Linear Momentum of Particle and Rod

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SUMMARY

The discussion focuses on the elastic collision between a small mass, m, and a rod of length L and mass M, where the mass initially moves at speed v0. After the collision, the mass continues with speed v, and the problem involves determining the ratio m/M for a second collision to occur. Key equations include the initial angular momentum (L/2)mv0 and the angular momentum after the collision (1/12)mL²ω + (L/2)mv. The origin of the reference frame remains at the center of the rod, allowing for consistent calculations of angular momentum before and after the collision.

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I don't need a solution to this problem, I just need some help understanding a part of it.

Homework Statement


The figure below shows a small mass, m, moving at an initial speed, v0 , colliding with a stick with
length, L, and mass, M. Both the mass and the stick lie on top of a table. The collision happens at the
tip of the stick. After the collision the mass continues in the same direction, but now with the speed, v.
For a specific value of the ratio m/M, the stick will collide with the small mass a second time. What is
this value, and how far will the small mass have traveled between collisions? Assume that the
collision is elastic.

Homework Equations


Initial Angular Momentum: (L/2)mv0
Angular Momentum After Collision: (1/12)mL2ω + (L/2)mv

The Attempt at a Solution


I know that the rod and the particle must be moving at same linear velocities for the second collision to happen. What I don't 100% understand is why the angular momentum for the particle after the collision is still (L/2)mv. Initially, the origin is at the center of the rod, but after the collision both the particle and the rod move to the right. I don't understand why, for the particle, the distance for the angular momentum is just (L/2); doesn't this place the origin at the center of mass of the rod, which is now moving and thus a noninertial frame?

Thanks for the help!
 
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The reference system has the origin at the centre of the rod in rest and stays the same after impact. With respect to that point, the angular momentum of the particle is mv0L/2 before the impact and mvL/2 after the impact (supposing that the velocity of the particle is perpendicular to the rod). Do not forget that the angular momentum is the vector product of the position vector with the linear momentum, so it is
mv*r*sin(angle between r and v)=mvL/2.

It is more convenient to use a frame of reference in rest then one connected to the centre of mass.

This is a very nice problem, really. It took me some time to imagine how the rod can hit the particle again. :smile:

ehild
 

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