 85
 4
 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 v_{b}^{2}/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_{b}^{2}/2 = M g 2L
v_{b} = 2√(g L)
m v^{2}/2 = m v^{2}/8 + k (Conservation Of Energy)
k = 3 m v^{2}/4
3 m v^{2}/4 = M v_{b}^{2}/2
v_{b}^{2} = 3 m v^{2}/(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?
M v_{b}^{2}/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_{b}^{2}/2 = M g 2L
v_{b} = 2√(g L)
m v^{2}/2 = m v^{2}/8 + k (Conservation Of Energy)
k = 3 m v^{2}/4
3 m v^{2}/4 = M v_{b}^{2}/2
v_{b}^{2} = 3 m v^{2}/(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?