# Homework Help: Momentum Between Two Colliding Balls

1. Nov 6, 2012

### cbasst

1. The problem statement, all variables and given/known data
A small ball of mass m is aligned above a larger ball of mass M = 0.63kg (with a slight separation), and the two are dropped simultaneously from height h = 1.8m. (Assume the radius of each ball is negligible relative to h.)
a. If the larger ball rebounds elastically from the floor and then the small ball rebounds elastically from the larger ball, what value of m results in the larger ball stopping when it collides with the small ball?
b. What height does the small ball then reach?

2. Relevant equations
Ug=mgh
K=1/2 mv2
p=mv

3. The attempt at a solution
Define time t=0 to be before the balls are dropped. Time t=1 is just before the large ball collides with the ground. Time t=2 is just after the large ball collides with the small ball.

K1 = Ug,0 = Mgh + mgh

To find momentum at t=1, velocity is needed. Velocity will be the same for both balls, since they are dropped form the same height. Considering only the larger ball:

K1 = Ug,0
1/2 Mv12 = Mgh
v1 = √(2gh)

V1 will also be the same for the smaller ball.

When the collision occurs, the momentum immediately before and after the collision (t1 and t2, respectively) should stay constant.

p1 = v1 (M + m) = √(2gh) * (M + m) = p2

After the collision, the velocity of the large ball is 0, so

p2 = mv2

To find v2, the kinetic energy equations are needed again. All the collisions are elastic, so the kinetic energy will be the same at t1 and t2.

K1 = Mgh + mgh = (M + m)gh
K2 = 1/2 mv22
K1 = K2
(M + m)gh = 1/2 mv22
v2 = √(2gh(M + m) / m)

So after some substitutions, it looks like

p1 = p2
(M + m)v1 = mv2
(M + m) * √(2gh) = m√(2gh(M + m) / m)

Everything is known except for m, so it seems like it should be able to be solved at this point. However, when I try to simplify this by squaring each side, I end up with M=0 or -m=M. Neither of those can be right, so I think I must have made a mistake in the physics part. I don't know where though. Can anyone spot the error?

2. Nov 7, 2012

3. Nov 7, 2012

### PeterO

In this arrangement, M is moving up at the same speed as m is moving down - before they collide [this comes about by the two phrases I have made red above]

In all collisions , the centre of mass of the bodies continues to move at the same velocity.

In an elastic collision, each separate body approaches and leaves the centre of mass at the same speed [not as each other, but separately]

eg, If M was travelling at +3 m/s, while m was travelling at -3 m/s and the centre of mass was moving at +1 m/s,
Then M is approaching the CofM at a relative speed of 2m/s, so will be moving away from the CofM at a relative speed of 2 m/s.
Also m is approaching the cofM at a relative speed of 4 m/s, so will be moving away from the CofM at a relative speed of 4 m/s after collision.

Clearly those figures don't match your situation, as it would have M moving down at 1m/s after collision - and they probably were not travelling at a speed of 3 m/s before collision either.
Perhaps you can adapt this idea however.

4. Nov 8, 2012

### cbasst

Are you suggesting I find the velocity of the CofM and use that along with the fact that the final velocity of the large ball is 0 in order to find the velocity of the smaller ball?

Before I show the velocity of the CofM, I think I want to redefine my time variables. The first definitions I gave are not working very nicely. to will be the time just before the large ball collides with the ground, t1 will be the time just before the large ball collides with the small ball, and t2 will be the time just after the large ball collides with the small ball. The velocity of the CofM can be described by
$$v_{CofM} = \frac{Mv_1 + m(-v_1)}{M + m}$$
m has a negative velocity because it is still going downwards, whereas M is positive because it is going upwards (but they both have the same speed; I think I may have missed the negative in my original work). Everything in that equation is known except for m, but from here I'm still a little lost. I understand your analogy I think, I'm just not sure how to implement the necessary strategy.

5. Nov 8, 2012

### PeterO

Just copyjng the original post in, as it looks like the red parts went missing???

