# Finding speed of objects after elastic collison

1. Nov 18, 2012

### joe426

1. The problem statement, all variables and given/known data

2. Relevant equations

1/2m1v12 + 1/2m2v22 = 1/2m1v1f + 1/2m2v2f
m1v1 + m2v2 = m1v1f + m2v2f
v1 - v2 = -(v1f - v2f)
3. The attempt at a solution
So I solved the momentum of conservation for the final velocity of object 1. I then plug that equation into the third equation listed above, as v1f. But the thing is the equation i get for v1f has v2f in it and that is what I am solving for and this is where I am stuck.

2. Nov 19, 2012

### joe426

I don't think any of the velocity equals 0 at any point. Initially they are moving in the same direction. Then since it says its perfectly elastic, one object will bounce back and have a velocity in the opposite direction. I think the 60g tennis ball will bounce back in the negative direction but I still can't figure out how to find the velocities after the collision :(

3. Nov 19, 2012

### ehild

ehild

4. Nov 19, 2012

### joe426

First use conservation of momentum
( m1v12 + m2v22 - m2v2f2 ) / m1 = v1f2

Then plug this v1f into this helper equation, the third equation listed in first post.
v2f = v1 - v2 + [( m1v12 + m2v22 - m2v2f2 ) / m1 ]1/2

And I'm stuck because I can't get both v2f to one side so i can solve for it. then once i get v2f i plug it back into the first equation, [( m1v12 + m2v22 - m2v2f2 ) / m1 ], and solve for v1f

5. Nov 19, 2012

### ehild

Use the last two equations. Plug in the numbers first. Express v1f from the third equation.

ehild

6. Nov 19, 2012

### ehild

Use the last two equations. Plug in the data first, express vif form the last equation, and substitute into the second one.

ehild

7. Nov 19, 2012

### joe426

v1f = -v1 + v2 - v2f

Plugged that into,
(m1v1 + m2v2 - m1v1f) / m2 = v2f

v2f = -.368m/s

8. Nov 19, 2012

### ehild

Check it.

9. Nov 19, 2012

### joe426

v1f = -v1 + v2 - v2f is really v1f = -v1 + v2 + v2f. ..

(m1v1 + m2v2 - m1v1f) / m2 = v2f turns into (m1v1 + m2v2 + m1v1f) / m2 = v2f

so v2f = .307m/s

10. Nov 19, 2012

### ehild

Check.

ehild

11. Nov 19, 2012

### ehild

Plug in the numbers for v1 and v2: v1f = v2f - 1.35.
Plugging in the data into the equation m1v1+m2v2=m1v1f+m2v2f and also substituting v2f - 1.35 for v1f: 0.2535 = 0.06(v2f - 1.35)+0.09v2f.
Expand and collect the terms with v2f, solve.

ehild

12. Nov 20, 2012

### Munzi5

Its 3.75 or 1.35?

Last edited: Nov 20, 2012
13. Nov 20, 2012

### ehild

What is 3.75 or 1.35?

ehild