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2D Elastic Collision Problem

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Let a particle of mass 2M have an initial velocity of v0i (the i merely indicating it is traveling on the x axis) and undergo an elastic glancing collision with a particle of mass M initially at rest. After the collision, the M particle moves off at an angle of 45° above the + x-axis.

    1. What is the speed of the M particle after the collision?

    2. What is the velocity (magnitude and direction) of the mass 2M particle after the collision?

    m1 = 2M
    m2 = M
    v1i = v0i
    θ2 = 45°

    2. Relevant equations

    pix = pfx (momentum conservation in the x direction)
    m1v1i = m1v1fcosθ1 + m2v2fcosθ2

    piy = pfy (momentum conservation in the y direction)
    0 = m1v1fsinθ1 + m2v2fsinθ2

    Ki = Kf (conservation of kinetic energy)
    m1v1i2 = m1v1f2 + m2v2f2

    3. The attempt at a solution

    It appears to be a 3 equations 3 unknowns problem (the unknowns being v2f, v1f, and θ1, which are the final velocities of the masses and the angle of mass 2M). I tried adding the two momentum equations to eliminate θ1 and get a new equation to use with the energy equation, that way I would have two equations and two unknowns. But I ended up finding a value for v1f to be -1.41v0i, meaning the ball of larger mass collided with the ball of half-mass, then started going in the opposite direction with a larger velocity, which makes no sense... Any help would be appreciated.
     
    Last edited: Apr 9, 2012
  2. jcsd
  3. Apr 10, 2012 #2

    ehild

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    The problem is wrong. The angle must be 30° or smaller.
    If you solve the problem symbolically assuming an angle θ, you get a quadratic equation with the discriminant non-negative when cosθ≥√3/2.

    ehild
     
  4. Apr 10, 2012 #3
    So are you saying my professor wrote the problem wrong? That the problem is unsolvable if the angle is greater than 30°? I don't see why the angle must be 30° or lower....

    Thanks for the reply by the way. :smile:
     
  5. Apr 10, 2012 #4

    ehild

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    Isolate v2fsin(θ2) and v2fcos(θ2) from the equations for momentum and substitute into the energy equation. Try to solve for v1 in terms of θ.

    ehild
     
  6. Apr 10, 2012 #5
    I don't understand. I need to isolate v2fsinθ2 and v2fcosθ2 in EACH momentum equation, then substitute into the energy equation? Am I isolating just v2f or do I include the trig functions in my isolation? And how am I supposed to substitute? Do I substitute the result from isolating v2fsinθ2 and v2fcosθ2 separately, giving me two different energy equations?
     
  7. Apr 10, 2012 #6

    ehild

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    Denote the components of the velocity of the smaller mass after the collision by v2x and v2y.
    The momentum equations:

    2Mv0=2Mv1f cos(θ)+Mv2x
    0=2Mv1f sin(θ)+Mv2y
    The energy equation:
    2Mv02=2Mv1f2+M(v2x2+v2y2)

    Isolating v2x and v2y from the first two equations:

    v2x=2v0-2v1fcos(θ)
    v2y=-2v1fsin(θ).

    Substituting into the energy equation:

    2v02=2v1f2+(2v0-2v1fcos(θ))2+(-2v1fsin(θ))2

    Expanding the squares and simplifying you arrive to the quadratic equation

    6v1f2-8v0v1fcos(θ)+2v02=0

    whis has real solution for v1f if cosθ≥√3/2.
     
  8. Jun 27, 2012 #7
    What would be the case, if the ball is collided with the plate. Just like the table tennis.
    Can we also consider the collision between the table tennis racket and ball as the 2D collision like collision between two ball ? ?
    Will the shape of the plate will effect the equations ?
     
  9. Jun 28, 2012 #8
    I am working on the 2D elastic collision problem.
    In the attachment find my calculations.
    First I solve the problem in forward direction and then I tried to solve the same problem with different angles.
    The main problem is, I require another condition to calculate my last parameter.
    Can anyone help me ...
     
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