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I found this problem in a Irodov's book. It's asked to find the momentum increment of the bullet-rod system. I wonder why the system momentum changes. They argue this:

The bullet-rod system is non-closed: apart from the counterbalancing forces, a horizontal component of the reaction force appears in the process of motion of the bullet at the point 0 of the rod. That component brings about the momentum increment of the system:

I don't understand it. Thank you for your help and time :)

This is the problem

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The case for momentum of a system to be conserved is that no external force should be acting on the system. This comes from newtons second law.

On alaysing the bullet rod system there are 2 forces, that acts on it::

Gravity- if we conserve momentum at the time just before the bullet hits the rod, to the momentum of the system just after the bullet collides with the rod, we can safely neglect gravity. This is because the time of collission is very small, subsequently the change in momentum of the system due to gravity is also very small. ( impulse or change in momentum is force times time.)

Hinge force - this is the force acting on the system at the point of contact of the rod with the horizontal wall. In the diagram it is a sort of triangular structure. It exerts force on the system, specifically normal and frictional force. Both these forces are impulsive and their effect on the net momentum of the system cannot be ignored. This is because, these contact forces take very large values at the time of collision, (thus they are impulsive) offsetting the small time period of collision. Hence momentum of the system cannot be conserved.

Contact forces between the bullet and rod are internal forces and cannot change the moment of the system. ( comes from newtons third law)

Finally to solve the question, you could try

Conserving angular momentum at the hinge of the rod. No force produces torque and so angular momentum is conserved.

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  • $\begingroup$ Thank you so much. I understood completely. Just one thing regarding conservation of angular momentum. Considering the hinge of the rod as the point of rotation, there is a force acting on it: the hinge force, but since there's no distance between them, the torque is 0. What about gravity? is it the same case as the one you mentioned? I wonder that since it seems to have a torque regarding the hinge of the rod once the rod is moving. I would appreciate your help one more time. Thank you so much. :) $\endgroup$ – Omar Jul 12 '15 at 5:11
  • $\begingroup$ I think there's a concept called angular impulse. As you said, the torque is applied during a brief tiem; therefore, it could be neglected. Is it correct? Thank you! $\endgroup$ – Omar Jul 12 '15 at 5:27
  • $\begingroup$ Yes that is correct, torque due to non impulsive forces like gravity is neglected. So there is no torque on the system, this angular momentum is conserved. Glad i could help :) $\endgroup$ – Sashurocks Jul 12 '15 at 7:35
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If you made the graphic of forces of the system, you could see that the angular momentum is conservated, you can use that to find the final angular momentum of the system when the bullet hit the rod, then the final velocity of the bullet.

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