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Transcript of Phy Bk2Ch2
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8/7/2019 Phy Bk2Ch2
1/15
2 Motion II
Practice 2.1 (p. 61)1 D
2 B
3 D
4 D
5 B
102
1030=
=v m s1
The velocity of the car at t= 2 s is 10 m s1.
6 C
7 (a) Total displacement
= 4 5 + (5) (7 5) = 10 m
The total displacement from the
staircase to her classroom is 10 m.
(b) Classroom C
8
9 (a) The object accelerates.
(b) The object first moves with a constant
velocity. Then it becomes stationary and
finally moves with a higher constant
velocity again.
(c) The object decelerates to rest, and then
accelerates in opposite direction to
return to its starting point.
(d) The object moves with uniform velocity
towards the origin (the zerodisplacement position), passes the origin,
and continues to move away from the
origin with the same uniform velocity.
10 (a) The object moves with a constant
velocity.
(b) The object moves with a uniform
acceleration from rest.
(c) The object moves with a uniform
deceleration, starting with a certain
initial velocity. Its velocity becomes
zero finally.
(d) The object first moves with a uniform
acceleration from rest, then at a constant
velocity, and finally moves with a
smaller uniform acceleration again.
(e) The object moves at a constant velocityand then suddenly moves at constant
velocity of same magnitude in the
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.
11 (a) The object moves with zero acceleration
(with constant velocity of 50 m s1).
(b) The object moves with a uniform
acceleration of 5 m s2.
(c) The object moves with uniform
deceleration of 5 m s2.
12 (a) It moves away from the sensor.
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(b)
13 (a)
(b) Total distance travelled
= area under the graph
=2
36)(12
= 27 m
(c) Average speed
=takentime
travelleddistancetotal
=3
27
= 9 m s1
14 (a) She moves towards the motion sensor.(b) The highest speed of the girl in the
journey is 3.5 m s1.
(c) The greatest rate of change in speed
2
5.30 =
= 1.75 m s2
(d) Total distance travelled
= area under the graph
=2
62
2
25.3 +
= 9.5 m
Practice 2.2 (p. 71)1 C
By v2 = u2 + 2as,2
3.6
290
= 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
CyclistXis moving at constant speed.
Time for cyclistXto reach finish line
= s305
150
time
ntdisplaceme==
For cyclist Y: u = 5 m s1, s = 250 m,
a = 2 m s2
By s = ut +21 at2,
250 = 5 t+2
1 2 t
2
t= 13.5 s or t=18.5 s (rejected)
Yneeds 13.5 s to reach finish line.
Therefore, cyclist Ywill 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.
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By v2 = u2 + 2as,
0 = 5002 + 2 (800 000) s
s = 0.156 m = 15.6 cm < 15.8 cm
The minimum thickness of the wall is
15.8 m.
4 CWhen the dog catches the thief at t = 5 s, its
total displacement is 30 m. The dog is sitting
initially, so u = 0.
By s =ut +2
1at
2,
30 = 0 +2
1a(5)2
a = 2.4 m s2
Its acceleration is 2.4 m s2.
5 D
6 a =t
uv
10
6.3
36
6.3
90
= = 1.5 m s2
By v2 = u
2 + 2as,
s =a
uv
2
22
=1.52
3.6
36
3.6
9022
= 175 m
The distance travelled by the motorcycle is
175 m and its acceleration is 1.5 m s2.
7 (a) Thinking distance= speed reaction time
=6.3
108 0.8 = 24 m
(b) Since the car decelerates uniformly,
braking distance
=2
uv +t
=2
06.3
108+
(3 0.8)
= 33 m
(c) Stopping distance
= thinking distance + braking distance
= 24 + 33 = 57 m
8 By v = u + at,
14 = u + 2 5
u = 4 m s1
By v2 = u2 + 2as,
142 = 42 + 2 2 s
s = 45 mThe displacement of the girl is 45 m.
9 (a) v = u + at= 0 + 20 0.3 = 6 m s1
The horizontal speed of the ball
travelling towards the goalkeeper is
6 m s1.
(b) By v2 = u2 + 2as,
a =0.82
60 22
= 22.5 m s2
The deceleration of the football should
be 22.5 m s2.
10 (a) The reaction time of the cyclist is
0.5 s.
