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Kinetic energy and momentum

  • Thread starter davon806
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  • #1
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Homework Statement


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.

Homework Equations




The Attempt at a Solution

 

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Answers and Replies

  • #2
You must give the original question.
 
  • #3
148
1
You must give the original question.
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
You say that A hits B, A rebounds with its original speed but B remains at rest? How is this possible.
 
  • #5
PeroK
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Homework Statement


Hi,I have been struggling on this problem for a whole day,looking for someone to sort it out!
Is this a problem you made up for yourself?
 
  • #6
SammyS
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Homework Statement


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.

Homework Equations



The Attempt at a Solution

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
148
1
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
148
1
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.
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 o_O
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
SammyS
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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 o_O
It is also quite confusing in the equation of KE and momentum,I assume they will yield the same answer without solving them simultaneously.
If v = -u, then you have not included conservation of momentum.

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

Added in Edit:
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:
  • #10
If A stops,by the conservation of momentum:
mu = MV ---> V = mu/M
What happened the momentum of B before the collision?
 
  • #11
SammyS
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To answer your last question first: "What's wrong?" Many things. Mostly, you can't have these cases for elastic collisions.

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
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 .
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)
If m < M, then A rebounds at a predictable speed, B moves forwards at another speed. (Both speeds depending upon the ratio m/M)
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 (A will rebound so it's not this case.)
If M<<m,V>>u
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.
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
This (the above) is the most extreme case of an inelastic collision, so none of it applies here.
Case 4:A stops,B moves forwards(= case 1)
mu^2 = MV^2
If m<<M,u>>V
If m>>M,u<<V
This is the case for m = M and only for m = M .
In this case, v = 0, and V = u . That's it .
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?
I hope that helps.
 

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