A bullet of mass m and speed v is fired at, hits and passes completely through a

Question

A bullet of mass m and speed v is fired at, hits and passes completely through a pendulum bob of mass M on the end of a stiff rod of negligible mass and length L. The bullet emerges with a speed of v/2 and the pendulum bob just makes it over the top of the trajectory without falling backward in its circular path. Determine an expression for the minimum initial speed the bullet must have in order for the pendulum bob to just barely swing through a complete vertical circle. (Use any variable or symbol stated above as necessary.)

Explanation / Answer

Let's consider the pendulum bob of mass M first then we'll look at the system as a whole. When the bullet strikes the bob, the bob now has this kinetic energy which then transfers it into g.pe. once it reaches to the top

Conservation of energy for the pendulum bob gives:

Ki + Ui = Kf + Uf
(1/2)Mv^2 + 0 = Mg(2l) + 0 <- Pendulum reaches the height of twice its radius at the top

v^2 = 4gl
v = Sqrt(4gl)

What happened to momentum? Well, let's look at it from the beginning. The pendulum initially stood still before the bullet was fired, but right after we see that that it started to move and the bullet also had its own speed

mv1 + Mv2 = mv1 +Mv2
mv1 + 0 = m(v/2) + M(Sqrt(4gl))
v = v/2 + M*Sqrt(4gl)/m
v/2 = M*Sqrt(4gl)/m
v = 2*MSqrt(4gl)/m = 4Sqrt(gl)*M/m

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