Two masses A and B of mass M and 2M respectiv | vedclass

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(C) The system is on a frictionless table,so there is no external force acting in the $x$-direction. The initial velocity of the system in the $x$-dir

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$-2 V_x \hat{i} + u \hat{k}$

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(C) The system is on a frictionless table,so there is no external force acting in the $x$-direction. The initial velocity of the system in the $x$-direction is zero,so the total momentum in the $x$-direction must remain zero throughout the motion.Let $v_{x_A}$ and $v_{x_B}$ be the $x$-components of the velocities of masses $A$ and $B$ respectively.Applying the Conservation of Linear Momentum in the $x$-direction:$M v_{x_A} + (2M) v_{x_B} = 0$$M v_{x_A} = -2M v_{x_B}$$v_{x_A} = -2 v_{x_B}$Given that the $x$-component of the velocity of mass $B$ is $V_x$,we have $v_{x_B} = V_x$.Therefore,$v_{x_A} = -2 V_x$.The velocity in the $z$-direction remains constant at $u \hat{k}$ for both masses because there are no external forces in the $z$-direction.Thus,the velocity of mass $A$ is $\vec{V}_A = -2 V_x \hat{i} + u \hat{k}$.

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