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Slipping block on a slipping ramp: final speed

  1. Apr 11, 2015 #1
    Hi, friends! A block having mass ##m## frictionlessly slips from height ##h## on a ramp of mass ##M##, which has an angle ##\theta## with the floor, where it slips with no friction.

    inclineWithTheta.gif

    I would like to prove that, as my book says, the ramp's speed when the block leaves it is

    ##V=\sqrt{\frac{2m^2gh\cos^2\theta}{(m+M)(M+m\sin^2\theta)}}.##​

    But I cannot reach this result. Because of the conservation of momentum in the direction parallel to the floor, which is a direction where no external force act on the system block-ramp, I would say that the speed ##v## of the block when it leaves the ramp is such that ##mv\cos\theta=M V## and, because of the conservation of mechanical energy, I would think that ##mgh=\frac{1}{2}mv^2+\frac{1}{2} MV^2=\frac{1}{2}\big(\frac{M^2V^2}{m\cos^2\theta} +MV^2\big)##, and therefore ##V=\sqrt{\frac{2m^2gh\cos^2\theta}{(m+M)(M^2+Mm\cos^2\theta)}}##
    which is a wrong result. Where am I wrong?

    I ##\infty##-ly thank you for any help!
     
  2. jcsd
  3. Apr 11, 2015 #2
    Where do you get the (m+M) term?
     
  4. Apr 11, 2015 #3

    haruspex

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    Relative to the ground, what direction will the block be moving?
     
  5. Apr 11, 2015 #4

    haruspex

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    I assumed that was a typo.
     
  6. Apr 11, 2015 #5

    ehild

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    See thread https://www.physicsforums.com/threads/wedge-and-block-initial-momentum.806992/. It is the same problem as yours.
    Decide what you denote by v. The block slides on the ramp which is moving horizontally with velocity -V. If the relative velocity of the block with respect to the ramp is u, the horizontal component is ucos(θ) and in the rest frame of reference it is ucos(θ)-V.
     
  7. Apr 12, 2015 #6
    Thank you all, friends!!! ehild's linked post answers the issue.
     
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