# Homework Help: Elastic collision -- Energy & Momentum

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1. May 21, 2017

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

Please see the attached photo. (down)
Hminitial= 1.5R
M = 2/3m
Perfectly elastic collision

What is the velocity of object m immidiatly after the collision? (by m,g,R)

2. Relevant equations
Conservation of energy
Conservation of momentum

3. The attempt at a solution
I assumed that because of the elastic collision the M object bounce back, so I considered positive velocity to the right and negative to the left (when we're looking exactly on the collision moments)

First I found the "before the collision" velocity of the falling ball M (with conservation of energy) which is (3Rg)

Now I used two equation to find the velocities after the collision :

Conservation of momentum: MVi + 0 = MVf +mvf
Conservation of energy: V0 + Vf = v0 + vf

After some work I get :
Vf = ⅓⋅√(3Rg)
vf = 4⋅√(3Rg) / 3

It seems make sense when the speed of the small ball got higher and the big ball, which bounced backward, slower .

But in the other hand I know that the M velocity after the collision has to be negative. because it bounced back!, perfectly elastic collision....
But I got 2 positive answers!

when I try to change the direction/sign (to minus) of the variable MVf on the first equation I get different and not logical answer (Vf = √(3Rg) ) , like the bigger ball didn't lost energy at all while it was giving energy to the smaller ball...

Thanks.

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Last edited: May 21, 2017
2. May 21, 2017

### Staff: Mentor

That's not true. They have to "bounce apart" (not stick together) and conserve kinetic energy, but the first mass needn't reverse direction. Imagine that first mass being 100 times more massive. Is it likely that hitting a small mass would make it move backwards?

3. May 21, 2017

You right. So basically the answers tell me that the bigger mass in this case didn't move backwards, It continued forward with third of its velocity?

4. May 21, 2017

### Staff: Mentor

Right!

5. May 21, 2017

Thank you very much.
For some reason I thought it needs to bounce backwards.

6. May 21, 2017

### haruspex

In a perfectly elastic head-on collision with a stationary object, the incomng object will bounce back if, and only if, it is lighter than the other. If they are equal it will come to rest.

7. May 22, 2017

### ehild

8. May 22, 2017

What am I missing?

9. May 22, 2017

### ehild

I do not know without seeing your work in detail.
The numerical values do not fulfill the equations for conservation of momentum and energy.

10. May 22, 2017

### Staff: Mentor

Yikes, ehild is right!

That equation for conservation of energy is incorrect.

Thus these results are incorrect.

Fix your conservation of energy equation and resolve.

Thanks to ehild for paying attention!

(Sorry for not checking your work earlier!)

11. May 22, 2017

### ehild

It is correct, if Vi stands for V0 and vi stands for v0 in
V0 + Vf = v0 + vf

Capital letters refer to the big ball, small case letters mean the velocities of the small ball.

12. May 22, 2017

### Staff: Mentor

Once again, ehild is correct (as always).

(That equation isn't a statement of energy conservation only, but is a result of energy conservation plus momentum conservation.)

@Deadawake: You must have made an algebra/computational error; try it once more.

13. May 22, 2017

### ehild

This is not true

14. May 22, 2017

Correct. this statement is result of conservation of energy and momentum equations.
Vbefore collision = √(3Rg)

1)
MV0 + 0 = MVf+mvf

⇒ 1.5mV0 = 1.5mVf +mvf
⇒ vf = 1.5⋅√(3Rg) - 1.5Vf

2)
V0 + Vf = v0 + vf
⇒ √(3Rg) +Vf = 0 +vf
⇒vf = √(3Rg) +Vf

⇒⇒⇒√(3Rg) +Vf = 1.5⋅√(3Rg) - 1.5Vf
⇒⇒⇒2.5Vf = 0.5⋅√(3Rg) →here was my error. I wrote 1.5 instead of 2.5
⇒⇒⇒Vf = 0.2√(3Rg)
⇒⇒⇒vf = √(3Rg) +0.2√(3Rg) = 1.2√(3Rg)

Last edited: May 22, 2017
15. May 22, 2017

### Staff: Mentor

Good. My answers agree with yours.

16. May 22, 2017

Some other question is - What height does the small mass ball reach after the collision? (hf)
(Just a reminder - we are talking about vertical standing circle/ring....)
So I have the initial speed of the small ball (after the collision) and I can use conservation of energy
½mv02 = mghf
½m⋅(1.2√3gh)2 = mghf
½⋅1.44⋅(3Rg) = g⋅hf
hf = 2.16⋅R

We know that the radius of the circle is R , so the maximum possible height is 2R.
Then the answer tells me that the small mass has enough energy to reach the top of the circle and a bit more.
The cunclusion is that the small ball will get to the maximum height which is 2R.
Am I right?

17. May 23, 2017

### Staff: Mentor

Sounds good to me.