Solving Momentum Problem: Max Height of m1 After Elastic Collision

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The discussion focuses on calculating the maximum height that block m1 reaches after an elastic collision with block m2, which is initially at rest. Participants emphasize the conservation of energy and momentum principles, using equations for kinetic and potential energy. The problem is broken down into three steps: the fall of m1, the collision, and the rise of m1, with energy conservation applied in each step. Clarification is provided on connecting initial and final velocities from the collision equations to the energy equations. Ultimately, the structured approach helps the original poster understand the solution process.
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Two blocks are free to slide along a frictionless wooden track ABC as shown in Figure P9.20. The block of mass m1 = 4.93 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2 = 9.60 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

The figure shows m1 on a curved ramp at a height of 5 m.

Since it is elastic, I know energy and momentum are conserved. So I have:

(1/2)m1*v1o^2+(1/2)m2*v2o^2 = (1/2)m1*v1f^2+(1/2)m2*v2f^2
and
m1*v1o+m2*v2o = m1*v1f+m2*v2f

m2 is initially at rest, so v2o=0. Now I am not sure how I am supposed to use these to find height, or anything at all for that matter. Can anyone give me a point in the right direction?
 
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Two equations, two unknowns. One of those unknowns will give you the energy of m1 immediately following the collision. Mechanical energy is conserved.
 
Well, mechanical energy is KE+PE, so

(1/2)m1*v1f+m1*g*hf = (1/2)m1*v1o+m1*g*ho

But I am not sure how to connect that to the other equations I have.

I am having a hard time getting my head around this one.
 
Since you didn't include the figure, I'm just guessing as to what it shows.

Elmon said:
Well, mechanical energy is KE+PE, so

(1/2)m1*v1f+m1*g*hf = (1/2)m1*v1o+m1*g*ho
Make sure you square those speeds in the KE terms. I assume v1o is the initial speed of m1 immediately after the collision. (That speed is called v1f in your collision equations.)

But I am not sure how to connect that to the other equations I have.
The collision equations will give you the speed of m1 after the collision.

There are three steps to this problem:
(1) The fall of m1 from point A to where it collides with m2
(2) The collision
(3) The rise of m1​

In each step, energy is conserved. In step 2, momentum is also conserved.
 
Okay, I got it, thanks a lot. When you put it into three steps, something clicked and everything came out. Thanks again.
 
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