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A shell of mass $200\, gm$ is ejected from a gun of mass $4\, kg$ by an explosion that generates $1.05\, kJ$ of energy. The initial velocity of the shell is .............. $\mathrm{ms}^{-1}$
$40$
$120$
$100$
$80$
Solution
$\begin{array}{l}
mv = Mv'\, \Rightarrow \,v' = \left( {\frac{m}{M}} \right)v\\
Total\,K.E.\,of\,the\,bullet\,and\,gun\\
\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}m{v^2} + \frac{1}{2}M{v^2}\\
Total\,K.E. = \frac{1}{2}m{v^2} + \frac{1}{2}M.\frac{{{m^2}}}{{{M^2}}}{v^2}
\end{array}$
$\begin{array}{l}
Total\,K.E. = \frac{1}{2}m{v^2}\left\{ {1 + \frac{m}{M}} \right\}\\
= \left\{ {\frac{1}{2} \times 0.2} \right\}\left\{ {1 + \frac{{0.2}}{4}} \right\}{v^2} = 1.05 \times 1000\,J\\
\Rightarrow \,\,\,\,{v^2} = \frac{{4 \times 1.05 \times 1000}}{{0.1 \times 4.2}} = {100^2}\\
\therefore \,\,\,\,\,v = 100m{s^{ – 1}}
\end{array}$
