Can Somebody check my center of mass answer?

AI Thread Summary
The discussion revolves around a physics problem involving a railroad flatcar and a man running on it. Participants confirm that the center of mass velocity remains constant due to the absence of external forces. The equations presented aim to relate the initial and final velocities of the car and the man, with a focus on converting relative velocities to absolute values. There is a correction regarding the use of an equal sign in the equations, which is acknowledged by the original poster. Overall, the calculations and reasoning provided are validated by peers in the thread.
toesockshoe
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Homework Statement


A railroad flatcar of mass Mc can roll without friciton along a straight horizontal track as shown in the diagram. Initially a man of mass Mm is standing on the car which is moving to the right relative to the ground with a speed of vi. What is the change in velocity of the car relative to the ground if the man runs to the left so that his speed relative to the car is given as vmc just before he jumps off at the left end?

the diagram is here: #5: http://nebula.deanza.edu/~Newton/4A/4AHWSet7.html

Homework Equations



x_{icm}=x_{fcm}

The Attempt at a Solution


Because there are no external forces, we can say that the center of mass velocity is always constant correct?[/B]
so v_{icm}=v_{fcm}

\frac{m_cv_{ci}+m_m+v_{ci}}{m_c+m+m} + \frac{m_cv_{cf}+m_mv_{mf}}{m_c+m_m}

we know that the initial velocity of the car and man are teh same. the unknowns in that equation are final velocity of the car and the final velocity of the man... we are given the final velocity of the man respect to the car, so we need to convert it to the final velocity of the man respect to the ground using the following mnemonic:

v_{mc} = v_{mg} + v_{gc} those should be vectors but idk how to do latex vectors
v_{mc} = v_{mg} - v_{cg} i swapped the subscripts
we need Vmg... so
v_{mg} = v_{mc}+v_{cg}
v_{cg} is the same as v_{cfinal} or v_{cf} for short...
plug this back into formula i put on the 2nd line (the velocity equation)
also the masses in the denominator cancel out... so substituting v_cg (or v_cf) in for the equation gives us

v_{cf} = \frac{v_{ci}(m_c+m_m) - m_mv_{mcf}}{m_c+m_m}

i know this isn't the change in velocity, but u can just subtract it from the given initial car speed. is this correct?
 
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toesockshoe said:

The Attempt at a Solution


Because there are no external forces, we can say that the center of mass velocity is always constant correct?[/B]
so v_{icm}=v_{fcm}

\frac{m_cv_{ci}+m_m+v_{ci}}{m_c+m+m} + \frac{m_cv_{cf}+m_mv_{mf}}{m_c+m_m}

I'm sure you meant to have an equal sign instead of the middle plus sign.

Your work looks correct. I agree with your result.
 
TSny said:
I'm sure you meant to have an equal sign instead of the middle plus sign.

Your work looks correct. I agree with your result.
yes yes, that should be an = sign. thanks!
 
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