WAEC 2018 · Paper 2 · Q14

(a)
A body is under the action of forces $M = (4\text{ N}, 030^\circ)$ , $N = (10\text{ N}, 300^\circ)$ and $P = (a\text{ N}, x^\circ)$ which keep it in equilibrium. Find the force $P$ .
Separate values with commas, e.g. 3, −2
(b)
The initial velocity of a particle of mass $0.2\text{ kg}$ is $40\text{ m s}^{-1}$ in the direction of the unit vector $\mathbf j$ . The velocity of the particle changed to $30\text{ m s}^{-1}$ in the direction of the unit vector $\mathbf i$ . Find the change in momentum.

Worked solution (try it first)

$M$ : $4\sin30^\circ = 2$ east and $4\cos30^\circ = 3.464$ north.
$N$ : $10\sin300^\circ = -8.660$ east and $10\cos300^\circ = 5$ north.
$M + N = -6.660\mathbf i + 8.464\mathbf j$ .
In equilibrium $P = -(M + N)$
$= 6.660\mathbf i - 8.464\mathbf j$ .
$|P| = \sqrt{44.36 + 71.64}$
$= \sqrt{116.0}$
$\approx 10.77\text{ N}$ .
$P$ points south-east: $\tan^{-1}\frac{6.660}{8.464} = 38.2^\circ$ east of south, a bearing of $180^\circ - 38.2^\circ = 141.8^\circ$ .
So $P \approx (10.77\text{ N}, 141.8^\circ)$ .

Change in momentum $= m\mathbf v - m\mathbf u$
$= 0.2(30\mathbf i) - 0.2(40\mathbf j)$
$= 6\mathbf i - 8\mathbf j$ .
Its magnitude is $\sqrt{36 + 64} = 10\text{ kg m s}^{-1}$ .