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Velocity of particle leaving wedge

  1. May 3, 2017 #1
    1. The problem statement, all variables and given/known data
    In the given figure, a wedge of mass 2m is lying at rest on a horizontal surface. The wedge has a cavity which is the portion of a sphere of radius R. A small sphere of mass m is released from the top edge of the cavity to slide down. All surfaces are smooth.Prove the maximum height acquired by the sphere after breaking off from the wedge is ##\frac{10R}{11}## (from the point B)
    question 2.jpg

    2. Relevant equations

    3. The attempt at a solution
    Let the velocity of the sphere at the point where the sphere leaves the wedge be ##v## and that of the wedge be ##v_w##, then conserving energy, we get $$mgR = \frac{1}{2}mv^2 + \frac{1}{2}2mv_{w}^{2} + mg\frac{R}{2}$$
    also conserving momentum along the x axis
    $$2mv_w = mvcos(\frac{\pi}{3})$$
    (Since the angle the tangent at point C makes with the horizontal is 60 degrees)
    solving these, I get $$v^2 = \frac{8}{9}gR$$
    and hence max height is ##\frac{4}{9}R + \frac{R}{2} = \frac{17}{18}R##
    where am I going wrong here?
    Last edited: May 3, 2017
  2. jcsd
  3. May 3, 2017 #2


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    Why is momentum conserved along the x-axis? Gravity is not the only external force on the sphere. The cavity surface exerts a variable normal force that has both x and y components.

    On edit: I misread the problem. I suppose that the horizontal surface on which the wedge lies is frictionless so that the horizontal momentum of wedge + ball is conserved. That was not clear after my initial reading of the problem.
    Last edited: May 3, 2017
  4. May 3, 2017 #3
    The velocity of sphere as it leaves the wedge is not vertical . Are you sure factor of 4/9 in the first term is correct ?
  5. May 3, 2017 #4
    You're right. The factor should be ##\frac{1}{3}R## yielding maximum height to be ##\frac{5}{6}R##
  6. May 3, 2017 #5


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    Which is what I get.
  7. May 3, 2017 #6
    I also get same answer . It might be possible , the answer 10R/12 is misprinted as 10R/11.
  8. May 3, 2017 #7
    It might be the case, or the sphere might not be leaving at 60 degrees at all, since the wedge is also moving backward with some velocity(acceleration?).
  9. May 3, 2017 #8
    Come to think of it, I don't really have a justification that the sphere will move tangentially to the arc at the point it leaves, save for an analogy with the case of uniform circular motion, which is certainly not happening here, but if not tangential, then which direction would the sphere take? Is the answer key simply wrong? Am I overthinking this?
  10. May 3, 2017 #9


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    It will be tangential to the arc, with the arc as frame of reference. But the arc is moving, so it will not be at 60 degrees to the horizontal.
  11. May 3, 2017 #10
    Brilliant !

    This gives you the correct result .
  12. May 3, 2017 #11


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    Of course!
  13. May 4, 2017 #12
    Thank you so much!
  14. May 4, 2017 #13
    Thanks Vibhor and kuruman
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