Velocity of particle leaving wedge

[WubbaLubba Dubdub]

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?

 

[kuruman]

... also conserving momentum along the x-axis ...

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.

 

[Vibhor]

hence max height is $\frac{4}{9}R + \frac{R}{2} = \frac{17}{18}R$

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 ?

 

[WubbaLubba Dubdub]

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 ?

You're right. The factor should be $\frac{1}{3}R$ yielding maximum height to be $\frac{5}{6}R$

 

[kuruman]

You're right. The factor should be $\frac{1}{3}R$ yielding maximum height to be $\frac{5}{6}R.$

Which is what I get.

 

[Vibhor]

You're right. The factor should be $\frac{1}{3}R$ yielding maximum height to be $\frac{5}{6}R$

I also get same answer . It might be possible , the answer 10R/12 is misprinted as 10R/11.

 

[WubbaLubba Dubdub]

I also get same answer . It might be possible , the answer 10R/12 is misprinted as 10R/11.

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?).

 

[WubbaLubba Dubdub]

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?

 

[haruspex]

I don't really have a justification that the sphere will move tangentially to the arc at the point it leaves

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.

 

[Vibhor]

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.

Brilliant !

This gives you the correct result .

 

[kuruman]

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.

Of course!

 

[WubbaLubba Dubdub]

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.

Thank you so much!

 

[WubbaLubba Dubdub]

Thanks Vibhor and kuruman