# 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. Commented May 22, 2018 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 Commented May 22, 2018 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. Commented May 22, 2018 at 19:17

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.

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