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2 Marking scheme: Worksheet (AS) 1

C

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B

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D

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B

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C

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6

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change in velocity time Acceleration is a vector. acceleration =

u = 0 v = 15 m s−1 v−u a= t 15 − 0 a= 0.30 a = 50 m s−2

a=?

[1] [1]

t = 0.30 s

[1] [1] [1]

a The object is travelling initially at 20 m s−1 and has a constant acceleration. b a = gradient of graph 35 − 20 a= 8 .0 a = 1.88 m s−2 ≈ 1.9 m s −2 c

Distance = area under the graph

a

b c

[1]

[1] [1]

u = 22 m s−1 v = 5.0 m s−1 a = ? t = 6.0 s v−u a= t 5.0 − 22 a= 6 .0 a ≈ −2.8 m s −2 (negative value → deceleration)

[1] [1] [1]

22 + 5 Average velocity = = 13.5 m s−1 2

[1]

distance = average velocity × time distance = 13.5 × 6.0 distance = 81 m

AS and A Level Physics

[1]

[1]

1 distance = area of ‘trapezium’ = (20 + 35) × 8.0 2 distance = 220 m

9

[1] [1]

[1] [1]

Original material © Cambridge University Press 2010

1

2 Marking scheme: Worksheet (AS)

u = 0 v = ? a = 9.81 m s−2 t = 2.3 s v = u + at v = 0 + 9.81 × 2.3 v = 22.6 m s−1 ≈ 23 m s −1 b s = ? u = 0 a = 9.81 m s−2 t = 2.3 s 1 s = ut + at 2 2 1 s = 0 + × 9.81 × 2.32 2 s ≈ 26 m

10 a

[1] [1] [1] [1] [1]

[1]

11 s = 9.0 m u = 4.0 m s−1 v = ? a = 0.45 m s−2 v2 = u2 + 2as v2 = 4.02 + (2 × 0.45 × 9.0) = 24.1 v2 = 24.1 ≈ 4.9 m s −1 12 s = 20 m u = 45 m s−1 v2 = u2 + 2as v 2 − u 2 0 − 452 a= = 2s 2 × 20 a ≈ −51 m s −2

[1] [1] [1]

v=0 a=?

13 a

[1]

[1] [1] 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 s = vt − (∆v)t 2 ∆v = at 1 1 hence s = vt − (at )t = vt − at 2 2 2 14 During free fall: s = 6.0 m u = 0 v = ? v2 = u2 + 2as v = 2 × 9.81 × 6.0 v = 10.85 m s−1

[1] [1] [1]

a = 9.81 m s−2 [1] [1] [1]

During landing on soft ground: s = 0.085 m u = 10.85 m s−1 v = 0 a = ? v 2 − u 2 0 − 10.852 = v2 = u2 + 2as ⇒ a = 2s 2 × 0.085 −2 a ≈ −690 m s

AS and A Level Physics

[1]

Original material © Cambridge University Press 2010

[1] [1]

2

2 Marking scheme: Worksheet (AS)

15 a

Distance = area under graph from 4 s to 8 s =

1 (12 + 6.0) × 4.0 2

distance = 36 m

[1]

b Acceleration = gradient of graph at 12.5 s ∆v 13 acceleration = = ∆t 20 acceleration = 0.65 m s−2 (allow ± 0.10 m s−2) c

16 a

[1] [1] [1]

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

uh =

s 2.5 = 4.17 m s−1 = t 0 .6

u2 = (uh2 + uv2) = 5.892 + 4.172 u = 7.21 m s−1

d tan θ =

[1] [1] [1] [1]

u v 5.89 = = 1.41 u b 4.17

[1]

θ = 55º above the horizontal

AS and A Level Physics

[1] [1]

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

[1]

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Original material © Cambridge University Press 2010

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C

[1]

2

B

[1]

3

D

[1]

4

B

[1]

5

C

[1]

6

7

8

change in velocity time Acceleration is a vector. acceleration =

u = 0 v = 15 m s−1 v−u a= t 15 − 0 a= 0.30 a = 50 m s−2

a=?

[1] [1]

t = 0.30 s

[1] [1] [1]

a The object is travelling initially at 20 m s−1 and has a constant acceleration. b a = gradient of graph 35 − 20 a= 8 .0 a = 1.88 m s−2 ≈ 1.9 m s −2 c

Distance = area under the graph

a

b c

[1]

[1] [1]

u = 22 m s−1 v = 5.0 m s−1 a = ? t = 6.0 s v−u a= t 5.0 − 22 a= 6 .0 a ≈ −2.8 m s −2 (negative value → deceleration)

[1] [1] [1]

22 + 5 Average velocity = = 13.5 m s−1 2

[1]

distance = average velocity × time distance = 13.5 × 6.0 distance = 81 m

AS and A Level Physics

[1]

[1]

1 distance = area of ‘trapezium’ = (20 + 35) × 8.0 2 distance = 220 m

9

[1] [1]

[1] [1]

Original material © Cambridge University Press 2010

1

2 Marking scheme: Worksheet (AS)

u = 0 v = ? a = 9.81 m s−2 t = 2.3 s v = u + at v = 0 + 9.81 × 2.3 v = 22.6 m s−1 ≈ 23 m s −1 b s = ? u = 0 a = 9.81 m s−2 t = 2.3 s 1 s = ut + at 2 2 1 s = 0 + × 9.81 × 2.32 2 s ≈ 26 m

10 a

[1] [1] [1] [1] [1]

[1]

11 s = 9.0 m u = 4.0 m s−1 v = ? a = 0.45 m s−2 v2 = u2 + 2as v2 = 4.02 + (2 × 0.45 × 9.0) = 24.1 v2 = 24.1 ≈ 4.9 m s −1 12 s = 20 m u = 45 m s−1 v2 = u2 + 2as v 2 − u 2 0 − 452 a= = 2s 2 × 20 a ≈ −51 m s −2

[1] [1] [1]

v=0 a=?

13 a

[1]

[1] [1] 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 s = vt − (∆v)t 2 ∆v = at 1 1 hence s = vt − (at )t = vt − at 2 2 2 14 During free fall: s = 6.0 m u = 0 v = ? v2 = u2 + 2as v = 2 × 9.81 × 6.0 v = 10.85 m s−1

[1] [1] [1]

a = 9.81 m s−2 [1] [1] [1]

During landing on soft ground: s = 0.085 m u = 10.85 m s−1 v = 0 a = ? v 2 − u 2 0 − 10.852 = v2 = u2 + 2as ⇒ a = 2s 2 × 0.085 −2 a ≈ −690 m s

AS and A Level Physics

[1]

Original material © Cambridge University Press 2010

[1] [1]

2

2 Marking scheme: Worksheet (AS)

15 a

Distance = area under graph from 4 s to 8 s =

1 (12 + 6.0) × 4.0 2

distance = 36 m

[1]

b Acceleration = gradient of graph at 12.5 s ∆v 13 acceleration = = ∆t 20 acceleration = 0.65 m s−2 (allow ± 0.10 m s−2) c

16 a

[1] [1] [1]

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

uh =

s 2.5 = 4.17 m s−1 = t 0 .6

u2 = (uh2 + uv2) = 5.892 + 4.172 u = 7.21 m s−1

d tan θ =

[1] [1] [1] [1]

u v 5.89 = = 1.41 u b 4.17

[1]

θ = 55º above the horizontal

AS and A Level Physics

[1] [1]

[1]

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

[1]

[1]

Original material © Cambridge University Press 2010

3

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