slow down. Its velocity will be zero
and then increases in the negative
direction (moves towards the left).
Revision exercise 1
Multiple-choice (p. 35)
1
C
2
D
3
B
2.4 m s–1.
6
C
7
C
8
C
9
B
10
A
Magnitude of displacement
= 2000 2 + 6000 2
= 6324.6 m
Magnitude of average velocity
6324.6
=
4 × 3600
= 0.439 m s–1
6000
tan θ =
2000
θ = 71.6°
His average velocity is 0.439 m s–1
(S 71.6° E).

11
(b) Displacement from Sheung Shui to Lok
C
Total time = 13 min = 780 s
840 × 2
= 2.15 m s −1
Average speed =
780
Ma Chau
1000
=
× 6.3
1
12
D
= 6300 m
13
(HKCEE 2003 Paper II Q3)
Magnitude of average velocity
6300
=
359
Conventional (p. 37)
1
= 17.5 m s–1
Total time left for the two players
= 4 × 60 + 9 + 5 × 60 + 16 = 565 s
(1M)
Total time they have been playing
= 6635 s (= 110 min 35 s = 1 h 50 min 35 s)
(1A)
(a) 50 m
(1A)
(b) Magnitude of average velocity of Kitty
50
=
(1M)
1× 60 + 15
= 0.667 m s −1
(c)
Average speed of the coach
5 + 50 + 5
=
1× 60 + 15
= 0.8 m s −1
3
(1A)
(1M)
(1A)
(a) Since she measures the time interval
based on 1 cycle of the pendulum, the
error (0.3 s) in measuring the cycle of
the pendulum accumulates.
(1A)
= 1500 + 40 × 1000 + 10 × 1000
Total time
= 2 × 3600 + 3 × 60 + 8
= 7388 s
Average speed
51 500
=
7388
= 6.97 m s–1
= 1.16 m s–1
Cycling:
Average speed
40 000
=
1 × 3600 + 1 × 60 + 53
Running:
cycles (e.g. 20) with the stop-watch and
Average speed
10 000
=
39 × 60 + 47
divide the time by the number of cycles.
= 4.19 m s–1
cycle, Jenny should time more pendulum
(1M)
(1A)
= 359 s (5 min 59 s)
(1A)
Average speed
1500
=
21 × 60 + 28
is from 8 to 14 s.
(1A)
(1M)
(b) Swimming:
= 10.8 m s–1
(a) Time required
7.4 × 1000
=
20.6
(1A)
(a) Total distance
The range of the time interval (10 cycles)
(b) When finding the time for one pendulum
4
(1M)
= 51 500 m
= 2 × 60 × 60 − 565
2
5
(1A)
His average speed was the highest in
(1M)
cycling.
(1A)

(1A)
(c)
(c)
Yes. Since the time interval of this
competition is quite long,
(1A)
= 3 min 47 s
= 3 × 60 + 47 = 227 s
v−u
a=
(1M)
t
431
−0
= 3.6
= 0.527 m s–2
(1A)
227
using stop-watch will not result in large
percentage error as the reaction time for
an average person is only 0.2 s.
6
(a) v = u + at
(1A)
(1M)
=0+6×4
= 24 m s–1
= 86.4 km h
The average acceleration of the train is
–1
0.527 m s–2.
(1A)
The maximum speed of the car is
8
(a) Total distance
–1
86.4 km h .
= 8000 + 4000 + 5000
(b) v = u + at
(1M)
= 17 000 m
= 24 + (–4) × 2
= 16 m s
Total time
–1
= 57.6 km h
= 1 × 3600 + 30 × 60 + 45 × 60
–1
(1A)
= 8100 s
–1
(c)
Total time = 5 min 45 s − 1 min 58 s
The final speed of the car is 57.6 km h .
v−u
a=
(1M)
t
16 − 0
=
6
= 2.67 m s–2
(1A)
Average speed
17 000
=
8100
(1M)
= 2.10 m s–1
(1A)
(b)
The average acceleration of the car is
2.67 m s–2.
7
(a) Average speed
30 000
=
8 × 60
= 62.5 m s–1
(1M)
(1A)
The average speed of the train is
62.5 m s–1.
(b) Maximum speed
430
=
= 119.4 m s−1 > average speed
3.6
(1A)
The average speed must be smaller than
the maximum speed because the train
needs to speed up from start and slows
down to stop during the trip.
(1A)
Magnitude of displacement
= 3000 2 + 4000 2 = 5000 m
Magnitude of average velocity
5000
=
= 0.617 m s–1
8100
4000
tan θ =
3000
(1A)
θ = 53.1°
His average velocity is 0.617 m s
(N 53.1° E).

(1A)
–1
9
(a) Distance travelled
10
(a) Total distance
= 10.5 × 3 × 60
(1M)
= (120 + 50) × 1000
(1M)
= 1890 m
(1A)
= 170 000 m
(1A)
(b) Circumference of the track
(b)
=2 r
= 2 (400)
∆XYZ is a
right-angled
triangle.
N
= 2513 m
Z
θ
50 km
The distance travelled by Marilyn is
3
1890 m which is about of the
4
circumference.
30°
Y
(1A)
α ψ
60°
120 km
X
Magnitude of displacement (from town
X to town Z)
= 120 000 2 + 50 000 2
= 130 000 m
120
tan θ =
50
θ = 67.4°
α = 60° − 22.6° = 37.4°
= 400 2 + 400 2
The total displacement of the car is
= 566 m
130 000 m (N 37.4° E).
Magnitude of average velocity
566
=
3 × 60
= 3.14 m s
400
tan θ =
400
θ = 45°
Her average velocity is 3.14 m s–1
(S 45° E).
(1A)
ψ = 90° − 67.4° = 22.6° Magnitude of displacement AB
–1
(1A)
(c)
(1A)
(1A)
Total time
170 000
=
= 10 200 s
60
3.6
(1A)
Magnitude of average velocity
130 000
=
10 200
(1M)
= 12.7 m s–1
Its average velocity is 12.7 m s
(N 37.4° E).

(1A)
–1
11
AC = 60 2 + 80 2 = 100 m
80
tan θ =
θ = 53.1°
60
(a)
(1M)
The total displacement of the athlete is
100 m (S53.1°W).
13
(a) The coin moves in the following
sequence: B A C C A
(Correct label of velocity with correct
direction (towards the left).)
Therefore, it is at A finally.
(1A)
= 15 cm
correct direction (towards the right).)
(b)
1
2
3
4
5
6
v / m s–1
–6
–4
–2
0
+2
+4
+6
(c)
= 15 + 30 + 30
(1M)
= 75 cm
(1A)
(i)
The car will slow down and
(1A)
its speed will drop to zero.
(1A)
= 0.0188 m s−1
(ii) Average speed
75 × 10 −2
=
8
After that the car will move towards the
right with increasing speed (uniform
acceleration).
12
(1A)
= 0.0938 m s−1
(a) Total distance travelled
= 60 + 80 + 80 + 60
(1M)
= 280 m
(1A)
(d) (i)
zero.
the coin is also zero.
= 420 m
(1A)
A
60 m
C
14
(1M)
(1A)
(a) Total distance
= πr
(1M)
= 5π
= 15.7 m
θ
(1A)
Therefore the average velocity of
Total distance travelled
(1M)
(1A)
(ii) The displacement of the coin is
(1A)
= 280 + 60 + 80
(1M)
The coin moves in the following
Therefore, it is at B finally.
The total displacement of the athlete is
(c)
(1A)
B A C C A B B
(1M)
160 m (west).
(1M)
sequence:
(b) Magnitude of total displacement
= 80 + 80 = 160 m
Total time = 2 s × 4 = 8 s
Average velocity
15 × 10 −2
=
8
(0.5A × 6)
(c)
(1A)
(b) Distance travelled by the coin
(1A)
0
(1M)
Displacement of the coin
(Correct label of acceleration with
Time / s
(1A)
(1A)
Total displacement
=5+5
(1M)
= 10 m
(1A)
80 m

The total displacement travelled by her
is 10 m.
(b) Jane’s statement is incorrect.
(1A)
Since both girls start at X and meet at Y,
they have the same displacement. (1A)
Betty’s statement is incorrect.
(1A)
Since both girls return to their starting
point, their displacements are zero. (1A)
Physics in articles (p. 40)
(a) From 19 January 2006 to 28 February 2007,
(1A)
It takes New Horizons spacecraft a total of
406 days to travel from the Earth to Jupiter.
(1A)
(b) (i)
Average speed
total distance travelled
=
total time of travel
=
8 × 108
406 × 24
= 8.21 × 104 km h−1
(ii) Average acceleration
change in velocity
=
total time of travel
=
(1A)
(1M)
(8.23 − 5.79)×10 4
406 × 24
= 2.50 × 104 km h−2
(c)
(1M)
July 2015
(1A)
(1A)

2
Motion II
Practice 2.1 (p. 61)
1
D
2
B
3
D
4
D
5
B
10
(a) The object moves with a constant
velocity.
(b) The object moves with a uniform
acceleration from rest.
(c)
deceleration, starting with a certain
30 − 10
= 10 m s–1
v=
2
initial velocity. Its velocity becomes
zero finally.
The velocity of the car at t = 2 s is 10 m s–1.
6
C
7
(a) Total displacement
The object moves with a uniform
(d) The object first moves with a uniform
acceleration from rest, then at a constant
velocity, and finally moves with a
= 4 × 5 + (−5) × (7 − 5) = 10 m
smaller uniform acceleration again.
The total displacement from the
(e)
staircase to her classroom is 10 m.
The object moves at a constant velocity
and then suddenly moves at constant
(b) Classroom C
velocity of same magnitude in the
8
opposite direction.
(f)
The object moves with uniform
deceleration from an initial velocity to
rest, and continue to move with the
uniform acceleration of the same
magnitude in opposite direction.
9
(a) The object accelerates.
11
(with constant velocity of 50 m s–1).
(b) The object first moves with a constant
(b) The object moves with a uniform
velocity. Then it becomes stationary and
acceleration of 5 m s–2.
finally moves with a higher constant
(c)
velocity again.
(c)
The object decelerates to rest, and then
accelerates in opposite direction to
(a) The object moves with zero acceleration
The object moves with uniform
deceleration of 5 m s–2.
12
(a) It moves away from the sensor.
return to its starting point.
(d) The object moves with uniform velocity
towards the origin (the zero
displacement position), passes the origin,
and continues to move away from the
origin with the same uniform velocity.

(c)
(b)
The greatest rate of change in speed
0 − 3.5
=
2
= –1.75 m s–2
(d) Total distance travelled
= area under the graph
3.5 × 2 2 × 6
=
+
2
2
= 9.5 m
Practice 2.2 (p. 71)
1
C
By v2 = u2 + 2as,
13
290
3.6
(a)
2
=0+2×1×s
s = 3240 m = 3.24 km < 3.5 km
The minimum length of the runway is
3.5 km.
2
B
Cyclist X is moving at constant speed.
Time for cyclist X to reach finish line
displacement 150
=
=
= 30 s
time
5
For cyclist Y: u = 5 m s–1, s = 250 m,
(b) Total distance travelled
a = 2 m s–2
1 2
at ,
2
1
250 = 5 × t + × 2 × t2
2
= area under the graph
(12 + 6) × 3
=
2
By s = ut +
= 27 m
(c)
Average speed
total distance travelled
=
time taken
27
=
3
= 9 m s–1
14
(a) She moves towards the motion sensor.
(b) The highest speed of the girl in the
journey is 3.5 m s–1.
t = 13.5 s or t = −18.5 s (rejected)
Y needs 13.5 s to reach finish line.
Therefore, cyclist Y will win the race.
3
B
Since the bullet start decelerates after fired
into the wall, we could just consider the
displacement of the bullet in the wall. To
prevent the bullet from penetrating the wall,
the bullet must stop in the wall.

By v2 = u2 + 2as,
8
2
14 = u + 2 × 5
0 = 500 + 2 × (−800 000) × s
u = 4 m s–1
s = 0.156 m = 15.6 cm < 15.8 cm
By v2 = u2 + 2as,
The minimum thickness of the wall is
142 = 42 + 2 × 2 × s
15.8 m.
4
s = 45 m
C
When the dog catches the thief at t = 5 s, its
total displacement is 30 m. The dog is sitting
The displacement of the girl is 45 m.
9
initially, so u = 0.
1
By s = ut + at2,
2
1
30 = 0 + a(5)2
2
travelling towards the goalkeeper is
6 m s−1.
(b) By v2 = u2 + 2as,
02 − 62
a=
= –22.5 m s−2
2 × 0.8
Its acceleration is 2.4 m s–2.
6
The deceleration of the football should
D
be 22.5 m s−2.
90 36
−
v−u
= 3.6 3.6 = 1.5 m s–2
a=
t
10
2
10
By v = u + 2as,
s=
2
v −u
=
2a
90
3.6
2
36
3.6
2 × 1.5
(b) Braking distance
(2.0 − 0.5)×15 = 11.25 m
=
2
2
−
= 175 m
Thinking distance
= 15 × 0.5 = 7.5 m
The distance travelled by the motorcycle is
–2
Stopping distance
= 11.25 + 7.5 = 18.75 m 20 m
175 m and its acceleration is 1.5 m s .
7
(a) Thinking distance
= speed × reaction time
108
=
× 0.8 = 24 m
3.6
(b) Since the car decelerates uniformly,
braking distance
v+u
=
×t
2
108
+0
= 3.6
× (3 − 0.8)
2
(c)
(a) The reaction time of the cyclist is
0.5 s.
2
2
(a) v = u + at = 0 + 20 × 0.3 = 6 m s−1
The horizontal speed of the ball
a = 2.4 m s–2
5
By v = u + at,
Therefore, the bicycle would not hit the
child.
11
2
By v = u2 + 2as,
0 = 32 + 2 × (–0.5) × s
s=9m8m
Therefore, the golf ball can reach the hole.
12
(a) (i)
By v = u + at,
0 = u + (–4)(4.75)
u = 19 m s–1
= 33 m
The initial velocity of the car is
Stopping distance
19 m s–1.
= thinking distance + braking distance
= 24 + 33 = 57 m

(ii) By v2 = u2 + 2as,
3
2
0 = 19 + 2 × (–4) × s
For option A, apply equation v2 = u2 – 2gs
s = 45.1 m
and take s = 0 (the ball returns to the second
The displacement of the car before
floor),
it stops in front of the traffic light
v = –u = –10 m s–1 (vertically downwards)
is 45.1 m.
This is the same velocity as the initial velocity
2
2
(b) By v = u + 2as,
of option B.
2
Therefore, in both ways the ball has the same
s = 48.2 m
vertical speed when it reaches the ground.
17 = 0 + 2 × 3 × s
The displacement of the car between
starting from rest and moving at 17 m s
4
–1
(a) By v2 = u2 + 2as,
v2 = 0 + 2 × 0.1 × 500
v = 10 m s–1
u = 150 m s–1
–1
His speed is 10 m s .
The speed of the bullet is 150 m s–1 when it is
(b) Consider the first section.
By v = u + at,
v−u
t=
a
10 − 0
=
0.1
= 100 s
Consider the second section.
1
By s = ut + at2,
2
1
800 = 10t + × 0.5t2
2
t = 40 s or t = –80 s (rejected)
Total time taken
= 100 + 40
= 140 s
It takes 140 s for Jason to travel
downhill.
Practice 2.3 (p. 83)
B
Take the upward direction as positive.
1
By s = ut + at2,
2
1
0 = u × 30 + × (−10) × 302
2
is 48.2 m.
13
C
fired.
5
Speed of
stone
v = u + at
Equation used
Distance
travelled by
the stone
1
s = ut + at 2
2
t=1s
10 m s–1
t=2s
20 m s
–1
20 m
t=3s
30 m s
–1
45 m
t=4s
40 m s–1
80 m
6
5m
1
By s = ut + at2,
2
1
10 = 0 + (10) t2
2
t = 1.41 s
v = u + at
= 0 + 10(1.41)
= 14.1 m s–1
1
D
It takes 1.41 s for a diver to drop from a 10-m
2
D
platform. His speed is 14.1 m s–1 when he
enters the water.

7
Take the upward direction as positive.
2
Besides, since Y spends a shorter time to
2
By v = u + 2as,
reach its highest point, it should be fired
2
4 = 0 + (2)(–10)s
s = 0.8 m
after X.
10
The highest position reached by the puppy is
0.8 m above the ground.
8
(a) Consider the boy’s downward journey.
t = 4.16 s or t = −5.76 s (rejected)
Take the downward direction as
It takes 4.16 s to reach the ground.
positive.
1
By s = ut + at2,
2
1
0.5 = 0 + (10) t2
2
(b) v = u + at = 8 + 10 × 4.16 = 49.6 m s–1
Its speed on hitting the ground is
49.6 m s–1.
11
Hang-time of the boy
= 6 – 2 – 1.2 = 2.8 m
= 0.316 × 2 = 0.632 s
(b) Let s be her vertical displacement when
(b) Take the upward direction as positive.
she jumps.
2
As the maximum jumping speed is
2
0 = u + 2 × (–10) × 0.5
8 m s–1, i.e. u = 8 m s–1.
u = 3.16 m s–1
By v2 = u2 + 2as,
v2 − u2
s=
2a
2
0 − 82
=
(upwards is positive)
2 × (−10)
By v = u + 2as,
The jumping speed of the boy is
3.16 m s–1.
9
Take the upward direction as positive.
(a) By v2 = u2 + 2as,
s = 3.2 m > 2.8 m
0 = u2 + 2(–10)(200)
Therefore, the indoor playground is not
u = 63.2 m s–1
The velocity of the firework X is
63.2 m s–1 when it is fired.
(b) By v = u + at,
0 = 63.2 + (–10)t
t = 6.32 s
(c)
(a) Distance between the ceiling and her
hands
t = 0.316 s
2
1 2
at ,
2
1
120 = 8t + × 10 × t2
2
(a) By s = ut +
12
safe for playing trampoline.
1
(a) By s = ut + at2,
2
1
132 = 0 × t + × 10 × t2
2
t = 5.14 s
It takes 6.32 s for the firework X to reach
The vehicle can experience a free fall in
that height.
the Zero-G facility for 5.14 s.
From (a) and (b), for firework Y to
explode at 130 m above the ground, the
speed of Y should be smaller than that of
(b) By v2 = u2 + 2as,
v2 = 02 + 2 × 10 × 132
v = 51.4 m s−1
X. Therefore, Y should be fired at a
The speed of the vehicle before it comes
lower speed.
to a stop is 51.4 m s−1.

