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.