A light and thin rod of length l is lying on smooth horizontal table. Two small balls of masses 3m and m are fixed on two ends of a massless rod. The rod is free to rotate about fixed vertical axel passing through midpoint of rod. A third ball of mass 2m is fastened to ball of mass m through a string of length l as shown in figure. The ball is projected horizontally with speed v as shown. The angular velocity of rod, just after the string becomes taut is

Angular momentum of the rod just before the string become taut is zero. Let us assume the angular velocity of rod just after the string become taut is $\omega$. Impulse imparted by string is J.

$\begin{array}{l}\sin \theta =\frac{4l}{5\times l}=\frac{4}{5}\text{}\\ \Rightarrow \theta ={53}^{\circ }\text{}\\ \text{Using concept of Angular Impulse,}\end{array}$

$\begin{array}{l}\left(J\cos {53}^{\circ }\right)\frac{l}{2}=\left(m+3m\right)\frac{l^2}{4}\omega \text{}\\ \Rightarrow \frac{3J}{5}=2ml\omega \\ \text{}\Rightarrow J=\frac{10}{3}ml\omega ----\left(1\right)\end{array}$

$\begin{array}{l}F\text{or }2m\text{ mass apply concept of Linear impulse,}\\ J=2m\left(v\cos {53}^0-v_1\right)---\left(2\right)\end{array}$

$\begin{array}{l}Using string constraint , \text{}\\ v_1=\left(\omega \frac{l}{2}\right)\cos {53}^{\circ }=\omega \frac{l}{2}\times \frac{3}{5}=\omega \frac{3l}{10}---\left(3\right)\text{}\\ from \left(1\right)\& \left(2\right) \text{ }\\ \text{So }\frac{10}{3}ml\omega =2m\left(\frac{3v}{5}-\frac{3\omega l}{10}\right)\left(\because from\left(3\right)\right)\end{array}$

$\begin{array}{l}\Rightarrow \frac{5l\omega }{3}=\frac{3v}{5}-\frac{3\omega l}{10}\text{}\\ \Rightarrow \frac{5\omega l}{3}+\frac{3\omega l}{10}=\frac{3v}{5}\\ \text{}\Rightarrow \frac{50\omega l+9\omega l}{30}=\frac{3v}{5}\\ \text{}\Rightarrow \omega =\frac{18v}{59l}\end{array}$