(c)
Take the upward direction as positive.
By v = u + at,
–v = v – gt
2v = gt
If the stone is projected with a speed of 2v, let
the new time of travel be t′.
(–2v) = (2v) – gt′
v
t′ = 4 ( )
g
= 2t
Its new time of travel is 2t.
6
Take the upward direction as positive.
1
s = ut + at2
2
1
= (10)(4) + (–10)(4)2
2
Revision exercise 2
Multiple-choice (p. 87)
1
B
D
By v2 = u2 + 2as,
= –40 m
0 = 102 + 2a(25 – 10 × 0.2)
The distance between the sandbag and the
a = –2.17 m s–2
ground is 40 m when it leaves the balloon.
His minimum deceleration is 2.17 m s–2.
2
D
7
D
3
B
8
C
Take the downward direction as positive.
Consider the rock released from the 2nd floor.
u = 200 m s–1, v = 5 m s–1, a = −20 m s–2
By v2 = u2 + 2as,
v2 = 2as
By v = u + at,
(as u = 0)
5 = 200 + (−20)t
Then consider the rock released from the 7th
t = 9.75 s
floor.
Note that s2 = 3.5s.
The rockets should be fired for at least 9.75 s.
(v2)2 = 2as2
Both C and D satisfy this requirement. But for
= 3.5(2as)
D, after firing for 10.2 s,
= 3.5v2
v = u + at
= 200 + (–20)(10.2)
v2 = 1.87v
= –4 m s–1
4
A
5
C
i.e. it flies away from the Moon with 4 m s–1
The stone returns to the ground with the same
upwards. It cannot land on the Moon.
speed (but in opposite direction).
Therefore, the correct answer is C.
9
D
10
D

11
(HKCEE 2006 Paper II Q1)
12
(HKCEE 2007 Paper II Q2)
13
(HKCEE 2007 Paper II Q33)
(b) (i)
Conventional (p. 89)
1
(a) The reaction time of the driver is 0.6 s.
(b)
v
a=
t
=
(1A)
(1M)
0 − 12
3.6 − 0.6
= –4 m s–2
(1A)
(Correct axes with label)
The acceleration of the car is –4 m s–2.
(c)
(A straight line with slope = 0.35 m s–1
The stopping distance of the car is the
area under graph.
(1A)
from t = 1.20 s to 1.25 s)
(1M)
(1A)
(A straight line with slope = –0.35 m s–1
Stopping distance
12 × (3.6 − 0.6)
=12 × 0.6 +
2
from t = 1.45 s to 1.50 s)
(1A)
Then, from t = 1.45 s to 1.50 s, the
(Correct axes with labels)
(1A)
object changes its moving direction and
(Correct graph with the acceleration of
−0.35 − 0.35
about
1.40 − 1.30
= 25.2 m
(ii)
(1A)
The stopping distance of the car is
shorter than 27 m. The driver will not be
charged with driving past a red light.
(1A)
2
(a) The object moves away from the motion
sensor with uniform velocity at
0.35 m s–1 from t = 1.20 s to 1.25 s.(1A)
From t = 1.25 s to 1.45 s, the object
moves with negative acceleration. (1A)
moves towards the motion sensor again
with a uniform velocity of –0.35 m s–1.
= –7 m s–2 at t = 1.30 s to 1.40 s) (1A)
(1A)
!

3
(b) Total displacement of the car
(a)
= area bound by the v−t graph and the
(c)
time axis
1
1
= (5 × 5) − (20 × 3)
2
2
(1M)
= −17.5 m
(1A)
Yes, the car moves 12.5 m forwards
from t = 0 to t = 5 s. Therefore, it hits
the roadblock.
5
(1A)
Take the upward direction as positive.
(a) From point A to the highest point:
(Correct axes with labels)
(1A)
(Correct shape of minibus’ graph)
(1A)
(Correct shape of sports car’s graph)
(1A)
(Correct values)
(1A)
By v2 = u2 + 2as,
0 = 42 + 2 (–10) s
s = 0.8 m
By v = u + at,
0 = 4 + (–10)t
(b) From the graph in (a), the two vehicles have
t = 0.4 s
the same velocity at t ≈ 2.3 s after passing the
traffic light.
(c)
The area under graph is the displacement of
(1M)
Consider their displacements at t = 3 s,
= –3.2 m
For the sports car:
1
s = × 15 × 3 = 22.5 m
2
3.2 m above the trampoline.
(b) Height of point A above the trampoline
(1A)
The minibus will take the lead 3 s after
passing the traffic light.
4
(1A)
The maximum height reached by him is
(1A)
For the minibus:
1
s = × (7 + 13) × 3 = 30 m
2
(1M)
From the highest point to the trampoline:
1
s = ut + at2
(1M)
2
1
= 0 + (–10)(1.2 – 0.4)2
2
(1A)
the cars.
(1M)
6
(1A)
= 3.2 – 0.8
(1M)
= 2.4 m
(1A)
(a) Initial velocity v
acceleration at −1 m s−2 from t = 0 s to
= 90 km h–1
90
=
m s–1
3.6
t = 5 s.
= 25 m s–1
(a) The car moves forward with uniform
(1A)
Its instantaneous velocity is 0 at t = 5 s.
Thinking distance
(1A)
=v×t
(1M)
Then the car changes its moving
= 25 × 0.2
direction. From t = 5 s to t = 8 s, it
=5m
moves backwards with a uniform
The thinking distance is 5 m.
−2
acceleration of −6.67 m s .
(1A)
(1A)
"

(b) By v2 = u2 + 2as,
v2 − u2
a=
2s
2
0 − 25 2
=
2 × (80 − 5)
(1M)
= −4.17 m s–2
(1A)
(c)
The slope of the graph is the magnitude
of the acceleration of the apple.
(1A)
speed / m s−1
7.75
Hence, the deceleration of the car is
4.17 m s–2.
(c)
By v2 = u2 + 2as,
2
(1M)
2
v −u
2a
0 2 − 25 2
=
2 × ( − 4.17 × 2)
s=
= 37.5 m
0
(1A)
(1M)
(d) The two graphs have no difference.
(1A)
= 37.5 + 5 = 42.5 m
(1A)
The driver could not stop before the
traffic light. Therefore, his claim is
incorrect.
8
(a) Take the downward direction as
positive.
By v2 = u2 + 2gs,
(1A)
v = u + 2 gs
= 0 2 + 2 × 10 × (40 − 3)
= 27.2 m s–1
positive.
(1M)
(1A)
cushion is 27.2 m s−1.
1
(b) (i) By s = ut + gt2,
2
1
40 – 3 = 0 + × 10 × t2
2
t = 2.72 s
(1M)
v = 2 × 10 × 3
The speed of the apple is 7.75 m s
(1A)
–1
when the apple just reaches the ground.
(1A)
The speed of the residents landing on the
The apple travels in air for 0.775 s.
= 7.75 m s−1
(1M)
2
(a) Take the downward direction as
(b) By v2 = u2 + 2as,
(2A)
(Straight line with a slope of 10 m s−2)
Stopping distance
1
By s = ut + gt2,
2
1
3 = 0 × t + × 10 × t2
2
3× 2
t=
= 0.775 s
10
time / s
(Correct labelled axes)
Braking distance = 37.5 m
7
0.775
(1M)
(1A)
The time of travel in air is 2.72 s.
u+v
(ii) By s =
t,
(1M)
2
2s
t=
u+v
2×3
=
t
27.2 + 0
= 0.221 s
(1A)
The time of contact is 0.221 s.

(b) Slope of the graph from t = 0
(c)
to t = 0.28 s
2.3 − 0
=
0.28 − 0
(1M)
= 8.21 m s–2
(1A)
The acceleration of the ball due to
gravity is 8.21 m s–2.
(c)
9
(Correct labeled axes)
(1A)
(Correct shape)
(1A)
(Correct values)
(1A)
(i)
(a) t = 2 s:
Displacement of the trolley
= 0.7 − 0.15 = 0.55 m
(1A)
t = 3.4 s:
Displacement of the trolley
= 1.175 − 0.15 = 1.025 m
(1A)
t = 4.9 s:
Displacement of the trolley
= 0.6 − 0.15 = 0.45 m
(1A)
(b) It moves away from the motion sensor
with a changing speed from t = 2 s to
t = 3.4 s.
(c)
a = −0.507 m s−2
(1M)
(1A)
11
(a) (i)
The ball is held 0.15 m from sensor
before being released. The ball hits
the ground which is 1.1 m from the
(1A)
Therefore, the ball drops a height
of 0.95 m.
(1A)
(ii) The ball rebounds to the positions
−0.507 m s−2.
(a) The motion sensor is protruded outside
the table to avoid the reflection of
ultrasonic signal from table.
(1A)
(1A)
sensor.
The acceleration of the trolley is
10
(1A)
by the transparent plastic plate.
(1A)
(1A)
(Correct shape)
since ultrasound will be reflected
Then it rests momentarily at t = 3.4 s.
sensor with a changing speed.
1
By s = ut + at2,
2
1
−0.1 = 0.7 × 2.9 + × a × (2.9)2
2
(1A)
(ii) The method does not work
(1A)
After that, it moves towards the motion
(Correct sign)
(1A)
which are 0.45 m, 0.65 m and
0.775 m from the sensor in its first
3 rebounds.

At the 1st rebound, the ball rises up
(1.1 − 0.45) = 0.65 m.
The average acceleration is 66.6 m s–2.
(c)
(1A)
v / m s−1
nd
At the 2 rebound, the ball rises up
(1.1 − 0.65) = 0.45 m.
(1A)
6.32
rd
At the 3 rebound, the ball rises up
(1.1 − 0.775) = 0.325 m.
(b) (i)
(1A)
t/s
t3
The ball hits the ground with
–1
velocities of 3.9 m s , 3.25 m s
t1
–1
t2
t4
t5
and 2.75 m s–1 in its first 3
rebounds.
(3A)
−6.32
(ii) Acceleration
= slope of graph =
3.9
(1M)
0.95 − 0.55
= 9.75 m s–2
12
(1A)
Take the downward direction as positive.
1
(a) By s = ut + gt2,
(1M)
2
1
2 = 0 × t + × 10 × t2
2
2×2
t=
= 0.632 s
(1A)
10
(3 straight lines)
(1A)
(Correct slopes)
(1A)
(Correct labels of time and velocity)(1A)
13
(a) Speed v = 70 km h–1
70
=
m s–1
3.6
= 19.4 m s–1
d
Reaction time =
v
6
=
19.4
It takes 0.632 s from t1 to t2.
(b) At t2,
= 0.309 s
v = u + at
(1A)
The reaction time of the man was
= 0 + 10 × 0.632
= 6.32 m s
(1M)
0.309 s.
–1
(1M)
–1
Shirley’s speed is 6.32 m s when she
lands on the trampoline at t2.
At t4, she leaves the trampoline at the
same speed. Therefore, from t3 to t4,
by v2 = u2 + 2as,
(b) By v2 = u2 + 2as,
v2 − u2
a=
2s
2
0 − 19.4 2
=
2 × 48
= –3.92 m s–2
(1M)
v2 − u2
2s
(−6.32) 2 − 0 2
=
2 × 0.3
3.92 m s–2.
(c)
(1A)
(1A)
The average deceleration of the car was
a=
= 66.6 m s–2
(1M)
Speed v
= 80 km h–1
80
=
m s–1
3.6
= 22.2 m s–1

Thinking distance
Take the upward direction as positive.
1
s = ut + at2
(1M)
2
1
= 7 × 1.75 + × (–10) × 1.752
2
= vt
= 22.2 × 0.309
= 6.86 m
2
(1A)
2
By v = u + 2as,
= –3.06 m (negative means the water
braking distance s
is below the spring board)
v2 − u2
=
2a
2
0 − 22.2 2
=
2 × (−3.92)
The spring board is 3.06 m above the
water.
Alternative method:
= 62.9 m
Consider the upward motion and
(1A)
downward motion separately.
Therefore, the stopping distance
For the upward motion, she takes 0.7 s
= 6.86 + 62.9
= 69.8 m
to reach the highest point from the
(1A)
spring board.
This stopping distance is greater than the
Take the upward direction as positive.
1
By s = ut + at2,
(1M)
2
1
s1 = 7 × 0.7 + × (–10) × 0.72
2
initial distance between the car and the
boy.
(1A)
Therefore, the car would have knocked
down the boy if the car had travelled at
= 2.45 m
80 km h−1 or faster.
For the downward motion, she takes
(d) A drunk has a longer reaction time.(1A)
1.05 s from the highest point to enter
This means that the thinking distance,
water.
and thus the stopping distance (sum of
Take the downward direction as
thinking distance and braking distance),
increases.
14
(1A)
positive.
(1A)
1 2
gt ,
2
1
s2 = 0 + × 10 × 1.052 = 5.51 m
2
By s = ut +
(a) Take the upward direction as positive.
By v = u + at,
(1M)
u = 0 − (−10) × 0.7
= 7 m s–1
Therefore the height of the spring board
(1A)
above the water
The speed of Belinda leaving the spring
–1
= s2 – s1
board is 7 m s .
= 5.51 – 2.45
(b) Total time taken from the spring board
to the water
= 0.7 + 1.05 = 1.75 s
(c)
= 3.06 m
(1A)
v = u + at
(1M)
= 0 + (−10) × 1.05
= −10.5 m s–1
(1A)
The speed of the diver entering the water
is 10.5 m s–1.

(d)
Deceleration of car Y
= slope of the graph during 0.5 s−8.5 s
=
0 − 19.4
= –2.43 m s–2
8.5 − 0.5
The deceleration of car Y is 2.43 m s–2.
(1A)
(c)
Thinking distance
= area under the graph during 0−0.5 s
= 19.4 × 0.5
= 9.7 m
(e)
(Correct shape)
(1A)
Braking distance
(Correct times)
(1A)
(Correct velocities)
(1A)
= area under the graph during 0.5 s−8.5 s
1
= × 19.4 × (8.5 – 0.5)
2
(See the figure in (d).)
= 77.6 m
(Correct slope - parallel to that in (d).)
distance are 9.7 m and 77.6 m
(Correct position – above that in (d).)
respectively.
(1A)
(d) The coloured area is equal to the
(a) Speed 70 km h–1
70
=
m s–1
3.6
= 19.4 m s
difference in the stopping distances
travelled by cars X and Y.
(e)
–1
(1M)
Since the distance between the cars is
travel in 2 s, the driver of car Y obeys
(1A)
= 9.7 + 77.6 – 48.5
(1M)
= 38.8 m < 50 m
(1M)
Since the difference in stopping
(b) Deceleration of a car is the slope of their
corresponding v–t graph.
Stopping distance of car X
Coloured area
greater than the distance that car Y can
the rule.
(1A)
= area under the graph during 0−5 s
1
= × 19.4 × 5 = 48.5 m
2
Distance travelled by car Y in 2 s
= vt = 19.4 × 2 = 38.8 m < 50 m
(1A)
The thinking distance and the braking
(1A)
15
(1A)
distances of the cars is smaller than the
(1M)
initial separation of the cars, the two cars
Deceleration of car X
do not collide with each other before
= slope of the graph during 0−5 s
they stop.
0 − 19.4
=
5−0
16
= –3.88 m s–2
The deceleration of car X is 3.88 m s–2.
(1A)
(a) From t = 0 s to t = 5 s, the car moves
with a uniform acceleration of
17 − 0
= 3.4 m s–2.
5
(1A)

(1A)
1 2
at
2
1
= 0 + × 17.5 × (8 × 60)2
2
From t = 5 s to t = 20 s, the car moves
(b) s = ut +
with a constant velocity of 17 m s–1.
(1A)
From t = 20 s to t = 28 s, the car moves
with a uniform acceleration of
0 − 17
= −2.125 m s–2.
28 − 20
(1A)
The Shuttle travels 2 016 000 m
(1A)
From t = 28 s to t = 30 s, the car remains
at rest.
= 2 016 000 m (2016 km)
(1M)
(2016 km) in the first 8 minutes.
19
(a) (i)
The cyclist is using first gear when
the acceleration is greatest before
(1A)
braking.
(b)
(1A)
(ii) The cyclist uses second gear for the
shortest time.
(1A)
(b) Distance travelled
= area under straight line PQ
(8 + 6) × 2
=
2
= 14 m
(1M)
(1M)
(1A)
The cyclist travels 14 m in second gear.
(c)
The acceleration during t = 18 s−20 s
0−9
=
(1M)
20 − 18
= −4.5 m s–2
(1A)
–2
The deceleration is 4.5 m s .
(c)
(Correct shape)
(1A)
(Correct time instants)
(1A)
(Correct accelerations)
(1A)
Yes.
(1A)
The car changes direction at t = 30 s.
(1A)
Its velocity changes from positive to
negative, showing a change in its
travelling direction.
17
(HKCEE 2002 Paper I Q8)
18
(a) v = u + at
(1A)
(1M)
= 0 + 17.5 × 8 × 60
= 8400 m s–1
(1A)
20
21
(HKCEE 2005 Paper I Q1)
1
(a) s = ut + at2
2
1
= 0 + × 10 × (500 × 10−3)2
2
= 1.25 m
(1M)
(1A)
Therefore the minimum height the
laptop must fall for it to be ‘saved’ is
1.25 m.
(b) v = u + at
(1M)
−3
= 0 + 10 × (500 × 10 )
= 5 m s−1
(1A)
The speed of the computer when it hits
the ground is 5 m s–1.
The speed of the Shuttle after the first 8
minutes is 8400 m s–1.

(c)
Most falls are likely to be from below
Physics in articles (p. 96)
this height,
(a) 2.45 m
(1A)
so the protection will not have taken
effect.
22
(a) Any one from:
(1A)
(1A)
By v2 = u2 + 2as,
2
(1M)
2
u = v − 2as
u = 5.35 m s−1
Displacement per unit time
(b) The velocity of a braking car is
(1A)
The vertical speed of Javier Sotomayor
(1A)
so the car has negative acceleration.(1A)
is 5.35 m s−1 when he leaves the ground.
(ii) Take the upward direction as positive.
Its displacement is (still) increasing with
Consider the upward journey.
time,
(1A)
so its velocity is (still) positive
(1A)
By v = u + at,
v − u 0 − 5.35
t=
=
= 0.54 s
a
− 10
In this case, the acceleration and
velocity are in opposite directions. (1A)
(c)
Take the upward direction as positive.
u2 = 0 − 2(−10)(2.45 + 0.07 − 1.09)
Rate of change of displacement
decreasing (with time)
(b) (i)
(1A)
(i)
(1M)
Consider the downward journey.
1
By s = ut + at2,
(1M)
2
1
− (2.45 + 0.07 − 0.71) = 0 + (− 10) t2
2
t = 0.60 s
The time that he stays in the air
= (0.54 + 0.60) = 1.14 s
(1A)
Alternative method:
(Correct graph)
(1A)
(ii) Vertical distance travelled
= area under the graph from 4.0 s
to 10.0 s
(
70 + 130)× 6
=
2
(1M)
= 600 m
(1A)
Take the upward direction as positive.
1
By s = ut + at2,
(1M)
2
(0.71 − 1.09) = 5.35t + 1 (− 10)t 2 (1M)
2
t = 1.14 s or t = −0.07 s (rejected)
(1A)
The time that he stays in the air is 1.14 s.
The vertical distance travelled by
the rocket between t = 4.0 s and t =
10.0 s is 600 m.

3
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
4
relative to the MTR train.
(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
man tends to move at his original
(a) Stretching a rubber band
direction, so he would move outwards
(b) Standing on the floor
(c)
Walking
(d) Exists in every object on the earth at any
relative to the MTR train.
7
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)
1
C
2
C
3
D
4
C
5
(a) No. Athletes would hit the wall of the
far out in space.
8
Joan moves on the ice surface with a constant
velocity.
Practice 3.3 (p. 122)
1
D
stadium if it is too close to the finishing
2
A
line.
3
B
4
A
5
D
(b) The mat is used to protect the athletes if
they hit the wall after passing the
finishing line.

6
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 θ 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 θ 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)
Resultant’s magnitude is 60 N and the
angle between the resultant and the
horizontal is 25°.
(d)
Horizontal component
= 40 + 30 cos 60° = 55 N
Vertical component
= 30 sin 60° = 26.0 N
Resultant = 55 2 + 26.0 2 = 60.8 N
Let θ be the angle between the resultant
and the horizontal.
26.0
θ = 25.3°
tan =
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°.

