Velocity of Bead Relative to Wedge on a Smooth Hemispherical Surface

[Sanchayan Dutta]

Homework Statement

A bead of mass m kept at the top of a smooth hemispherical wedge of mass M and radius R is gently pushed towards right.As a result,the wedge slides due left.Find the magnitude of velocity of bead relative to the wedge.

Homework Equations

$$MV=m(v\cos(\theta)-V)$$
and,

$$mgR(1-cos(\theta))=(1/2)mv^2+(1/2)MV^2$$ i.e.
$$mgR(1-cos(\theta))=(1/2)mv^2+(1/2)M(\frac{mvcos(\theta)}{m+M})^2$$

The Attempt at a Solution

On solving the two equations I get $$v=\sqrt{\frac{2gR(1-cos(\theta))(m+M)^2}{(M+m)^2+Mm(cos(\theta))^2}}$$

But this answer is wrong according to my book.Where am I going wrong?

 

[mfb]

How do you define v? You seem to use two different definitions in the first two equations.

 

[Sanchayan Dutta]

How do you define v? You seem to use two different definitions in the first two equations.

Oh that's my mistake.So in the second equation i should use $$\sqrt{(vcos(\theta)-V)^2+(vsin(\theta))^2}$$ instead of v.Right?

 

[mfb]

That would make it consistent, yes.