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Kinetic energy and momentum

  1. May 27, 2015 #1
    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
     

    Attached Files:

  2. jcsd
  3. May 27, 2015 #2
    You must give the original question.
     
  4. May 27, 2015 #3
    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...
     
  5. May 27, 2015 #4
    You say that A hits B, A rebounds with its original speed but B remains at rest? How is this possible.
     
  6. May 27, 2015 #5

    PeroK

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    Is this a problem you made up for yourself?
     
  7. May 27, 2015 #6

    SammyS

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    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.
     
  8. May 27, 2015 #7
    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?
     
  9. May 27, 2015 #8
    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.
     
  10. May 27, 2015 #9

    SammyS

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    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: May 27, 2015
  11. May 27, 2015 #10
    What happened the momentum of B before the collision?
     
  12. May 27, 2015 #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.

    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.
     
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