# Homework Help: Elastic Collision Of Uneqal Masses

1. Oct 3, 2009

### Warmacblu

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

Two particles of masses m and 8m move toward each other along the x-axis with the same initial speeds of 9.69 m/s. Mass m is traveling to the left and mass 8m to the right. They undergo a head-on elastic collision, and each rebounds along the same line as it approached.

Find the final speed of the heavier particle.

2. Relevant equations

p = mv
ke = 1/2mv2

3. The attempt at a solution

1/2 * m * -9.692 = 46.95mJ
1/2 * 8m * 9.692 = 375.58mJ

total ke = 375.58 + 46.95 = 422.53

p = m * -9.69 = 9.69
p = 8m * 9.69 = 77.52

total p = 77.52 - 9.69 = 67.83

so ...

p = mv1 + 8mv2 = 67.83

ke = 1/2 * mv12 + 1/2 8mv22 = 422.53

The m's can cancel and I get

v1 + 8v2 = 67.83
v1 = 67.83 - 8v2

Then I do not know where to go from here. Do I plug it into the ke equation and solve for v2? Which would give me my answer of the velocity of the heavier particle.

Thanks for any help.

2. Oct 3, 2009

### w3390

Are these v's you're using the velocities after the collision?

3. Oct 3, 2009

### Warmacblu

Yeah, that is what I am trying to solve for.

4. Oct 4, 2009

### Warmacblu

Could anyone advise me if my math is correct?

5. Oct 4, 2009

### w3390

Yes your math looks correct. So just substitute the v1 into the kinetic energy equation and solve for v2. For the record, it makes it easier to keep your initial and final v's separate by indicating the final velocities as v'(v prime). I was confused in your math because of this, but I think you did it correctly and should work out.

6. Oct 4, 2009

### Warmacblu

Okay, I will try it either tonight or tomorrow and will let you know. Thanks.

7. Oct 5, 2009

### Warmacblu

Okay, so my v1 value is:

v1 = 67.83 - 8v2

I then plug that into

Ke = 1/2 * v12 + 1/2 * 8v22 = 422.63

I dropped the m's because I believe they can cancel.

I then get this expression:

(1/2)(67.83 - 8v2)2 + (1/2)(8v2)2 = 422.63

I am unsure on what to do from here; do I square (67.83 - 8v2) to get rid of the square? I am pretty positive I can do this:

(1/2)(67.83 - 8v2)2 + 4v22 = 422.63

Then divide both sides by 4

(1/2)(67.83 - 8v2)2 + v22 = 105.6575

Then multiply both sides by 2 to get rid of the (1/2)

(67.83 - 8v2)2 + v22 = 211.315

I am then stuck here.

Any help is appreciated,
Thanks

8. Oct 6, 2009

### Warmacblu

Anyone? How can I get rid of the square in the (67.83 - 8v2) term?

9. Oct 6, 2009

### rl.bhat

(1/2)(67.83 - 8v2)2 + (1/2)(8v2)2 = 422.63

You can rewrite this step as

(67.83 - 8v2)2 + (8v2)2 = 422.63*2
Expand the first term and write down the quadratic in v2 and solve for v2.

10. Oct 6, 2009

### Warmacblu

Okay, how does this look?

(67.83 - 8v2)(67.83 - 8v2) + 64v22 = 845.26

FOIL ...

4600.9089 - 542.64v2 - 542.64v2 + 64v22 + 64v22 = 845.26

128v22 - 1085.28v2 + 3755.6489 = 0

I plugged this into my calculators program "PolySmlt" and I got two unreal answers. However, if I change the 3755.6489 to a negative then I get:

x1 = -2.639
x2 = 11.118

X2 does not make sense but the -2.639 does because it is traveling in the negative direction at a slower speed then it started.

11. Oct 6, 2009

### rl.bhat

This step is wrong. It should be
(67.83 - 8v2)(67.83 - 8v2) + 8v22 = 845.26

12. Oct 7, 2009

### Warmacblu

Alright, I corrected my mistake and solved the quadratic for v2. Thanks for all the help.