(d) Resultant = 40 2 + 30 2 = 50 N
8
Hence, the angle between the two 5-N forces
Let θ be the angle between the resultant
is 120°.
and the horizontal.
30
tan =
θ = 36.9°
40
Alternative method:
the resultant 5-N force form an equilateral
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
(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 θ = 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 θ
Horizontal component Fy
= 5 − 5 cos θ = 5 × (1 − cos θ)
(magnitude of the resultant)2 = Fx2 + Fy 2
52 = (5 sin θ)2 + [5 × (1 − cos θ)]2
1 = sin θ + 1 − 2 cos θ + cos θ
2
2
2T cos 45° = W
2 × 20 × cos 45° = W
cos θ = 0.5
W = 28.3 N
θ = 60°

12
(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
6
= T cos 10°
C
By v2 – u2 = 2as,
= 1440 × cos 10°
0 – u2 = 2a(20)
= 1420 N
−u2 = 40a
u2
a=−
40
The component of the force that pulls
the car is 1420 N.
13
(a)
Resistance = ma = 12 × −
7
u2
= –0.03u2
40
‘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
a= =
= 2 m s–2
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°
= 20 cos 30°
6945 N.
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

11
Take the upward direction as positive.
14
Weight = mg
Take the downward direction as positive.
Let R be the reading of the balance.
5
(a) By F = ma,
6
R − mg = 0
= 3 × 10 × 10
= 3 × 10 N
Net force = ma
R = 20 N
5
= 3 × 10 × 12
The reading of the balance is 20 N.
6
= 3.6 × 10 N
(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
6
= 3.6 × 10 + 3 × 10
6
R = 17 N
6
= 6.6 × 10 N
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
(b) The friction acting on the box is 3 N.
(c)
By F = ma,
F 5−3
m=
=
= 1 kg
a
2
The reading of the balance is 21 N.
15
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 θ = (2)(10)sin 10° = 3.47 N
(b) By F = ma,
mg sin θ − f = ma
Weight, air resistance
(2)(10) sin 30° − 3.47 = 2a
(ii) Weight
(iii) Weight, air resistance
(b)
a = 3.27 m s–2
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.
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

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
constant velocity as no net force acts on
it. During 5–10 s, the object is moving
on B by A.
7
(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,
the object is moving with an acceleration
of 1 m s–2 as a net force of 5 N acts on it.
Therefore the skater moves backwards.
(b) When the diver pushes the platform, the
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
it.
Practice 3.5 (p. 148)
1
D
2
C
Force acting on A by B and force acting
the diver gains speed and dives.
(c)
When we push ourselves against the side
of the pool, the pool exerts an equal but
opposite force on us. Therefore we
accelerate forwards.

(d) When the runner exerts a force on the
(c)
starting block, the block exerts an equal
but opposite force on the runner.
(a) (i)
2.5 m s–2.
Trolley A’s weight component
(ii) By F = ma,
down the plane
= mg sin θ
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
By Newton’s third law,
= 10.3 N – T (down the plane)
(b) (i)
force acting on B by A
Trolley B’s weight component
down the plane
= mg sin θ
= 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
By F = ma,
20
F
a= =
= 2.5 m s–2
m 3+5
The accelerations of the blocks are
Therefore the runner moves forwards.
8
(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
!

Practice 3.6 (p. 165)
1
(b)
A
Moment of force
= 30 sin 15° × 0.7 = 5.44 N m (clockwise)
2
(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
(In opposite directions and acting at
(a) Torque = 5 N × 0.5 m = 2.5 N m
different positions.)
"

(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.
Take moment about A.
t = 2.4 = 6 tE = 2.45 tE
4
C
5
D
6
B
7
A
8
B
9
D
platform of the stand.
10
D
11
B
Revision exercise 3
12
A
13
D
14
(HKCEE 2006 Paper II Q31)
15
(HKCEE 2007 Paper II Q6)
16
(HKCEE 2007 Paper II Q30)
17
(HKCEE 2007 Paper II Q27)
m × 10 × 0.1 = 2.4 × 10 × 0.06
m = 1.44 kg
The maximum mass that the system can
withstand is 1.44 kg.
(b) Use a G-clamp to fix the stand on the
bench or add a heavy weight on the
Multiple-choice (p. 170)
1
A
Moment of force about O
=F×d
= 8 sin 45° × 0.4
= 2.26 N m (clockwise)
2
A
By F = ma,
Conventional (p. 173)
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
On the Earth:
1
By s = ut + at 2 ,
2
1
2 = 0 + (10)t E 2
2
(1A)
Thus, it takes more time for the block on
Mars to reach the ground.
2
(a) (i)
(1A)
normal
force
tE = 0.4
tension
T1 from
m1
M
weight

tension
T2 from
m2
(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
(ii)
(1A)
(1M)
normal
force
tension
T1 from
m1
500 × 3 = 700 × 1 + Y × 4
Y = 200 N
M
friction
(1M)
(1A)
Besides,
tension
T2 from
m2
net force = 0
(1M)
X + Y = 700 + 500
X + 200 = 1200
X = 1000 N
weight
(2 tensions with T1 > T2)
(1A)
5
(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.
82.5 m.
(c)
in a way similar to that in (b)(i) but
(1A)
along the path of the block.
of the system will be smaller. (1A)
3
Any one of the following:
Add a layer of oil / polystyrene beads
the magnitude of the acceleration
(1A)
(1A)
The displacement of the box is
If T1 > T2 + f, the masses will move
remain at rest.
(1M)
The acceleration of the box is 5 m s–2.
1
(1M)
(b) s = ut + at 2
2
1
= 4 × 5 + (5)(5)2 = 82.5 m
(1A)
2
Mass m1 accelerates downwards,
If T1 = T2 + f, the masses will
(1A)
Use air cushion.
6
(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
= 10 × 1
= 10 N m (anticlockwise) (1A + 0.5A)
(Weight of Joan)
(1A)
(Reaction from the balance on Joan)
(1A)

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
speed
Lift slows
down and
stops
(c)
< 500 N
lighter
(1A)
(1A)
(Correct force)
(1A)
(Correct label)
(1A)
(Correct force)
(1A)
(Correct label)
(1A)
Let R be the normal reaction acting on
Joan by the balance (the reading of the
scale) and W be the weight of Joan.
Take the upward direction as positive.
(i)
By F = ma
(1M)
(c)
For the 30-kg box,
R − W = ma
R − 500 = 50 × 3
R = 650 N
(1A)
The reading of the scale is 650 N.
(ii) Since acceleration is 0 and, by
Let T be the tension in the string.
By F = ma,
(1M)
T = 30 × 1.25 = 37.5 N
(1A)
The tension in the string is 37.5 N.
(d) Net force = 50 − T
= 50 − 37.5
F = ma, the reading of the scale
R = W = 500 N.
(iii) By F = ma,
12.5 N.
(e)
R − 500 = 50 × (−2)
(1A)
The reading of the scale is 400 N.
(1A)
The net force acting on the 10-kg box is
(1M)
R − W = ma
R = 400 N
= 12.5 N
(1A)
Her statement is not correct.
(1A)
When the string breaks, the net force
acting on the 30-kg box is zero.
(1A)
By Newton’s first law of motion, the
box will continue to move and its
velocity will be constant.

(1A)
8
(a)
v−u
t
2 −1
=
= 20 m s–2
0.05
a=
F = ma
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.
(1A)
(Correct forces)
(1A)
(Correct labels)
(1A)
(Correct forces)
(1A)
(Correct labels)
(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
The reaction of m1 (R) and the force
friction-compensated, such that the
weight component of the box along the
plane balances the friction acting on the
box.
(1A)
Therefore, the net force acting on the
box along the plane is zero and the box
acting on pan A by m1 (R′) form an
action-and-reaction pair.
(1A)
(b) The pans and masses would move
up/down at constant speed
(1A)
or remain at rest.
(1A)
will move along the plane with a
uniform speed after pushing the box

11
v / m s−1
(a)
t/s
(Correct forces)
(1A)
(Correct labels)
(1A)
(Axes with correct labels)
(The speed of the pot increases at a
decreasing rate.)
(b) The weight of Jackie is constant. (1A)
The air resistance acting on her increases
13
(a) (i)
(1A)
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
(1A)
(a) I do not agree with Gloria.
(1A)
(1A)
The air resistance acting on the flower
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,
(1A)
the downward net force acting on the pot
is always greater than or equal to zero.
By F = ma, the pot will not slow down.
(1A)
(b) Take the downward direction as
positive.
(ii) Take the direction down the plane
as positive.
1 2
at ,
2
1
2 = 0 + × a × 42
2
a = 0.25 m s–2
By s = ut +
(1M)
(1A)
The acceleration of the trolley is
0.25 m s–2.
(iii) F = ma
(1M)
= 1 × 0.25 = 0.25 N
The resultant force acting on the
trolley is 0.25 N (down the plane).
(1A)

(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 θ.
(1A)
The student is wrong.
(1A)
normal reaction
(Axes with correct labels)
(1A)
(The graph decreases linearly from A to
friction
B.)
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
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)
By v 2 − u 2 = 2as ,
2
(1A)
(1M)
(1M)
2
80 − 0 = 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
food parcel and thus the acceleration
(1M)
5
= 2.5 × 10 × 2
decreases (from point A to point B).
= 5 × 105 N
(1A)
(c)
I would adjust the thrust to balance the
Eventually, the air resistance balances
air resistance and the weight of the
the weight of the food parcel. The net
aeroplane.
force acting on the food parcel and thus
16
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
–1
(50 m s ).
(1A)
(1A)
(a) 0–4 s: The object moves with an
the acceleration becomes zero (from
(1A)
(1A)
acceleration of –6 m s–2.

(1A)
(1A)
(1A)
(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 θ =
1
During 4–8 s:
F = ma = 2 × 0 = 0
(1A)
θ = 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)
(Correct slopes : during 0−9 s,
slope = 3 m s–2; then slope = 0; final part
(Forces F normal to the wings)
(1A)
(Weight)
(1A)
(b)
F θ
F θ
steeper than the first part with negative
(3 × 1A)
slope)
θ
(v = 0 at the starting point and the end
point)
(1A)
–1
(v = 27 m s at t = 9 s)
(1A)
(b) The magnitude of the maximum
acceleration of the train is 4 m s–2. (1A)
18
(a) (i)
Net force along vertical direction
=3N–2N
= 1 N (downwards)
(1A)
(ii) Net force along horizontal
Consider the forces in the vertical
direction.
2F × cos θ = mg
(1A)
The aeroplane does not fly with uniform
velocity.
(1A)
This is because a net force, 2F sin θ, acts
on the aeroplane towards the left. By
F = ma, the aeroplane has an
direction
acceleration.
= 10 N – 5 N
= 5 N (towards the right)
weight
(1A)

(1A)
20
(a)
21
(a)
(Weight of passenger)
(1A)
(2 tensions)
(1A)
(Reaction from the platform to the
(Weight)
(1A)
passenger)
(b) The net force acting on the picture is
zero.
(b) Take the upward direction as positive.
(i)
Total time = 24 s
(1M)
Average speed =
80°
2T × cos
= 1 × 10
2
T = 6.53 N
Total distance during initial rise
= 50 – 2 = 48 m
Consider the vertical components.
2T cos θ = mg
(1A)
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, θ will be
smaller.
Since T =
(1A)
mg
, T decreases with θ.
2 cos θ
(1A)
Therefore, the tension in a longer string
is smaller and it is harder for the string
to break.
(1A)

23
Let P be the pulling force.
(a) The net force acting on the case is 0.
By F = ma,
(1A)
P + mg = ma
(b) Let T be the tension.
(1M)
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.
(c)
(1A)
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.
24
The acceleration of the balloon is
–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 = anticlockwise (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 = anticlockwise (1M)
moment
moment
resistance increases. As a result, the
acceleration of the sandbag decreases.
200 × 2 = 600 × d
(1A)
d = 0.667 m
!

(1A)
27
When the plank begins to tip, the
25
(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
2
2
2
2
(1A)
Then the ball-bearing moves with a
−2
constant speed called terminal speed.
By v − u = 2as,
(1M)
(1A)
(b)
0 − u = 2 × (−0.4) × 2.965
u = 1.54 m s
−1
(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.
When an aeroplane travels at a high
(1A)
(1A)
"

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)
The student must use a
In order to improve the speed of an
friction-compensated runway to carry
aeroplane, the body of the aeroplane
should be streamlined so that air can
(1A)
out this experiment.
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)
The mass of the weights is 0.244 kg.
(b) (i)
(ii) The acceleration of the trolley is
equal to the slope of the graph in
Figure w.
The acceleration is 1.24 m s−2.(1A)
(c)
According to Newton’s second law
(F = ma),
(1A)
the tension pulling the force sensor and
the trolley T = ma.
T = (0.333 + 0.718) × 1.24 = 1.30 N(1A)
This theoretical result is not close to the
result in (b)(i). The results are not in
accordance with Newton’s second law.
(1A)
(Force on boat by air)
(1A)
velocity
(1A)
towards the direction it is pushed.
of the string is 1.93 N
(1A)
(1A)
(ii) The boat moves with a constant
By the data in Figure v, the tension
(t = 1.1 s to 1.5 s).
(Weight)
(1A)
(b) (i)
As fans B and C blow air
backwards, an action force acts on
the air by the fans.
(1A)
Thus, an equal and opposite
reaction force acts on the fans by
the air.
(1A)
Therefore, the boat moves forwards.
(1A)
(ii) By F = ma,
0.2 + 0.2 = 1a
a = 0.4 m s–2
(1M)
v = u + at
(1M)
= 0 + (0.4)(5) = 2 m s
–1
Its speed is 2 m s .

–1
(1A)
(iii) Any one of the following:
(c)
(i)
(ii) The reading on the balance is not
(1A)
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
By F = ma,
(1M)
F
−6000
a= =
= −0.706 m s–2
m 5.5 × 103 + 3000
land on the ground and will be
stopped by the friction.
By v2 – u2 = 2as,
Control both fans B and C to blow
(1M)
2
0 – u = 2(–0.706)(30)
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
downwards. Therefore, an action force
acts on the air by fan.
(1A)
Force acting on truck Q by truck P
= force acting on truck P by truck Q (but
Then, an equal and opposite reaction
in opposite direction)
force acts on the fan of the toy by the air.
= –719 000 N
(1A)
Such upward reaction force is larger
than the weight of the toy. Therefore, the
toy can go up in mid-air.
(c)
(c)
(i)
(1A)
(1A)
(1A)
(d) Net force acting on truck Q
= force acting on truck Q by truck P +
friction
= –719 000 + (−3000)
Minimum upward force
= –722 000 N
(1M)
= weight of the toy
50
=
× 10
1000
By F = ma,
(1M)
= 0.5 N
–722 000 = 3000a
(1A)
a = –240.7 m s–2

(b) (i)
By v = u + at,
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)
(ii) The weight and the normal reaction
(1A)
(iii) The force acting on the plank by
the square object and the normal
provide an anticlockwise moment
force acting on the square object by
about X.
the plank.
(1A)
On the other hand, the clockwise
moment is zero.
(1A)
(iv) (1) F should be the force acting on
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.
(1A)
(1A)

1
fon:e and Motio
ha
Therefore. F is equal to the
.Y -:;
magnitude. which is 10 N.(IA)
10 )( I
( lA)
act.
(lA)
(2)
The result on the torque is not the
(v)
A
(lA)
same.
so the nonnat reaction acting on
the square object is nOI equal to
60 N
the weight of the square object
(JO ~) in mOlgnitude.
the torque ( = F x d).
36
(a )
i~
(lA)
" '""
3.0
><
(IM)
=-0.80
(I IKCEE 2006 Paper I Q-i)
Cvlomem) F
(lA)
(Correct arrow)
(I A)
Thereibre. F is dilTcrcnt and so
8
r '
The square object is accelerating.
3.75 m s-~
[)Cmcndicular di stance
(b)
(IA+IA)
from pivot
(i)
( lA)
Air resistance increases with SI)C{.-d.
(I A)
(lA)
1.501
so the net force (= forward thms! -
Assumption: unifonnlregular beam.
air resistance) decreases. and the
(I A)
(H)
(I A)
weight of the body seems to
=\O\lm
(i)
0.6 m
point on which the (entire)
= Fxd
( b)
( I M)
(iii) ( I ) The centre of gravity is the
(2) Torque
(a)
force and Motlo
40><x "" 60><(I - x)
"eight of the square object in
=
;)
(lA)
acceleration decreases.
(I)
(ii)
A
to m
c
B
When the air resistance is equal to
the forward thrust in magnitude.
(lA)
fzs
the net force and hence the
acceleration is lero. so the car
60N
reaches a constant specd.
(Pi\ot hct\\ccn the cube of
(c)
weight and the centre of the
beam)
(2)
(i)
( lA)
The ,"clocity ofthc ear decreases al
( lA)
a decreasing mtc.
(I A)
Lcl x be thc distance between
The car eventually travels at a
the new position of tile pivot
constant \elocity.
(ii)
and the original position.
(lA)
When the parachute is opened. the
Take moment about the joint.
ai r resistance becol11es grC31cr than
In equilibrium. clockwise
the thrust in magnitude. As a rcsuil .
momem
i~
the velocity decreases.
equal to the
(lA)
antic lock" i5e moment. (IM)
New Senior Secondary Physics at Work
70
Cl Oxford University Press 2009
Force and Mot
ha t@f' 3
Air resistance dc:;;fe.h......, "ith speed.
(b)
(I)
I
~e
2
2
(l A )
No net force .
(" )
(b)
(0)
(I M)
4 O+.!.(IO) r
the <:J!
tralclsatconqam\dc>\"lf!o_
37
.
Bp III +- ar.
and is fin all~ equal ILl the thru~ t in
magnilUde. Th..:rc::
force and Motio
1- 0.8945
(l A )
( i)
1\·\X\""l\X - ":.:
(l A )
( ii)
R = w -I\ - .
11.\)
(lA)
The man takes 0.894 s 10 HIli from Ihe
first floor to the ground \\ ithout the air
bag.
(I)
(ii)
0.5 5+ 0.1 s = 0.6 s
(I A)
The lime inle"31 bcl\\een the man
O.Sm
jumps lmd the bag is fully inflated is
!
H:;:::J=b=======::j
pC]
, 'I.
(Hi) From (i) and (iil.the time for air
,
be r<.>duced to a small value.
I,
m
15
1
3m R
0H·
0.6
'I
0.5m
rc"i"t;mcc actin!! on the man is ~
Q
w
BO N
~.
shon and the Iclocity orthc man cannot
20 N
(lA)
In (hi:. case. the man is mainly proK'Cted
(Do\\ O\\ ard for":':.1
(l A)
by the thick special cu"hion or the air
(Upward for.:e)
(l A)
bag when he reache:. the ground.
(Ill)
(Correcl dislance of all dOlI nIl ard
rorce ~ )
( ii)
(l A)
Tal-. e mome-m aboUllhe Pl\Ot.
In equil ibrium.
II' X
0.1 =l'iO
.:!n
...
11 \I)
(L\ )
11 "'600'
(COITCCI perpendicular distances of
the force s from the plI01.)
(1 ~1)
(Hi) R = 600 +80 -.:!O"" -OO '
(l A)
Physics in articles (p, 183)
(a)
When the air bag inflates. the air re-~lstance
acting on the air bag and the peThOn mcreases.
( l A)
When the air resistance is grealer than the
weight orthc person. the person \\ ill
down
(F ~
( I .\)
II/a).
1I is easier for a person 10 land
~Io\\
\I
jl h
falling speed.
New Senior Secondary Physics at Work
a 10\\ er
(I \ )
71
, Oxford University Press 2009
4
Work, Energy and Power
Practice 4.1 (p. 196)
9
(a) Component of force in the direction of
1
D
motion = 25 cos 50° N
2
B
Work = (F cos θ)s
3
B
= 25 cos 50° × 10
4
C
= 161 J
5
(a) Work = Fs = 375 × 10 × 1 = 3750 J
The work done by John is 161 J.
(b) The chemical energy of John converts to
The work done on the water skier by the
the kinetic energy of the sledge.
tension is 3750 J.
(b) Since the skier moves at uniform speed,
10
(a) Work = Fs = 10 × 3 = 30 J
The work done by F on the block is 30 J.
the energy gained by the skier is zero.
(b) Work = fs = –4 × 3 = –12 J
The work done on the skier by the
tension is used to overcome the gravity
The work done by f on the block is
and the friction/air resistance acting on
–12 J.
(c)
the skier.
The chemical energy of the source of the
6
The work done is 0.
force converts to the kinetic energy of
7
Work = Fs
the block.
500 = 10 × 10 × s
11
(a) Work = Fs = 10 × 10 × 0.8 = 80 J
The work done by the man is 80 J.
s=5m
8
The depth of the well is 5 m.
The chemical energy of the man
(a) Work = Fs
converts to the gravitational potential
30 = F × 1.5
energy of the box.
(b) No, he has not done work in this process.
F = 20 N
Yes, the man feels tired.
The size of the force is 20 N.
(b)
F = ma
Practice 4.2 (p. 202)
20 − 1 × 10 = 1(a)
a = 10 m s
–2
1
–2
The acceleration of the box is 10 m s .
(c)
2
2
By v – u = 2as,
Let mt and vt be the mass and the speed of the
thief respectively, and
2
v – 0 = 2(10)(1.5)
v = 5.48 m s
A
mp and vp be the mass and the speed of the
–1
The velocity of the box is 5.48 m s
–1
when it reaches 1.5 m above the ground.
policeman respectively.
1
1
m p v p 2 = mt vt 2
2
2
vp
mt
1
=
=
vt
mp
2