(b) Braking distance
=( )
25.112
155.00.2=
m
Thinking distance
= 15 0.5 = 7.5 m
Stopping distance
= 11.25 + 7.5 = 18.75 m 20 mTherefore, the bicycle would not hit the
child.
11 By v2 = u2 + 2as,
0 = 32 + 2 (0.5) s
s = 9 m 8 m
Therefore, the golf ball can reach the hole.
12 (a) (i) By v = u + at,
0 = u + (4)(4.75)
u = 19 m s1
The initial velocity of the car is
19 m s1.
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(ii) By v2 = u2 + 2as,
0 = 192 + 2 (4) s
s = 45.1 m
The displacement of the car before
it stops in front of the traffic light
is 45.1 m.(b) By v2 = u2 + 2as,
172 = 0 + 2 3 s
s = 48.2 m
The displacement of the car between
starting from rest and moving at 17 m s1
is 48.2 m.
13 (a) By v2 = u
2 + 2as,
v2 = 0 + 2 0.1 500
v = 10 m s1
His speed is 10 m s1.
(b) Consider the first section.
By v = u + at,
t=a
uv
=1.0
010
= 100 s
Consider the second section.
By s = ut+ 2
1
at
2
,
800 = 10t+2
1 0.5t
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)1 D
2 D
3 C
For option A, apply equation v2 = u
2 2gs
and take s = 0 (the ball returns to the second
floor),
v = u = 10 m s1 (vertically downwards)
This is the same velocity as the initial velocityof option B.
Therefore, in both ways the ball has the same
vertical speed when it reaches the ground.
4 B
Take the upward direction as positive.
By s = ut +2
1at
2,
0 = u 30 +2
1 (10) 302
u = 150 m s1
The speed of the bullet is 150 m s1 when it is
fired.
5
Speed of
stone
Distance
travelled by
the stone
Equation used atuv = 2
2
1atuts +=
t= 1 s 10 m s1 5 m
t= 2 s 20 m s1 20 m
t= 3 s 30 m s1 45 m
t= 4 s 40 m s1 80 m
6 By s =ut +2
1at
2,
10 = 0 +2
1(10) t2
t= 1.41 s
v = u + at
= 0 + 10(1.41)
= 14.1 m s1
It takes 1.41 s for a diver to drop from a 10-m
platform. His speed is 14.1 m s1 when he
enters the water.
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7 Take the upward direction as positive.
By v2 = u
2 + 2as,
42 = 0 + (2)(10)s
s = 0.8 m
The highest position reached by the puppy is
0.8 m above the ground.8 (a) Consider the boys downward journey.
Take the downward direction as
positive.
By s =ut +2
1at
2,
0.5 = 0 +2
1(10) t2
t= 0.316 s
Hang-time of the boy
= 0.316 2 = 0.632 s(b) Take the upward direction as positive.
By v2 = u2 + 2as,
0 = u2 + 2 (10) 0.5
u = 3.16 m s1
The jumping speed of the boy is
3.16 m s1.
9 Take the upward direction as positive.
(a) By v2 = u2 + 2as,
0 = u2 + 2(10)(200)
u = 63.2 m s1
The velocity of the fireworkXis
63.2 m s1 when it is fired.
(b) By v = u + at,
0 = 63.2 + (10)t
t= 6.32 s
It takes 6.32 s for the fireworkXto reach
that height.
(c) From (a) and (b), for fireworkYto
explode at 130 m above the ground, the
speed ofYshould be smaller than that of
X. Therefore, Yshould be fired at a
lower speed.
Besides, since Yspends a shorter time to
reach its highest point, it should be fired
afterX.
10 (a) By s = ut +2
1at
2,
120 = 8t+ 2
1
10
t
2
t= 4.16 s or t= 5.76 s (rejected)
It takes 4.16 s to reach the ground.
(b) v = u + at= 8 + 10 4.16 = 49.6 m s1
Its speed on hitting the ground is
49.6 m s1.
11 (a) Distance between the ceiling and her
hands
= 6 2 1.2 = 2.8 m
(b) Let s be her vertical displacement whenshe jumps.
As the maximum jumping speed is
8 m s1, i.e. u = 8 m s1.
By v2 = u2 + 2as,
s =a
uv
2
22
=10)(2
80 22
(upwards is positive)
s = 3.2 m > 2.8 m
Therefore, the indoor playground is not
safe for playing trampoline.
