Conservation of linear/angular momentum in a ballistic pendulum

[Quadrat]

I did a problem a coule of weeks ago with a vertical rod, frictionless hanging around the upper end, length L and mass m. Then a bullet with the same mass, m speed v is fired horizontally at the lowest point of the rod and becomes embedded in the rod. So I calculated the maximum angle it would make etc.

So I know that I can relate the angular momentum Lmv with the torque I_total*omega. I'm wondering if one can think of the situation as: linear momentum is conserved as angular momentum just before the collision and just after (before any external torque is acting on the system)? Or how can one break down the conservation of momenta in this case? I get that mechanical energy is lost due to heat/sound. But the momentum is not as clear for me. The task is solved, I just wanted some additional questions that arose answered. Any help is much appreciated!

 

[Quadrat]

Never mind, I get it now. Since I chose the pivot point as my origin any horizontal forces contributes to any torque around that point and therefore angular momentum is conserved.

 

[Jun Pan]

When bullet hits the pendumlum,just before and at the moment, momentum conservation is applied for bullet to transfer momentum to pendulum and bullet and at the same time cause the horizontal force to the pendulum and bullet and so the torque for the pendulum and bullet around the pivot. After then is the angular momentum conservation process of pendulum and bullet.