A shell at rest on a smooth horizontal surfac | vedclass

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Question

(C) Initially,the shell is at rest,so the initial momentum is $0$. By the law of conservation of linear momentum,the final momentum must also be $0$.L

Vedclass · Accepted Answer

$\frac{1}{2} m_1 u^2\left(1+\frac{m_1}{m_2}\right)$

Vedclass · Answer

(C) Initially,the shell is at rest,so the initial momentum is $0$. By the law of conservation of linear momentum,the final momentum must also be $0$.Let $v$ be the speed of mass $m_2$. Then $m_1 u = m_2 v$,which gives $v = \frac{m_1 u}{m_2}$.The work done by internal forces is equal to the change in kinetic energy of the system.$W = K_f - K_i = \left(\frac{1}{2} m_1 u^2 + \frac{1}{2} m_2 v^2\right) - 0$.Substituting $v = \frac{m_1 u}{m_2}$:$W = \frac{1}{2} m_1 u^2 + \frac{1}{2} m_2 \left(\frac{m_1 u}{m_2}\right)^2 = \frac{1}{2} m_1 u^2 + \frac{1}{2} \frac{m_1^2 u^2}{m_2}$.$W = \frac{1}{2} m_1 u^2 \left(1 + \frac{m_1}{m_2}\right)$.

$A$ shell at rest on a smooth horizontal surface explodes into two fragments of masses $m_1$ and $m_2$. If just after explosion $m_1$ moves with speed $u$,then the work done by internal forces during the explosion is:

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