What are the final velocities of two balls after a head-on elastic collision?

[perfectionist17]

Problem:
"A ball with the mass of .440 kg moving east (+x direction) with a speed of 3.70 m/s collides head-on with a .220 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each object after the collision?

Relevant equations.

Conservation of momentum:

m* vi of b1 + m* vi of b2= m* vf of b1 + m* vi of b2

and conservation of kinetic energy...

1/2mvf^2= 1/2 mvi^2

Attempt at Solution:

.440(3.70) + .220(0)= .440(vf1) + .220(vf2)

1.63= .44(vf1) + .220(Vf2)

and then...

1/2 (.440)(3.7^2)= 1/2 (.440)(Vf^2) + 1/2(.220)(vf^2)

3.01= " "

**** Where do I go from here? Also, do I have to use the component form (as in, momentum of initial x component= momentum of final x component), etc? What about angles? I'm so confused because I feel like I should be using angles if its 2-D collisions, but none are given in the problem and I'm not sure if they'll go off a 90 degree angle or not?

 

[hotvette]

You are quite close actually. Here are a couple of hints.

1. Re angles, think of playing pool and using the cue ball for an absolute straight in shot to the pocket

2. Re kinetic energy, you need to consider the total kinetic energy of the system before and after the collision

 

[perfectionist17]

thank you very much!

so does that mean I don't consider components or angles at all?

since there are two unknowns in each equation, do I simply solve for one say in the kinetic equation then substitute it back into my conservation of momentum?

and when do you consider angles? i thought this was 2-D motion and that's when angles have to be taken into account, but maybe I'm wrong?

 

[hotvette]

Well, if it had been labeled 1D you wouldn't need to think as much! Actually, can't distance + time be considered 2D?

The clue to me was that the 2nd ball was at rest. If it had been initially moving in a direction that was non-colinear with the direction of the 1st ball, then angles would be involved.

Yep, I see this as a 2-equation in 2-unknown problem.