12 (a) By s = ut +2
1at
2,
132 = 0 t+2
1 10 t
2
t= 5.14 s
The vehicle can experience a free fall in
the Zero-G facility for 5.14 s.
(b) By v2 = u
2 + 2as,
v2 = 02 + 2 10 132
v = 51.4 m s1
The speed of the vehicle before it comes
to a stop is 51.4 m s1.
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(c)
Revision exercise 2Multiple-choice (p. 87)
1 DBy v2 = u2 + 2as,
0 = 102 + 2a(25 10 0.2)
a = 2.17 m s2
His minimum deceleration is 2.17 m s2.
2 D
3 B
Consider the rock released from the 2nd floor.
By v2 = u2 + 2as,
v2 = 2as (as u = 0)
Then consider the rock released from the 7th
floor.
Note that s2 = 3.5s.
(v2)2 = 2as2
= 3.5(2as)
= 3.5v2
v2 = 1.87v
4 A
5 C
The stone returns to the ground with the same
speed (but in opposite direction).
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
t= 4 )(g
v
= 2t
Its new time of travel is 2t.
6 B
Take the upward direction as positive.
s = ut +2
1at
2
= (10)(4) +21 (10)(4)2
= 40 m
The distance between the sandbag and the
ground is 40 m when it leaves the balloon.
7 D
8 C
Take the downward direction as positive.
u = 200 m s1, v = 5 m s
1, a =20 m s2
By v = u + at,
5 = 200 + (20)t
t= 9.75 s
The rockets should be fired for at least 9.75 s.
Both C and D satisfy this requirement. But for
D, after firing for 10.2 s,
v = u + at
= 200 + (20)(10.2)
= 4 m s1
i.e. it flies away from the Moon with 4 m s1
upwards. It cannot land on the Moon.
Therefore, the correct answer is C.
9 D
10 D
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!
11 (HKCEE 2006 Paper II Q1)
12 (HKCEE 2007 Paper II Q2)
13 (HKCEE 2007 Paper II Q33)
Conventional (p. 89)
1 (a) The reaction time of the driver is 0.6 s.(1A)
(b)t
va = (1M)
=6063
120
..
= 4 m s2 (1A)
The acceleration of the car is 4 m s2.
(c) The stopping distance of the car is the
area under graph. (1M)
Stopping distance
=12 0.6 +2
606312 )..(
= 25.2 m (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 s1 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)
Then, from t= 1.45 s to 1.50 s, the
object changes its moving direction and
moves towards the motion sensor again
with a uniform velocity of 0.35 m s1.
(1A)
(b) (i)
(Correct axes with label) (1A)
(A straight line with slope = 0.35 m s1
from t= 1.20 s to 1.25 s) (1A)
(A straight line with slope = 0.35 m s1
from t= 1.45 s to 1.50 s) (1A)
(ii)
(Correct axes with labels) (1A)
(Correct graph with the acceleration of
about30.140.1
35.035.0
= 7 m s2 at t= 1.30 s to1.40 s) (1A)
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"
3 (a)
(Correct axes with labels) (1A)
(Correct shape of minibus graph) (1A)
(Correct shape of sports cars graph) (1A)
(Correct values) (1A)(b) From the graph in (a), the two vehicles have
the same velocity at t 2.3 s after passing the
traffic light. (1A)
(c) The area under graph is the displacement of
the cars. (1M)
Consider their displacements at t= 3 s,
For the sports car:
s =2
1 15 3 = 22.5 m (1A)
For the minibus:
s =2
1 (7 + 13) 3 = 30 m (1A)
The minibus will take the lead 3 s after
passing the traffic light. (1A)
4 (a) The car moves forward with uniform
acceleration at 1 m s2 from t= 0 s to
t= 5 s. (1A)
Its instantaneous velocity is 0 at t= 5 s.
(1A)
Then the car changes its moving
direction. From t= 5 s to t= 8 s, it
moves backwards with a uniform
acceleration of6.67 m s2. (1A)
(b) Total displacement of the car
= area bound by the vtgraph and the
time axis (1M)
= ( ) ( )3202
155
2
1
= 17.5 m (1A)
(c) Yes, the car moves 12.5 m forwards
from t= 0 to t= 5 s. Therefore, it hits
the roadblock. (1A)
5 Take the upward direction as positive.
(a) From pointA to the highest point:
By v2 = u2 + 2as,
0 = 42 + 2 (10) s
s =0.8 m (1M)
By v = u + at,
0 = 4 + (10)t
t=0.4 s (1M)
From the highest point to the trampoline:
s = ut +2
1at
2 (1M)
= 0 +2
1(10)(1.2 0.4)2
= 3.2 m (1A)
The maximum height reached by him is
3.2 m above the trampoline.
