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Internal Force

A shell at re...

Question

A shell at rest on a smooth horizontal surface explodes into two fragments of masses $m_{1}$ and $m_{2}$ . If just after explosion $m_{1}$ move with speed $u$ , then work done by internal forces during explosion is

\frac{1}{2} (m_{1} + m_{2}) \frac{m_{2}}{m_{1}} u^{2}

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\frac{1}{2} (m_{2} + m_{1}) u^{2}

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\frac{1}{2} m_{1} u^{2} (1 + \frac{m_{1}}{m_{2}})

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\frac{1}{2} (m_{2} - m_{1}) u^{2}

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Solution

The correct option is C $\frac{1}{2} m_{1} u^{2} (1 + \frac{m_{1}}{m_{2}})$
According to the question shell is exploding due to its internal force
No external force is acting on the shell
So momentum will be conserved
Using momentum conservation, $m_{1} u = m_{2} v$ ----------(1)

Now using work energy theorem,

$W = \frac{m_{1} u^{2}}{2} + \frac{m_{2} v^{2}}{2}$
From equation 1
$W = \frac{m_{1} u^{2}}{2} (\frac{m_{1} + m_{2}}{m_{2}})$
Hence work done is $\frac{1}{2} m_{1} u^{2} (1 + \frac{m_{1}}{m_{2}})$

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A shell at rest on a smooth horizontal surface explodes into two fragments of masses m1 and m2. If just after explosion m1 move with speed u, then work done by internal forces during explosion is

A shell at rest on a smooth horizontal surface explodes into two fragments of masses $m_{1}$ and $m_{2}$ . If just after explosion $m_{1}$ move with speed $u$ , then work done by internal forces during explosion is