AP Physics Question: Conservation of Energy?

Question

A small block of mass 2m initially rests on a track at the bottom of the circular, vertical loop-the-loop, which as radius = r. The surface contact between the block and the loop is frictionless. A bullet of mass m strikes the block horizontally with initial velocity v and remains embedded in the block as the block and bullet circle the loop. Determine each in terms of m, v, r, and g

a. the speed of the block and bullet immediately after impact
b. the kinetic energy of the block and bullet when they reach a point on the loop-the-loop (stays on the track), which is half the height of the loop-the-loop. The height at this point = r.
c. the minimum initial speed of the bullet if the block and bullet are to successfully execute a complete circuit of the loop.

Answer 1

a) speed, s using conservation of momentum
m*v=3*m*s
s=v/3

b) The KEt of the combined at the top of the loop is the KE after collision less the PE due to the rise of 2*r
.5*3*m*v^2/9=3*m*g*2*r+KEt
KEt=3*m*(v^2/18-g*2*r)

c) The speed at the top of the track, vt, must be sufficient so that the centripetal force balances the force of gravity:

3*m*vt^2/r=3*m*g
vt^2=r*g

from above
KEt=3*m*(v^2/18-g*2*r)
which is expressed as
.5*3*m*vt^2=3*m*(v^2/18-g*2*r)
simplify and plug in vt^2
v=sqrt(g*r*45)

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