2
(c)
A
1
By KE = mv 2 ,
2
1
2.17 × 10–18 = × 9.1 × 10 −31 × v 2
2
this sport. This is because a short
weightlifter needs to move the barbell
for a shorter displacement in the
direction of the force applied. Therefore,
v = 2.18 × 106 m s–1
less work done is required.
Its speed is 2.18 × 106 m s–1.
3
10
B
(a) Change in the gravitational potential
energy of each worker
Her gain in gravitational PE
4
= loss in the gravitational potential
= mgh = 50 × 10 × 30 = 15 000 J
1
KE of the ball = mv 2
2
=
1
57
246.2
×
×
2 1000
3.6
energy of each worker
= mgh = 75 × 10 × (–3.5) = –2625 J
2
(b) The gain in KE of each worker = 0
(c)
1
mv 2 ,
2
1 14.9
521 = ×
× v2
2 1000
= gain in the kinetic energy of each
By KE =
worker + work done against tension
Since the platform is lowered in a
uniform speed, there is no gain in KE of
v = 264 m s–1
each worker. Then, the loss in
The speed of the bullet fired is 264 m s–1.
6
7
)
(
9
worker is equal to the work done against
)
Her gain in gravitational PE
= mgh = 50 × 10 × 72 = 36 000 J
8
gravitational potential energy of each
Gain in KE
1
1
= m v2 − u2 = (0.2 ) 30 2 − 5 2 = 87.5 J
2
2
(
tension.
Practice 4.3 (p. 212)
Speed of the passenger
1400
=
= 1.94 m s–1
12 × 60
1
C
2
C
3
B
KE of the passenger
1
1
= mv 2 = × 75 × 1.94 2 = 141 J
2
2
4
B
(a) (i)
Loss in the gravitational potential energy
of each worker
= 133 J
5
A short weightlifter has an advantage in
Work done
= Fs = 176 × 10 × 1.8 = 3168 J
(ii) Minimum force that each of his
arms acted on the barbell
176 × 10
=
= 880 N
2
(b) Work done
Work done against friction = loss in KE
1
Fs = m v 2 − u 2
2
(
1
9000s = × 1500 ×
2
)
72
3.6
2
−
36
3.6
2
s = 25 m
The distance travelled by the car when it
slows down is 25 m.
= Fs = 176 × 10 × 2 = 3520 J

5
(a) The chemical energy of the weightlifter
(b) By the law of conservation of energy, all
is converted to the gravitational potential
kinetic energy of the stone is converted
energy of the barbell.
to its potential energy when it reaches
(b) The electrical energy is converted to the
6
the highest point. Therefore, the
gravitational potential energy of the
potential energy gain is 0.5 J when it is
passengers and the lift.
at the highest point.
(a) PE = mgh = 80 × 10 × 6.14 = 4912 J
(c)
The gain in his gravitational potential
ground level be 0.
energy at the highest point is 4912 J.
PE = mgh
0.5 = 0.01 × 10 × h
(b) By conservation of energy,
kinetic energy when he left the ground
h=5m
= gain in his gravitational potential
The maximum height the stone reaches
energy at the highest point
is 5 m.
(d) By the law of conservation of energy, all
= 4912 J
7
(a) PE = mgh = 0.4 × 10 × 5 = 20 J
potential energy of the stone at its
The potential energy before it falls is
highest point is converted back to kinetic
20 J.
energy when it falls. Its kinetic energy is
(b) By the law of conservation of energy,
the potential energy of the ball is
(c)
0.5 J on hitting the ground again.
9
(a) Take the potential energy of the bob at
converted to its kinetic energy when it
the lowest level be 0.
falls. Therefore, its KE on hitting the
PE = mgh = 0.01 × 10 × 0.1 = 0.01 J
ground is 20 J.
1
KE = mv2
2
1
20 = × 0.4 × v2
2
2 × 20
v=
= 10 m s–1
0.4
Its potential energy gain after being
Its velocity on hitting the ground is
8
Take the potential energy of the bob at
10 m s−1.
1
1
(a) KE = mv2 = × 0.01 × 102 = 0.5 J
2
2
The kinetic energy of the stone is 0.5 J
when it is thrown from the ground.
raised is 0.01 J.
(b) By conservation of energy, the potential
energy of the bob in (a) is converted to
its kinetic energy at its lowest position.
Therefore, its kinetic energy is 0.01 J as
it passes its lowest position.
1
By KE = mv2,
2
1
0.01 = × 0.01 × v2
2
v = 1.41 m s−1
The speed is 1.41 m s−1 as it passes its
lowest position.

(c)
New total energy
1
= mv12 + original PE
2
1
= × 0.01 × 12 + 0.01
2
4
The potential energy of the bob at the
Usual output power
E mgh (60 × 10)× 10 × (3 × 20)
= =
=
= 18 kW
t
t
20
E
By P = ,
t
E 45 × 10 × 2000
t= =
= 45 000 s = 12.5 hr
P
20
other end = mgh = 0.015 J
The climbing time of Jack is 12.5 hours.
5
= 0.015 J
0.01 × 10 × h = 0.015
6
By P = Fv,
h = 0.15 m
10
Let v be Alex’s maximum speed.
The height of the bob above its lowest
P = mgv
point at the other end is 0.15 m.
30 = 65 × 10 × v
v = 0.0462 m s−1
(a) Gravitational PE of the carts and the
His maximum speed is 0.0462 m s−1.
passengers at A
= mgh = 5000 × 10 × 85 = 4.25 × 106 J
7
Let F be the force against friction.
P = Fv
(b) By the law of conservation of energy,
10 × 103 = F ×
loss in gravitational PE = gain in KE
1
mg(hA – hB) = mv2
2
Force against friction = 1200 N
Since the block moves at a constant velocity,
v = 2 × 10 × (85 − 53)
the net force acting on the block is zero and
= 25.3 m s−1
(c)
At B, the actual kinetic energy of the
carts and the passengers
1
1
= mv2 = × 5000 × 202 = 106 J
2
2
the friction = F = 1200 N.
8
15 × 103 = (120 + 70n) × 10 × 2
loss in gravitational PE
n=9
= gain in KE + work done against
The maximum number of people who can be
friction
= 5000 × 10 × (85 – 53) − 106
= 6 × 105 J
Let n be the maximum number of people that
the lift can raise at 2 m s−1.
mgh
P=
t
By the law of conservation of energy,
work done against friction
30
3.6
raised at 2 m s−1 is 9.
9
Work done by the engines
= gain in kinetic energy of the cars
For car A:
Practice 4.4 (p. 219)
Work done by car A’s engine
1
D
=
2
B
P=
3
D
E 60 000
=
= 500 W
t
2 × 60
100
1
1
mv2 = × m ×
2
2
3.6
2
= 386m J
Power of car A’s engine
work done 386m
=
=
= 64.3m W
time
6

v = 2 gh
For car B:
Work done by car B’s engine
100
1
1
= mv2 = × m ×
2
2
3.6
Therefore, if the block moves at 2v at B, the
2
height of the block should be 4h.
= 386m J
4
Power of car B’s engine
work done 386m
=
= 96.5m W
=
time
4
C
A:
= kinetic energy of the ball bearing at A
1
1
= mv2 = × 0.1 × 52 = 1.25 J
2
2
Therefore, car B’s engine can output more
power.
10
Work done by Stephen
B:
(a) Loss in PE of water per second
By the law of conservation of energy,
the kinetic energy at A is converted to
= mgh = 4000 × 10 × 500 = 2 × 107 J
potential energy at B. i.e.
1
mv2 = mgh
2
In each second, water of 4000 kg loses
potential energy of 2 × 107 J.
v = 2 gh
(b) Since all the potential energy is assumed
C:
to be converted into electrical energy,
down from B to A.
the power output of the generator is
7
2 × 10 W.
(c)
energy is lost in heating up the wire,
5
A
(1) By the law of conservation of energy,
generator, driving the turbine, etc.
potential energy at the highest point, i.e.
total energy
= mgh = M × 10 × 0.1 = M J
(2) By the law of conservation of energy,
B
the potential energy of the bob is
PE gained by the load
5
= mgh = 50 × 10 ×
= 2.5 J
1000
converted to kinetic energy at its lowest
D
Work done by the force
30
= Fs = 20 × 2π ×
= 37.7 J
100
3
energy.
total energy of the bob is equal to its
Multiple-choice (p. 222)
2
It is true by the law of conservation of
moving the movable parts of the
Revision exercise 4
1
D:
Not all potential energy of the water is
converted into electrical energy because
It is not true when the ball bearing rolls
C
By the law of conservation of energy, the
potential energy at A is converted to kinetic
1
energy at B. i.e. mgh = mv2
2
position. i.e.
1
mv2 = mgh
2
v = 2 gh
which is independent of the mass of the
bob.
(3) By conservation of energy, the bob will
move up to a point at the same level as A,
whether there is a pin at C or not.

6
A
Work done by the braking force
ball at A is equal to the kinetic energy of
= change in KE of the car
1
Fd2 = 0 – mv2
2
the ball at B.
Fd2 = 0 –
(1) By the law of conservation of energy,
the gravitational potential energy of the
1
mv2
2
1
0.1 × 10 × h = × 0.1 × 42
2
mgh =
−247m
F
−61.7 m −247m
d1 : d2 =
:
=1:4
F
F
The ratio of the braking distance of d1 to d2 is
(2) By the law of conservation of energy,
kinetic energy of the ball at B
= gravitational potential energy of the
ball at C + kinetic energy of the ball at C
1
1
mvB2 = mghC + mvC2
2
2
1
1
× 42 = 10 × 0.5 + × vC2
2
2
vC = 6 m s−1
(3) When the ball arrives at B and D, it has
the same gravitational potential energy
1 : 4.
9
D
10
(HKCEE 2006 Paper II Q5)
11
(HKCEE 2007 Paper II Q4)
12
(HKCEE 2007 Paper II Q31)
13
(HKCEE 2007 Paper II Q32)
Conventional (p. 224)
1
(1A)
gravitational potential energy.
D has the same kinetic energy, hence the
energy
= mgh
= change in KE of the car
1
Fd1 = 0 – mv2
2
40
1
m
2
3.6
−61.7 m
d1 =
F
(1M)
= 50 × 10 × 1860
Work done by the braking force
Fd1 = 0 –
(1A)
(b) His change in gravitational potential
same velocity.
A
(1A)
Then his kinetic energy converts to his
same level. Therefore, the ball at B and
8
(a) His chemical energy converts to
kinetic energy.
which is zero because B and D are at the
A
2
d2 =
h = 0.8 m
7
80
1
m
2
3.6
2
= 9.3 × 105 J
2
(1A)
distance
(a) Speed of Dora =
time
50
=
40
= 1.25 m s–1
1 2
mv
2
1
= × 60 × 1.252
2
KE =
= 46.9 J
(1M)
(1A)
The kinetic energy of Dora is 46.9 J.

T = 12.2 °C
(b) Dora loses all her kinetic energy over
(1A)
the last 1 m from side B. The loss of her
The temperature of water at the bottom of the
kinetic energy is due to the work done
waterfall is 12.2 °C.
4
against the decelerating force.
(a) As the man moves at a constant speed,
Let F be the decelerating force and s be
the tension acting on the man is equal to
the distance of travel.
the weight of the man. Hence, the
Fs = –46.9
(1M)
tension acting on the man is 700 N.(1A)
F × 1 = –46.9
(b) Work done by the tension
F = –46.9 N
(1A)
= Fs
(1M)
Therefore, the decelerating force is
= 700 × 15
46.9 N.
= 10 500 J
(1A)
should exert a force of the same size as
(Correct labelled axes)
(1A)
the decelerating force but in opposite
(Correct graph)
(1A)
direction, so that net force acting on her
(Correct values)
(1A)
(c)
Alternative method:
2
2
By v = u + 2as,
a=
v 2 − u 2 0 − 1.25 2
=
= −0.781 m s–2
2s
2 ×1
(1M)
The average deceleration is 0.781 m s–2.
F = ma = 60 × 0.781 = 46.9 N
(1A)
The average decelerating force is
46.9 N.
(c)
In order to swim at uniform speed, Dora
5
is zero.
Power of Dora = Fv
(1M)
= Fs
(1A)
= 8.128 × 106 J
3
the ground to the top floor
508
=
= 30.2 s
bottom of the waterfall and m be the mass of
water reaching the bottom.
1010
By the law of conservation of energy,
60
loss in gravitational potential energy
mgh = mc∆T
(1A)
(b) Time required to transport the load from
Let T be the temperature of water at the
= gain in internal energy
(1M)
= 16 000 × 508
= 46.9 × 1.25
= 58.6 W
(a) Work done on the load by the lift
P=
(1M)
(2M)
=
10 × 100 = 4200 × (T – 12)
E
t
(1M)
8.128 × 10 6
30.2
= 269 kW
(1A)
The power of the light in (a) is 269 kW.
!

(c)
(ii) By the law of conservation of
The actual power of the lift is larger than
that in (b).
energy,
(1A)
When calculating the actual power of the
loss in KE of the cylinder
lift, besides the maximum capacity of
= gain in PE of the cylinder + work
done against friction
the lift, the weight of the lift needs to be
6
taken into account as well.
1
(a) KE = mv2
2
1
72
= × 1500 ×
2
3.6
Loss in KE = mgh + fs
(1A)
8 = 1 × 10 × h + 5h
(1M)
h = 0.533 m
2
= 3 × 105 J
the cylinder is 0.533 m.
(1A)
(b) Let v be the minimum initial speed of
the metal cylinder to win the game.
3 × 105 J.
Loss in KE of the cylinder
= gain PE of the cylinder
+ work done against friction
1
× 1 × v2 = 1 × 10 × 3 + 5 × 3
2
(1M)
Work done against friction
= Fs = 500 × 1200 = 6 × 105 J
(c)
cylinder is 9.49 m s−1.
72
3.6
= 10 kW
(c)
72
0 −
3.6
Use a heavier metal cylinder.
Increase the friction between the
(1M)
cylinder and the support.
7
(a) (i)
Move the pivot of the plank closer to A.
2
8
= 2 × (−4) × s
s = 50 m
(a) The ball should be released at a position
1 m above the ground.
(1A)
(1A)
(b) Let H be the height that the ball should
Initial KE of the metal cylinder
1
= mv2
(1M)
2
1
= × 1 × 42
2
=8J
(2 × 1A)
Put the bell higher.
By v2 − u2 = 2as,
2
Any two of the following and other
reasonable methods:
(1A)
(d) Acceleration when braking
F −6000
= =
= −4 m s−2
m 1500
(1A)
The minimum initial speed of the metal
(1M)
= 500 ×
(1M)
v = 9.49 m s−1
(1A)
Power of the car engine
= fv
(1A)
The maximum height reached by
The kinetic energy of the car is
(b) Distance travelled in 60 s
72
= vt =
× 60 = 1200 m
3.6
(1M)
be released.
By conservation of energy,
mgH = mgh + 10% × mgH
(1M)
H = 1 + 0.1H
= 1.11 m
(1A)
(1A)
The ball should be released at a height of
1.11 m above the ground.
"

(c)
The energy loss of the ball due to the
Work done by Chris
work done against friction is converted
= 400 × 2.5 × 50 = 50 000 J
into the internal energy of the rail and
Chris is the winner.
the ball.
(1A)
She will not win.
(1A)
(b) (i)
kinetic energy of the ball as zero.
the kinetic energy of ball at A will not be
(1A)
(1A)
Power of Chris
50 000
=
= 833 W
60
zero. Therefore, the ball will not stop at
(1A)
(a) After the car starts its engine, it
accelerates.
(1A)
Power of Billy
20 000
=
= 1000 W
20
If she pushes the ball at the beginning,
9
Power
(ii) Power of Alan
6000
=
= 600 W
10
The calculation in (b) takes the initial
A and she will not win.
(1A)
(1A)
The real winner is Billy.
(1A)
11
(a)
It continues to accelerate because the
output power is greater than the power
against the frictional force (= fv). (1A)
Once the velocity increases to a value
such that the power against the frictional
force is equal to the output power, (1A)
the car moves with constant velocity.
(1A)
(b) For maximum speed v,
output power = power against friction
(1M)
80 × 1000 = 1600 × v
v = 50 m s−1
(1A)
(Labelled 4 forces: weight, tension,
friction, normal reaction)
(b) By F = ma,
a = 2 m s–2
−1
50 m s .
(1M)
(1A)
1
By s = ut + at 2 ,
2
1
10 = 0 + (2)t 2
2
(1A)
t = 3.16 s
(a) Work done by Alan
= 30 × 10 × 2 × 10 = 6000 J
Work done by Billy
= 80 × 10 × 25 = 20 000 J
(1M)
80 – 10 – 10 × 10 × sin 30° = 10a
The maximum speed of the car is
10
(4 × 1A)
(1M)
(1A)
The man takes 3.16 s to pull the block to
the top.

