# Another Collision problem

1. May 30, 2014

### BrainMan

1. The problem statement, all variables and given/known data
A ball of mass = 3-kg moving to the right at 3 m/s collides elastically with a ball of mass m2 = 2 kg moving at 4 m/s to the left. Find (a) the final speed of each object and (b)
the kinetic energy of each ball before and after the collisions.

2. Relevant equations
V1i + V1f = V2f + V2i
V1i - V2i = - (V1f - V2f)

3. The attempt at a solution
I plugged the numbers into the above formulas
3 + V1f = -4 + V2f
V2f = 7 + V1f

3 - (-4) = -(V1f - V2f)
7 = -V1f + V2f

substitute V2f
7 = -V1f + 7 + V1f
I got 0 = 0 the answer should be V1f = -2.60 ms, V2f = 4.40 m/s

2. May 30, 2014

### Nathanael

Where'd you get those equations from? Conservation of momentum states m$_{1}$V$_{1i}$+m$_{2}$V$_{2i}$=m$_{1}$V$_{1f}$+m$_{2}$V$_{2f}$

3. May 30, 2014

### BrainMan

In this problem since it is perfectly elastic kinetic energy is conserved so 1/2m1v1i2 + 1/2m2v2i2 = 1/2m1v1f2 + 1/2m2v2f2
with some mathematical manipulation you can acquire the above formulas.

4. May 30, 2014

### Nathanael

Ah, I see now. Still you seem to have made a mistake with your manipulations since there's no possible way to manipulate out of the equation m1 and m2

Last edited: May 30, 2014
5. May 30, 2014

### BrainMan

I got these equations directly from my physics book.

6. May 30, 2014

### Nathanael

Well don't trust your physics book. If it comes from conservation of KE then do the manipulations yourself.

And let me know if you figure it out, because I don't think it's possible to turn
into:
The simplest I can make those equations is:
m$_{1}(V_{1i}+V_1f)(V_{1i}-V_{1f})=m_{2}(V_{2f}+V_2i)(V_{2f}-V_{2i})$

Last edited: May 30, 2014
7. May 30, 2014

### Nathanael

But using either
m$_{1}(V_{1i}+V_1f)(V_{1i}-V_{1f})=m_{2}(V_{2f}+V_2i)(V_{2f}-V_{2i})$
or
m$_{1}$V$_{1i}$+m$_{2}$V$_{2i}$=m$_{1}$V$_{1f}$+m$_{2}$V$_{2f}$

It's just that the equations you used (from your textboook) are wrong for this situation.

8. May 30, 2014

### CAF123

No, those equations are correct for a perfectly elastic collision.

Brainman, you got 0=0 because your 'two' equations are not independent. One is simply a rearrangement of the other. You can use the result you got in the OP, together with conservation of momentum. This will give the result because the first equation uses the fact that the collision was perfectly elastic, so provides additional information.

Last edited: May 30, 2014
9. May 30, 2014

### BrainMan

What was my mistake then?

10. May 30, 2014

### CAF123

I just edited my post. See above.

11. Jun 2, 2014

### BrainMan

OK I did that and I am still not getting the correct answer. I am pretty sure I did something wrong when I substituted. Here are my calculations.

equation 1
m$_{1}$V$_{1i}$+m$_{2}$V$_{2i}$=m$_{1}$V$_{1f}$+m$_{2}$V$_{2f}$
3(3) + 2(-4) = 3(v1f) + 2(v2f)

equation 2
V1i + V1f = V2f + V2i
3 + v1f = v2f + 4
v2f = 7 + v1f

substitute v2f
1 = 3(v1f) + 2(7 + V1F)
1= 3V1F + 14 + 2V1f
1- 3v1f = 14 + 2v1f
-3v1f = 13 + 2v1f
-v1f = 13
v1f = -13

the answer should be - 2.60 m/s

12. Jun 2, 2014

### CAF123

Small sign error there.

13. Jun 2, 2014

### BrainMan

I am still getting the wrong answer

1 = 3v1f + 2v1f - 2
3/5 = v1f

there must be something else I'm doing wrong.

14. Jun 2, 2014

### CAF123

From conservation of momentum, you correctly obtained the equation 1 = 3v1f + 2v2f.

From the equation derived by manipulating the equations for conservation of kinetic energy and that of momentum (do you understand the derivation?) you obtain, by fixing the sign error in the previous post, -7 = v1f - v2f

Solve these two equations for v1f and v2f.

15. Jun 2, 2014

### BrainMan

I am having trouble deriving that formula. What I did was
V1i + V1f = V2f + V2i
3 - vif = -4 -v2f // the final velocities are negative because they will both switch directions.
-1(3 - vif = -4 -v2f)
-3 + vif = 4 +v2f
-7 + vif = v2f

I then tried to substitute this into the other equation.

1 = 3v1f + 2v2f
1 = 3v1f + 2v1f - 14
15 = 5v1f
v1f = 3
and I did not get the right answer

16. Jun 2, 2014

### SammyS

Staff Emeritus
Do what CAF123 says above.

One way to do this is by elimination: Multiply the 2nd equation by 2, then add the equations to eliminate v2f .

17. Jun 2, 2014

### BrainMan

What did I do wrong in my previous post?

18. Jun 2, 2014

### SammyS

Staff Emeritus
Don't change signs on these velocities. You didn't on the other equation.

Algebra will take care of the signs! (That's one of Algebra's most endearing characteristics.)

Last edited: Jun 2, 2014
19. Jun 3, 2014

### BrainMan

OK I see what I did and now got the right answer. Thanks!