A uniform rod of length $l$ and mass $2 m$ rests on a smooth horizontal table. A point mass $m$ moving horizontally at right angle to the rod with velocity $v$ collides with one end of the rod and sticks to it, then

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angular velocity of the system after collision is $\frac{v}{2 l}$ .

the loss in kinetic energy of the system as a whole as a result of the collision is $\frac{7 m v^{2}}{24}$ .

the loss in kinetic energy of the system as a whole as a result of the collision is $\frac{m v^{2}}{6}$ .

angular velocity of the system after collision is $\frac{v}{l}$ .

answer is A, C.

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Detailed Solution

$M V (\frac{l}{3}) = [3 M {(\frac{l}{3})}^{2} + \frac{2 M L^{2}}{12} + 2 M {(\frac{l}{6})}^{2}] ω . . . . . . . . . . . . . . (1) M V = 3 m V_{C M} . . . . . . . . . . . . . . (2)$ Solving $(1) \infty = \frac{v}{l}$

Solving $(2) v_{C M} = \frac{v}{3}$ and from (1), $ω = \frac{3 V}{5 l}$

${(K . E .)}_{i} = \frac{1}{2} M V^{2}, {(K . E .)}_{f} = \frac{1}{2} I ω^{2} + \frac{1}{2} (3 M) v_{C M}^{2}, {(K . E .)}_{f} = \frac{1}{2} [\frac{M l^{2}}{3}] ω^{2} + \frac{1}{2} (3 M) (\frac{v^{2}}{a}) {(K . E .)}_{f} = \frac{M l^{2}}{6} ω^{2} + \frac{M v^{2}}{6} = \frac{M V^{2}}{6} + \frac{M V^{2}}{6} = \frac{M V^{2}}{3} ∆ K . E . = {(K . E .)}_{f} - {(K . E .)}_{i} = \frac{M V^{2}}{6}$