Rod and particle

Let $v_0$ be the velocity of particle before collision. And $\omega$ the angular velocity of road after collision and $v_x$ and $v_y$ the component of particle just after collision along and perpendicular the rod respectively

From the equation of $e$ along the line of impact:

$e = 1 = - \displaystyle \dfrac{\frac l4 \times \omega - v_y}{ 0 - \frac 45 \times v_0}$

$\rightarrow \frac l4 \times \omega - v_y = \frac 45 \times v_0$

Now conserving angular momentum form the point where rod is attached to the axis:

$m \times \frac {4v_0}{5} \times \dfrac {l}{4} = mv_y \times \frac l4 + I \omega$ , where $I$ is the moment of inertia of rod $( I = \frac{ml^2}{12})$

Now substituting the value of $v_y$ we get $\omega = \boxed{6}$

i first did by a different method but ended with a wrong answer ,

i'll be glad if u tell my mistake ,first i found the velocity of particle in direction perpendicular to the rod let it be v' ,

v be the initial perpendicular velocity of the particle which is $\sqrt (2gh)$ $\cos 37$ the net impulse on rod is m(v+v')

Now using angular impulse - momentum theorem $\frac{Ml^{2}}{12}$ angular velocity = m(v+v') l/4

after solving i get 5.3333

can u tell what is wrong in my approach ?

i firstly didn't feel like conserving momentum as impulse is acting at the point of contact and hinge will also apply an impulse.

Just conserve angular momentum and put the condition of elasticity.