Conservation of Momentum in Perfectly Elastic Collisions

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Homework Help Overview

The problem involves a perfectly elastic collision between two balls of different masses and initial velocities. The original poster seeks to determine the final speeds and directions of both balls after the collision, applying the conservation of momentum and kinetic energy principles.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use conservation of momentum and kinetic energy equations but struggles with the resulting equations due to having two unknowns. Some participants suggest showing full calculations to identify errors, while others propose substituting variables to simplify the problem.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem. There is recognition of the complexity involved in solving the equations, and multiple interpretations of the results are being explored, particularly regarding the potential outcomes from the quadratic formula.

Contextual Notes

Participants note the assumptions of no friction and the conditions of a perfectly elastic collision. The original poster expresses uncertainty about the final velocities and whether the results obtained are valid.

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Homework Statement


A .3kg ball, moving with a speed of 2.5 m/s has a head on collision with a .6kg ball initially moving away from it at a speed of 1 m/s. assuming a perfectly elastic condition, what is the speed and direction of each ball after the collision?


Homework Equations


conservation of momentum m1vi+m2vi=m1vf+m2vf


The Attempt at a Solution

I assumed there's no friction and assumed that the final KE is equal to the initial KE. I tryed using the conservation of momentum equation and KEi =KEf but I am not getting anywhere with that...
 
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hi pb23me! :wink:

that's the right approach …

show us your full calculations, and then we'll see what went wrong, and we'll know how to help! :smile:
 
(.3)(2.5)+(.6)(1)=(.3)vf+(.6)vf
1/2(.3)(2.5)2+1/2(.6)(1)2=1/2(.3)(vf)2+1/2(.6)(vf)2
i have two unkowns in both equations. i don't know either of the final velocitys. when i try solving for final velocity of one and plug that into the other equation it becomes a huge mess! then i can't figure out how to get vf by itself. am i missing another equation?
 
that's ok (apart from missing out the 2 and writing vf twice! :wink:) …

just substitute from the first equation, and you should get an easy quadratic equation :smile:
 
ok i had to use the quadratic formula and came up with two answers .466 m/s and 2.53 m/s...so which is it?
 
pb23me said:
ok i had to use the quadratic formula and came up with two answers .466 m/s and 2.53 m/s...so which is it?

difficult to say without seeing your calculations :redface:,

but I'm guessing that your 2.53 is the original 2.5, ie the solution if the balls miss! :biggrin:
 

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