WAEC 2019 · Paper 2 · Q6

(a)
The force of attraction, $F$ , between two bodies varies directly as the product of their masses, $m_1$ and $m_2$ , and inversely as the square of the distance, $d$ , between them. Given that $F = 20\text{ N}$ when $m_1 = 25\text{ kg}$ , $m_2 = 10\text{ kg}$ and $d = 5\text{ m}$ , find: (i) an expression for $F$ in terms of $m_1$ , $m_2$ and $d$ ; (ii) the distance $d$ when $F = 30\text{ N}$ , $m_1 = 7.5\text{ kg}$ and $m_2 = 4\text{ kg}$ .
(b)
The diagram is a pentagon with interior angles $x$ , $(x + 20^\circ)$ , $(x + 40^\circ)$ , $(x + 80^\circ)$ and $(x + 60^\circ)$ . Find the value of $x$ .

Worked solution (try it first)

$F$ varies directly as the product $m_1m_2$ (on top) and inversely as $d^2$ (underneath): $F = \dfrac{km_1m_2}{d^2}$ .
Put in $F = 20$ , $m_1 = 25$ , $m_2 = 10$ , $d = 5$ : $20 = \dfrac{k \times 25 \times 10}{25}$
$= 10k$ , so $k = 2$ .
The expression is $F = \dfrac{2m_1m_2}{d^2}$ .

Put in $F = 30$ , $m_1 = 7.5$ , $m_2 = 4$ : $30 = \dfrac{2 \times 7.5 \times 4}{d^2}$
$= \dfrac{60}{d^2}$ .
So $d^2 = \dfrac{60}{30} = 2$ and $d = \sqrt2 \approx 1.41$ m.

The interior angles of a pentagon add up to $(5 - 2) \times 180^\circ = 540^\circ$ .
So $x + (x + 20) + (x + 40) + (x + 80) + (x + 60) = 540$ , which gives $5x + 200 = 540$ , $5x = 340$ and $x = 68^\circ$ .