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A $0.5\, kg$ ball moving with a speed of $12\,m/s$ strikes a hard wall at an angle of $30^o $ with the wall. It is reflected with the same speed at the same angle. If the ball is in contact with the wall for $0.25$ seconds, the average force acting on the wall is ........... $N$
$96$
$48$
$24$
$12$
Solution
$\begin{array}{l}
\,\,\,\,\,\,\,\,\,Components\,of\,momentum\,parallel\\
to\,the\,wall\,add\,each\,other\,and\,commponents\\
of\,momentum\,in\,the\,prependicular\,to\,\\
the\,wall\,are\,opposite\,to\,each\,other.\,\\
Therefore\,change\,of\,momentum\,is\,final\\
momentum\, – \,initial\,momentum\\
i.e,.\,(mv\sin \theta \,after\,collision\\
– ( – mv\sin \theta )\,before\,collision)
\end{array}$
$\begin{array}{l}
\,\,\,F \times t = change\,in\,momentum\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,2mv\sin \theta \\
\therefore \,\,F = \frac{{2mv\sin \theta }}{t}\\
\,\,\,\,\,\,\,\,\,\, = \frac{{2 \times 0.5 \times 12 \times \sin {{30}^ \circ }}}{{0.25}} = 48 \times \frac{1}{2} = 24\,N
\end{array}$
