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miss_story
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Conversation of Momentum (2 problems)
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:
mA = mass of bullet = 20.0 g = .02 kg
mB = mass of block = 2.50 kg
VBi = initial velocity of the block = 0
d = distance block travels after collision = 2.50 m
u = mu or the coefficient of kinetic friction = .600
Vf = final velocity of masses = 0
Conservation of momentum when the collision is inelastic
mAVAi + mBVBi = (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)(mA + mB)Vf^2 - (1/2)(mA + mB)V^2 = -Fkd
where Fk is the force of kinetic friction, and Fk = uN and N = the normal force.
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:
VAi = (.02 + 2.50)(sqrt(2*.60*9.8*2.50))/.02
VAi = 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!
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Problem #2
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:
Vmf = the velocity of the bullet after collision = (1/3)Vmi
Since the pendulum barely remains along a circular path, the velocity of the block at the top of the circular path, VMtop = 0.
Cons of Momentum:
mVmi + MVMi = mVmf + MVMf
Cons of Energy (this system has potential gravitational energy and kinetic energy), applied after collision to the block:
(1/2)M(VMf)^2 + Mghi = (1/2)M(VMtop)^2 + Mghf
I set up my coordinate system so that y = 0 when the pendulum was at rest initially. So, that means hi = 0, and hf = 2L.
I assumed the initial velocity of the block before collision is zero because I assumed the block was at rest. So, VMi = 0
So, my equations become:
Cons of Momentum:
mVmi + MVMi = mVmf + MVMf
mVmi = mVmf + MVMf
and mVmf = (1/3)Vmi, so the equation now becomes:
mVmi = (1/3)Vmi + MVMf
(2/3)mVmi = MVMf
Vmi = (3/2)(MVMf)/m
Solving the cons of energy equation for VMf gives VMf = 2sqrt(gL)
Subbing that into the equation from the cons. of momentum gives:
Vmi = 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.
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:
mA = mass of bullet = 20.0 g = .02 kg
mB = mass of block = 2.50 kg
VBi = initial velocity of the block = 0
d = distance block travels after collision = 2.50 m
u = mu or the coefficient of kinetic friction = .600
Vf = final velocity of masses = 0
Homework Equations
Conservation of momentum when the collision is inelastic
mAVAi + mBVBi = (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)(mA + mB)Vf^2 - (1/2)(mA + mB)V^2 = -Fkd
where Fk is the force of kinetic friction, and Fk = 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:
VAi = (.02 + 2.50)(sqrt(2*.60*9.8*2.50))/.02
VAi = 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:
Vmf = the velocity of the bullet after collision = (1/3)Vmi
Since the pendulum barely remains along a circular path, the velocity of the block at the top of the circular path, VMtop = 0.
Homework Equations
Cons of Momentum:
mVmi + MVMi = mVmf + MVMf
Cons of Energy (this system has potential gravitational energy and kinetic energy), applied after collision to the block:
(1/2)M(VMf)^2 + Mghi = (1/2)M(VMtop)^2 + Mghf
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 hi = 0, and hf = 2L.
I assumed the initial velocity of the block before collision is zero because I assumed the block was at rest. So, VMi = 0
So, my equations become:
Cons of Momentum:
mVmi + MVMi = mVmf + MVMf
mVmi = mVmf + MVMf
and mVmf = (1/3)Vmi, so the equation now becomes:
mVmi = (1/3)Vmi + MVMf
(2/3)mVmi = MVMf
Vmi = (3/2)(MVMf)/m
Solving the cons of energy equation for VMf gives VMf = 2sqrt(gL)
Subbing that into the equation from the cons. of momentum gives:
Vmi = 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.
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