Conservation of Momentum in Perfectly Elastic Collisions

[pb23me]

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 m₁v_i+m₂v_i=m₁v_f+m₂v_f

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 KE_i =KE_f but I am not getting anywhere with that...

 

[tiny-tim]

hi pb23me!

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!

 

[pb23me]

(.3)(2.5)+(.6)(1)=(.3)v_f+(.6)v_f
1/2(.3)(2.5)²+1/2(.6)(1)²=1/2(.3)(v_f)²+1/2(.6)(v_f)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 v_f by itself. am i missing another equation?

 

[tiny-tim]

that's ok (apart from missing out the ² and writing v_f twice! ) …

just substitute from the first equation, and you should get an easy quadratic equation

 

[pb23me]

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?

 

[tiny-tim]

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 ,

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