2 Force and Motion
Chapter 2 Motion (II)
Practice 2.3 (p.78)
8
Take upwards as positive. Consider the
1
D
upward journey.
2
C
(a)
3
A
By v2 = u2 + 2as,
Consider the downward journey. Take
u = v 2  2as
= 0  2(9.81)3
downwards as positive.
0 .5
Time taken =
= 0.25 s
2
= 7.67 m s1
The whale’s velocity is 7.67 m s1
Its initial velocity is zero.
1
1
s = ut + at 2 = 0 + (9.81)0.252 = 0.307 m
2
2
4
upwards.
(b)
C
Take downwards as positive.
By v = u + at,
v  u 5  200
t=
=
= 9.75 s
 20
a
The time of firing the rocket must be longer
Its final velocity is zero.
By v = u + at,
v  u 0  7.67
t=
=
= 0.782 s
a
9.81
The whale takes 0.782 s to move to the
highest position.
9
Take upwards as positive.
(a)
than this value.
The final velocity of X is zero.
For option D,
By v2 = u2 + 2as,
v = u + at = 200 + (20)10.2 = 4 m s1
u = v 2  2as
The spacecraft would move upwards.
= 0  2(9.81)200
 The answer is C.
= 62.64 m s1
5
B
 62.6 m s1
6
B
The speed of X is 62.6 m s1 when it is
Consider the upward journey. Take upwards
fired.
as positive.
30
Time taken =
= 15 s
2
(b)
By v = u + at,
v  u 0  62 .64
t=
=
= 6.39 s
9.81
a
Its final velocity is zero.
X takes 6.39 s to reach the highest point.
By v = u + at,
(c)
u = v  at = 0  (9.81)15 = 147 m s1
7
initial speed should be smaller than that
Take upwards as positive.
1
By s = ut + at 2,
2
1
3 = 8t + (9.81)t 2
2
 4.905t  8t + 3 = 0
2
 t = 0.584 s
or
1.05 s (rejected)
Since Y explodes at a lower height, its
of X.
Besides, since Y takes a shorter time to
reach its highest point, it should be fired
after X.
10
Take upwards as positive.
(a)
Consider the downward journey.
The ball will hit the ceiling after 0.584 s.
New Senior Secondary Physics at Work (Second Edition)
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1
2 Force and Motion
Time taken =
Chapter 2 Motion (II)
1 .2
= 0.6 s
2
The slope of the graph is the
instantaneous velocity of the apple.
His initial velocity is zero.
1
s = ut + at 2
2
1
= 0 + (9.81)0.62
2
(d)
before reaching the ground. This is
because all objects accelerate at the same
rate under gravity in the absence of air
= 1.7658 m
resistance.
 1.77 m
12
The maximum height is 1.77 m above
the trampoline.
(b)
Consider the journey from A to the
Take downwards as positive.
1
(a) By s = ut + at 2,
2
1
132 = 0 + (9.81)t 2
2
highest point.
t = 5.19 s
By v2 = u2 + 2as,
v2  u 2
s=
2a
0  42
=
2(9.81)
The vehicle experiences free fall for
5.19 s.
(b)
= 0  2(9.81)132
= 50.9 m s1
Height of A above trampoline
The maximum speed is 50.9 m s1.
= 1.7658  0.8155 = 0.950 m
13
Take downwards as positive.
(a)
Take downwards as positive.
(a)
By v = u + at,
vu 90
t =
=
= 0.917 s
a
9.81
0.917 s.
By v2 = u2 + 2as,
v2  u 2
92  0
s=
=
= 4.13 m
2a
2  9.81
= 0.785 m
The apple is 4.13 m above the ground
His maximum displacement is 0.785 m
before it falls.
(c)
Consider the downward journey.
0 .8
Time taken =
= 0.4 s
2
His initial velocity is zero.
1
s = ut + at 2
2
1
= 0 + (9.81)0.42
2
The apple travels through the air for
(b)
By v2 = u2 + 2as,
v = u 2  2as
= 0.8155 m
11
The two apples have the same speed just
above the ground.
s/m
(b)
Consider the downward journey.
v = u + at = 0 + 9.81  0.4 = 3.92 m s1
4.13
His speed is 3.92 m s1 when he falls
back to the ground.
0
2
0.917
t/s
New Senior Secondary Physics at Work (Second Edition)
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2 Force and Motion
(c)
Chapter 2 Motion (II)
v / m s1
3.92
0
0.4
0.8
t/s
3.92
a / m s2
9.81
0
0.8
t/s
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
3