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4-1.Newton's Laws of Motion
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A body of mass $3\, kg$ hits a wall at an angle of $60^o$ and returns at the same angle. The impact time was $0.2 \;.s$ Calculate the force exerted on the wall
A
$100\,N$
B
$50\sqrt 3 \,N$
C
$150\sqrt 3 \,N$
D
$75\sqrt 3 \,N$
(AIIMS-2013) (AIPMT-2000)
Solution
$\begin{array}{l}
Change\,in\,momentum\,along\,the\,wall\\
= mv\cos {60^ \circ } – mv\cos {60^ \circ } = 0\\
Change\,in\,momentum\,perpendicular\,\\
to\,the\,wall\\
= mv\sin {60^ \circ } – \left( { – mv\sin {{60}^ \circ }} \right)\\
= 2mv\,\sin {60^ \circ }\\
\therefore Applied\,force
\end{array}$
$\begin{array}{l}
= \frac{{Change\,in\,momentum}}{{Time}}\\
= \frac{{2\,mv\sin {{60}^ \circ }}}{{0.20}}\\
= \frac{{2 \times 3 \times 10 \times \sqrt 3 }}{{2 \times 0.20}} = 50 \times 3\sqrt 3 \\
= 150\sqrt 3 \,newton
\end{array}$
Standard 11
Physics
