Minimum Velocity for Pendulum to Clear Arc

[warfreak131]

Homework Statement

A pendulum consists of mass M hanging at the bottom of a mass-less rod of length L with a frictionless pivot at its top. A bullet of mass m and velocity v collides with M and gets embedded. What is the smallest velocity v of the bullet sufficient to cause the pendulum with the bullet to clear the top of the arc.

Homework Equations

mv=(m+M)v'
KE=\frac{1}{2}(m+M)v'^2
PE=(m+M)gh

The Attempt at a Solution

First I found out the velocity of the moving block.

mv=(m+M)v'
\frac{mv}{(m+M)}=v'

and KE_{bottom}=PE_{top}

\frac{1}{2}(m+M)v'^2=(m+M)g(2L)

plug in v'

\frac{1}{2}(m+M)\frac{{m^2}{v^2}}{(m+M)^2}=(m+M)2gL

cancel out (m+M)

\frac{1}{2}\frac{{m^2}{v^2}}{(m+M)}=(m+M)2gL

cross multiply

{m^2}{v^2}={(m+M)^2}2gL

divide by m^2

v^2=\frac{{(m+M)^2}2gL}{m^2}}

v=\frac{(m+M)\sqrt{4gL}}{m}

v=\frac{(m+M)\sqrt{4}\sqrt{gL}}{m}

v=\frac{(m+M)2\sqrt{gL}}{m}

 

[rl.bhat]

At the top of the arc, the velocity of M is not equal to zero because it clears the top.

 

[warfreak131]

but is it more or less correct?

 

[rl.bhat]

The total energy at the top is
(m + M)*2gL + 1/2*(m + M)*v"^2
At the top acceleration v"^2/L = g. Or v"^2 = Lg. Sustitute this value in the above expression.
One more mistake.
Check the step from v^2 to v.