Part (a): This was attempted by many candidates but few failed this technically by writing
- three forces can be represented by the three sides of a triangle taken in order.
- Given three forces they can be represented in magnitude and direction by the three sides of a triangle taken in order.
While the first was incorrect outright, the second failed to include the necessary condition that the three forces must be in equilibrium.
Part (b): This did not pose a problem to most responding candidates. Performance was above average.
Part (c): This was avoided by majority of the candidates. Many got the shapes wrong and a few candidates used separate axes which was contrary to the instruction. Performance was poor.
Part (d): (i-iii) was done correctly by most candidates however, a few candidates could not calculate (i-iii) because they applied the wrong formula.
Part(iv) posed a challenge to some of the respondent candidates. They approached it thus:
Ft = (Mp + MQ)v. This was not a problem of impulse but that of conservation of momentum i.e
MpUp + MQUQ = (Mp + MQ)v.
Performance was poor.
The expected answers are:
a) Triangle law of vector addition
If three forces are in equilibrium, they can be represented in magnitude and direction by the
three sides of a triangle taken in order
(b) Physical quantities associated with equations of motion
e.g - distance/displacement
- speed/velocity
- acceleration
- time.
(c)
Both axes correctly distinguished
Correct shape for potential energy
Correct shape for kinetic energy
(d) (i) W = ½ ke2
10 = ½ x k x( )2
k = 8000 Nm-1
(ii) F = ke
= 8000 x
= 400N
(iii) F = ma a = ω2A = A
400 = 2.0 x a OR = x 5 x 10-2
a = 200 ms-2 = 200 ms-2
(iv) Initial velocity of block P = Up = 200 x 0.25 = 50ms-1
Mp Up + MQUQ = (Mp+MQ)v
Where v is the common velocity after collision
(2 x 50) + (4.0 x 0) = (2 + 4) v
100 = 6v
v = 16.7 ms-1