(b) Height of pointA above the trampoline
= 3.2 0.8 (1M)
= 2.4 m (1A)
6 (a) Initial velocity v
= 90 km h1
=6.3
90m s1
= 25 m s1
Thinking distance
= vt (1M)
= 25 0.2
= 5 m (1A)
The thinking distance is 5 m.
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(b) By v2 = u2 + 2as, (1M)
a =s
uv
2
22
=5)(802
25022
=
4.17 m s
2
(1A)Hence, the deceleration of the car is
4.17 m s2.
(c) By v2 = u2 + 2as, (1M)
s =a
uv
2
22
=2)4.17(2
250 22
= 37.5 m (1M)
Braking distance = 37.5 m
Stopping distance
= 37.5 + 5 = 42.5 m (1A)
The driver could not stop before the
traffic light. Therefore, his claim is
incorrect. (1A)
7 (a) Take the downward direction as
positive.
By s = ut+2
1gt
2, (1M)
3 = 0
t+ 2
1
10
t
2
t=10
23= 0.775 s (1A)
The apple travels in air for 0.775 s.
(b) By v2 = u2 + 2as, (1M)
v = 3102
= 7.75 m s1 (1A)
The speed of the apple is 7.75 m s1
when the apple just reaches the ground.
(c) The slope of the graph is the magnitude
of the acceleration of the apple. (1A)
(Correct labelled axes) (2A)
(Straight line with a slope of 10 m s2)
(1A)
(d) The two graphs have no difference.
(1A)
8 (a) Take the downward direction as
positive.
By v2 = u2 + 2gs, (1M)
v = gsu 22 +
= 3)(4010202 +
= 27.2 m s1 (1A)
The speed of the residents landing on the
cushion is 27.2 m s1.
(b) (i) By s = ut +21 gt2, (1M)
40 3 = 0 +2
1 10 t
2
t= 2.72 s (1A)
The time of travel in air is 2.72 s.
(ii) By s =2
vut, (1M)
t=vu
s
+
2
=
02.27
32
+
t
= 0.221 s (1A)
The time of contact is 0.221 s.
speed / m s1
time / s0 0.775
7.75
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(c)
(Correct labeled axes) (1A)
(Correct shape) (1A)
(Correct values) (1A)
9 (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. (1A)
Then it rests momentarily at t= 3.4 s.
(1A)
After that, it moves towards the motion
sensor with a changing speed. (1A)
(c) By s = ut +2
1at
2, (1M)
0.1 = 0.7 2.9 +2
1a (2.9)
2
a = 0.507 m s2 (1A)
The acceleration of the trolley is
0.507 m s2.
10 (a) The motion sensor is protruded outside
the table to avoid the reflection of
ultrasonic signal from table. (1A)
(b) Slope of the graph from t= 0
to t= 0.28 s
=0280
032
.
.(1M)
= 8.21 m s2 (1A)
The acceleration of the ball due to
gravity is 8.21 m s2.
(c) (i)
(Correct sign) (1A)
(Correct shape) (1A)
(ii) The method does not work (1A)
since ultrasound will be reflected
by the transparent plastic plate.
(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
sensor. (1A)
Therefore, the ball drops a height
of 0.95 m. (1A)
(ii) The ball rebounds to the positions
which are 0.45 m, 0.65 m and0.775 m from the sensor in its first
3 rebounds.
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At the 1st rebound, the ball rises up
(1.1 0.45) = 0.65 m. (1A)
At the 2nd rebound, the ball rises up
(1.1 0.65) = 0.45 m. (1A)
At the 3rd rebound, the ball rises up
(1.1 0.775) = 0.325 m. (1A)(b) (i) The ball hits the ground with
velocities of 3.9 m s1, 3.25 m s1
and 2.75 m s1 in its first 3
rebounds. (3A)
(ii) Acceleration
= slope of graph =0.550.95
3.9
(1M)
= 9.75 m s2 (1A)
12 Take the downward direction as positive.
(a) By s = ut+2
1gt
2, (1M)
2 = 0 t+2
1 10 t
2
t=10
22= 0.632 s (1A)
It takes 0.632 s from t1 to t2.