A small ball of mass m is aligned above a larger ball of mass M = 0.63kg (with a slight separation), and the two are dropped simultaneously from height h = 1.8m. (Assume the radius of each ball is negligible relative to h.)
a. If the larger ball rebounds elastically from the floor and then the small ball rebounds elastically from the larger ball, what value of m results in the larger ball stopping when it collides with the small ball?
b. What height does the small ball then reach?

I believe those red bits mean that your to,t1 & t2 are all the same time - an idealisation but I believe that is what is wanted: not that it makes any difference to the way you are working it out.

You said that everything in that equation - except m - was known - so why don't you just solve for m
OR
Did you really mean you don't know vCoM either.

6. Nov 9, 2012

### cbasst

Sorry - I should have specified: I know everything except m on the right hand side of that equation. vCofM is also unknown. So the equation has two unknowns: m and vCofM.

Yes, I had also figured that each of the times was the same time, but representing them as different times can help me keep my work a little more organized.

Anyway, so how should I find vCofM? Or do I need to set the equation for vCofM to be equal to something else (maybe relate it to the speed of one or both of the balls)? How would I do that?

7. Nov 10, 2012

### PeterO

Remember that bit I mentioned about objects involved in an elastic collision.

If you watch the collision from the C of M frame of reference, each object approaches at some speed [not necessarily equal], then bounces off at exactly the same speed.
That concept can lead to the speed of the centre of mass, since the large M STOPS in the Earth frame of reference.

to run you through an example again:

Suppose a 3 kg mass is travelling North at 5 m/s, while a 1 kg mass is travelling South at 3 m/s, and they have a head on, elastic collision,
When you do your standard analysis of that you will find that AFTER collision, the 3 kg mass will be travelling North at 1 m/s, while the the 1 kg mass will be travelling North at 9 m/s

Throughout, the C of M is travelling at 3 m/s North
Ʃp = 3x5 + 1x(-3) = 12 Ns
Ʃm = 3 + 1 = 4 kg

Velocity of CofM = Ʃp/Ʃm = 12/4 = 3 and positive 3 means 3 m/s North.

Look at it from the frame of reference of the CofM

You are travelling at 3 m/s, while the 3 kg mass is travelling at 5 m/s --> it is approaching at 2 m/s
After collision, you are travelling at 3 m/s and the 3kg mass is travelling at 1 m/s --> it is moving away at 2 m/s

Also

You are travelling at 3 m/s, while the 1 kg mass is travelling at -3 m/s --> it is approaching at 6 m/s
After collision, you are travelling at 3 m/s and the 1 kg mass is travelling at 9 m/s --> it is moving away at 6 m/s.

8. Nov 10, 2012

### cbasst

I think I may understand what you are getting to. Are you suggesting that I come up with two equations for the vCofM, one for just before the collision, and one for just after? If I do that, I get $$v_{CofM} = \frac{v_1(M - m)}{M + m} = \frac{M * 0 + mv_2}{M + m}$$ I still have two unknowns, since both m and v2 are unknown. However, if I set the initial kinetic energy equal to the final kinetic energy, I can solve for either v2 or m. I won't show that work here, but after I combine the above equations for vCofM and the solution for v2 using the kinetic energy equations, I get $$v_1(M - m) = m \sqrt{\frac{2Mgh}{m} + gh}$$ which can then simplify to m2 - 6Mm + 2M2 = 0. After substituting 0.63 kg in for M, m is 0.22 kg. Should I be worried if the answer the textbook gives is 0.21 kg? As far as I can tell, my methods are correct, so I'm not sure why this difference would exist. Thanks for the help!

9. Nov 11, 2012

### PeterO

The numeric answer is precisely 0.21, so if you get 0.22 you are slightly wrong.

I think you have still not noticed the value of the Velocity of the Centre of Mass after the mass M bounces.

Here we go.

after hitting the ground, M is traveling up at some speed V. m is still travelling down at speed V

Ʃp = MV - mV,
Ʃmass = M + m

VCofM = (MV - mV) / (M + m) = V(M - m) / M + m

BUT VCofM = V/2

So V(M - m) / M + m = V/2
--> 2(M - m) = M + m
--> 2M - 2m = M + m
--> M = 3m

since M = 0.63, we get m = 0.21

Now how did I get VCofM = V/2

before the balls collided, M was moving up at speed V - approaching the Centre of Mass at some speed.
After the balls collided, M was stationary - but leaving the Centre of Mass at that same speed.

