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Two balls each of mass $M$ moving in opposite directions with equal speeds $v$ undergo a head-on collision. the velocity of the two balls after the collision will be {fill regular} and ____.

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Detailed Solution

Two balls each of mass M moving in opposite directions with equal speeds v undergo a head-on collision. the velocity of the two balls after the collision will be

v_{1} = - v

and

v_{2} = v

.
Given
Two ball mass are

M 1 = M 2 = M

;

u 1 = v and u 2 = –v

Final velocity of the ball 1 is given by

v_{1} = \frac{(M_{1} - M_{2}) u_{1} + 2 M_{2} u_{2}}{M_{1} + M_{2}}

\Rightarrow v_{1} = \frac{(M - M) v + 2 M {(- v)}_{2}}{M_{} + M}

\Rightarrow v_{1} = \frac{(0) v + 2 M (- v)}{M + M}

\Rightarrow v_{1} = \frac{- 2 M (v)}{2 M}

\Rightarrow v_{1} = - v

Final velocity of the ball 2 is given by

v_{2} = \frac{(M_{2} - M_{1}) u_{1} + 2 M_{1} u_{1}}{M_{1} + M_{2}}

\Rightarrow v_{2} = \frac{(M - M) v + 2 M {(v)}_{}}{M_{} + M}

\Rightarrow v_{2} = \frac{(0) v + 2 M (v)}{M + M}

\Rightarrow v_{2} = \frac{2 M (v)}{2 M}

\Rightarrow v_{2} = v

Thus, after the collision, the two balls bounce back with equal speeds v

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