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Momentum Conservation

  1. Jan 13, 2017 #1
    1. In the figure shown , a ball of mass m collides perpendicularly on a smooth stationary wedge of mass M , kept on a smooth horizontal plane. If the coefficient of restitution is e , then determine the velocity of the wedge after collision.
    a6aa4078_ac2c_44b7_b2b0_9221d740fd9b.jpg https://postimage.org/][/PLAIN] [Broken]

    Given
    mass of ball = m
    mass of wedge = M
    coefficient of restitution = e
    velocity of wedge after collision = ?
    answer to problem =
    (1+e)mv sinθ / M+m sin^2 θ



    2.
    mv = mv1 + Mv2
    v1 = velocity of m after collision
    v2 = velocity of M after collision
    e = relative velocity after collision / relative velocity before collision


    3. The attempt at a solution

    So I tried conservation of momentum along the common normal , ie at the line of impact but since the wedge can only move in x direction I am stumped and cannot get an answer after solving. Tried center of mass approach and still arriving at weird solution. Any advice will be appreciated. Thanks.
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Jan 13, 2017 #2
    When can we conserve momentum?
     
  4. Jan 13, 2017 #3
    momentum is always conserved that i know of.
     
  5. Jan 13, 2017 #4
    Nope, look up Netwon's second law, momentum is conserved only in the absence of external forces. Now in which direction are there no external forces?
     
  6. Jan 13, 2017 #5
    perpendicular to the line of impact?
    or along the slide of wedge
     
  7. Jan 13, 2017 #6
    Ok, let me rephrase it - which is the external force on the system (ball + wedge)?
     
  8. Jan 13, 2017 #7
    Normal force from the ground?
    So we conserve momentum in x direction
     
  9. Jan 13, 2017 #8
    Yes that is one. What are the others?
     
  10. Jan 13, 2017 #9
    No other external forces.
    Rest of the forces are from within the system.
     
  11. Jan 13, 2017 #10
    What about gravity, is it not external?
     
  12. Jan 13, 2017 #11
    yes sorry it is.
     
  13. Jan 13, 2017 #12
    It's ok. So the Normal force balances the wedge's weight. But the ball's weight is not balanced by anyone, so there is a net force in the direction of the ball's weight. So in which direction is there no external force.?
     
  14. Jan 13, 2017 #13
    x direction
     
  15. Jan 13, 2017 #14
    Correct, so momentum can be conserved in the x-direction. So conserve momentum along x-axis, use the equation for e and I think you're done.
     
  16. Jan 13, 2017 #15
    Ok so I did this
    mv sinθ = Mv1 + mv2x ------------1
    mv2
    x = velocity of m in x direction
    e = (v1 + v2) / v
    ev = v1 + v2
    v2 = ev - v1
    Now velocity in x direction for m will be v2 sinθ
    v2 sinθ = (ev - v1) sinθ -------2
    substituting 2 in equation 1
    mv sinθ = Mv1 + m sinθ (ev-v1)
    v1 = mv sinθ (1-e) / (M-m sinθ)

    This is not the answer , what am I doing wrong?
     
  17. Jan 13, 2017 #16
    Well your e-equation in wrong. The e-equation is valid only for velocities along the line of impact.
     
  18. Jan 13, 2017 #17
    Is this correct?
    ev = v1 sinθ + v2
    then substitute as
    v2 sinθ = (ev-v1 sinθ) sinθ
     
  19. Jan 13, 2017 #18
    I think so, does that get you your answer?
     
  20. Jan 13, 2017 #19
    Well almost ,
    It seems as I have to take
    ev = v1 sinθ - v2
    As by assigning each velocity a direction . Silly mistake by me.
    Thank you for the help.
     
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