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Ball hitting an inclined plane

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data
    http://www.sumoware.com/images/temp/xzafcttptkeoehmo.png [Broken]
    An inclined plane is put on a smooth floor. The inclined plane is hit (collided) by an elastic ball moving horizontally before the collision. The ball bounce from the inclined plane and land again right at the point of first collision. If the inclination angle is Θ, determine the ratio between the ball mass and the inclined plane mass

    2. Relevant equations
    px = px'
    E = E'

    3. The attempt at a solution

    Actually I don't know what to do, so I just draw the condition after collision
    http://www.sumoware.com/images/temp/xzknipxthmfhalxb.png [Broken]
    Vp is the velocity of inclined plane

    p = p
    mb vb = mb vbx' + mp vp'

    Then, I don't know what to do
    Please help me
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Feb 10, 2015 #2

    ehild

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    Energy is conserved, as the ball is elastic and the ground is smooth.
    The ball reaches the plane at the first collision point again. What does it mean for the horizontal distances travelled by the plane and by the ball? What information you have about the vertical and horizontal velocity components of the ball?
     
  4. Feb 10, 2015 #3
    Hmm... When the ball reaches the plane at the first collision point, it means that the horizontal distance travelled by the ball is the same as travelled by the plane.
    So
    p = p
    mb vb = mb vbx' + mp vp'
    mb vb = (mb + mp)v'x

    The vertical component of the ball is v'by = v cos Θ
    Horizontal component is v'bx = v sin Θ

    (1/2)mvb2 = (1/2)mvb'x2 + (1/2)mvp'x2
    (1/2)mvb2 = mvb'x2
    vb2 = 2vb'x2
    vb2 = 2(vb sin Θ)2
    vb = √(2(vb sin Θ)2)
    Oops.. It seems like a dead-end to me.. Please help me
     
  5. Feb 10, 2015 #4

    ehild

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    Correct.
    Why??????
     
  6. Feb 10, 2015 #5
    Because if theta increases, the y component of the velocity of ball after collision decreases, right ? So, it's cosine function, right ?
    And if theta increases, the x component of the velocity of ball after collision increases, right ? So, it's sinus function, right ?
    But, I don't know what it will be if the 'launch' angle has nothing to do with the angle theta
     
  7. Feb 10, 2015 #6

    haruspex

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    There are trig functions besides cos and sine.
    Let the impulse the block delivers to the ball be J. Which way does it point? What is its component in the x direction?
     
  8. Feb 10, 2015 #7
    http://www.sumoware.com/images/temp/xzaciotcpqkxexdr.png [Broken]
    J = mg cos Θ Δt
    Jx = m v'bx - m vb

    But, what is v'bx ?
     
    Last edited by a moderator: May 7, 2017
  9. Feb 10, 2015 #8

    ehild

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    The plane can exert force only in the perpendicular direction, as you drew, but the impulse is not connected to gravity. The parallel (with the plane) component of the initial velocity of the ball does not change during the collision. Only the perpendicular component changes. Conservation of energy provides a relation among the speed of the slope Vp, initial speed of the ball vb, and the perpendicular component after collision, vm.

    Both the perpendicular and parallel components of the velocity after collision have horizontal components, depending on the angle theta. You can get an expression for v'bx in terms of theta.
    You know that v'bx=Vp, and conservation of momentum yields Vp=mbvb/(mb+mp). There are enough equation to get the mass ratio in terms of theta.
     
  10. Feb 10, 2015 #9

    haruspex

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    As ehild says, g doesn't enter into it. Δt is very small.
    Deduce J from Jx (simple trig), and deduce Jy from that.
    Or use ehild's observation that "The parallel (with the plane) component of the initial velocity of the ball does not change during the collision", which I was trying to avoid assuming. The two methods amount to the same.
     
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