WAEC 2008 · Paper 2 · Q18✱

(a)
An object is thrown up a smooth plane inclined at an angle of $30^\circ$ to the horizontal. If the plane is $15\text{ m}$ long and the object comes to rest at the top, find the time it takes to reach the top. (Take $g = 10\text{ m s}^{-2}$ .)
(b)
Forces of magnitudes $5\text{ N}$ , $5\sqrt3\text{ N}$ , $10\text{ N}$ , $5\sqrt3\text{ N}$ and $5\text{ N}$ act on a body $P$ of mass $5\text{ kg}$ as shown in the diagram. Find the: (i) magnitude of the resultant force; (ii) acceleration of the body.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

On a smooth plane the only force along the slope is the component of the weight, $mg\sin30^\circ$ , acting down the slope.
So the deceleration is $g\sin 30^\circ = 10 \times \frac12$
$= 5\text{ m s}^{-2}$ .
The object stops at the top, so work backwards from rest: $s = \frac12 at^2$ gives $15 = \frac12(5)t^2$ .
$t^2 = 6$ , so $t = \sqrt6 \approx 2.45$ seconds.

Take the 10 N force along the $x$ -axis.
The others make $60^\circ$ , $30^\circ$ , $-30^\circ$ and $-60^\circ$ with it.
Vertical components: $5\sin60^\circ + 5\sqrt3\sin30^\circ - 5\sqrt3\sin30^\circ - 5\sin60^\circ = 0$ .
They cancel.
Horizontal components: $2(5\cos60^\circ) + 2(5\sqrt3\cos30^\circ) + 10 = 5 + 15 + 10$
$= 30$ .
The resultant is $30\text{ N}$ along the direction of the 10 N force.