That can only happen if the C of M is moving at V/2

When the Mass is moving at V, it is approaching the C of M at speed V/2
When the mass is stationary, the C of M is moving away from it at speed V/2

10. Nov 11, 2012

### PeterO

Just in case you want some numeric examples.

Let's say that M bounces with a velocity of 6 m/s. Then lets consider various different C of M velocities.

1: VCofM = 1 m/s
Here the mass M is travelling at 5 m/s faster that the C of M, so is approaching at 5 m/s
After collision, it will be moving away from the C of M at 5 m/s.
Only a velocity of -4 m/s meets that requirement - so if the VCofM = 1 m/s, the mass M will recoil at 4 m/s [but we are told it merely stops.

2: VCofM = 2 m/s
Here the mass M is travelling at 4 m/s faster that the C of M, so is approaching at 4 m/s
After collision, it will be moving away from the C of M at 4 m/s.
Only a velocity of -2 m/s meets that requirement - so if the VCofM = 2 m/s, the mass M will recoil at 2 m/s [but we are told it merely stops].

3. VCofM = 5 m/s
Here the mass M is travelling at 1 m/s faster that the C of M, so is approaching at 1 m/s
After collision, it will be moving away from the C of M at 1 m/s.
Only a velocity of 4 m/s meets that requirement - so if the VCofM = 5 m/s, the mass M will continue on at 4 m/s [but we are told it stops].

4. VCofM = 4 m/s
Here the mass M is travelling at 2 m/s faster that the C of M, so is approaching at 2 m/s
After collision, it will be moving away from the C of M at 2 m/s.
Only a velocity of 2 m/s meets that requirement - so if the VCofM = 4 m/s, the mass M will continue on at 2 m/s [but we are told it stops].

5. VCofM = 3 m/s
Here the mass M is travelling at 3 m/s faster that the C of M, so is approaching at 3 m/s
After collision, it will be moving away from the C of M at 3 m/s.
Only a velocity of 0 m/s meets that requirement - so if the VCofM = 3 m/s, the mass M will stop. We are told that happens.

Note that if VCofM = 3 m/s, then the Centre of Mass is moving at exactly 1/2 the speed of the Mass M [which was traveling at 6 m/s here].

11. Nov 11, 2012

### cbasst

Okay, that makes a lot of sense. The explanation for why vCofM = v/2 was very useful. So just to clarify that I understand this, let's say that instead of M stopping after the collision, it continued moving at, say, 2 m/s. Then the vCofM would be given by $$v_{CofM} = \frac {\sqrt{2gh} - 2 m/s}{2}$$ Is that correct? And I am curious: what was wrong about the way I solved the problem? I think that if I know what I was doing wrong that will help me understand these sorts of problems better in the future.

12. Nov 11, 2012

### PeterO

The stuff I made red is certainly correct - I actually used the first part in my solution.

It turns out that M would stop no matter what height this pair was dropped from [in reality it wouldn't, since M would fall a little more than m due to their physical sizes.

As far as the VC of M is concerned, it is just the average of the Velocity of M before and after its collision with m. It is also the average of the Velocity of m before and after - though we weren't given any specific value for the velocity of m after collision.

BUT: we now know VC of M was travelling up at V/2, while m was travelling down at V - a closing speed of 3V/2, so after it will be separating at 3V/2 so will be traveling at 2V.
If you want to find a numeric value of that, you need to evaluate V by finding how fast an object will be travelling once it has dropped 1.8m

In the example you mentioned - where M is still travelling up at 2 m/s, you would have to first calculate that is reaches the ground, and therefore bounces back up, at about 4 m/s (just over)
[use v2 = u2 + 2as for an object falling 1.8 m]

The average of 4 and 2 is 3, so VC of M = 3 m/s

13. Nov 11, 2012

### cbasst

Okay, that makes sense. Thanks for the help!