WAEC 2011 · Paper 2 · Q16

(a)
An object $P$ of mass $6.5\text{ kg}$ is suspended by two light inextensible strings $AP$ and $BP$ . The strings make angles $50^\circ$ and $60^\circ$ respectively with the downward vertical. (i) Express the forces acting on $P$ in component form. (ii) If $P$ is at rest, write down the vector equation connecting all the forces. (iii) Calculate, correct to one decimal place, the tensions in the strings. $[\text{Take } g = 10\text{ m s}^{-2}]$
Separate values with commas, e.g. 3, −2
(b)
A particle of mass $5\text{ kg}$ moves with initial velocity $\begin{pmatrix} -1 \\ 2 \end{pmatrix}\text{ m s}^{-1}$ and final velocity $\begin{pmatrix} 3 \\ 4 \end{pmatrix}\text{ m s}^{-1}$ . Find the magnitude of its change in momentum.

Worked solution (try it first)

The strings make $40^\circ$ ( $AP$ ) and $30^\circ$ ( $BP$ ) with the horizontal.
The weight is $6.5 \times 10 = 65\text{ N}$ .
$\mathbf T_1 = \begin{pmatrix} -T_1\cos40^\circ \\ T_1\sin40^\circ \end{pmatrix}$ , $\mathbf T_2 = \begin{pmatrix} T_2\cos30^\circ \\ T_2\sin30^\circ \end{pmatrix}$ and $\mathbf W = \begin{pmatrix} 0 \\ -65 \end{pmatrix}$ .

Across: $T_1\cos40^\circ = T_2\cos30^\circ$ , so $T_2 = 0.8846\,T_1$ .
Up: $T_1\sin40^\circ + T_2\sin30^\circ = 65$ , so $T_1(0.6428 + 0.4423) = 65$ .
$T_1 = \dfrac{65}{1.0851}$
$\approx 59.9\text{ N}$ and $T_2 \approx 53.0\text{ N}$ .

Change in momentum $= m(\mathbf v - \mathbf u)$
$= 5\begin{pmatrix} 3 + 1 \\ 4 - 2 \end{pmatrix}$
$= \begin{pmatrix} 20 \\ 10 \end{pmatrix}$ .
Its magnitude is $\sqrt{400 + 100} = 10\sqrt5$
$\approx 22.36\text{ kg m s}^{-1}$ .