(b) At t2,
v = u + at
= 0 + 10 0.632
= 6.32 m s1 (1M)
Shirleys speed is 6.32 m s1 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, (1M)
a =s
uv
2
22
=3.02
0)32.6( 22
= 66.6 m s2 (1A)
The average acceleration is 66.6 m s2.
(c)
(3 straight lines) (1A)
(Correct slopes) (1A)
(Correct labels of time and velocity)(1A)
13 (a) Speed v = 70 km h1
=6.3
70m s1
= 19.4 m s1
Reaction time =v
d(1M)
=4.19
6
= 0.309 s (1A)
The reaction time of the man was
0.309 s.
(b) By v2 = u2 + 2as, (1M)
a =s
uv
2
22
=482
4.190 22
= 3.92 m s2 (1A)
The average deceleration of the car was
3.92 m s2.
(c) Speed v
= 80 km h1
=6.3
80m s1
= 22.2 m s1
t3
6.32
v / m s1
t / s
6.32
t1 t2 t4 t5
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Thinking distance
= vt
= 22.2 0.309
= 6.86 m (1A)
By v2 = u
2 + 2as,
braking distance s
=a
uv
2
22
=)92.3(2
2.220 22
= 62.9 m (1A)
Therefore, the stopping distance
= 6.86 + 62.9
= 69.8 m (1A)
This stopping distance is greater than the
initial distance between the car and the
boy. (1A)
Therefore, the car would have knocked
down the boy if the car had travelled at
80 km h1 or faster.
(d) A drunk has a longer reaction time.(1A)
This means that the thinking distance,
and thus the stopping distance (sum of
thinking distance and braking distance),
increases. (1A)
14 (a) Take the upward direction as positive.
By v = u + at, (1M)
u = 0 (10) 0.7
= 7 m s1 (1A)
The speed of Belinda leaving the spring
board is 7 m s1.
(b) Total time taken from the spring board
to the water
= 0.7 + 1.05 = 1.75 s
Take the upward direction as positive.
s = ut +2
1at
2 (1M)
= 7 1.75 +2
1 (10) 1.752
= 3.06 m (negative means the water
is below the spring board)
The spring board is 3.06 m above the
water. (1A)
Alternative method:
Consider the upward motion and
downward motion separately.
For the upward motion, she takes 0.7 s
to reach the highest point from the
spring board.
Take the upward direction as positive.
By s = ut +2
1at
2, (1M)
s1 = 7 0.7 +2
1 (10) 0.7
2
= 2.45 m
For the downward motion, she takes
1.05 s from the highest point to enter
water.
Take the downward direction as
positive.
By s = ut +2
1gt
2,
s2 = 0 +2
1 10 1.05
2 = 5.51 m
Therefore the height of the spring board
above the water
= s2 s1
= 5.51 2.45
= 3.06 m (1A)
(c) v = u + at (1M)= 0 + (10) 1.05
= 10.5 m s1 (1A)
The speed of the diver entering the water
is 10.5 m s1.
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(d)
(Correct shape) (1A)
(Correct times) (1A)
(Correct velocities) (1A)
(e) (See the figure in (d).)
(Correct slope - parallel to that in (d).)
(1A)
(Correct position above that in (d).)
(1A)
15 (a) Speed 70 km h1
=6.3
70m s1
= 19.4 m s1
Distance travelled by car Yin 2 s
= vt= 19.4 2 = 38.8 m < 50 m (1M)
Since the distance between the cars is
greater than the distance that car Ycan
travel in 2 s, the driver of car Yobeys
the rule. (1A)
(b) Deceleration of a car is the slope of their
corresponding vtgraph. (1M)
Deceleration of carX
= slope of the graph during 05 s
=05
4190
.
= 3.88 m s2
The deceleration of carXis 3.88 m s2.
(1A)
Deceleration of car Y
= slope of the graph during 0.5 s8.5 s
=5.05.8
4.190
= 2.43 m s2
The deceleration of car Yis 2.43 m s2.
(1A)
(c) Thinking distance
= area under the graph during 00.5 s
= 19.4 0.5
= 9.7 m (1A)
Braking distance
= area under the graph during 0.5 s8.5 s
=2
1 19.4 (8.5 0.5)
= 77.6 m (1A)
The thinking distance and the braking
distance are 9.7 m and 77.6 m
respectively.
