A uniform rod of mass M and length 2L lies on a smooth horizontal table. There is a smooth peg O fixed on the table. One end of the rod is kept touching the peg as shown in the figure. An impulse J is imparted to the rod at its other end. The impulse is horizontal and perpendicular to the length of the rod. Find the magnitude of impulse experienced by the peg.

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$\frac{2 J}{3}$

$\frac{J}{2}$

$\frac{J}{4}$

$\frac{3 J}{4}$

answer is C.

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Detailed Solution

Let the impulse applied by the peg be J₀

${J+J}_{0} = M V_{0} \dots \dots \dots . . (i)$
Angular impulse about COM $= (J - J_{0}) L$

Mol about an axis through COM $: I_{CM} = \frac{1}{12} M (2 L)^{2} = \frac{{ML}^{2}}{3}$
$∴ \frac{{ML}^{2}}{3} ω = (J - J_{0}) L \Rightarrow ω = \frac{3 (J - J_{0})}{ML}$

Velocity of end O of the rod immediately after the impulse is applied is zero.
$∴ V_{0} = ωL ∴ \frac{J + J_{0}}{M} = 3 \frac{(J - J_{0})}{M} \Rightarrow J + J_{0} = 3 J - 3 J_{0} \Rightarrow J_{0} = \frac{J}{2} \Rightarrow J + J_{0} = 3 J - 3 J_{0} \Rightarrow J_{0} = \frac{J}{2}$