Linear Momentum - Bullet hitting pendulum

AI Thread Summary
The discussion focuses on solving a physics problem involving a bullet hitting a pendulum, using both linear momentum and energy conservation principles. The linear momentum calculations yield a velocity formula of v = 4M√(gL)/m, which is confirmed as correct. However, the energy conservation approach leads to a different result, indicating a potential mistake in the calculations. The participant acknowledges the discrepancy and seeks clarification on where the error might have occurred in the energy conservation method. The conversation highlights the complexities of applying different physical principles to the same problem.
EEristavi
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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 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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EEristavi said:
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?
 
good point.
Maybe conservation of momentum made me confused here...
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...

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