A thin uniform rod of mass m and l length is free to rotate about its upper end. When it is at rest, it receives an impulse J at its lowest point, normal to its length. Immediately, after impact

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The angular velocity of rod is $\frac{3 J}{m l}$

The kinetic energy of rod is $\frac{3 J^{2}}{2 m}$

The angular momentum of rod is $J l$

The linear velocity of mid point of rod is $\frac{3 J}{2 m}$

answer is A, B, C, D.

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Detailed Solution

About point of contact

$\begin{array}{r} L = (J) l . \dots \dots \dots . . (1) \\ Jl = (\frac{{ml}^{2}}{3}) ω \\ W = \frac{3 J}{ml} \dots \dots \dots (2) \\ K . E = \frac{1}{2} {Iω}^{2} = \frac{3 J^{2}}{2 m} . \end{array}$

And linear velocity of mid point of rod is given by

$ω \times \frac{l}{2} = \frac{3 J}{ml} \times \frac{l}{2} = \frac{3 J}{2 m}$

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