(c)
Fs
t
80 × 10
=
3.16
Power of the man =
= 253 W
(d) By v2 – u2 = 2as,
(d) The work done against the force acting
(1M)
on Fanny (between A and her lowest
position) is equal to her kinetic energy at
A.
(1A)
Let F be the minimum average force
(1M)
acting on her after passing A.
v2 – 0 = 2(2)(10)
v = 6.32 m s–1
Gain in KE
1
= m(v12 – v02)
2
1
= × 10 × (6.322 – 0)
2
F = 658 N
Let v be the required speed.
output power = power against friction
The minimum average decelerating
13
(HKCEE 2004 Paper I Q7)
14
(HKCEE 2005 Paper I Q2)
15
In this question, g is taken to be 9.8 m s−2.
(a) (i)
(1M)
∆E = mg∆h
= 88 J
−1
(1A)
(a) Loss in PE
= mgh
(1A)
(iii) Work done = Fs
(1M)
(1M)
= 5260 J
(1A)
108 = F × 0.4
(1M)
(1A)
= loss in PE − work done
(1M)
= 88 − 20
= 68
(1M)
1
KE = mv 2
2
1
68 = × 18 × v 2
2
(1A)
5260 J.
Fanny’s mechanical energy is not
conserved.
F = 270 N
(b) Gain in KE
At point A, Fanny’s kinetic energy was
(c)
v = 2.75 m s−1
(1A)
(c)
This is because her potential energy loss
at A is greater than her gain in kinetic
energy.
(1A)
(ii) 88 + 20 = 108 J
= 50 × 10 × 12
= 6000 J
1
(b) KE = mv2
2
1
= × 50 × 14.52
2
(1M)
= (16.8 + 1.2) × 9.8 × 0.5 (1M)
253 = 10 × v
12
(1A)
force acting on her was 658 N.
(1A)
v = 25.3 m s
(1M)
5260 = F × 8
(1M)
= 200 J
(e)
KE = Fs
(1M)
(1A)
Fanny’s mechanical energy is not
conserved because she has to do work
against the air resistance acting on her.
(1A)

(1M)
(1A)
(Graph starts at origin and forms a full
rounded peak.)
(1A)
(Exactly two cycles, i.e. 4 peaks, shown
but not arches.)
(1A)
(Height of peaks decreases and peaks
approximately equally spaced.)
16
(HKCEE 2006 Paper I Q3)
17
(a) PE required
(1A)
= mgh = 75 × 10 × 1.6 = 1200 J
(1M)
KE required
1
1
= mv 2 = × 75 × 0.80 2 = 24 J
2
2
(1M)
Total energy required
= 1200 + 24 = 1224 J
(1A)
(b) Let v be the minimum speed.
1
mv 2 = 1224
2
1
× 75 × v 2 = 1224
2
(1M)
v = 5.71 m s−1
(1A)
−1
The minimum speed is 5.71 m s .
Physics in articles (p. 228)
(a) A manhole gained the maximum potential
energy when it reached its maximum height,
which is 10 m as stated in the passage.
Maximum potential energy gained by a
manhole cover in the explosion
= mgh
(1M)
= 20 × 10 × 10
= 2000 J
2
(1A)
2
(b) By v – u = 2as,
(1M)
2
v – 0 = 2(10)(10)
v = 14.1 m s–1
(1A)
The speed of a manhole cover when it fell
back to ground was 14.1 m s–1.
(c)
The air resistance is assumed to be negligible.
(1A)

5
Momentum
Practice 5.1 (p. 238)
1
Momentum of runner A = mA vA = p
1
2
KE of runner A = m A v A = E
2
D
Take the upward direction as positive.
Momentum of runner B = 2mA vA = 2p
1
KE of runner B = (2m A )v A 2 = 2E
2
Change in momentum of the ball
= mv − mu
= 0.5 × 6 – 0.5 × (−8)
The momentum and kinetic energy of runner
= 7 kg m s–1
2
B are 2p and 2E respectively.
C
Momentum of the tennis ball
57
= mtvt =
× 60 = 3.42 kg m s–1
1000
7
(a) Momentum of the object before the
force acts
= mu = 2 × 5 = 10 kg m s–1
(b) Momentum of the object after the force
Momentum of the football
has acted
= momentum of the tennis ball
= mv = 2 × 10 = 20 kg m s–1
= 3.42 kg m s–1
(c)
mfvf = 3.42
= mv − mu = 20 − 10 = 10 kg m s–1
400
× v f = 3.42
1000
vf = 8.55 m s
Gain in momentum
(d) Force acting on the object
change in momentum 10
=
=
=2N
5
time of impact
–1
The velocity of the football is 8.55 m s–1.
8
(a) The cushion in the glass column can
3
B
4
B
reduce the force of impact acting on the
5
Magnitude of momentum of the boy
peanut by lengthening the time of impact.
= mv = 60 × 4 = 240 kg m s–1
Therefore, the peanut does not break.
(b) The cushion in an envelope can reduce
Magnitude of momentum of the girl
= mv = 40 × 6 = 240 kg m s–1
the force of impact acting on the fragile
The magnitudes of momenta of the boy and
items by lengthening the time of impact.
the girl are the same.
KE of the boy
1
1
= mv 2 = × 60 × 42 = 480 J
2
2
KE of the girl
1
1
= mv 2 = × 40 × 62 = 720 J
2
2
6
9
(a) Take the travelling direction of the car
as positive.
mv − mu
F=
t
80 × 0 − 80 × 20
=
1
= −1600 N
The girl has larger kinetic energy.
The force acting on the driver is 1600 N
Let the mass of runner A be mA and the
opposite to the travelling direction of the
velocity of runner A be vA.
car.

(b) Since the force acting on the driver is
10
For football Q:
huge (1600 N), the driver will be
Take the initial travelling direction as
seriously hurt or even dead if he/she
positive.
does not wear a seat-belt.
Net force =
mv − mu
t
m × (−15) − m × 20
=
0.5
(a) By v2 = u2 + 2as,
v = 2 gs = 2 × 10 × 10 = 14.1 m s–1
When the dry cell hits the ground, its
= −70m N
speed is 14.1 m s–1.
mv − mu 0.02 × 14.1 − 0
= 70.5 N
(b) F =
=
t
4 × 10 −3
Force from the ground + weight = net force
Force from the ground = −70m − 10m
= −80m N
The net force acting on the cell is
Football Q experiences a force of −80m N.
70.5 N.
11
Therefore, football Q experiences a larger
(a) Take the initial travelling direction of
water as positive.
mv − mu 0 − 15 × 25
=
= −375 N
F=
t
1
force.
13
(a) The shaded area represents the impulse
of the force acting on the trolley and
impulse = Ft, where F is the force on
The water experiences a force of 375 N,
trolley and t is the time of impact.
in a direction opposite to its initial
The area is also equal to the change of
travelling direction, by the wall.
momentum of the trolley,
(b) By Newton’s third law, the force exerted
i.e. impulse = Ft = mv – mu
by the water on the wall is equal to the
where m is the mass of the trolley, and u
force exerted by the wall on water.
and v are the velocities of trolley before
Therefore, the force exerted by the water
on the wall is 375 N, in the initial
travelling direction of the water.
12
and after impact respectively.
(b) Time of impact = 0.3 s
(accept 0.3−0.4 s)
For football P:
Area under curve = Ft = 0.47 N s
0.47
F=
= 1.57 N
0.3
Take the initial travelling direction as
positive.
mv − mu
t
m × (− 15) − m × 20
=
0.5
Net force =
= −70m N
(accept 1.175−1.57 N)
The force acting on the force sensor is
1.57 N.
(c)
From the v–t graph, the velocities of the
Force from the wall = net force on P
trolley before and after impact are
Football P experiences a force of −70m N
0.36 m s−1 and −0.35 m s−1 respectively.
from the wall.
Note that the direction of the initial
velocity of the trolley is positive.

By Ft = m × (v − u),
−0.47
Ft
m=
=
= 0.662 kg
v − u − 0.35 − 0.36
The mass of the trolley is 0.662 kg.
(b) This is because the mass of the earth is
huge and the motion of the earth is not
noticeable.
7
Let v be the common velocity of the bullet
and trolley.
Practice 5.2 (p. 257)
By conservation of momentum,
1
B
2
total momentum before impact
C
3
= total momentum after impact
A
4
(a) This is an inelastic collision. Some of
mbulletubullet + mtrolleyutrolley = (mbullet+ mtrolley) × v
0.006 × ubullet + 0.8 × 5 = (0.006 + 0.8) × 8.5
the kinetic energy of the bullet is
converted into internal energy of the
The velocity of the bullet before the impact is
apple.
475 m s−1.
(b) This is an inelastic collision. The kinetic
energy of the car is converted to sound
(c)
ubullet = 475 m s–1
8
By conservation of momentum,
energy and internal energy of the wall
total momentum before hitting the ball
and the car.
= total momentum after hitting the ball
This is an elastic collision. The kinetic
0 = mKathyvKathy + mballvball
energy of the puck does not change.
5
0 = 3 + 0.3 × vball
Take the moving direction of the shell as
vball = –10 m s–1
positive.
The velocity of the volleyball after
By conservation of momentum,
impact is –10 m s−1.
3
mv − mu
(b) F =
=
= 12 N
t
0.25
total momentum before firing the shell
= total momentum after firing the shell
0 = mshellvshell + mcannonvcannon
The average force acting on Kathy is
0 = 5 × vshell + 8000 × (–0.08)
12 N.
vshell = 128 m s
9
–1
−1
The velocity of the shell is 128 m s .
6
(a) Take backwards as positive.
(a) It does not contradict the law of
conservation of momentum. We
consider the system which includes the
ball only. There is force acting on the
ball by the ground during the impact and
also there is weight acting on the ball,
(a) By conservation of momentum,
total momentum before collision
= total momentum after collision
mwhiteuwhite + mblueublue
= mwhitevwhite + mbluevblue
0.135 × uwhite + 0
= 0.135 × 0.2 + 0.135 × 0.5
uwhite = 0.7 m s–1
providing external net forces on the
The speed of the white ball when it hits
system. Therefore, the total momentum
the blue ball is 0.7 m s−1.
of the system is not conserved.

(b) Total KE before the collision
1
1
= mwhite(uwhite)2 + mblue(ublue)2
2
2
1
= × 0.135 × 0.72 + 0
2
(b) Before collision:
Velocity of trolley A, uA
= slope of the graph
0.9 − 0.2
=
2.6 − 0.6
= 0.0331 J
= 0.35 m s–1 (accept 0.34−0.39 m s−1)
Total KE after the collision
1
1
= mwhite(vwhite)2 + mblue(vblue)2
2
2
1
1
= × 0.135 × 0.22 + × 0.135 × 0.52
2
2
Velocity of trolley B, uB = 0
After collision:
Velocity of trolleys A and B
= slope of the graph
1.5 − 0.9
=
6 − 2.6
= 0.0196 J
Since there is loss of total kinetic energy,
= 0.176 m s–1 (accept 0.16−0.19 m s−1)
the collision is inelastic.
10
(c)
(a) Let v be the velocity of the boat after
= m Au A + m Bu B
dropping water.
= 0.5 × 0.35 + 0
By conservation of momentum,
= 0.175 kg m s–1
total momentum before dropping water
Total momentum after collision
= total momentum after dropping water
= (mA + mB)v
mboatvboat = (mboat + mwater) × v
= (0.5 + 0.5) × 0.176
0.45 × 1 = (0.45 + 0.01) × v
= 0.176 kg m s–1
v = 0.978 m s−1
Momentum is conserved within limits of
The velocity of the boat after dropping
water in it is 0.978 m s −1.
(b) Total KE before dropping the water
1
1
= mu2 = × 0.45 × 12 = 0.225 J
2
2
Total KE after dropping the water
1
= mv2
2
1
= × (0.45 + 0.01) × 0.9782
2
= 0.220 J
(c)
Since the total kinetic energy of the boat
and water before ‘collision’ is not equal
to that after ‘collision’; and the boat and
water ‘stick’ together after the collision,
Total momentum before collision
experimental error.
12
Let u and v be the speeds of the white ball and
the red ball after the collision respectively.
Along the direction in which the white ball
travels before collision:
By conservation of momentum,
0.135 × 1.8 = 0.135 × u cos 30° +
0.135 × v cos 60°
3
1
1.8 =
u+ v
2
2
v = 3.6 − 3u …………(1)
Along the direction perpendicular to which
the white ball travels before collision:
the ‘collision’ is inelastic.
11
(a) It is a completely inelastic collision.

5
By conservation of momentum,
D
0 = 0.135 × u sin 30° − 0.135 × v sin 60°
1
3
0= u−
v
2
2
All choices (A−D) follow the conservation of
u = 3v …………(2)
and the total KE of the balls after collision is
Substitute (2) into (1):
greater than that before collision.
momentum.
D violates the law of conservation of energy
v = 3.6 − 3 × 3v
v = 0.9 m s
6
−1
C
Along the direction in which fragment P
−1
Substitute v = 0.9 m s into (2):
u = 3v = 1.56 m s
travels:
−1
Take the direction in which fragment P
travels be positive.
The speeds of the white ball and the red ball
−1
after the collision are 1.56 m s and 0.9 m s
−1
By conservation of momentum,
respectively.
vP + vR cos 45° = 0
3 + vR cos 45° = 0
Revision exercise 5
vR = −4.24 m s−1
Multiple-choice (p. 261)
The speed of fragment R is 4.24 m s−1.
1
C
7
A
2
B
8
A
By Ft = mv – mu,
1.5 = 0.024 × v − 0.024 × (−15)
9
(HKCEE 2007 Paper II Q29)
10
(HKCEE 2005 Paper II Q45)
v = 47.5 m s
3
4
–1
The speed of the ball is 47.5 m s–1 when it
Conventional (p. 262)
leaves the racket.
1
(a) Change in momentum
= area under F–t graph
1
= × 0.15 × 18 000
2
(1M)
mv − mu
By F =
, the largest net force acting
t
on the object is represented by the steepest
= 1350 N s
(1A)
slope of the graph.
The change in momentum of the ball is
D
1350 N s.
D
110
96
− −
3.6
3.6
÷ 0.0013 = 44000 N
B: F =
120
98
− −
3.6
3.6
÷ 0.0018 = 33600 N
C: F =
130
112
− −
3.6
3.6
÷ 0.0021 = 32000 N
140
109
D: F =
− −
3.6
3.6
÷ 0.0015 = 46100 N
A: F =
(b) Average force on the ball
impulse
=
time of impact
=
(1M)
1350
0.15
= 9000 N
(1A)
The average force experienced by the
ball is 9000 N.

2
The velocity of trolley B was 2.5 m s−1
(a) Take the initial direction of the bullet be
positive.
mv − mu
F=
t
3 × 10 −3 × (−500 − 500)
=
0.01
after collision. Its direction was the same
as the travelling direction of trolley A
(1M)
before collision.
4
= −300 N
(a) Initial momentum of the jet fighter
= mfighterufighter
(1A)
= 8000 × 100
The force on each bullet is 300 N, in the
= 800 000 kg m s–1
same direction as the final travelling
= mmissilevmissile
(b) By Newton’s third law of motion,
= 10 000 kg m s–1
= −(average force acting on the
total momentum before firing
= total momentum after firing
total time
× (−500 − 500)
60
= mfightervfighter + (mmissilevmissile) × 5
(1A)
800 000 + 0
The average force acting on Superman is
= 7900 × vfighter + 10 000 × 5
2.5 N, in the initial direction of the
vfighter = 94.9 m s–1
bullets.
(1A)
The velocity of the jet fighter after firing
(a) By v2 = u2 + 2as,
(1M)
five missiles is 94.9 m s–1.
v = u 2 + 2as
(c)
= 0 + 2× 2× 4
=4ms
(1M)
mfighterufighter + mmissileumissile
= 2.5 N
3
(1M)
By conservation of momentum,
bullets)
= − total change in momentum of bullets (1M)
50 × 3 × 10
(1M)
= 20 × 500
average force on Superman
=−
(1A)
(b) Momentum of each missile when fired
direction of the bullet.
−3
(1M)
By conservation of momentum,
total momentum before firing
–1
The velocity of trolley A was 4 m s
(1A)
= total momentum after firing
−1
mfighterufighter + mmissileumissile
(1M)
= mfightervfighter + (mmissilevmissile) × 5
before collision.
7900 × 94.9 + 0
(b) Take the travelling direction of trolley A
before collision as positive.
= 7800 × vfighter − 10 000 × 5
By conservation of momentum,
vfighter = 103 m s–1
total momentum before collision
The velocity of the jet fighter after firing
= total momentum after collision (1M)
another five missiles is 103 m s–1.
5
mAuA + mBuB = mAvA + mBvB
vB = 2.5 m s
–1
(a) Magnitude of impulse
= Ft
1 × 4 + 2 × 0 = 1 × (−1) + 2 × vB
(1A)
(1A)
(1M)
= 80 × 0.10
=8Ns
(1A)

