- #1
DubbTom
- 3
- 0
Homework Statement
A bullet of mass m and speed v passes completely through a pendulum bob of mass
M. The bullet emerges on the other side of the pendulum bob with half its original
speed. Assume that the pendulum bob is suspended by a stiff rod 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 cycle
Homework Equations
m1v1i2 + m2v2i2 = m1v1f2 + m2v2f2
w=Δk+Δp
k = 1/2mv2
p = mgh
The Attempt at a Solution
initial kinetic energy of bullet = potential energy of bob @ max height + final kinetic energy of bullet
1/2mv2 = Mg2L + 1/2 m(1/2v)2
1/2mv2 - 1/8 mv2 = 2MgL
3/8mv2=2MgL
v2 = (16MgL)/(3m)
v = 4[(MgL)/(3m)]^(1/2)
Answer is [4M(gL)^(1/2)]/m
The key sets
1/2MVb2 = Mg2L
if Vb = velocity of the bob, then at max height the bob is not moving hence velocity is 0 and all of the energy that the bob carried at the bottom is now potential energy.
and then solves via the equation:
mv = MVb + mv/2
initial momentum = momentum of bob + momentum of bullet after going through bob
If everything that I have said about the professors method is true, then I feel like I am beginning to understand why the professors way works, but why doesn't my method come to the same conclusion.
Last edited: