# Dynamics Questions // Elastic Collision with different Vo

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1. Feb 17, 2017

### sunnnystrong

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

A 2.3 kg object traveling at 6.1 m/s collides head-on with a 3.5 kg object traveling in the opposite direction at 4.8 m/s. If the collisions is perfectly elastic, what is the final speed of the 2.3 kg object?

2. Relevant equations

Conservation of Energy:
1/2 m*vi^2 = 1/2 m*vf^2

Conservation of Momentum:
Pix = Pfx

3. The attempt at a solution

Pix For 2.3 kg Object
(2.3kg)(6.1m/s) = (2.3kg)(Vf2) + (3.5kg)(Vf1)

Solving for Vf1 -->
[14.03 Ns - (2.3kg)(Vf2)]/3.5kg = Vf1

KE
1/2(2.3kg)(6.1m/s)^2 = 1/2(2.3kg)(Vf1)^2 + 1/2(3.8kg)(Vf2)^2
42.7915 = 1.15(Vf1)^2 + 1.9(Vf2)^2

Plugging in Vf1
42.7915 = 1.15(0.431837(Vf2)^2-5.26841Vf2 + 16.0686) + 1.9(Vf2)^2
Simplify and you get a quadratic:
0=2.39661 (Vf2)^2 - 6.05867 (Vf2) -24.3126
Solve for Vf2 = 4.69 m/s

Plug back into:
[14.03 Ns - (2.3kg)(Vf2)]/3.5kg = Vf1

Solve for Vf1 = 0.96m/s (which is definitely wrong)

I'm really confused & not sure how to solve for Vf1?

2. Feb 17, 2017

### Staff: Mentor

Total momentum of the system is conserved. Don't forget the initial momentum of the other mass. (Careful with signs!)

3. Feb 17, 2017

### sunnnystrong

So going back -->
(2.3kg)(6.1m/s) + (3.5kg)(-4.8m/s) = (2.3kg)(Vf2) + (3.5kg)(Vf1)
Solve for Vf1 :
Vf1 = (-2.77Ns -2.3kg(Vf2))/(3.5kg)
Square it for later use:
0.431837(Vf2)^2 + 1.04016(Vf2) + .6264 = (Vf1)^2

Looking at total system -->
1/2(2.3kg)(6.1m/s)^2 + 1/2(3.5kg)(-4.8m/s)^2 = 1/2(2.3kg)(Vf1)^2 + 1/2(3.8kg)(Vf2)^2
Simplify this:
42.7915+40.32 = 1.15(Vf1)^2 + 1.9(Vf2)^2
Plug in again you get a quadratic:
83.1115 = 1.15(0.431837(Vf2)^2 + 1.04016(Vf2) + .6264) + 1.9(Vf2)^2
Solve
Vf2 = 5.62 m/s :((
I'm confused because the answer is 6.6m/s

4. Feb 17, 2017

### sunnnystrong

Do you think you could show me how to set it up?

5. Feb 17, 2017

### Staff: Mentor

Your set up looks good to me. (Now that you've corrected the momentum equation.) I didn't check all the steps you did, but if you're getting a different answer than expected double check your arithmetic.

6. Feb 17, 2017

### sunnnystrong

I'm not sure anymore. i went through and combed through my work & can't find where i went wrong.

If you have time, could you look at my last response? I'd really appreciate it as my teacher is useless haha

7. Feb 17, 2017

### TomHart

You used 3.8 kg instead of 3.5 kg. I don't know if that will solve all of your problem, but it is at least part of it.

8. Feb 17, 2017

### sunnnystrong

I went back and did it carefully in my notebook but this time got 9m/s which is also wrong XD
at this point i just don't know what went wrong... do you have any ideas? does my set up look right?

9. Feb 17, 2017

### TomHart

When I work these perfectly elastic collision problems, it is easy for me to make a mistake. But what I got for an answer was as follows:
The 2.3 kg object initially had a velocity of +6.1 m/s. Its final velocity is -7.055 m/s.
The 3.5 kg object initially had a velocity of -4.8 m/s. Its final velocity is 3.845 m/s.

But like I said, it is easy for me to make mistakes because of the numerous math steps involved.

I thought your set-up looked right. I will try to go through your solution to see if I can find the problem.

10. Feb 17, 2017

### sunnnystrong

I'd really appreciate it. this is for my exam on tuesday. This problem & another one has been causing me troubles haha.
I used the same process throughout ^^ However if some of my numbers are weird above ignore those as i went through and did it again and got an answer around 9m/s for the car.

11. Feb 17, 2017

### Staff: Mentor

12. Feb 17, 2017

### TomHart

It looks like in the first equation you have the 3.5 kg object's velocity defined as Vf1, and in the second equation you have the 2.3 kg object defined as Vf1. That could be causing a problem.

13. Feb 17, 2017

### sunnnystrong

I will go back and find my math error because I set it up right ^^^ but made a mistake

14. Feb 18, 2017

### haruspex

In perfectly elastic collisions there is a shortcut that can avoid quadratics.
Newton's experimental law says that for two given materials (not necessarily perfectly elastic) in a straight line collision the relative velocity afterwards is -R times the initial relative velocity, where R is a constant (coefficient of restitution) depending only on the materials. For a perfectly elastic collision R=1.
Thus, in the present problem, you could write down that the relative velocity after the collision is 6.1+4.8=10.9 m/s (but reversed). Combining that with conservation of momentum gives you the answer without any quadratics.
The law, for the perfectly elastic case, can be deduced fromconservation of energy and momentum.

15. Feb 18, 2017