# Kinetic energy and momentum

1. May 27, 2015

### davon806

1. The problem statement, all variables and given/known data
Hi,I have been struggling on this problem for a whole day,looking for someone to sort it out!

My question:(please see the attached diagram as well)
Consider 2 spheres A and B,having different masses (m and M) but equal in size.
A is moving to the right at a speed u, B is stationary(and at the right hand side of A).
The floor is smooth and the collision is elastic.

First,consider the KE of the system.Since KE is conserved,
0.5mu^2 = 0.5mv^2 + 0.5MV^2
mu^2 = mv^2 + MV^2

v = final speed of A
V = final speed of B
The max value of v = u (where A rebounds at it original speed),(i.e.velocity = -u)
in this case,

mu^2 = m(-u)^2 + MV^2
V = 0
B remains at rest and this makes sense.

The min value of v = 0,that is,
mu^2 = MV^2
V = √(mu^2/M)

On the other hand,the momentum of the system is conserved.
mu = mv + MV

If v = -u,
2mu = MV

And if M =m, V= 2u BUT NOT V = 0

If v = 0,
mu =MV
V = mu/M Which contradicts the above result again!

I know there must be something wrong leading to these inconsistent solutions.
Thx for everyone's help.
2. Relevant equations

3. The attempt at a solution

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2. May 27, 2015

### rolotomassi

You must give the original question.

3. May 27, 2015

### davon806

My question is,why would there be two different answers of V provided v is a constant(see my working).
No matter which equation you are using,it should yield the same result?However things just don't work here...

4. May 27, 2015

### rolotomassi

You say that A hits B, A rebounds with its original speed but B remains at rest? How is this possible.

5. May 27, 2015

### PeroK

Is this a problem you made up for yourself?

6. May 27, 2015

### SammyS

Staff Emeritus
Those are not solutions to the general problem you have here. They are just solutions for extreme cases.

Yes, it's true that if the collision is elastic and sphere B remains stationary, then sphere A rebounds with the same speed it had prior to the collision. Of course for this case momentum is not conserved. Some external agent must be keeping sphere B from moving.

On the other hand. if you apply momentum conservation, and the spheres have equal mass, and furthermore sphere A rebounds with velocity -u, then sphere B will have final velocity 2u. You will also find that the final kinetic energy of the system is 5 times the initial kinetic energy.

What you need to do for this problem is to solve the two equations (one from each conservation law) simultaneously.

By the way, I assume that the collision is head-on.

7. May 27, 2015

### davon806

Sorry for being inarticulate,as this question was made up by myself while I was helping with my sister's homework.
The original question gave me the following conditions:
1.A collides with B (head-on collision)
2.A is at a speed u to the right.B is at rest
3.The collision is elastic
4.The floor is smooth
5.A has a mass m,B has a mass M,you don't know which one is bigger but it doesn't matter(in that question)

This is the background of my question:
So I am thinking of the final velocity of B (V) under several circumstances:

When A collide with B,
there are 4 outcomes,
which can be understood by solving the following equation:
0.5mu^2 = 0.5mv^2 + 0.5MV^2 ..........(conservation of KE)
mu^2 = mv^2 + MV^2

Case 1:A rebounds at an unknown speed,B moves forwards at another speed
That is:
mu^2 = mv^2 + MV^2
0<|a|<|v|,
MV^2 < mu^2
The value of V would depend on the size of M,
If M>>m, V<<u ,
If M<<m,V>>u, (V and u are in same direction)

Case 2:A moves forwards at a speed v.v < u , and B moves forwards at a speed V > u
That is:
mu^2 = mv^2 + MV^2
0<v <u <V
MV^2<mu^2 (conservation of KE)
Again,V depends on M,
If M>>m,V<<u
If M<<m,V>>u

Case 3:They stick together and move forwards at a speed V < u
mu^2 = (m+M)V^2 = mV^2 + MV^2
MV^2 < mu^2
If M>>m,V<<u
If M<<m,V>>u (rejected,V<u)
Therefore if M>>m,case 3 is possible

Case 4:A stops,B moves forwards(= case 1)
mu^2 = MV^2
If m<<M,u>>V
If m>>M,u<<V

So let's forget those trashy details above,(I hope to provide some more information to you guys)
When I consider the momentum in this system,I found something weird,
take Case 4 as an example:
If A stops,by the conservation of momentum:
mu = MV ---> V = mu/M
On the other hand,V = √(mu^2/M) by equating Case 4.

What's wrong?

8. May 27, 2015

### davon806

Yes,I did equate both equations:
mu^2 = mv^2 + MV^2 --> MV^2 = mu^2 - mv^2 = m(u+v)(u-v)
and
mu = mv + MV ----> MV= mu - mv

Hence,MV^2 = MV(u+v)
V = u+v
if v = -u (rebounds at it original speed) , V = 0,B remains stationary(it seems sensible if B is very heavy)
if v = 0,V = u, so B moves at u

The problem is that it is quite difficult to visualize what's going on with those symbols....Especially when m and M are different.How could you accept V = u+v
It is also quite confusing in the equation of KE and momentum,I assume they will yield the same answer without solving them simultaneously.

9. May 27, 2015

### SammyS

Staff Emeritus
If v = -u, then you have not included conservation of momentum.

(I'll respond to your previous post shortly.)

That is to say, if v = -u, AND V =0, then you have conserved KE, but not conserved momentum.

On the other hand, if v = -u, AND you conserve momentum, then $\displaystyle \text{V}=\frac{2m}{M}u\ .\$ Of course, then KE is no longer conserved.

Last edited: May 27, 2015
10. May 27, 2015

### rolotomassi

What happened the momentum of B before the collision?

11. May 27, 2015

### SammyS

Staff Emeritus
To answer your last question first: "What's wrong?" Many things. Mostly, you can't have these cases for elastic collisions.

If you combine this with conservation of momentum
m⋅u = m⋅v + M⋅V ,​
then you have essentially 2 equations in two unknowns. Treat u as a known, else solve the equations for v/u and V/u .

So there is only one set of solutions - only that one case.

For a collision of two otherwise isolated objects, the Law of Conservation of Momentum always holds true !

If the collision is elastic, that's a very special requirement. Most if not all of your cases try to impose this condition while ignoring the condition which always holds, namely the Law of Conservation of Momentum .
If m < M, then A rebounds at a predictable speed, B moves forwards at another speed. (Both speeds depending upon the ratio m/M)
If m > M, then A moves forwards at a speed v, also B moves forwards at a speed V.
In this case, both of the following are always true: v < u and V > u .

You made a statement regarding M >> m. However, that makes A rebound, so it's not this case.
This (the above) is the most extreme case of an inelastic collision, so none of it applies here.
This is the case for m = M and only for m = M .
In this case, v = 0, and V = u . That's it .
I hope that helps.