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    susus's Avatar
    susus Posts: 138, Reputation: 2
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    #1

    Dec 17, 2010, 11:45 AM
    Bat hitting a Ball
    Bat hitting a Ball
    A ball and bat, approach one another each with the same speed of 2.70 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)


    Find the factor by which the KE of the ball increases due to the collision.
    harum's Avatar
    harum Posts: 339, Reputation: 27
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    #2

    Dec 19, 2010, 09:24 PM
    Quick, equation-less approach: If the bat is much havier than the ball, then you can safely assume that the ball is colliding with a moving wall. In the frame of reference associated with the ground and with the positive direction where the wall moves, the wall moves with the speed v, and the ball's speed is -v. If you switch to a new frame of reference in which the wall is at rest, then in this frame of reference the ball is moving toward the wall with the speed -2v. The switching is done by subtracting v from the speeds of both the wall and the ball: v - v = 0 (the wall is at rest in the new reference frame; -v - v = -2v (the ball's speed in the new frame). The ball bounces elastically off the wall, that is without a change in the magnitude of the speed, the direction of the motion is reversed. The new ball's speed is 2v. It was -2v and became 2v. This all in the frame of reference associated with the wall. To go back to the frame of reference associated with the ground, add v back to both speeds. That is, the ball's speed after collision in the frame of reference associated with the ground is 3v (2v + v) and the wall's/ bat's speed has remained v. Before collision, the ball's energy was 1/2*m*v^2. After collision it became 1/2*m*(3v)^2.
    harum's Avatar
    harum Posts: 339, Reputation: 27
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    #3

    Dec 19, 2010, 11:23 PM
    If you prefer equations, use the two conservation laws (momentum and energy) in the frame of reference associated with the ground. Momentum:

    (1) M*v - m*v = M*v1 + m*v2

    Energy:

    (2) M*v^2 + m*v^2 = M*v1^2 + m*v2^2

    Here, M - mass of the bat, m - mass of the ball, v is the speed before collision and v1 and v2 - after.
    Solve (1) for v1 and plug into (2), then divide what you get by M and get rid of all terms containing (m/M)^2 to simplify your equation. This is because the bat is much havier than the ball. Solve a simple quadratic equation.
    susus's Avatar
    susus Posts: 138, Reputation: 2
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    #4

    Dec 20, 2010, 10:49 AM
    Comment on harum's post
    even though it look strange and long haha , it was so easier than using the equations thanks :):):)

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