# 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?

• Yes, that is what linear momentum means. If you’re in doubt you can just check the dimensions, which match up. – knzhou May 22 '18 at 18:41
• my problem is also that getting the result, as linear momentum is not conserved in the case, and the angular momentum equation only contains 3 terms – Morten Sode May 22 '18 at 18:50
• Angular momentum is conserved about the pivot using which you can find the final angular velocity and hence the final translational velocity of the centre of mass of the rod and bullet. – Farcher May 22 '18 at 19:17