What is the maximum height reached by two masses after an elastic collision?

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Homework Help Overview

The problem involves two masses, one initially at rest, sliding down a frictionless quadrant of a vertical circle and colliding elastically with another mass at rest. The goal is to determine the maximum height each mass reaches after the collision, considering the conservation of energy and momentum principles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need to calculate the speed of the first mass before the collision using conservation of energy, while others suggest using momentum conservation for the collision itself.
  • There is confusion regarding the correct application of energy and momentum equations, with participants questioning the validity of their approaches and the assumptions made about energy conservation.
  • Some participants express uncertainty about the relationship between height and radius in the context of the problem.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants have provided guidance on using conservation laws, while others are clarifying misconceptions about the equations involved. There is no explicit consensus on the best method to proceed, and multiple interpretations are being considered.

Contextual Notes

Participants are navigating through the complexities of elastic collisions and energy conservation, with some expressing confusion about the necessity of calculating specific speeds versus focusing on energy alone. The problem setup includes specific masses and a radius, which are central to the calculations being discussed.

  • #31
Ah, ok.
I calculate the ratio of the heights, h2/h1 = (2m1/(m1-m2))2. That agrees with your final answer. It follows that for m1 to climb higher than m2 it needs to have at most 1/3 of the mass of m2.
Consider the case where the masses are the same. m1 would stop dead and m2 would climb to height h.
 
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  • #32
haruspex said:
Ah, ok.
I calculate the ratio of the heights, h2/h1 = (2m1/(m1-m2))2. That agrees with your final answer. It follows that for m1 to climb higher than m2 it needs to have at most 1/3 of the mass of m2.
Consider the case where the masses are the same. m1 would stop dead and m2 would climb to height h.

but...
i don't get it .. if u divide 16 kg by 3 .. you get 5.333 kg
and mine is 6 kg ?
 
  • #33
Lolagoeslala said:
but...
i don't get it .. if u divide 16 kg by 3 .. you get 5.333 kg
and mine is 6 kg ?
Quite so. 6k is greater than one third of 16k, so the 16k ball will climb higher. For the light ball to climb higher it must be less than one third the mass of the other.
 
  • #34
haruspex said:
Quite so. 6k is greater than one third of 16k, so the 16k ball will climb higher. For the light ball to climb higher it must be less than one third the mass of the other.

Seriously? :O But i have always been though that the smaller mass will go higher... why would the bigger mass travel further?
 
  • #35
Lolagoeslala said:
Seriously? :O But i have always been though that the smaller mass will go higher... why would the bigger mass travel further?
The scenario is not symmetric. As I suggested, consider two equal masses. The first (initiating) mass will stop dead and transfer all its KE to the other. Now make the first mass a bit smaller. It will bounce back some, but still most of the energy will be transferred. It won't achieve its 'fair' share of energy (i.e. in proportion to mass) until it's down to one third the mass of the other. And to get equal energy it would have to be smaller still.
 

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