Linear Momentum of Particle and Rod

In summary, the problem presents a scenario where a small mass collides with a stick on a table, causing the mass to continue in the same direction with a changed speed. The question asks for the specific value of the mass ratio at which the stick will collide with the mass again and the distance the mass travels between collisions. The solution involves understanding the angular momentum before and after the collision and using a frame of reference at the center of the rod. This is a challenging problem that requires careful visualization.
  • #1
KDawgAtsu
12
0
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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  • #2
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
 

FAQ: Linear Momentum of Particle and Rod

1. What is linear momentum?

Linear momentum is a measure of an object's motion, specifically its inertia and velocity. It is a vector quantity, meaning it has both magnitude and direction.

2. How is linear momentum calculated?

Linear momentum is calculated by multiplying an object's mass by its velocity. The formula for linear momentum is p = m * v, where p is momentum, m is mass, and v is velocity.

3. What is the conservation of linear momentum?

The conservation of linear momentum states that in a closed system, the total momentum remains constant. This means that the total momentum before a collision or interaction is equal to the total momentum after.

4. How is linear momentum different from angular momentum?

Linear momentum is the measure of an object's motion in a straight line, while angular momentum is the measure of an object's rotational motion. Angular momentum takes into account an object's mass, velocity, and distance from the axis of rotation.

5. How is linear momentum applied in real-world situations?

Linear momentum is applied in many real-world situations, such as in sports like football or baseball, where players use their momentum to throw or hit a ball. It is also important in calculating the motion of vehicles, projectiles, and other moving objects.

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