Momentum Between Two Colliding Balls

[cbasst]

Homework Statement

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?

Homework Equations

U_g=mgh
K=1/2 mv²
p=mv

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 = U_g,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 = U_g,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?

 

[Spinnor]

Study the following problem (which is similar to yours) and see if it makes sense.

http://answers.yahoo.com/question/index?qid=20110525095138AA6Z6zL

Found from,

https://www.google.com/webhp?hl=en#...5ec0820d464c81&bpcl=37643589&biw=1093&bih=491

 

[PeterO]

Homework Statement

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?

Homework Equations

U_g=mgh
K=1/2 mv²
p=mv

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 = U_g,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 = U_g,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?

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 traveling at +3 m/s, while m was traveling 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 traveling at a speed of 3 m/s before collision either.
Perhaps you can adapt this idea however.

 

[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. t_o 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.

 

[PeterO]

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. t_o 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.

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 t_o,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 v_CoM either.

 

[cbasst]

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

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 v_CofM? Or do I need to set the equation for v_CofM to be equal to something else (maybe relate it to the speed of one or both of the balls)? How would I do that?

 

[PeterO]

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

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 v_CofM? Or do I need to set the equation for v_CofM to be equal to something else (maybe relate it to the speed of one or both of the balls)? How would I do that?

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 traveling North at 5 m/s, while a 1 kg mass is traveling 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 traveling North at 1 m/s, while the the 1 kg mass will be traveling North at 9 m/s

Throughout, the C of M is traveling 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 traveling at 3 m/s, while the 3 kg mass is traveling at 5 m/s --> it is approaching at 2 m/s
After collision, you are traveling at 3 m/s and the 3kg mass is traveling at 1 m/s --> it is moving away at 2 m/s

Also

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

 

[cbasst]

That concept can lead to the speed of the centre of mass, since the large M STOPS in the Earth frame of reference.

I think I may understand what you are getting to. Are you suggesting that I come up with two equations for the v_CofM, 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 v_CofM 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!

 

[PeterO]

I think I may understand what you are getting to. Are you suggesting that I come up with two equations for the v_CofM, 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 v_CofM 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!

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 traveling down at speed V

Ʃp = MV - mV,
Ʃmass = M + m

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

BUT V_CofM = 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 V_CofM = 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

 

[PeterO]

Now how did I get V_CofM = V/2

Just in case you want some numeric examples.

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

1: V_CofM = 1 m/s
Here the mass M is traveling 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 V_CofM = 1 m/s, the mass M will recoil at 4 m/s [but we are told it merely stops.

2: V_CofM = 2 m/s
Here the mass M is traveling 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 V_CofM = 2 m/s, the mass M will recoil at 2 m/s [but we are told it merely stops].

3. V_CofM = 5 m/s
Here the mass M is traveling 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 V_CofM = 5 m/s, the mass M will continue on at 4 m/s [but we are told it stops].

4. V_CofM = 4 m/s
Here the mass M is traveling 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 V_CofM = 4 m/s, the mass M will continue on at 2 m/s [but we are told it stops].

5. V_CofM = 3 m/s
Here the mass M is traveling 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 V_CofM = 3 m/s, the mass M will stop. We are told that happens.

Note that if V_CofM = 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].

 

[cbasst]

Okay, that makes a lot of sense. The explanation for why v_CofM = 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 v_CofM 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.

 

[PeterO]

I think I may understand what you are getting to. Are you suggesting that I come up with two equations for the v_CofM, 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 v_CofM 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!

Replying here, so I can highlight parts of your solution.

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 V_{C 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 V_{C of M} was traveling up at V/2, while m was traveling 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 traveling once it has dropped 1.8m

In the example you mentioned - where M is still traveling 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 V_{C of M} = 3 m/s

 

[cbasst]

Okay, that makes sense. Thanks for the help!