Solving Momentum Problem: Max Height of m1 After Elastic Collision

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Homework Help Overview

The problem involves two blocks sliding along a frictionless track, where one block (m1) is released from a height and collides elastically with another block (m2) that is initially at rest. The goal is to determine the maximum height to which m1 rises after the collision.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of energy and momentum in elastic collisions, with the original poster expressing uncertainty about how to apply these principles to find the height after the collision. Others suggest breaking the problem into steps and connecting kinetic and potential energy equations.

Discussion Status

Some participants have provided guidance on the steps involved in the problem, emphasizing the conservation laws and the need to connect the equations for energy and momentum. The original poster indicates a moment of clarity after this breakdown, suggesting productive progress in understanding the problem.

Contextual Notes

The original poster did not include a figure that is referenced in the problem, which may affect the clarity of the discussion. There is also a mention of mechanical energy conservation, but the specific details of the setup are not fully described.

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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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