# 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?
Homework Equations
p = mV

K = mV^2/2
Solving using Linear Momentum:

M vb2/2 = M g 2L
vb = 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 vb2/2 = M g 2L
vb = 2√(g L)

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

3 m v2/4 = M vb2/2
vb2 = 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?

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#### PeroK

Science Advisor
Homework Helper
Gold Member
2018 Award
Solving using Linear Momentum:

M vb2/2 = M g 2L
vb = 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 vb2/2 = M g 2L
vb = 2√(g L)

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

3 m v2/4 = M vb2/2
vb2 = 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...

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