AS and A Level Physics Original material © Cambridge University Press 2010 1

2 Marking scheme: Worksheet (AS) 1 C [1]

2 B [1]

3 D [1]

4 B [1]

5 C [1]

6 time

yin velocit changeonaccelerati = [1]

Acceleration is a vector. [1]

7 u = 0 v = 15 m s−1 a = ? t = 0.30 s

tuva −

= [1]

30.0015 −

=a [1]

a = 50 m s−2 [1]

8 a The object is travelling initially at 20 m s−1 and has a constant acceleration. [1]

b a = gradient of graph [1]

0.82035 −

=a [1]

a = 1.88 m s−2 2s m 9.1 −≈ [1]

c Distance = area under the graph [1]

distance = area of ‘trapezium’ = ( ) 0.8352021

×+ [1]

distance = 220 m [1]

9 a u = 22 m s−1 v = 5.0 m s−1 a = ? t = 6.0 s

tuva −

= [1]

0.6220.5 −

=a [1]

2s m 8.2 −−≈a (negative value → deceleration) [1]

b 22 5Average velocity

2+

= = 13.5 m s−1 [1]

c distance = average velocity × time distance = 13.5 × 6.0 [1] distance = 81 m [1]

2 Marking scheme: Worksheet (AS)

AS and A Level Physics Original material © Cambridge University Press 2010 2

10 a u = 0 v = ? a = 9.81 m s−2 t = 2.3 s v = u + at [1] v = 0 + 9.81 × 2.3 [1] v = 22.6 m s−1 1s m 23 −≈ [1]

b s = ? u = 0 a = 9.81 m s−2 t = 2.3 s 2

21 atuts += [1]

s = 0 + 21 × 9.81 × 2.32 [1]

m 26≈s [1]

11 s = 9.0 m u = 4.0 m s−1 v = ? a = 0.45 m s−2 v2

= u2 + 2as [1]

v2 = 4.02 + (2 × 0.45 × 9.0) = 24.1 [1]

v2 = 1s m 9.41.24 −≈ [1]

12 s = 20 m u = 45 m s−1 v = 0 a = ? v2

= u2 + 2as [1]

202450

2

222

×−

=−

=suva [1]

2s m 51 −−≈a [1]

13 a

Line of positive slope [1]

Correct labels on axes [1]

b s = area under the graph s = area of ‘larger’ rectangle − area of shaded triangle [1]

tvvts )(21∆−= [1]

∆v = at [1]

hence 2

21)(

21 atvttatvts −=−= [1]

14 During free fall: s = 6.0 m u = 0 v = ? a = 9.81 m s−2 v2 = u2 + 2as [1]

0.681.92 ××=v [1] v = 10.85 m s−1 [1]

During landing on soft ground: s = 0.085 m u = 10.85 m s−1 v = 0 a = ?

v2 = u2 + 2as ⇒ 085.0285.100

2

222

×−

=−

=suva [1]

2s m 690 −−≈a [1]

2 Marking scheme: Worksheet (AS)

AS and A Level Physics Original material © Cambridge University Press 2010 3

15 a Distance = area under graph from 4 s to 8 s = 21 (12 + 6.0) × 4.0 [1]

distance = 36 m [1] b Acceleration = gradient of graph at 12.5 s [1]

acceleration = 2013

=∆∆

tv [1]

acceleration = 0.65 m s−2 (allow ± 0.10 m s−2) [1]

c Constant acceleration of 1.5 m s−2 from 0 to 10 s [1] Acceleration gradually decreasing to zero after 10 s [1]

16 a uh = 6052

.

.ts= = 4.17 m s−1 [1]

b Vertically final velocity is zero as ball passes over cross-bar. vv = uv + at 0 = u – 9.81 × 0.60 [1] uv = 5.89 m s−1 [1]

c u2 = (uh2 + uv

2) = 5.892 + 4.172 [1] u = 7.21 m s−1 [1]

d tan θ =174895

b

v

.

.uu

= = 1.41 [1]

θ = 55º above the horizontal [1]