gce o level paper answers for 2009

GCE ‘O’ Level October/November 2009 Suggested Solutions

Physics (5058 /01)
#

Ans

1

version 1.1

Workings/Remarks

C

End correction = 0.2cm
Actual length of object X = 1.1 + 0.2 = 1.3cm

2

D

Force, Acceleration, Velocity and Displacement are vectors. Work, Time and Mass are scalars. 3

D

Gradient of speed-time graph = acceleration
Changing gradient (curve) = non uniform acceleration
Constant gradient (straight line) = uniform acceleration

4

C

… 9 dots …
20 cm

… 9 dots …

… 9 dots …

… 9 dots …

… 9 dots …

Noting that dots are shared across consecutive interval, we calculate based on the spacings between consecutive dots.
In each 20 cm interval, there are 11 dots ⇒ 10 spacings
⇒ Between 50 dots, there’s a total of 49 spacings.
× 10s

Since 50 dots/49 spacings are moved in 1 sec, time taken for 1 spacing =
Time taken to move 20 cm/10 spacings =
∴Speed of tape =

distance time =

20

1
×10
49

1

49

= 98 cm/s ≈ 100 cm/s

5

B

Resultant force = 10 – 2 = 8 N

6

C
A

Density is inversely proportional to volume (Density = Mass / Volume)

8

A

s

Acceleration = (30 – 10) / 16 = 1.25 m/s2

7

1

49

Clockwise moment = 1.5 × (50 − 30) = 30 N cm

The weight of 5N (towards centre of earth/downwards on trolley) is not included in the calculation because it is not acting in the same direction as the motion of trolley (along a level bench).
F = ma = 12000 × 1.25 = 15000N
Minimum volume = Maximum density and vice versa
Anticlockwise moment = 2 × (30 −15) = 30 N cm
Resultant moment is the difference between clockwise and anticlockwise moment.

9

B

Pressure = Force / Area = Weight / Area
A: P = 30/100

= 0.3 N/cm2

B: P = 500/150

= 3.33 N/cm2

C: P = 750 / 300

= 2.5 N/cm2

D: P = 10000 / 4000 = 2.5 N/cm2
For tuition, exam papers & Last-Minute Buddha Foot Hugging Syndrome treatment

[phone] / [e-mail]