Linear Momentum - Bullet hitting pendulum

[EEristavi]

Homework Statement

bullet of mass m and speed v
passes completely through a
pendulum bob of mass M. The
bullet emerges with a speed
of v/2. The pendulum bob is
suspended by a stiff rod (not a
string) of length L, and negligible
mass. What is the minimum
value of v such that the pendulum bob will barely
swing through a complete vertical circle?

Relevant Equations

p = mV

K = mV^2/2

Solving using Linear Momentum:

M v_b²/2 = M g 2L
v_b = 2√(g L)

m v = m v/2 + M (2√(g L) )
v = 4 M √(g L) / m

Note: I see from the answers - that this is correct.

--------------

Next, I tried to solve it via Energy conservation point of view.

M v_b²/2 = M g 2L
v_b = 2√(g L)

m v²/2 = m v²/8 + k (Conservation Of Energy)
k = 3 m v²/4

3 m v²/4 = M v_b²/2
v_b² = 3 m v²/(4 M)

3 m v^²/(4 M) = 4 g L
v = 4 √(M g L) / (3 m)Answers are different.
Considering that linear momentum answer is correct - I must have made a mistake, when solving via conservation of energy.
Can anyone tell - where I'm making mistake?

 

[PeroK]

Solving using Linear Momentum:

M v_b²/2 = M g 2L
v_b = 2√(g L)

m v = m v/2 + M (2√(g L) )
v = 4 M √(g L) / m

Note: I see from the answers - that this is correct.

--------------

Next, I tried to solve it via Energy conservation point of view.

M v_b²/2 = M g 2L
v_b = 2√(g L)

m v²/2 = m v²/8 + k (Conservation Of Energy)
k = 3 m v²/4

3 m v²/4 = M v_b²/2
v_b² = 3 m v²/(4 M)

3 m v^²/(4 M) = 4 g L
v = 4 √(M g L) / (3 m)Answers are different.
Considering that linear momentum answer is correct - I must have made a mistake, when solving via conservation of energy.
Can anyone tell - where I'm making mistake?

Why do you think energy is conserved?

 

[EEristavi]

good point.
Maybe conservation of momentum made me confused here...