2
Force and Motlo
( b)
ha ter 5
Assume the direction ofthc forc c
(c)
applied is as shown in the figure below.
Momentu
Let v be the velocity of the vehicles just
after :he collision.
Along the direction towards the north:
o
Take the direction 10\\ ards thc nonh as
positive.
F
B} conscr"\ ation of lllomentmn.
Along the direction in \\ hich the puck
IOtal momentum before collision
travcls before impact:
,., tolalmomcnlum aflercollision (lM)
1ll0111enlJlll before impact + impulse
3000 x 18 sin 45" + 2500 )( 21.6 sin 45"
(IM)
= momentum alter impact
= (3000
0.25 x 15 + (- 8)cos 0 '" 0
v = 13.9m ~-'
0 = 62.00
(c)
+ 2500) )( v
( lA)
The velocity of the vehicles is 13.9111
Along the direction perpendicular to
IOwanls the north just after the
colli~ion.
which the puck travels before impact:
(I A)
velocity of the puck alter impact
= velocity bctore impact
7
+ impulse( I M)
(a)
Let
I'
be the \ elocity of can Yafter
col[ision.
0 + 8 sin 62.0°
= 7.07
n: s'
1.2 m 5-,1/
( lA)
2m5- 1
1"
m
--
"-
Along the original moving direction of
== 18ms '
cart X:
The velocity of the mi ni bus is 18 m
Si
By conservation ofmomemurn.
towards lhe Ilortheasljust before
75 )( 2 = 75 x 1.2 cos 45 ° + 150)( vcos 0
(I A)
collision.
0 = 0.576 ......... (1)
l'COS
Along Ihe direction lowards the cast:
Along the direction perpendicular to the
Take the direction towards the cast as
original moving direction of cart X:
positivc.
By conservmion of momentum.
By conservation of momentum.
o
lotal momentum before collision
o
o
lI,ot= 21.6ms ·1
In
s i
New Senior Secondilry PhYSiCS at Work
sin 0
( lA)
36.4"
rrorn ( 1):
the nortlmcst jusl before
collision.
I'
(2) + (1)'
0.424
tan O" " - 0.576
3000 )( 18 cos 45° - 2500 x II..,., COS 45°
The vclocilY oflhe car is 21.6
75)( 1.2 sin 45" - 150)(
\' sin 0= 0.424 ........ (2)
( I M)
= lotalmomenlutn after collision
10\\ ards
(45
v "
(IM)
1I + 1fI
= 22 + (2»(2
(b)
"
,,
7.07015 ' .
C')
,
0-=-"----0:10
,
The speed of the puck after impact is
6
S- l
I'
(lA)
cos 36.4° "., 0.576
1' : 0.7[6111 5
89
I
(lA)
Oxford University Press 2009
1:
force and Motio
ha ter 5
After the coll ision, cart )'moves at a
(b)
9
(a)
When the plunger is released. clastic
velocity of 0.716 m s I at an angle of
potential energy ofthc spring
36.40 to thc originAl moving direction of
is convertcd into sound energy :md (I A)
cart X.
kinetic energy of trolleys.
(I A)
By conservat ion of energy,
( I A)
Bcfore the collision. total KE of carts X
and Y
I
,I
=- m lll\~ - 2
.
2
(b)
,
sound energy + K E of trolleys
I
,
,
5 x 0.7 = - (11/ ,1'1- + 1118 " 8· ) . ....... ( I)
2
=
111111)·
2
2
(I A)
= 150 J
(lA )
After the col lision, total KE of cam.\'
and }'
I
,I
(c)
,
By conservation of momentum,
(I A)
III 11'1 = 11/81'8 • ....... (2)
(I A)
From (2).
v~ =
= -II/\I'r- + -III}l'r-
0.5 x
I
_
, I
,
= - x 7";) x (Ut +- x 150 x 0.7162
2
Substitute
2
~
(lA)
elaSlic potential energy oflhe spring
=..!..x 75 x 21 +..!.. x 150 x 0 2
2
92.4 J
I"H ""'
collision. the collision is incla~tic. (I A)
~
change in momentum
By/' =
(IM)
lime of impact
2=
1'1
Xl' S
1'1 = ~31 '1I
X
into (I),
(~31'8 )~ +
1.08 ms
[.5
X I's 2]
I
""-3 x 1.08 =-3.24
111
Si
The veloc ity of trolley A is 3.24 m 5
(Iowards the left).
0.Sx(0.5 - u)
1I =-2ms ·1
(Iowards the right).
S-I
before
10
il
(:I)
I
( I i\)
The velocity of trolley /J is I.OS
I
The speed of 1\ ater is 2111
- 1.5
5 x 0.7 =..!. [0.5
2
(lA)
Since the carts lose kinet ic energy in the
(h)
Momentu
III 5 I
(lA)
Take the travelling direction o r lhe bullet
hits Victor.
(I A)
after collision as positive.
I don '! agrce with him.
( lA)
Let A denote the bullet and /J denote the
This is because friction and nomlal
block.
reaction acts on thc fcet of Victor when
By conservalion of momentum.
he takes shower. The friction and nonna[
lolal momentum before collision
reaction balances the force acting on him
::0:
by the water (zero net force).
( lA)
lotal momentum after coll ision (IM)
11/ ,Ill
+- IJIsll8 = (11/ , + II/R) x
I'
0.0511 , + 0 = (0.05 + I) x 5
When external nct force acts on objects.
11 1 =
conservation of momentum is not valid.
105
I11S-'
(lA)
However. it is correct in the absence of
The speed oflhe hullet is 105 III 5 I just
external net force.
before the collision.
New Senior Secondary Physics at Work
(lA)
90
" Oxford University Press 2009
2
Force .nd Hatlo
ha ter 5
(b) Average force
change in momentum
(b)
Change in KE of the team
I
,
= -(600 + 70 x22)x6-
( lA)
0.2
The average force acting on the block by
(a)
2
the bullet is 25 N. in the travelling
= 38520J
direction of the bullet.
A, er.lge POI' er
increase in KE
By
,
,
+ 2(/:;.
t'- = 111/
2
(IM)
23.3 m
3
arrives at the sur/acc of the cushion is
= 12840W
23.3 m S·I .
The average power of the team in the
Impulse =
=
(c)
50x (23.3 - 0)
The impulse acting on the
parkcd at the pier
is
By thc conscnatiOI1 of momentum.
Average forcc acting on the person
impulse
=
( lA)
timc of impact
when athletes mOlled forwards and
stepped on the pier. the boat would
Since the cushion can lengthen Ihe lime
the same magnitude.
of impact.
Therefore. they need a rope to lix the
moved hackwards \\ ith mOlllentum of
( lA)
13
(3)
Take the direl:tion lowards the right as
cushion reduces the chance of injury;
positive.
therefore it saves peoplc.
(i)
( lA)
Nct force on the waler ejected
1111 ' - 111//
According to Newton's third law of
motion.
( ")
=
whcn the men paddle. an action force
\\ hich is in backwards di rCClion acts on
the water by the paddles.
(I M )
0.5 x (1O - 0)
= 5N
(I A)
By Newton's 3rd law. the net force
A reaction forcc which is in forwards
(thrust) acting on thc rocket is 5 N
direction then acts on the paddles by the
water.
(lA)
position of the boat before landing.
falling on the cushion is reduced so the
(" )
(I A )
1165 N s.
the average force acting on the person
12
The tOlal momentum of the boat and the
athletes \\as zero when the boat was
( lA)
per~on
( lA)
first 3 seconds is 12840 W.
(IVI)
1111' - /J/l1
= 1165Ns
(c)
(I M )
" ~3,,g =52:::0
( lA)
S i
The velocity of the person when he just
(b)
(I M)
lime taken
== 0 + 2( IOX30 - 2.75)
1'=
1
thc men travel at 6 m 5. 1.
Ix(5 - 0) = 25 N
11
1' = /I "' ul=0 + 2x3 = 6ms
After aecclerating for 3 s.thc boat and
(lM)
time of impact
Homentu
(towards the left).
(I A)
(I AI
Therefore, the boat can movc forwards
by paddling.
New Senio- Secondary Physics at Work
91
Cl Oxford university Press 2009
(ii)
By conservation of momentum.
(0)
Air cushions lengthen the time of im[h1ct
(I A)
momentum be fore launching
for people falling from a height.
'" momentum after launch ing
so il can reduce the fo rce acting on the
o~ II/, V, + 111" 1',,
0 = (2 - 0.5) x
I'r
people when they fall on the cushion and
(1:-'-1)
+5
reduce Ihe chance of injury.
1',=-3.33ms · 1
(lA)
(lA)
The rodet mme!> aI3.33 m s'
15
(a ) The inilial vc1ocityof B is - 10 III
5-
(I A)
towards the lefl.
(b)
Since the thrust is less than the weight of
and its final velocity is -3.5 m s r. (I A)
Ihe rocket. the rocket cannot ny up in air.
(b)
By conservation of momentum.
total mOlllentum before collision
(I. \)
Any onc of the following modifications:
= 10lal momenlum after collision (I M )
(I A)
+ 11/8118
1111111
More water can be ejectcd.
(" )
( i)
" I'""-6ms
( lA)
case~
- 6 illS
arc the sa111e.( I A)
(0)
I,
11/
x ( 1'.1 -
Ill)
ball just bc lore colliding \\ ilh Ihe
· 0.2< (-6 -20)
platc depends on the height of
=
~am e
in
-5.2 kg m S-1
0.5 s.
after collision is the same in both
~
orce ""
(I AI
(ii) Force acting on an object F
= change in momentum
time of impJcl
(lA)
(lA)
- 10.4 N
A fo rce of -\ 0.4 N acts on block A
during collision.
16
for both plates.
I
F~
and
time of impact
(,)
( ;)
Take the d irection oflhe final
\ elocity of the trolley as positive.
Change in momentum
I
'=
FIwIl :F""r.""-'0.1 0.1
New Senior Secondary PhySiCS at Work
(lM)
- 5.2
the ball after colli~ion b the same
"" 1 : I
change in momenlum
lime of imp..1cl
._0.5
Since the change in momentum of
I
(lA)
From Figure h. the lime of impact is
both cases. And the \ eIOCII) j U!>1
cases.
( lA)
Change in momentum of block A
This is because Ihe \("Iocit) oflhe
release. h. \\ hich i!> the
l
The \ clocity of block A after collision is
Yes. the momentum change of the
ba1! in both
11/ rl' l ~ 1118 " 1/
,--, 0.21' 1+ 0.8 x (- 3.5)
The harder the plate. the shon("r the lime
ofimpacl.
(h)
=
0.2 x 20.,. 0.8 x (- 10)
Water can be ejected al a higher speed.
I.
1
1111' - (- 1111')
"" 21111'
(IM)
(I A)
(lA)
92
, Oxford University Press 2009
2
Force and Motio
ha ter 5
17
Change in KE
I
2
=' -1/11'
2--I IIII' ~~
2
Average force
(IM)
(b)
(lA)
(c)
(lA)
2
(I M)
A\erage net force
= change in momentum
- 350
1.2
(I M)
N
(I A)
The:1\ erage net force acting on him
\\ hen he reaches the ground is - 292 N.
time of impact
11/1'
(d)
(I A)
~ -
2,
trol1cy~wall
( lA)
~--=-292
Average force
= change in momcntum
Momentum of the
Change in momentum
time of impact
I
~
-1111'- = -- IIII'~
2
- 292
system is
not conserved in both (a)
(I A)
:md (b).
(I A)
net force = force by the ground
\\eight
(IM)
=
F..,.. 700
r '"'--992N
(I A)
The average force acting on him by the
ground is -992 N.
Since the wall is fixed on the ground. the
ground exens a force on the wall in the
(c)
l Ie bends his knecs to increasc the lime
collisions and the momentum of the
of impact and hence reduce the loree
trolley-\\all system is not cOlbervcd.
acti ng on him \\ hen he reaches the
(lA)
(I A)
ground.
The momentum ofa mechanical system
18
(a)
Before Ihe impact. the velocity orthe
(I A)
is conserved only ifno external force
exerts on the system.
(d)
lTl.
= 70 x (0-5)
(I A)
Change in KE
1 ,
(c)
(lA)
= -350 l\' s
~ 0 - (-1/I1') = 1II1'
(ii)
1.25 m
'" m(l· - /I)
Change in momentum
0-
(IM)
Gilbert jumps from a height of U5
21111'
=
- Il = 20s.
.\.=
time of impact
(i)
\ ,2
(lA)
= change in momentum
(b)
By
5 ~ - 0 = 2(10)5
~ O
(ii)
Take the downward direction as positive.
(a)
(IM)
Momentu
After the impact. the \'c locity of the ball
The collision in (a) is clastic and
( lA)
the collision in (b) is inelastic.
(I A)
l~ 0.36ms
(b)
(lA)
1.
Impulse = IJIV-1/I1/
(IM)
Thi s is because the trolley in (a) does not
= 0.2 x (- 0.36
lose kinetic energy in collision while the
= -O.156Ns
trolley in (b) loses kinet ic energy in the
collision.
(c)
(lA)
Avemge force
=
0.42)
impulse
time of impact
(lA)
(IM)
-0. 156
1.75 1.6
=-I.04N
New Senior Secondary Physics at Work
93
(lA)
() Oxford University Press 2009
2
ha
Force and Motio
I.
(a)
er 5
Momentu
By conservation ofmomcntulTI.
(The impact timc of the driver and the
momcntum before the actor reaches the
windscreen becomes OA5 s. three times
m
the original impact time.)
=
21
momentum after the actor reaches the
car
(a)
1' =
Assume their velocity after the collision
is as shown belo\\.
(I")
1000 x 20 = (1000 + 75)
(1 A)
y
;,
I'
IS.6ms I
(I A)
The speed of the car is [8.6 m S-I when
the actor reaches the car.
(b)
A Sms "
"---'-"'-'----1' - (}- - - - - - - .... x
Increase in KE
I
,
,
4
=-x 75 x(,,--18.6-)= 10
2
,, =:: 24.8 ms- I
(lM)
Impulse =
( I")
Ill\> - 11111
60 kg
( lA)
8 70 kg
Along .\"-a.\is:
= 75 x (24.8 - 18.6)
= 465Ns
By consenation of momentum.
( lA)
60 x 5 = (60 + 70) x I'COS 0
The impulse on the actor is 465 N s.
20
(u)
I'COS
Area under the graph
,
I
=-
x 0.15 x 25 000
30
0 =- ..
13
..(1)
Alon£y-axis:
=1875Ns
(b)
(I M)
By conservation of momentum.
(I A)
70 x 6 = (60 + 70) x "sin 0
The area under graph in (a) is the
I'
impulse experienced by the driH'r. (l A)
(Or it is the change in momentum oflhe
sin 0 =_
42 ......... (2)
13
(2) + (1),
42
30
dri,'er.)
tan 0 =-
,,, (,)
0 = 54.5 0
"
(I A)
From (I):
1"
30
cos 54.5 °
13
3.97 m S-I
I'
(lA)
Their velocity is 3.97 m S·l, at an angle
tI ' "J
of 54. 5° to thc original direction orA.
(b)
2
(The largest force experienced b~ the
driver becomes
8330~.
Total loss in kineti c energy
=c x60x5 +
onc-third of the
original force.)
-G
" A)
=
New Senior Secondary Physics at Work
94
986 J
~ X70 X 6 2 )
1
x 130 X3.97 )
(I M)
(lA)
10 Oxford University Press 2009
2
22
ha ter 5
Force and Motio
(:1)
(i)
(ii)
Momentu
The shaded area is the impulse
It is smaller than that by the steel plunger
acting on the force sensor.
in (a)( ii).
(lA)
Let F be the average force acting
I f a man is knocked down by a car. with
on the force sensor.
the same change in momentum of the
From the graph.
man. the average force acting on the
man \,ould be smaller if the bumper is
t ime of impact
=
3.021 - 3.012
(lA)
softer.
"'- 9x10 1 s
A::. a result. the injury caused to the Illan
(lM)
Impulse = arl!a under F- I graph
can be reduced and a softer bumper is
Impulse = FI
safer to the public.
0.24457 = F x 9 x 10
23
.1
F = 27.2N
(a)
(i )
Before the impact (i.e. whell
1<
(I A)
(1 A)
2.50 s). trolley IJ moves towards
An average force 01'27.2 N
the len \\ ith an average speed of
(towards the right) acts on the force
about0,42ms- l .
sensor by the trolley during the
During the impact (i.e. from
impact.
r= 2.50 s to f '" 2.75 s). trolley B
(lA)
decelerates. It becomes
(iii) The average force acting on the
momentarily at rest at
force sensor by the trolley and the
f ~
2.66 s
(I A)
average lorce acting on thc trolley
and then ren'rses its travelling
by the force sensor lonn an
action-and-reaction pair.
(h)
( lA)
(I A)
direction.
By Newton's third law.lhese
Afterthe impact (i.e. when
forces have the same magnitude
f
but in opposite direction. i.e. the
about -0.15 m s
average force acting on the trolley
(The acceptable range o f time of
by the force sensor is also 27.2 N
impact is from 1 = 2.50- 2.52 s to
(towards the Jell).
1=
Figures nand
0
(lA)
(ii)
show that soil/elastic
> 2.75 s). trolley B trtl\ els at
l.
2.75- 2.8 s)
From Figure r. the time ofill1pact
of trolley is (2.75 s- 2.5 s) = 0.25 s.
materials tend to have a longer time of
( IM)
imp..1ct and a smaller maximum force of
A \ crage force
change in momentum
(lA)
impact.
For mbbcr plungcr.
time of impact
avcmge force acting on the iorce sensor
change in momentum
I ime of
(lA )
I .38x (- 0.15 - 0.42)
0.25
impact
=-3.15 N
6S,=
• ,,"",0'0-2:c7-,
4c2.570 - 2.550
=
13.7 N
New Senior Secondary Physics at Work
(I A)
95
Cl Oxford University Press 2009
2
Force and Motio
ha ter 5
The average force acting on trolley
(b)
(i)
B is 3.15 N (towards the right).
the force sensor with a velocity of
0.43 m S-I and rebounds with
In Figure q, the velocity of trolley A
changes from 0.55 m
S ·1
The trolley accelerates from rest
down the run way. It collides with
(lA)
(b)
Momentum
to - 0.55 ms- I .
- 0.36 ms-I .
"crage forcc acting 011 A by B
= changc in momentum
(lA)
The collision is inelast ic because
A
the trolley rebounds at a smaller
timeofimpact
(lA)
velocity.
O.69x(- 0.55 - 0.55)
(ii)
0 .25
Time of impact
..,. 1.65 - 1.60 = 0.05 s
=-3.04 N
Average force
area undcr the curve
The avcragc fo rce acting on trolley A is
3.04 N (towards the left).
(lA)
time of impact
Within c.\.pcrimental error. the a\ ('rage
0.44
~--
force acting
011
A by 13 has the same
0.05
- 8.8 N
magnitude but in the opposite direction
as the average force acting on B
Therefore. the experimental
b~"!.
re~ult
(ii i) As compared with the collision
(lA)
with plasticinc. the collision with
i!> in
spring has a longer time of impact
closer to an clastic collision. (2A)
(I A )
(:1)
(i)
25
The trollcy accelerates from rest
Perpendicular to the initial travelling
direction of the ball:
sensor at 0.50 m S- I and SIOp~.{].-\ )
By conservation of momentum.
(I M)
0 = M/lVlI sin 12.0" + MI'VI' sin 36.0"
lrol1e~
0 = 5.90 x VB sin 12.0"-
and the force sensor is ineLhllc.
, lA)
(ii)
(3)
down the runwa). I1 hits the force
The collision bet\\ een the
JIf" x 3.30 si n 36.0"
The shaded area rcpresenb the
VlI = 1.58MI' ...... ..( I)
impulse of the force .
Along the initialtr:lvelling direc tion of
'1.-\ )
(iii) From the F-I graph. time ofimpacl
the ball:
= 1.544 - 1.531
= 0.013
s
By consen'ation ofll1omentum.
, I \1)
Jl-f/jV = MRVB cos 12.0"
A verage force
5.90 x 3.00 = 5.90
area under the cun e
X
+ MpVp cos 36.0°
V/I cos t 2.0 0 +
'\(" x 3.30 cos 36.0"
time of impac t
17.7 = 5.77V8 + 2.67MI'
0.31
... (2)
Substitute (I) into (2):
0.013
= 23.8
<l
and a smaller average force. It is
accordance with Newton· s third la\\.
24
( lA)
N
New Senior Secondary Physics at Work
17.7 = 5.77(I.S8M,,) - 2.67M,.
( 1.\)
Mp = 1.S0kg
96
(lA)
Cl Oxford University Press 2009
2
Force and Motio
(b)
ha ter 5
Substitute Mp = 1.50 kg into (I):
(Hi) Mean resistance force
time
~ 60,(0-3.67)
1.58 x 1.50
= 2.37ms
(0)
_ change in momentum
(IM)
VB = 1.58M"
=
(lA)
I
= -55. 1 N
them is 55.1 N towards the left .
(i\') It is invalid because the external
force is not negligible .
27
(a)
24.7 J
(IM)
(b)
Since the total kinetic cnergy ofthc ball
By conservation of momentum.
1'/ +
I 'B=
0)
If Ihey stick together on coHision,
v/ =
1'8.
Then.
1'.0/-0. VB
(a)
(lA)
:::>
The IOtal momentum of the system is
conserved.
(ii)
(lA)
(b)
(I)
(lA)
Final KE
2
1
2
Mean force
change in momentum
timeo!" impact
(IM)
,
1
.'
1
2
.'
25111
=
III S-l
(Correct calculations for both KE)
(1 M)
Fraction of original KE lost
(IM)
~20'(367-8)
(0)
0.50
= - 173 N
A
... -mvIJ
= - IIIX)- + - IIIX)-
"" 3.7 m S- l (I A)
( ii)
,
I
= -1111'
(IM)
1' = 3.67
(lA)
Original KE
IOtal momentum after collision
20 x 8 + 40 x 1.5 = 601'
. m s.,
= I 'B = )
1
Total momentum before collision
=
1'..1
= 10
=~lIn(IO"=50111
provided that there is no external nct
force acting on the system .
( l A)
10
that after collision, the collision is
26
(lA)
10 = 1//1'/
...
11/1'8
(lA)
11/ X
and pin before collision is not cqualto
inelastic.
(lA)
The mean resistive force acting on
Total kinetic energy aftcr collision
I
,I
,
= 2MBVs- + M I'VI'2
_ I
?
~
I
_
.2
--x 5.90 x_ .37 + - xl. )0x3.30
2
2
(Correct calculations)
( IM)
4.0
Total kinetic energy before collision
I
~
1
~
_
= - M !lV- = - x 5.90x 3.00- = 26.)5 J
2
2
=
Momentum
(lA)
50111- 25111
I
50111
2
0)
1'1
(ii)
The total KE is the same before
=
0,
VB=
10 m S-l
and aftcr collision.
The mean force acting on the left
(lA)
(lA)
(lA)
hand trolley is 173 N towards the
lcft.
New S enior Secondary Physics at Wo rk
97
© Oxford University Press 2009
2 force and Motlo
ha teT 5
(iii) Steel is 'hard" so collisions arc
Package Yhas a longer time of impact.
( lA)
elastic.
( I A)
A passes on momentum to B. then
With the same change in momentum in
B to C. Cto D. 1) toE. In each
each case.
si ngle collision . all the momentum
the force experienced by the television
(lA)
is transferred.
Therefore. A. B. C and D are left
stat ionary. and E gets the
momentum and s\\ ings.
28
(11)
(i)
( lA )
(lA)
set in Y is smaller.
( lA)
(For efTecli\ e communication)
(IC)
29
(IIKCEE 2007 Paper I Q9)
30
(:I)
(i)
Impulse or change inmornentum.
Total momentum before collision =
total momentum after collision
0.60 x 40 = 0.60 x 35 + 0.045v
( lA )
(ii)
Momentu
Change in momentum
=
shaded area
=
1.6x 1.7 = 1.361\'
(1 M)
)' =
,
s
66.7 III Si
(lA)
The velocity oflhe ball after the
(1:0.1)
coll ision is 66.7 m s
Since the kinetic encrg) of the ball
I
towards the
right.
after the collision is the !)arlle as it
(i i)
was before collision. the
The Principle of Conservation of
Momentum is nOI applicable if
magniludes of momentum before
there is an external net force acting
and after collision arc the same.
on the system (the ball and
(] \1 )
cl ub-head).
Initial momentum o rlhe ball
(lA)
During the collision. the golfcr
,
= ..!.-xI.36 = 0.68," s
!lA)
applies force on his club. \\ hich is
an external force to the system.
(ii i)
momentum
( lA)
(b )
Tota l KE before collision
I
,
=-- xO.60 x402
ot - - - - - - 0.5
1.0
t 5
= 480 J
2.0
Total KE after collision
""
(Not linear)
(I
,I
2
2
= 468 J
A.J
kinetic energy is lost in the collision .
(I A)
li"el~ to
New Senior Secondary Physics at Work
(lA)
than that before collis ion. Therefore.
1I AI
survive without damage.
,
The total KE aftcr collision is smaller
same magnitude bd'ore and after
(b) Television set in ) is more
I
=- x O.60 x35- +-x O.045 x 66.7·
(From negati\ e to posili\e. \\ ilh
collision)
( lA)
f 1.\ )
98
o Oxford
University Press 2009
2
Force and Motio
(c)
ha ter 5
By Newton's second 1,1\\.
Momentu
(lA)
momentum (IM)
mean "lorce = chan\!em
,
lime laken
~ O.045 x(66. -- 0)
1.5x 10-.1
( lA)
= 2001 "'"
Physics in articles (p. 272)
(a)
li lengthens the time of impact during
collision.
(lA)
This reduces the force acting on the
passcngcr.
(b)
The mass of the object and
(lA)
the time of impact.
(lA)
(For estimation. the final velocity of the
object is usually taken to be 7ero. Ifmore
precise result is needed. the final velocity
should also be known.)
(c)
(i)
11
= 331 km h- t = 92.2 III s
I
F = m(v - II)
(I M)
(IM)
I
0.l x (O _331)
3.6
0.1
= - 92.1 N
(I A)
The average force acting on the object is
92.2 N.
(H)
The air bag bursts at a very high speed.
When it hits the baby. the force on the
baby is so large tha! may severely injure
it.
New Senior Secondary PhySiCS at Work
(lA)
99
e
Oxford University Press 2009
6
Projectile Motion
Practice 6.1 (p. 286)
1
In the vertical direction:
1
1
sy = gt 2 = × 10 × 0.5 2 = 1.25 m
2
2
D
In the horizontal direction:
s
10
t = x = =2 s
u
5
The arrow drops 1.25 m over this range.
(b) The archer should aim above the target.
In the vertical direction:
(c)
vy = gt = 10 × 2 = 20 m s−1
1
1 s
sy = gt 2 = g x
2
2
u
2
=
1
1
gs x 2
2
u
2
Its vertical speed is 20 m s−1.
The vertical distance fell is inversely
2
C
proportional to the square of the release
3
C
speed. When the release speed is
4
In the horizontal direction:
v x = u x = 20 m s−1
reduced by half (from 60 m s−1 to
30 m s−1), the distance fell is four times
In the vertical direction:
as calculated in (a), i.e. 1.25 × 4 = 5 m.
vy = gt = 10 × 4 = 40 m s
−1
Speed of the object after 4 s
2
2
= v x + v y = 20 2 + 40 2 = 44.7 m s−1
5
(a) The aeroplane and the bomb travel in the
same horizontal velocity. Therefore, at
Practice 6.2 (p. 295)
1
D
2
A
3
(a) Greatest height
u 2 sin 2 θ 25 2 × sin 2 30°
=
=
= 7.81 m
2 × 10
2g
the moment when the first bomb hits the
ground, the aeroplane is right above the
(b) By sy = (u sin θ ) t −
impact point.
In vertical direction:
1
1
sy = gt 2 = × 10 × 15 2 =1125 m
2
2
1
0 = (25 sin 30°) t − × 10 × t 2
2
5t 2 − 12.5t = 0
Therefore, at that moment, the aeroplane
t = 0 (rejected) or 2.5 s
is 1125 m right above the impact point.
The time of flight of the object is 2.5 s.
(b) Distance between the successive impact
points of the bombs on the ground
720
=
× 1 = 200 m
3.6
6
(a) In the horizontal direction:
s
30
t= x =
= 0.5 s
u 60
1 2
gt ,
2
(c)
4
The time need for the object to reach its
1
highest point = × 2.5 = 1.25 s
2
(a) Best possible distance
u 2 25 2
=
=
= 62.5 m
10
g

