Another momentum problem

  • Thread starter pb23me
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  • #1
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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 theres 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 im not getting anywhere with that...
 

Answers and Replies

  • #2
tiny-tim
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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
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(.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 dont 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 cant figure out how to get vf by itself. am i missing another equation?
 
  • #4
tiny-tim
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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:
 
  • #5
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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?
 
  • #6
tiny-tim
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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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