2005 3E Sc(Phy) CA2 Suggested Answers:
Section A:
1.D 7.D 13.B
2.D 8.A 14.C
3.B 9.D 15.C
4.D 10.A
5.B 11.D
6.B 12.A
Section B:
16.(a)
Average speed = Total distance / Total time
= 1.8 / 1.5
= 1.2 m s-1
(b) Distance travelled = Area under graph
= ½ v (1.5)
= 0.75 v
= 1.8 / 0.75
= 2.4 m s-1
(c) Acceleration = Δv / Δt
= (0 – 2.4) / (0 – 1.5)
= 1.60 m s-2
17. (a) Work done = Force x Distance moved
= 12 x 4.0
= 48.0 J
(b) a = F / m
= 12 / 3
= 4 m s-2
Distance moved = Area under graph
= ½ (4x) x
= 2x2
= 2.0
= 1.414 s
Speed = 4x
= 4 (1.414)
= 5.66 m s-1
18. Acceleration = (50 – 20) / 3
= 10 m s-2
F = ma
= 100 x 10
= 1000 N
19.(a) First second because acceleration is higher in the first second.
(b) Eleventh second because area under the speed-time graph is greater for the eleventh second than for first second.
Or
Because the average speed for the eleventh second is higher.
20. (a) W = mg
= 20 x 10
= 200 N
(bi) 80N since the box moved at a constant speed, implying zero net force on the box.
(ii)(FIGURE CANNOT BE SHOWN HERE)
IF U WAN..PLS TELL SIAO HUAN..SHE WILL SENT TO U PPL=)
(ci) F = ma
= 20 x 1.5
= 30 N
(ii) Friction= 80 – 30
= 50 N
revise!!! jyjy~~ CA2=)
good luck ppl..don eat snake arhs.(mr.leow)
signin` off
-huan-
perserverance yield success ! PYSS ,
