A smooth sphere of mass M moving with velocit | vedclass

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(C) For a one-dimensional elastic collision,the final velocity $v_2$ of the second body (mass $m_2$) is given by the formula:$v_2 = \left( \frac{m_2 -

Vedclass · Accepted Answer

$\frac{2u}{1 + \frac{m}{M}}$

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(C) For a one-dimensional elastic collision,the final velocity $v_2$ of the second body (mass $m_2$) is given by the formula:$v_2 = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) u_2 + \left( \frac{2m_1}{m_1 + m_2} \right) u_1$Here,$m_1 = M$,$m_2 = m$,$u_1 = u$,and $u_2 = 0$.Substituting these values into the formula:$v = \left( \frac{m - M}{M + m} \right) (0) + \left( \frac{2M}{M + m} \right) u$$v = \frac{2Mu}{M + m}$Dividing the numerator and denominator by $M$:$v = \frac{2u}{\frac{M}{M} + \frac{m}{M}} = \frac{2u}{1 + \frac{m}{M}}$Thus,the correct option is $C$.

$A$ smooth sphere of mass $M$ moving with velocity $u$ directly collides elastically with another sphere of mass $m$ at rest. After collision,their final velocities are $V$ and $v$ respectively. The value of $v$ is

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