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Question

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A small block of mass 2m initially rests on a track at the bottom of the circular. Vertical loop-the-loop shown above. which has a radius r. The surface contact between the block and the loop is frictionless. A bullet of mass m strikes the block horizontally with speed v0 and remains embedded in the block as the block and bullet circle the loop. Determine each of the following in terms of m, vo r, and g. The speed of the block and bullet immediately after impact The kinetic energy of the block and bullet when they reach point P on the loop The minimum initial speed vmin of the block and bullet are to successfully execute a complete circuit of the loop.

Explanation / Answer

we have,
initial speed of bullet = vo
mass of bullet = m
mass of block = 2m
initial speed of block = 0
let final speed of the system be v
From conservation of momentum,
mvo + 2m x 0 = (m+2m) x v
mvo = 3mv
v = vo/3

a. Final Speed of block and bullet = v = vo/3

b. Now we have,
Initial Kinetic Energy of block and bullet KEi= 0.5(2m+m)(v)^2
= 0.5 x 3m x vo^2/9
= mvo^2/6
Let Kinetic Energy at point P be K.
we have from Work Energy Theorem,
Change in Kinetic Energy = Net Work Done
K.E. - KEi = - (m+2m)gr (Potential Energy at height r)
K.E. - mv0^2/6 = -3mgr

Kinetic Energy at Point P = mv02/6 - 3mgr

c. To complete a successful loop, at top of circuit, we should have Normal Force N> = 0

(2m+m)g = mv2min/r

3g = v2min/r

vmin = (3gr)1/2

   Therefore minimum velocity = vmin = (3gr)1/2

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