Conservation of Momentum in Perfectly Elastic Collisions

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In a discussion about a perfectly elastic collision between a 0.3 kg ball moving at 2.5 m/s and a 0.6 kg ball moving at 1 m/s, participants focused on applying the conservation of momentum and kinetic energy equations. The initial equations set up were m1vi + m2vi = m1vf + m2vf and KEi = KEf, but complications arose due to having two unknown final velocities. One participant struggled to isolate vf and ended up using the quadratic formula, yielding two potential solutions: 0.466 m/s and 2.53 m/s. The community suggested that the 2.53 m/s value likely represented the original speed of the first ball if the balls did not collide. The discussion emphasized the importance of careful substitution and calculation to resolve the equations correctly.
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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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