A rod of mass M and length L is lying on a ho | vedclass

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(B) Let $v$ be the velocity of the center of mass of the rod and $\omega$ be the angular velocity of the rod about its center of mass after the collis

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $v$ be the velocity of the center of mass of the rod and $\omega$ be the angular velocity of the rod about its center of mass after the collision.$1$. Conservation of linear momentum:$mu = Mv \implies v = \frac{mu}{M} \quad \dots(i)$$2$. Conservation of angular momentum about the center of mass of the rod:$mu \left(\frac{L}{2}ight) = I \omega = \left(\frac{ML^2}{12}ight) \omega$$\implies \omega = \frac{6mu}{ML} \quad \dots(ii)$$3$. Coefficient of restitution for elastic collision $(e=1)$:$e = \frac{	ext{Velocity of separation}}{	ext{Velocity of approach}} = 1$$1 = \frac{v + \omega(L/2)}{u}$$u = v + \frac{\omega L}{2} \quad \dots(iii)$Substituting $(i)$ and $(ii)$ into $(iii)$:$u = \frac{mu}{M} + \left(\frac{6mu}{ML}ight) \left(\frac{L}{2}ight)$$u = \frac{mu}{M} + \frac{3mu}{M} = \frac{4mu}{M}$$1 = \frac{4m}{M} \implies \frac{m}{M} = \frac{1}{4}$Comparing with $\frac{m}{M} = \frac{1}{x}$, we get $x = 4$.

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