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(A) The impulse $J = Mv$ is applied at one end of the rod. This impulse imparts a linear momentum $p = Mv$ to the system at that point.Let the center

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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 imparts a linear momentum $p = Mv$ to the system at that point.Let the center of mass of the system be at point $O$,which is the midpoint of the rod. The distance of each ball from the center of mass is $L/2$.The initial angular momentum $L_i$ of the system about the center of mass $O$ is given by the impulse multiplied by the perpendicular distance from the axis of rotation:$L_i = J 	imes (L/2) = (Mv) 	imes (L/2) = \frac{MvL}{2}$.The moment of inertia $I$ of the system about the center of mass $O$ is:$I = M(L/2)^2 + M(L/2)^2 = 2M(L^2/4) = \frac{ML^2}{2}$.The final angular momentum $L_f$ of the system is $I\omega$,where $\omega$ is the angular velocity.By the law of conservation of angular momentum about the center of mass:$L_i = L_f$$\frac{MvL}{2} = I\omega$$\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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