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(C) Let the length of the rod be $L$. The impulse $J$ applied at $A$ creates a linear impulse and an angular impulse about the center of mass $C$ of t

Vedclass · Accepted Answer

$\frac{J}{m}$

Vedclass · Answer

(C) Let the length of the rod be $L$. The impulse $J$ applied at $A$ creates a linear impulse and an angular impulse about the center of mass $C$ of the system.Since the masses are equal,the center of mass $C$ is at the midpoint of $AB$.The linear velocity of the center of mass $v_{cm}$ is given by $J = M_{total} v_{cm} = (2m) v_{cm}$,so $v_{cm} = \frac{J}{2m}$.The angular impulse about $C$ is $J 	imes \frac{L}{2} = I_{cm} \omega$,where $I_{cm} = m(\frac{L}{2})^2 + m(\frac{L}{2})^2 = \frac{mL^2}{2}$.Thus,$\frac{JL}{2} = \frac{mL^2}{2} \omega$,which gives $\omega = \frac{J}{mL}$.The velocity of particle $A$ is $v_A = v_{cm} + \omega r_A = \frac{J}{2m} + (\frac{J}{mL})(\frac{L}{2}) = \frac{J}{2m} + \frac{J}{2m} = \frac{J}{m}$.

Two particles of equal mass $m$ at $A$ and $B$ are connected by a rigid light rod $AB$ of length $L$ lying on a smooth horizontal table. An impulse $J$ is applied at $A$ in the plane of the table and perpendicular to $AB$. Then the velocity of the particle at $A$ is:

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