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Homework Help: Find the factor by which the KE of the ball increases

  1. Dec 18, 2009 #1
    Hi,
    Can anyone help me in this question?
    1) A ball and bat, approach one another each with the same speed of 1.30 m/s, collide. Find the speed of the ball after the collision. (Assume the mass of the bat is much much larger than the mass of the ball and that this is an elastic collision with no rotational motion).
    (in m/s)
    2)Find the factor by which the KE of the ball increases due to the collision.



    I have tried to solve it but I did not know how to continue.

    Sorry for my broken english & thanks in advance.
     
  2. jcsd
  3. Dec 18, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Momentum

    Show what you have tried.

    Hint: View things from a frame in which the bat is at rest.
     
  4. Dec 18, 2009 #3
    Re: Momentum

    *mass of the ball= m
    mass of the bat= M
    *speed of the ball after the collision= u1
    speed of the bat after the collision= u2
    *speed of the ball before the collision= v1= 1.30m/s
    speed of the bat before the collision=v2= 1.30 m/s

    *mv1 + Mv2 = mu1 + Mu2
    1.30m + 1.30M = mu1 + Mu2

    *0.5m*(1.30)^2 +0.5M*(1.30)^2 = 0.5m*(u1)^2 +0.5M*(u2)^2
    1.69m + 1.69M = m*(u1)^2 + M*(u2)^2

    And then I did not know how to continue.
     
  5. Dec 18, 2009 #4
    Re: Momentum

    I have tried this now:

    (1)m(1.30-u1) = M(u2-1.30)
    (2)m(1.69-u1^2) = M(u2^2 - 1.69)

    (2)/(1) --> (1.69-u1^2)/(1.30-u1) = (u2^2 -1.69)/(u2-1.30)

    (1.30-u1)(1.30+u1)/(1.30-u1)= (u2-1.30)(u2+1.30)/(u2-1.30)
    1.30+u1 = u2 +1.30
    u1=u2


    ??
     
  6. Dec 18, 2009 #5
    Re: Momentum

    I think the responder tried to make it easier for you by assuming the bat was at rest. To make the problem the same for this situation what would the velocity of the ball be?

    Makes it a whole lot easier.
     
  7. Dec 20, 2009 #6

    Doc Al

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    Staff: Mentor

    Re: Momentum

    You can certainly solve the problem from the 'ground' frame, it's just harder. To do that, first combine the equations for conservation of momentum and energy to deduce a relationship about relative velocities before and after the collision.

    Hint: Since the ball is assumed to be much heavier than the ball, M >> m, what can you say about the final speed of the bat to a good approximation?

    My point was that it might be easier to solve this from a frame in which the bat is at rest. In that frame, the bat is like a wall. If you bounce a ball elastically off a wall, what is its final velocity? Get the answer in that frame, then transform back to the original frame.

    Careful. While the speed of ball and bat is the same, their direction of motion is not. They move in opposite directions, so they have different velocities.
     
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