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WubbaLubba Dubdub
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Homework Statement
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)
[Paraphrased]
Homework Equations
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?
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