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 $\frac{1}{2}mv_1^2 = K/9$,which gives $v_1 = \sqrt{2K/(9m)} = \frac{1}{3}\sqrt{2K/m} = u_1/3 = p/(3m)$.Since the collision is one-dimensional,we apply the law of conservation of linear momentum:$p_{initial} = p_{final}$$p = -mv_1 + p_B$$p_B = p + mv_1$Substituting $v_1 = p/(3m)$:$p_B = p + m(p/(3m)) = p + p/3 = 4p/3$.

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