Phy Bk2Ch2

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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.


(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).


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


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)



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)


(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)


(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


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.


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 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 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:


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.

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