Find the velocity of the centre of mass $C$ of the system constituting 'the rod plus the particle'.

\frac{2 M v}{M - m}

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\frac{2 m v}{M + m}

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\frac{M v}{M + m}

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\frac{m v}{M + m}

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Solution

The correct option is D $\frac{m v}{M + m}$
Let the center of mass(C) of the rod moves with a linear velocity $v^{'} .$

As linear momentum is conserved, i-e, $P_{i} = P_{f}$

$m v + 0 = (M + m) v^{'}$

$⟹ v^{'} = \frac{m v}{M + m}$

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Find the velocity of the centre of mass C of the system constituting 'the rod plus the particle'.

The correct option is D mvM+mLet the center of mass(C) of the rod moves with a linear velocity v′.As linear momentum is conserved, i-e, Pi=Pfmv+0=(M+m)v′⟹v′=mvM+m

Find the velocity of the centre of mass $C$ of the system constituting 'the rod plus the particle'.

The correct option is D $\frac{m v}{M + m}$
Let the center of mass(C) of the rod moves with a linear velocity $v^{'} .$

As linear momentum is conserved, i-e, $P_{i} = P_{f}$

$m v + 0 = (M + m) v^{'}$

$⟹ v^{'} = \frac{m v}{M + m}$