Revision exercise 6
1
(b) sx = × 62.5 = 31.25 m
2
u 2 sin 2θ
By sx =
,
g
Multiple-choice (p. 298)
1
A
Along the vertical direction:
1
By sy = u y t − gt 2 ,
2
1
−10 = 0 − × 10 × t 2
2
252 × sin 2θ
31.25 =
10
sin 2θ = 0.5
2θ = 30° or 150°
t = 1.414 s
θ = 15° or 75°
Minimum speed
The angle of elevation is 15° or 75°.
5
= minimum ux =
Along the vertical direction:
1
By sy = uyt − gt 2 ,
2
1
0 = uy × 5 − × 10 × 5 2
2
lower bridge is 14.1 m s−1.
The minimum speed of the car to reach the
uy = 25 m s−1
u = u x 2 + u y 2 = 9 2 + 25 2 = 26.6 m s−1
uy
25
tan θ =
=
= 70.2°
ux
9
2
A
3
(HKALE 2004 Paper II Q3)
4
(HKALE 2006 Paper II Q3)
Conventional (p. 299)
1
(a) Range =
70 =
His initial velocity is 26.6 m s–1 at an angle of
u 2 sin 2θ
g
(a) Maximum height =
=
2
u sin θ
+ 1.8
2g
2
20 × sin 30°
+ 1.8
2 × 10
3.14°.
(b) From R =
= 6.8 m
The maximum height that the volleyball
g
2u 2 cos 2 θ
−1.8 = (tan 30°)s x −
u 2 sin 2θ
, when u increases
g
and R needs to be kept constant,
(1A)
θ should be decreased.
(1A)
(c)
can reach is 6.8 m.
(b) By sy = (tan θ )s x −
(1A)
The angle of projection θ should be
2
2
(1M)
80 2 × sin 2θ
10
θ = 3.14° or 86.9° (rejected)
70.2° to the ground.
6
20
= 14.1 m s−1
1.414
Along the horizontal direction:
s
45
ux = x =
= 9 m s−1
t
5
sx2 ,
10s x 2
2 × 20 2 × cos 2 30°
0.016 67sx2 − 0.5774 sx − 1.8 = 0
Solving the quadratic equation, we have
sx = 37.5 m or −2.88 m (rejected)
Player B releases the arrow at a higher
The horizontal distance AB is 37.5 m.
position than player A.

(1A)
(d) Greatest height H
u 2 sin 2 θ
=
+ 30
2g
With all conditions being the same,
player B’s arrow would travel a longer
horizontal distance (sx) when reaching
the same level of the target as shown
above.
2
=
(1A)
10 2 × sin 2 30°
+ 30
2 × 10
To reduce sx, the angle of projection θ
= 31.25 m
should be smaller, i.e. he should aim at a
The greatest height H above the ground
lower position above the target.
reached by the stone is 31.25 m.
(a) Along the vertical direction:
1
By sy = u y t − gt 2 ,
2
1
−30 = 10 sin 30° × t − × 10 × t 2
2
(1A)
3
t = 3 s or −2 s (rejected)
(1A)
Consider the projectile motion afterwards.
g
By sy = (tan θ ) s x − 2
sx 2 ,
(1M)
2
2u cos θ
10
d2
−1 = (tan 0 ) d −
2 × 8.94 2 × cos 2 0
(1M)
(1A)
d = 4.00 m
The distance R of point P from the foot
of the cliff is 26.0 m.
4
(1A)
(a) By sy = (tan θ )s x −
Horizontal component of vP
= ux = 10 cos 30° = 8.66 m s−1
(1M)
(1M)
Velocity vP
(1A)
25
tan θ =
8.66
10 × 12 2
2u 2 cos 2 30°
(1A)
positive.
Along the horizontal direction:
vx = u x
= 12.2 cos 30°
= 10.6 m s
(1A)
The velocity vP of the stone is 26.5 m s−1
at 70.9° to the horizontal.
s x 2 ,(1M)
(b) Take the direction to the right as
= 10 sin 30° − 10 × 3
θ = 70.9°
2u 2 cos 2 θ
u = 12.2 m s−1
= uy − gt
= 8.66 2 + (− 25)2 = 26.5 m s−1
g
2.43 − 2 = (tan 30°)(12) −
Vertical component of vP
= −25 m s−1
(1M)
v = 2 gh = 2 × 10 × (5 − 1) = 8.94 m s−1 (1M)
= 10 cos 30° × 3
= 26.0 m
Let v be the speed when the boy is projected.
loss in PE = gain in KE
1
mgh = mv 2
2
(b) Along the horizontal direction:
R = u xt
(1A)
By the law of conservation of energy,
(1M)
5t2 − 5t − 30 = 0
(c)
(1M)
(1M)
−1
(1A)
Take the upward direction as positive.
Along the vertical direction:
vy2 − uy2 = 2asy
2
(1M)
2
vy − (12.2 sin 30°) = 2 × (−10) × (−2)
vy = −8.79 m s−1 (1A)

The vertical velocity is 8.79 m s−1
By sy = (tan θ ) s x −
(downwards) and the horizontal velocity
(c)
is 10.6 m s−1 (towards right).
−97.2 = (tan 30°) s x
Along the vertical direction:
−
By vy = uy − gt,
vy − u y
t=
−g
−8.79 − 12.2 sin 30°
=
− 10
= 1.49 s
Distance XY = 168 + 92.5
= 260.5 m
Along the horizontal direction:
(1A)
vx = u x
= 19.44 × cos 30°
= 16.8 m s
(1A)
(1M)
−1
(1A)
the ball falls within the court on the
Take the downward direction as
opposite side.
positive.
(1A)
(a) Speed of the ball when it is released
Along the vertical direction:
vy2 − uy2 = 2asy
= speed of the car
–1
= 70 km h
70
=
= 19.44 m s−1 = 19.4 m s−1
3.6
2
(1M)
2
vy − (19.44 sin 30°) = 2(−10)(−97.2)
vy = 45.1 m s−1 (1A)
(1A)
The horizontal velocity is 16.8 m s−1
(b) Height of the ball
(towards left) and the vertical velocity is
= vertical distance travelled by the car
= 19.44 × sin 30° × 10
(1M)
= 97.2 m
(1A)
The height of the ball above the ground
45.1 m s−1 (downwards).
6
(HKALE 2000 Paper I Q1)
7
(HKALE 2003 Paper I Q7)
when it is released is 97.2 m.
Horizontal distance travelled before
released
= 19.44 × cos 30° × 10 = 168 m
(1A)
(d) Take the direction to the left as positive.
by the volleyball (15.8 m) is shorter than
(c)
10
sx 2
2 × 19.44 2 × cos 2 30°
sx = 92.5 m or −59.7 m (rejected) (1M)
Since the horizontal distance travelled
5
sx2 ,
Solving the quadratic equation, we have
Horizontal distance travelled
the distance AD (21 m),
2u cos 2 θ
0.0176 sx2 − 0.577 sx − 97.2 = 0
(1A)
= uxt = 10.6 × 1.49 = 15.8 m
g
2
(1M)
Consider the motion after the ball is
released.
Along the vertical direction:

7
Uniform Circular Motion
Practice 7.1 (p. 308)
Practice 7.2 (p. 324)
1
D
1
D
2
A
2
B
3
Angular speed
θ 1.5
= =
= 0.2356 = 0.236 rad s−1
t
20
Centripetal force =
Weight of the aircraft = mg
centripetal force mv 2
1
=
×
weight
r
mg
Linear speed
= rω = 28 × 0.2356 = 6.60 m s−1
4
(a) The angular speeds of Peter, Paul and
=
Mary are the same.
Angular speed
2
=
= 7.27 × 10−5 rad s−1
24 × 60 × 60
(b) Linear speed of Peter
3
=
= (6400 × 10 ) × (7.27 × 10 )
= 465 m s−1
= rω
10 × (10 × 10 3 )
v2
= 0.4
gr
θ = 21.8°
4
D
Corresponding to the dry road, we have:
= (6400 × 103 ×cos 60°) × (7.27 × 10−5)
= 233 m s
200 2
A
tan θ =
−5
Linear speed of Paul
v2
gr
= 0.4
3
= rω
mv 2
r
fmax =
−1
Linear speed of Mary
mv max 2
r
The maximum frictional force is
= rω
= 0 × (7.27 × 10−5)
road.
=0
5
f max mvmax 2 1
=
× =
2
r
2
(a) Angular velocity ω
2
=
= 8.25 × 10−4 rad s−1
127 × 60
= [(1740 + 200) × 103] × (8.25 × 10−4)
Centripetal acceleration required
1600 2
v2
=
= 1.32 m s−2
=
r (1740 + 200)× 10 3
vmax
2
2
r
road is wet
30
= 21.2 m s−1
=
2
= rω
(c)
m
Therefore, the maximum safe speed when the
(b) Linear velocity v
= 1600 m s−1
f max
on wet
2
5
(a) Vertical component of tension
= weight of the mass
T cos θ = mg
T cos 20° = 0.8 × 10
T = 8.51 N

(b) Frictional force
The tension of the string is 8.51 N.
(b) Centripetal force on the mass
(c)
= T sin 20° = 8.51 sin 20° = 2.91 N
mv 2
Centripetal force =
= 2.91
r
0.8 × v 2
= 2.91
1.5 sin 20°
(c)
v = 1.37 m s−1
θ = 48.4°
The speed of the mass is 1.37 m s−1.
6
The angle that the motorcycle makes
Horizontal component which provides the
centripetal force:
mv 2
R sin α =
………(1)
r
with the vertical is 48.4°.
Revision exercise 7
Vertical component which balances the
Multiple-choice (p. 327)
weight:
1
B
R cos α = mg ………(2)
2
D
(1) ÷ (2):
3
D
tan α =
tan 30° =
Centripetal force = mrω2
2
v
rg
For A and B, the values of m and ω are both
2
the same, and rB = 2rA.
v
20 × 10
v = 10.7 m s
Therefore, centripetal force for B is twice that
−1
for A.
−1
7
The maximum speed is 10.7 m s .
Centripetal for A = T1 − T2
Angular speed of the pendulum
Centripetal for B = T2
= 2π × 0.5 = π rad s
−1
Then we have:
T2 = 2 × (T1 − T2)
Horizontal component which provides the
centripetal force:
3T2 = 2T1
T1 3
=
T2 2
T sin α = mrω
2
T sin α = 0.4 × (2 sin α) × π2
4
(HKALE 2000 Paper II Q11)
Vertical component which balances the
5
(HKALE 2006 Paper II Q4)
weight:
6
(HKALE 2007 Paper II Q4)
T = 7.90 N
T cos α = mg
7.90 cos α = 0.4 × 10
α = 59.6°
8
= centripetal force
mv 2 500 × 15 2
=
=
= 5625 N
r
20
f
tan θ =
N
5625
tan θ =
5000
(a) Normal reaction
= weight of the rider and the motorcycle
= mg = 500 × 10 = 5000 N
Conventional (p. 328)
1
(a) Angular speed of the Earth
2
=
365.26 × 24 × 60 × 60
= 1.99 × 10−7 rad s−1

(1M)
(1A)
(b) Linear speed of the Earth
4
= rω
(1M)
−7
11
= (1.50 × 10 ) × (1.99 × 10 )
= 2.99 × 104 m s−1
(c)
(1A)
Centripetal force required
= mrω2
r = 7710 m
(1M)
= 3.55 × 1022 N
(b) Let θ be the angle of banking.
(1A)
tan θ =
(a) The tension in the string provides the
centripetal acceleration of mass m. (1A)
mv 2
(b) Centripetal force =
=T
r
mv 2
= Mg
(1A)
r
Mgr
v=
(1A)
m
(c)
3
No
The angle of banking is 80.5°.
(c)
The apparent weight of the pilot is the
normal reaction R acting on him by the
seat.
R cos θ = mg
(1M)
R = 3940 N
(1M)
(1A)
The apparent weight of the pilot in the
turn is 3940 N.
5
−3 2
= 400 × 50 × (3.49 × 10 )
Centripetal force = m1rω2
For maximum value of r (= rmax):
(1M)
m2g + µsm1g = m1rmaxω2
For the cart at the lowest point of the circle:
(1M)
m × 10 + 0.5 × m × 10 = m × rmax × 6
F − mg = 0.244
2
10 + 5 = 36rmax
F − 400 × 10 = 0.244
rmax = 0.417 m
(1A)
m2g − µsm1g = m1rminω2
point of the circle is 4000.244 N (upwards).
m × 10 − 0.5 × m × 10 = m × rmin × 6
For the cart at the highest point of the circle:
rmin = 0.139 m
(1A)
The force supporting the cart at the highest
point of the circle is 3999.756 N (upwards).
(1M)
2
10 − 5 = 36rmin
mg − F = 0.244
400 × 10 − F = 0.244
(1A)
For minimum value of r (= rmin):
The force supporting the cart at the lowest
F = 3999.756 N
(1A)
R cos 80.5° = 65 × 10
= mrω2
F = 4000.244 N
(1M)
θ = 80.5°
Centripetal force required
= 0.244 N
v2
gr
tan θ = 6
(1A)
Angular speed
2
=
= 3.49 × 10−3 rad s−1
0.5 × 60 × 60
(1A)
The minimum radius is 7710 m.
= (5.98 × 1024)(1.5 × 1011) (1.99 × 10−7)2
2
(a) Let r be the minimum radius of the path.
v2
= 6g
(1M)
r
(2 × 340)2 = 6 × 10
r
6
(a) When there is no friction,
v2
tan θ =
gr
tan 30° =
(1M)
v2
10 × 100
v = 24.0 m s−1

(1A)
(1A)
(b) (i)
To avoid overturning, the motorcyclist has to
lean inwards as shown in the free-body
(Weight)
(1A)
(Normal reaction)
(1A)
(Friction)
(1A)
diagram below.
(1A)
(Correct diagram)
(1A)
(ii) Along vertical direction:
N cos 30° − f sin 30° − mg = 0
mg
………(1)
N = f tan 30° +
cos 30°
(1M)
In equilibrium,
v2
tan θ =
gr
Along radial direction:
N sin 30° + f cos 30° =
2
mv
…(2)
r
(1M)
According to the equation above, the higher
the speed v, the larger the angle θ, i.e. the
Substitute (1) into (2):
mg
)sin 30°
(f tan 30° +
cos 30°
mv 2
+ f cos 30° =
r
motorcyclist leans closer to the ground. (1A)
8
(a) Consider v = rω. Since B and C have the
same v and rB < rC, ωB > ωC.
f × (tan 30° sin 30° + cos 30°) =
(1A)
Meanwhile, the angular speed of A and
B are the same. Therefore, the angular
700 × 48.0 2
− 700 × 10 × tan 30°
100
speed of C is slower than that of A. This
f = 10 500 N
means that A will overtake C.
(1A)
The value of the frictional force f is
10 500 N.
7
(1A)
(b) Consider a = rω . Since rA > rB and ωA =
ωB, A has a higher centripetal
When a motorcyclist is turning around a
acceleration than B.
corner, frictional force acting on the wheels
On the other hand, consider a =
provides the centripetal force and also gives a
turning moment on the motorcycle.
(1A)
(1A)
2
(1A)
2
v
.
r
Since rB < rC and vB = vC, B has a higher
centripetal acceleration than C.
(1A)
Therefore, A has the highest acceleration
(1A)
and C has the lowest.

