A body A moving with momentum P collides one- | vedclass

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(C) Let the mass of both bodies be $m$. Initially,body $A$ has momentum $P$,so its velocity is $v_A = P/m$. Body $B$ is stationary,so $v_B = 0$.During

Vedclass · Accepted Answer

$(b)$ and $(c)$ are correct.

Vedclass · Answer

(C) Let the mass of both bodies be $m$. Initially,body $A$ has momentum $P$,so its velocity is $v_A = P/m$. Body $B$ is stationary,so $v_B = 0$.During the impact,body $A$ exerts an impulse $J$ on body $B$. By Newton's third law,body $B$ exerts an equal and opposite impulse $J$ on body $A$.Statement $(b)$ is correct because the impulse exerted by $B$ on $A$ is equal in magnitude to the impulse exerted by $A$ on $B$.After the impact,the momentum of $B$ is $P_B = J$,so its velocity is $v_B' = J/m$.The momentum of $A$ becomes $P_A' = P - J$,so its velocity is $v_A' = (P - J)/m$.The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach:$e = \frac{v_B' - v_A'}{v_A - v_B} = \frac{\frac{J}{m} - \frac{P-J}{m}}{\frac{P}{m} - 0} = \frac{2J - P}{P} = \frac{2J}{P} - 1$.Thus,statement $(c)$ is also correct.Therefore,statements $(b)$ and $(c)$ are correct.

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