(d) The coloured area is equal to the
difference in the stopping distances
travelled by carsXand Y. (1A)
(e) Stopping distance of carX
= area under the graph during 05 s
=
2
1 19.4 5 = 48.5 m
Coloured area
= 9.7 + 77.6 48.5 (1M)
= 38.8 m < 50 m (1M)
Since the difference in stopping
distances of the cars is smaller than the
initial separation of the cars, the two cars
do not collide with each other before
they stop. (1A)
16 (a) From t= 0 s to t= 5 s, the car moves
with a uniform acceleration of
4.35
017=
m s2. (1A)
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From t= 5 s to t= 20 s, the car moves
with a constant velocity of 17 m s1.
(1A)
From t= 20 s to t= 28 s, the car moves
with a uniform acceleration of
2028170
= 2.125 m s2. (1A)
From t= 28 s to t= 30 s, the car remains
at rest. (1A)
(b)
(Correct shape) (1A)
(Correct time instants) (1A)
(Correct accelerations) (1A)
(c) 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. (1A)
17 (HKCEE 2002 Paper I Q8)
18 (a) v = u + at (1M)
=0 + 17.5 8 60
= 8400 m s1 (1A)
The speed of the Shuttle after the first 8
minutes is 8400 m s1.
(b) s = ut +2
1at
2 (1M)
= 0 +2
1 17.5 (8 60)
2
= 2 016 000 m (2016 km) (1A)
The Shuttle travels 2 016 000 m
(2016 km) in the first 8 minutes.
19 (a) (i) The cyclist is using first gear when
the acceleration is greatest before
braking. (1A)
(ii) The cyclist uses second gear for the
shortest time. (1A)
(b) Distance travelled
= area under straight line PQ (1M)
=2
2)68( +(1M)
= 14 m (1A)
The cyclist travels 14 m in second gear.
(c) The acceleration during t= 18 s20 s
=1820
90
(1M)
= 4.5 m s2 (1A)
The deceleration is 4.5 m s2.
20 (HKCEE 2005 Paper I Q1)
21 (a) s = ut +2
1at
2 (1M)
= 0 +2
1 10 (500 103)2
= 1.25 m (1A)
Therefore the minimum height the
laptop must fall for it to be saved is
1.25 m.
(b) v = u +at (1M)
= 0 + 10 (500 103)
= 5 m s1 (1A)
The speed of the computer when it hitsthe ground is 5 m s1.
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(c) Most falls are likely to be from below
this height, (1A)
so the protection will not have taken
effect. (1A)
22 (a) Any one from: (1A)
Rate of change of displacementDisplacement per unit time
(b) The velocity of a braking car is
decreasing (with time) (1A)
so the car has negative acceleration.(1A)
Its displacement is (still) increasing with
time, (1A)
so its velocity is (still) positive (1A)
In this case, the acceleration and
velocity are in opposite directions. (1A)
(c) (i)
(Correct graph) (1A)
(ii) Vertical distance travelled
= area under the graph from 4.0 s
to 10.0 s (1M)
=( )
2
613070 +
= 600 m (1A)
The vertical distance travelled bythe rocket between t= 4.0 s and t=
10.0 s is 600 m.
Physics in articles (p. 96)(a) 2.45 m (1A)
(b) (i) Take the upward direction as positive.
By v2 = u2 + 2as, (1M)
u2 = v2 2as
u2
= 0 2(10)(2.45 + 0.07 1.09)u = 5.35 m s1 (1A)
The vertical speed of Javier Sotomayor
is 5.35 m s1 when he leaves the ground.
(ii) Take the upward direction as positive.
Consider the upward journey.
By v = u +at, (1M)
54.010
35.50=
=
=
a
uvt s
Consider the downward journey.
By s = ut +2
1at
2, (1M)
( ) ( )102
1071.007.045.2 +=+ t2
t= 0.60 s
The time that he stays in the air
= (0.54 + 0.60) = 1.14 s (1A)
Alternative method:
Take the upward direction as positive.
By s = ut +2
1at
2, (1M)
( ) ( ) 2102
135.509.171.0 tt += (1M)
t= 1.14 s or t=0.07 s (rejected)
(1A)
The time that he stays in the air is 1.14 s.