WAEC 2013 · Paper 2 · Q17

(a)
A particle is under the action of forces $\mathbf P = (4\text{ N}, 030^\circ)$ and $\mathbf R = (10\text{ N}, 300^\circ)$ . Find the force that will keep the particle in equilibrium.
Separate values with commas, e.g. 3, −2
(b)
A train travelling at $45\text{ m s}^{-1}$ is brought to rest after covering a distance of $1500\text{ m}$ . Find the: (i) time taken to come to rest; (ii) uniform retardation.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

$\mathbf P$ : $4\sin30^\circ = 2$ east and $4\cos30^\circ = 3.464$ north.
$\mathbf R$ : $10\sin300^\circ = -8.660$ east and $10\cos300^\circ = 5$ north.
The balancing force is $-(\mathbf P + \mathbf R) = 6.660\mathbf i - 8.464\mathbf j$ .
Its size is $\sqrt{6.660^2 + 8.464^2} \approx 10.77\text{ N}$ .
It points south-east: $\tan^{-1}\frac{6.660}{8.464} = 38.2^\circ$ east of south, a bearing of $141.8^\circ$ .

(b)(i)

$s = \frac12(u + v)t$ : $1500 = \frac12(45)t$ , so $t = 66\frac23\text{ s}$ .

(ii)

$v^2 = u^2 + 2as$ : $0 = 2025 + 3000a$ , so $a = -0.675$ : a retardation of $0.675\text{ m s}^{-2}$ .

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