A ball A of mass m moving along the positive | vedclass

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(C) Initial kinetic energy of ball $A$ is $K = p^2 / (2m)$,so $p = \sqrt{2mK}$.Initial velocity of ball $A$ is $u_1 = p/m = \sqrt{2K/m}$.After collisi

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$4p/3$

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(C) Initial kinetic energy of ball $A$ is $K = p^2 / (2m)$,so $p = \sqrt{2mK}$.Initial velocity of ball $A$ is $u_1 = p/m = \sqrt{2K/m}$.After collision,ball $A$ moves in the negative $x$-direction with kinetic energy $K' = K/9$.Let $v_1$ be the final velocity of ball $A$. Then $K' = 1/2 m v_1^2 = K/9$,which gives $v_1 = \sqrt{2K/(9m)} = (1/3) \sqrt{2K/m} = p/(3m)$.Since ball $A$ moves in the negative $x$-direction,its final velocity is $v_1 = -p/(3m)$.By the law of conservation of linear momentum: $p_{initial} = p_{final}$.$p = m v_1 + p_B$,where $p_B$ is the final momentum of ball $B$.$p = m(-p/(3m)) + p_B$.$p = -p/3 + p_B$.$p_B = p + p/3 = 4p/3$.

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