Linear momentum rod hit by bullet I am trying to solve the question which I have attached. My attempt at a solution is that the angular momentum is conserved thereby we have 
$$
mva = \frac{ML^2\Omega}{3} + a^2m\Omega
$$
My problem is understanding what is meant by linear momentum in this case. Is this $\left(m+M\right)v_f$ or something else?



 A: Linear momentum before the impact is obviously $$p_{\rm bullet} = m v$$ But after impact, it is the sum of the two masses multiplied by the velocities of their center of mass
$$ p_{\rm rod} = m ( a \Omega) + M ( \tfrac{L}{2} \Omega ) $$
But these two are not equal because the pivot is also imparting momentum (an impulse) which we name $J$
$$ m  v = \underbrace{ m ( a \Omega) + M ( \tfrac{L}{2} \Omega )}_{p\text{ in answer}} + J $$
But you cannot use the equation above along to solve for $\Omega$ and $J$. You also need conservation of angular momentum, which you have correctly stated as
$$ a (m v) = m \Omega a^2 + \tfrac{M}{3} \Omega  L^2 $$
Solve the above for $\Omega$ and plug it
$$ p =m ( a \Omega) + M ( \tfrac{L}{2} \Omega ) $$ to get your answer.
A: Apply angular momentum conservation about an axis passing from hinge and perpendicular to the plane and find angular velocity then find the combimed center of mass and find velocity of CM about hinge. Multiply it by total mass you will get the answer. Check the solution here -
Solution
