Body of mass M is much heavier than the other | vedclass

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Question

From conservation of momentum,

$M v+m 	imes 0=M v_{1}+m v_{2}$

$\Rightarrow M\left(v-v_{1}ight)=m v_{2} \ldots$ $(i)$

Again, from the cons

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$2 v$

Vedclass · Answer

From conservation of momentum,

$M v+m 	imes 0=M v_{1}+m v_{2}$

$\Rightarrow M\left(v-v_{1}ight)=m v_{2} \ldots$ $(i)$

Again, from the conservation of kinetic energy (as collision is of elastic nature)

$\frac{1}{2} M v^{2}+\frac{1}{2} m 	imes 0=\frac{1}{2} M v_{1}^{2}+\frac{1}{2} m v_{2}^{2}$

$\Rightarrow M\left(v^{2}-v_{1}^{2}ight)=m v_{2}^{2} \ldots($ ii $)$

On solving equations $(i)$ and $(ii),$ we get

$\frac{M\left(v-v_{1}ight)}{M\left(v+v_{1}ight)\left(v-v_{1}ight)}=\frac{m v_{2}}{m v_{2}^{2}}$

$v_{2}=v+v_{1} \ldots$ $(iii)$

Now, solving equations $(i)$ and $(iii)$, we get

$v_{1}=\frac{(M-m) v}{(M+m)}$ and $v_{2}=\frac{2 M v}{(M+m)}$

As $M \gg m$

$So , v_{1}=v \Rightarrow v_{2}=v+v=2 v$

Body of mass $M$ is much heavier than the other body of mass $m$. The heavier body with speed $v$ collides with the lighter body which was at rest initially elastically. The speed of lighter body after collision is

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