(1A)
(c)
Linear speed of B = 5 m s−1
(1A)
Angular speed of B
v
5
= B =
= 0.2 rad s−1
rB 25
(1A)
9
(a)
Acceleration of B
=
vB 2 52
=
= 1 m s−2
25
rB
(1A)
Angular speed of A
= angular speed of B = 0.2 rad s−1 (1A)
Linear speed of A
= rAωA
(Upthrust)
(1A)
(Air resistance)
(1A)
(Weight)
(1A)
Acceleration of A
(Pushing force)
(1A)
= rAωA
upthrust = weight
(1A)
air resistance = pushing force
(1A)
= (25 + 11) × 0.2
= 7.2 m s
−1
(1A)
2
= (25 + 11) × 0.2
= 1.44 m s
2
−2
(1A)
Linear speed of C
= linear speed of B = 5 m s
The horizontal component of
upthrust provides the centripetal
−1
Angular speed of C
v
5
= c =
= 0.139 rad s−1
rc 25 + 11
(1A)
vc 2
52
=
= 0.694 m s−2
25 + 11
rc
(1A)
(1A)
(d) Centripetal force
= mass × centripetal acceleration
Along radial direction:
mv 2
U sin θ =
………(2)
r
(2) ÷ (1):
200 2
v2
tan θ =
=
gr 10 × (15 ×103 )
(1A)
(e)
(1A)
The centripetal forces are provided by
the frictional force between the feet of
the athletes and the ground.
(1M)
(1A)
The angle of banking is 14.9°.
Since the athletes have the same mass,
centripetal force of A > B > C.
(1A)
U cos θ = mg ………(1)
θ = 14.9°
From (b), we have centripetal
acceleration of A > B > C.
acceleration of the aircraft.
(ii) Along vertical direction:
Acceleration of C
=
(b) (i)
(1A)

10
(iii) Centripetal force = mrω2 = 121
(a)
(1M)
7 × (1.2 sin 60°) × ω = 121
2
ω = 4.08 rad s−1
(1A)
The angular speed of the steel ball
is 4.08 rad s−1.
(iv) We have:
T cos θ = mg………(1)
T sin θ = mrω2
= m(L sin θ)ω2
(Correct direction of ∆v )
T = mLω2………(2)
(1A)
Substitute (2) into (1):
(Correctly using tip-to-tail method)(1A)
(mLω2)cos θ = mg
g
cos θ = 2
ω L
The direction of acceleration of the steel
ball is the same as the direction of ∆v in
the figure above, pointing towards the
centre.
(1M)
(1M)
θ increases as the metal ball rotates
(1A)
faster and faster, i.e. ω increases, θ
(b)
increases.
11
(1A)
(a) For the same angular distance on the
curved tracks, the outer tracks are longer
than the inner ones since their radii of
curvature are larger.
(1A)
If the starting lines are all aligned,
athletes will need to run different lengths
(c)
(Tension)
(1A)
(Weight)
(1A)
(i)
Along vertical direction:
T cos θ = mg
(1M)
T cos 60° = 7 × 10
T = 140 N
(1A)
(ii) Centripetal force
= radial component of tension
= T sin θ
(1M)
= 140 × sin 60°
= 121 N
of track. This makes the race unfair.(1A)
(b) Let r1 and r2 be the radius of curvature
from the midde of tracks 1 and 2 to the
centre of the circular path, respectively.
πr1 = 100
100
r1 =
r2 = r1 + 1 =
100
+1
θr2 = 100
θ=
(1A)
(1M)
100
= 3.05 rad = 175°
100
+1
The centripetal force acting on the
steel ball is 121 N.

(1M)
(1M)
(1A)
(c)
13
For the athlete running on track 1:
v
ω1 = 1
(1M)
r1
=
(a) Angular velocity
2
=
10
= 0.6283 rad s−1 = 0.628 rad s−1
10
100
= 0.314 rad s−1
(1M)
(1A)
The angular velocity of the player is
0.628 rad s−1.
(1A)
(b)
For the athlete running on track 2:
v
ω2 = 2
(1M)
r2
=
10
100
+1
= 0.305 rad s−1
12
(1A)
While stopping along the straight line, the
motion of the car can be described by
v2 − u2 = 2as
where v = 0, u = v0 and s = r
−v 2
a= 0
2r
(c)
(Force exerted by the chain)
(1A)
(Weight)
(1A)
The centripetal force is provided by the
horizontal component of the force
exerted by the chain.
Therefore, the frictional force between the
mv 2
(1A)
road and the car is 0 .
2r
(d) (i)
(1A)
Vertical component which balances
the weight:
While turning at the corner along a circular
T cos θ = mg………(1)
path, the centripetal force is provided by the
Horizontal component which
frictional force.
provides the centripetal force:
Centripetal force required by the car
mv0 2 mv0 2
=
>
r
2r
(1A)
T sin θ = mrω2
(1M)
(1M)
T sin θ = m(l sin θ)ω
2
(1A)
T = mlω2………(2)
(1M)
The centripetal required is larger than the
Substitute (2) into (1):
frictional force between the road and the car.
mlω2 cos θ = mg
g
10
cos θ = 2 =
ω l 0.62832 × 28
The car will skid and be in danger. Therefore,
the man should not attempt to take the turn.
θ = 25.2°
(1A)
(1A)
(ii) Tangential speed
= rω
(1M)
= (28 sin 25.2°)(0.6283)
= 7.49 m s−1
!

(1A)
When angle θ is 30°,
10
cos 30° = 2
ω × 28
The tangential speed of the player
−1
is 7.49 m s .
(iii) Centripetal force required
= mrω2
ω = 0.642 rad s−1
(1M)
Therefore, the angular speed should not
= (70)(28 sin 25.2°)(0.6283)2
= 329 N
be higher than 0.642 rad s−1.
(1A)
The centripetal force required to
From the two constraints, we can
keep the player in circular motion
conclude that the maximum angular
speed that can be used is 0.642 rad s−1.
is 329 N.
(e)
(1M)
No, the student is wrong.
(1A)
(1A)
14
From (d)(i), we have derived that:
g
cos θ = 2
(1A)
ω l
(a)
The angle θ is independent of the mass
of the seat-player system.
(1A)
Since the angular velocity ω and the
length l are the same for all seats, all
seats including the empty one will make
the same angle θ with the post.
(f)
Recall the equation (2) in (d)(i):
T = mlω2
Along vertical direction:
(1M)
(N1 + N2)cos 35° − mg −
When the tension is 1400 N,
1400 = (60 + 10) × 28 × ω2
ω = 0.845 rad s−1
(f1 + f2)sin 35° = 0
(1M)
(N1 + N2)cos 35° − mg −
(1A)
(µN1 + µN2)sin 35° = 0
mg
N1 + N2 =
………(1)
cos 35° − µ sin 35°
The maximum angular speed that can be
used is 0.845 rad s−1.
(g) At frequency 0.11 cycles per second,
Along horizontal direction:
angular speed
(N1 + N2)sin 35° +
= 0.11 × 2π
= 0.691 rad s−1
(f1 + f2)cos 35° =
(1M)
Therefore, the angular speed should not
mv 2
r
(1M)
(N1 + N2)sin 35° +
be higher than 0.691 rad s−1.
From (d)(i), we have
g
cos θ = 2
ω l
(1M)
(µN1 + µN2)cos 35° =
(1M)
N1 + N2 =
"

mv 2
r
mv 2
…(2)
r (sin 35° + µ cos 35°)
Substitute (2) into (1):
mv 2
r (sin 35° + µ cos 35°)
mg
=
cos 35° − µ sin 35°
v=
=
Substitute (3) into (1):
1.6f1 cos 35° − f1 sin 35° − mg = 0
f1
g
=
= 13.6
m 1.6 cos 35° − sin 35°
(1M)
Substitute (3) into (2):
gr (sin 35° + µ cos 35°)
cos 35° − µ sin 35°
1.6f1 sin 35° − f1 cos 35° =
(10)(18)(sin 35° + 0.8 cos 35°)
v=
cos 35° − 0.8 sin 35°
= 24.8 m s−1
mv 2
r
f1
(1.6 sin 35° − cos 35°)
m
= 13.6(1.6 sin 35° − cos 35°)
(1A)
The car would start to skid outwards at
= 1.16 m s−1
−1
24.8 m s .
The car would start to overturn inwards
(b)
at 1.16 m s−1.
(c)
The range of speed for turning this
corner safely is from 1.16 m s−1 to
24.8 m s−1.
15
(HKALE 2001 Paper I Q1)
16
(HKALE 2003 Paper I Q1)
17
(HKALE 2004 Paper II Q1)
Just before overturning occurs, N2 = 0.
Along the vertical direction:
N1 cos 35° − f1 sin 35° − mg = 0 ……(1)
(1M)
Along the horizontal direction:
mv 2
N1 sin 35° − f1 cos 35° =
……(2)
r
(1M)
Take moment about c.g.
clockwise resultant torque =
anticlockwise resultant torque
2.5
N1 ×
= f1 × 2
2
N1 = 1.6f1 ………(3)
(1A)
(1M)

8
Gravitation
Practice 8.1 (p. 339)
1
The mass of the rocket is 316 kg.
5
A
Gravitation force
Gm1 m 2
=
r2
=
(6.67 ×10 )(5.98 ×10 )(250)
(6.37 ×10 + 3600 ×10 )
−11
24
3 2
6
= 1000 N
2
A
aM =
aE =
GM M
rM 2
FAC =
GM E
rE
2
With the information given, we have:
1
aM = a E
6
GM M 1 GM E
= ×
6 rE 2
rM 2
=
the Earth is about 1 : 4.
3
−11
0.5 2
=
−11
82
= 3.13 × 10−12 N
To find resultant force F, resolve FAC into
=
The magnitude of the gravitational force they
−7
act on each other is 9.60 × 10 N.
Gm1 m 2
Gravitational force =
r2
6.67 × 10 −11 5.98 × 10 24 (m 2 )
1800 =
2
6.37 × 10 6 + 2000 × 10 3
m2 = 316 kg
(6.67 ×10 )(3)(1)
Magnitude of F
= 9.60 × 10 N
(
10 2
8
= 1.06 × 10−12 N
10
6
FAC sin θ = 1.33 × 10−12 × = 7.98 × 10−13 N
10
(6.67 ×10 )(60)(60)
(
−11
FAC cos θ = 1.33 × 10−12 ×
−7
4
(6.67 ×10 )(2)(1)
FAC cos θ and FAC sin θ.
Gravitational force
Gm1 m 2
=
r2
=
rCA 2
= 1.33 × 10−12 N
Gm B mC
FBC =
rCB 2
rM
M
1
1
= 6× M = 6×
= 0.272 ≈
rE
ME
81
4
The ratio of the radius of the Moon to that of
Gm A mC
)(
)
)
=
(FBC + F AC cos θ )2 + (F AC sin θ )2
(3.13 × 10
−12
+ 1.06 × 10 −12
) + (7.98 × 10 )
2
= 4.27 × 10−12 N
F AC sin θ
tan φ =
FBC + F AC cos θ
=
7.98 × 10 −13
3.13 × 10 −12 + 1.06 × 10 −12
φ = 10.8°

−13 2
The resultant gravitational force acting on C
by A and B is 4.27 × 10
−12
7
N at an angle of
10.8° to BC.
Gravitational force = centripetal force
GM E m mv 2
=
r
r2
v=
Practice 8.2 (p. 353)
1
C
2
B
3
D
4
Gravitational field strength
GM N
=
RN 2
=
GM E
r
(6.67 ×10 )(5.98 ×10 )
−11
(6370 + 1600)×10 3
= 7070 m s−1
Its linear speed is 7070 m s−1.
8
−11
30
approximation:
3 2
13
= 3.32 × 10 N kg
When the satellite is close to the Earth’s
surface, we can take the following
(6.67 ×10 )(100 ×1.99 ×10 )
=
(20 ×10 )
Gravitational force = mg
−1
Then we have:
Gravitational force = centripetal force
The gravitational field strength at the surface
13
mg = mrω2
g
ω=
r
−1
of the neutron star is 3.32 × 10 N kg .
GM M
5
g initial (3R )2
1
=
=
GM
9
g final
M
=
R2
6
24
The ratio is 1 : 9.
2
ω=
T
2
=
11.9 × 365 × 24 × 60 × 60
−8
= 1.674 × 10 rad s
3
T=
=
−1
=
2
ω
2
1.253 × 10 −3
= 5015 s (= 1 hr 23 min 35 s)
GM S
ω2
3
Its period is 1 hr 23 min 35 s.
9
(a) Gravitational force = centripetal force
GM S M M M M v 2
=
r
r2
v=
(6.67 ×10 )(1.99 ×10 )
(1.674 ×10 )
−11
6370 × 10 3
= 1.253 × 10−3 rad s−1
Gravitational force = centripetal force
GM S M J
= M J rω 2
r2
r=
10
30
−8 2
= 7.80 × 1011 m
The distance of Jupiter from the Sun is
=
GM S
r
(6.67 ×10 )(1.99 ×10 )
1.5 × (1.50 × 10 )
−11
30
11
= 2.43 × 104 m s−1
The speed of Mars is 2.43 × 104 m s−1.
7.80 × 1011 m.

(b) T =
=
=
2
4
ω
2 r
v
2 × 1.5 × 1.50 × 1011
(
2.43 × 10
4
A
Gravitational force = centripetal force
GMm mv 2
=
r
r2
)
= 5.82 × 107 s
T=
= 674 days
The Mars orbits around the Sun once in
T=
674 days.
T=
Multiple-choice (p. 356)
C
Weight on the planet
GM P m
=
RP 2
Weight on the Earth
GM E m
=
RE 2
RP = 2 ×
A
2 r
GM
r
= 2π
r3
GM
(HKALE 2003 Paper II Q11)
6
(HKALE 2004 Paper II Q7)
7
(HKALE 2005 Paper II Q28)
1
Gravitational force = centripetal force
GM E m mv 2
=
r
r2
v=
MP
× RE
ME
=
GM E
r
(1M)
(6.67 ×10 )(5.98 ×10 )
−11
2
24
(6370 + 7000)× 103
= 5460 m s−1
= 2 RE
3
2 r
………(2)
v
5
= 2 × 2 × RE
D
ω
Conventional (p. 356)
weight on the Earth
eight on the planet =
2
GM P m 1 GM E m
= ×
2
RP 2
RE 2
2
2
Substitute (1) into (2):
Revision exercise 8
1
GM
………(1)
r
v=
(1A)
The linear speed of the satellite is 5460 m s−1.
GM M
(a) gM =
rM 2
G×
=
1
ME
10
1
rE
2
2
(1M)
2 GM E
= ×
5 rE 2
=
2
g
5

(1A)
(b) Weight on Mars
= mgM
(b) (i)
(1M)
2
= m×
g
5
2
= (mg )
5
2
= × weight on Earth
5
2
= × 700
5
= 280 N
(c)
(
(1M)
= (3700 − 2200) × 3.35 × 10
7
= 5.0 × 1010 kg
(1A)
(ii) Since point A is long way from the
granite rock, we assume the granite
(1A)
rock does not affect the
As the astronaut is further away from
gravitational field strength at A.
Mars, his weight (the gravitational force)
(1M)
(1A)
Difference between gravitational
Therefore, the force supporting the
field strength
G∆M
= 2
r
astronaut by the chair, which balances
the astronaut’s weight to maintain
constant speed, becomes smaller. (1A)
=
−2
The value 10 m s is the accepted value of
the acceleration due to gravity of the Earth
(1M)
(6.7 ×10 )× (5.0 ×10 )
(0.40 ×10 )
−11
10
3 2
= 2.09 × 10−5 N kg−1
(1A)
(1M)
(1A)
(iii)
at positions very close to the surface of the
Earth.
(1M)
= difference in density × volume
chair will be decreasing as the spacecraft
3
)
Difference in mass
The force acting on the astronaut by the
becomes lighter.
(1M)
= 3.35 × 107 m3
(1A)
is leaving Mars.
Volume of granite
4 3
=
r
3
3
4
=
0.2 × 103
3
(1A)
When involving celestial body in the space,
the distances used in calculations are different
from the radius of the Earth and gravities due
to different bodies have to be considered.(1A)
Therefore, this value is invalid for those
(Correct shape always below
calculations.
4
(a) The radius/diameter of the planet. (1A)
The mass/density of the planet.
(1A)
original curve)
5
(a) Gravitation field strength =
(1A)
GM
r2
(1M)
Gravitation field strength due to the
Earth
GM E
=
3.6 × 108
(

)
2
(1M)
v−u
t
0 − 3.4
=
2.13
Gravitation field strength due to the
(ii) a =
Moon
=
GM M
(4.0 ×10
8
G
=
=
(4.0 ×10
8
− 3.6 × 10
(1M)
)
8 2
ME
81
= −1.60 m s−2
(1M)
(1M)
(1A)
The acceleration due to gravity on
− 3.6 × 10
the Moon is 1.60 m s−2.
)
8 2
(iii) The other time
GM E
(3.6 ×10 )
8 2
= gravitation field strength due to the
Earth
= 4.26 − 0.90
(1M)
= 3.36 s
(1A)
(iv)
(1A)
The gravitational fields due to the Moon
and the Earth are equal in magnitude at
point P.
(b)
(Correct diagram)
(3 × 1A)
(v) This method is not valid on the
Earth
(Correct direction)
(c)
6
because there is atmosphere on the
(1A)
Earth which gives air resistance to
After P, the rocket will encounter a net
attraction to the Moon.
magnitude only)
(1A)
while velocity is a vector (is described
by both magnitude and direction). (1A)
1
(b) (i) By s = (u + v)t,
(1M)
2
1
4.26
3.6 = (u + 0)
(1M)
2
2
u = 3.4 m s
affect the motion of the projectile.
(1A)
(a) Speed is a scalar (is described by
−1
(1A)
The initial vertical velocity of the
−1
projectile is 3.4 m s .
(1A)
(1A)
(c)
(i)
Resultant initial velocity
= 2.0 2 + 3.4 2
= 3.94 m s−1
3.4
tan θ =
2.0
θ = 59.5°
(1A)
(1A)
The resultant initial velocity of the
projectile is 3.94 m s−1 and the
angle between the resultant and the
horizontal is 59.5°.
(Correct calculating method for
both values)
(1M)
(ii) The projectile will land on the
moving Moon vehicle.

(1A)
Throughout the motion of the
projectile, the only force acting on
it is the gravitational force along
the vertical direction. Therefore,
the net force in horizontal direction
is zero, and it moves with constant
speed of 2.0 m s−1, which is the
same as that of the Moon vehicle.
(1A)
Physics in articles (p. 359)
(a) The space debris orbits the Earth in circular
motion (or elliptical motion).
(1A)
The gravitational forces acting on them
provide the centripetal forces
(1A)
instead of pulling them down to the Earth.
(1A)
(b) Gravitational force
GM E m
=
r2
=
(1M)
(6.67 ×10 )(5.98 ×10 )(260)
(6370 ×10 + 1000 ×10 )
−11
24
3
= 1910 N
3 2
(1M)
(1A)
The gravitational force acting on the satellite
by the Earth is 1910 N.
(c)
Collision with satellites or spacecrafts. (1A)
Return to the Earth and cause damage. (1A)
