Calculating Bullet Speed for Pendulum Swing?

[miss_story]

Conversation of Momentum (2 problems)

Homework Statement

The problem:
"A 20.0 gram bullet is fired into a 2.50 kg wooden block initially at rest on a horizontal surface. After impact the block slides 2.50 m before coming to a rest. If the coefficient of kinetic friction between the block and the surface is .600, what was the speed of the bullet immediately after impact? Derive the necessary equations."

Given:
m_A = mass of bullet = 20.0 g = .02 kg
m_B = mass of block = 2.50 kg
V_Bi = initial velocity of the block = 0
d = distance block travels after collision = 2.50 m
u = mu or the coefficient of kinetic friction = .600
V_f = final velocity of masses = 0

Homework Equations

Conservation of momentum when the collision is inelastic
m_AV_Ai + m_BV_Bi = (mA + mB)V

Conservation of energy in the presence of a non-conservative force

Since this system only has kinetic energy, the formula is:

(1/2)(m_A + m_B)V_f^2 - (1/2)(m_A + m_B)V^2 = -F_kd

where F_k is the force of kinetic friction, and F_k = uN and N = the normal force.

The Attempt at a Solution

Here are links to the work I did:
Pg. 1
http://www.box.net/shared/8n4dx0bq6b
Pg. 2
http://www.box.net/shared/39qk8an8ae

I don't have the calculation in the written work, but this is what I did:
V_Ai = (.02 + 2.50)(sqrt(2*.60*9.8*2.50))/.02
V_Ai = 683 m/s

However, the answer that my teacher has the correct answer is 68.3, so my answer is off by a power of 10, but I don't know what I did wrong...I've done the problem twice (and if you look at my solution, it takes me awhile each time heh) and I still can't figure it out. Any help is appreciated!

-----------------------------------------------------------------------------------------------------------------------------------------------
Problem #2

Homework Statement

The problem:
"A bullet of mass m is fired into a wooden block of mass M, suspended from a rod (considered massless) of length L. The bullet passes through the block and comes out with a third of its original speed. What must be the minimum speed of the bullet if the pendulum is to swing 180 degrees, barely remaining on a circular path?"

Given:
V_mf = the velocity of the bullet after collision = (1/3)V_mi
Since the pendulum barely remains along a circular path, the velocity of the block at the top of the circular path, V_Mtop = 0.

Homework Equations

Cons of Momentum:
mV_mi + MV_Mi = mV_mf + MV_Mf

Cons of Energy (this system has potential gravitational energy and kinetic energy), applied after collision to the block:
(1/2)M(V_Mf)^2 + Mgh_i = (1/2)M(V_Mtop)^2 + Mgh_f

The Attempt at a Solution

I set up my coordinate system so that y = 0 when the pendulum was at rest initially. So, that means h_i = 0, and h_f = 2L.

I assumed the initial velocity of the block before collision is zero because I assumed the block was at rest. So, V_Mi = 0

So, my equations become:

Cons of Momentum:
mV_mi + MV_Mi = mV_mf + MV_Mf
mV_mi = mV_mf + MV_Mf

and mV_mf = (1/3)V_mi, so the equation now becomes:

mV_mi = (1/3)V_mi + MV_Mf
(2/3)mV_mi = MV_Mf
V_mi = (3/2)(MV_Mf)/m

Solving the cons of energy equation for V_Mf gives V_Mf = 2sqrt(gL)

Subbing that into the equation from the cons. of momentum gives:
V_mi = 3Msqrt(gL)/m

The answer my teacher has is (3/2)Msqrt(5gL)/m. I have no idea how he got 5gL under the sqrt root.

 

[aim1732]

Though I was too lazy to check your first problem---I think it's a calculation error----you did not correctly find the required velocity for pendulum to make a complete rotation in the second problem. You can not find out this speed by only conserving energy----the correct limiting condition is for tension in the thread and not the velocity of the pendulum to vanish. That gives you sqrt(5gl).
Hint:Try using a little dynamics as well as energy conservation.