Exam Style Answers 2 Asal Physics CB

 

 

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded may be different.

 Coursebook answers Chapter 2 Exam-style questions care to change 200 km h −1 into m s−1  and 1.4 km into 1400m.)  [1]

1

A (Tak  (Takee

2

C 

[1]

3

C 

4

Using v2 = u2 + 2as   distance s =  distance =

 

−

u2 )

2a

(302 − 10 2 )

2 × 4.0 = 100 m   5

(v 2

 

 

time taken

(50 − 25 )

 

[1]

[1]

 

average speed = 

[1]

 

[1]

c

 

−

( 25 + 50 ) 2

 37.5 m s 1 

[1]   distance tra travelled velled = 37.5 × 20 = 750 m  [1] −

=

B must travel travel an extra 50 m; its additional speed is 10 m s 1  [1] −

50

[1]

 final velocity = 50 − 0.50 × 100 [1] = 0  final

[1]

s = ut +  1 at2  2

[1]

8

[1]

  d

so, time required =  10  = 5 s  [1] Consider car A: it tra travels vels at 40 m s 1 for 25 s  [1]

 

total distance travelled 40 × 25 = 1000 m  [1]

a

vertical component component of velocity = v sin 30° [1]

 

=

−

 5.6 sin 30° = 2.8 m s 1 

[1]

−

[1]

 

Using

1 s = ut +  2 at2 with a = − g  and  and

Train  T rain slows to rest and covers a distance of 2500 m  [1]

 

t =  2.8

 = 0.57 s 

b

horizontal component of velocity =  v cos 30° 

[1]

 5.8 cos 30° = 4.85 m s 1 ≈ 4.9 m s 1 

 2500 m 

a

Using s = ut +  12 at2  2

[1] 2

[1]

s = 20t − 0.5

[1]

 

=

−

[1]

b

Substituting values of t in the equation gives  [1]

 

horizontal distance = speed × time 

[1]

 

after 2.0 s, displacement = 20.4 m ≈ 20 m [1]

  a

=

[1] [1]

 

after 6.0 s, displacement = −56.4 m ≈  −56 m 

 

t = 0.202

b

i

×

 9.8t  = 20t − 4.9t  

4.9

s = 0 [1  [1]]

 

[1]

c  

Substituting s = 0 gives  0 = 20t − 4.9t2 

[1]

 

t =  20  

 

t = 4.08

a

distance travelled at constant speed = speed × time 

 

1

=

change in speed

 

[1]

=

7

 

[1]

distance  d istance tra travelled velled =  50 × 100 − 0.5 × 0.50 × 1002 

6

acceleration = 

 = 1.25 m s 2  [1] 20 distance tra travelled velled = average speed × time taken  [1]

Using v = u + at   Using Using

b

[1]

4.9

s ≈ 4.1 s 

 40 × 20 = 800 m 

=

9

−

 4.85 × 0.57 = 2.77 m ≈ 2.8 m  0.2 =  12   × 9.81 × t2  s ≈ 0.20 s 

v2 = u2 + 2as;

[1]

2.912 = 1.922 + 2a × 

0.25 

[1]

m s 2 ≈ 9.6 m s 2 

[1]

 

a = 9.56

[1]

 ii ii

Air resistance 

[1]

[1]

 

Acts in the opposite direction to the velocity and so reduces the acceleration 

[1]

[1]

−

−

[1]

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020

   

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

10 a

i

 ii ii

 iii iii

Ball travels upwards (or reverses direction) on bouncing. 

12 a [1]

In both cases, the ball is accelerating due to gravity only.  [1] Initial height of the ball above the ground.  [1]

v2 = u2 + 2as;

[1]

 

23.7 ≈ 24 m s 1 

b

 = 0.39 s  [1] 60 The reactio reaction n time is approximately 0.3 s, so the driver was alert.   [1]

 

[1]

−

t = 

v s

 =

23.7

 iv iv

−

Ball does or not(Kinetic) bounce as high is aslost initial position. energy (as heat/internal energy) during the bounce.  [1]

0 = v2 − 2 × 2 × 140 

1

=

100000

  60 × 60

1

−

=

c

100 km h1   ≈ 28 m s  

 

The driver was not speeding, as the speed of 24 m s 1 is less than the speed limit.   [1]

−

   27.8 m s

[1]

−

b

i

v2 = u2 + 2as leading v2 = 2

×

to

 9.81 × 1.2 

 

v = 4.85

 ii ii

v2 = 2

 

v = 3.96

×

m s 1 ≈ 4.9 m s 1  −

−

 9.81 × 0.8 

13 a

[1]

b

[1]

m s 1 ≈ 4.0 m s 1  −

[1]

11 a

2

1.55 (±0.05) s 

 ii ii

Area under graph calculated between t = 0 and t = 1.55 s  [1] 1.55 = 15 ×   = 11.6 ≈ 12 m  [1] 2 Area between t = 1.55 s and t = 4.1 s  [1]

 

to 4.85 = −3.96 + a × 0.16 

[1]

2

 

a = 55.1

 55 m s  

[1]

 

Upwards direction 

[1]

−

Tangent drawn at t = 0.7 s and gradient of graph determined  [1] (±0.2) m s 2 

 iii iii

 

c

a = 0.8

b

Acceleration is constant from t = 0 to about t = 0.5 s 

[1]

 

Acceleration decreases from t = 0.5 s 

[1]

 

Gradient constant from t = 0 to t = 0.5 s and decreases from t = 0.5 s  [1]

c

Area under the graph used 

 

Correct method, e.g. trapezium rule or  squares counted  [1]

−

[1]

 

distance = 0.20 ± 0.01 m 

[1]

d

Random errors: errors: the points are either side of the line  [1]

 

Systematic errors: the whole line is shifted up or down  [1]

i

[1]

31.8 ≈ 32 m; accept error carried forward from time in i 

[1]

The initial speed of the ball or the hot-air balloon is 15 m s 1 

[1]

−

[1]

 

[1]

i

[1]

−

 iii v = u + at leading iii

≈

Constant gradient 

 ii ii

14 a

The acceleration is in the opposite direction to the initial speed of the ball. or The acceleration acce leration due to gravity is downwards and the ball initially rises.  [1]

v2 = u2 + 2as;

202 = 0 + 2 × 9.81 × s 

 20 m 

[1] [1]

 

s = 20.4

b

v = u + at;

 

t = 2.04

c

distance = 80 × 2.04 = 163 m ≈ 160 m  [1]

≈

≈

20 = 0 + 9.81 × t 

 2.0 s 

[1] [1]

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020