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(A) The impulse $J = Mv$ is applied at one end of the rod. This impulse provides an angular momentum about the center of mass $(CM)$ of the system.The

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$v/L$

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(A) The impulse $J = Mv$ is applied at one end of the rod. This impulse provides an angular momentum about the center of mass $(CM)$ of the system.The center of mass of the two-ball system is at the midpoint of the rod.The angular momentum $L_{CM}$ imparted by the impulse $J$ about the center of mass is given by $L_{CM} = J 	imes r$,where $r = L/2$ is the distance from the center of mass to the point of application of the impulse.$L_{CM} = (Mv) 	imes (L/2) = \frac{MvL}{2}$.The moment of inertia $I$ of the system about an axis passing through the center of mass and perpendicular to the rod is $I = M(L/2)^2 + M(L/2)^2 = 2M(L^2/4) = \frac{ML^2}{2}$.Using the relation $L_{CM} = I\omega$,we have:$\frac{MvL}{2} = \left( \frac{ML^2}{2} ight) \omega$.Solving for $\omega$:$\omega = \frac{MvL/2}{ML^2/2} = \frac{v}{L}$.

Consider a body,shown in the figure,consisting of two identical balls,each of mass $M$,connected by a light rigid rod of length $L$. If an impulse $J = Mv$ is imparted to the body at one of its ends,what would be its angular velocity?

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