A mass M moving with a certain speed V collid | vedclass

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Question

(D) The collision is elastic,so both linear momentum and kinetic energy are conserved.According to the conservation of linear momentum:$M V = M V^{\pr

Vedclass · Accepted Answer

$v=V+V^{\prime}$

Vedclass · Answer

(D) The collision is elastic,so both linear momentum and kinetic energy are conserved.According to the conservation of linear momentum:$M V = M V^{\prime} + m v \implies M(V - V^{\prime}) = m v \dots (i)$According to the conservation of kinetic energy:$\frac{1}{2} M V^2 = \frac{1}{2} M V^{\prime 2} + \frac{1}{2} m v^2 \implies M(V^2 - V^{\prime 2}) = m v^2 \dots (ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{M(V^2 - V^{\prime 2})}{M(V - V^{\prime})} = \frac{m v^2}{m v}$$\frac{(V - V^{\prime})(V + V^{\prime})}{(V - V^{\prime})} = v$$V + V^{\prime} = v$Thus,option $(d)$ is correct.

$A$ mass $M$ moving with a certain speed $V$ collides elastically with another stationary mass $m$. After the collision,the masses $M$ and $m$ move with speeds $V^{\prime}$ and $v$,respectively. All motion is in one dimension. Then,

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