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