Can Somebody check my center of mass answer?

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SUMMARY

The discussion focuses on calculating the change in velocity of a railroad flatcar when a man runs to the left on it. The initial conditions include a flatcar mass (Mc) and a man mass (Mm) with an initial velocity (vi). The key equations used involve the conservation of momentum and the relationship between the velocities of the man and the flatcar. The final consensus confirms that the approach to solving for the final velocity of the flatcar (v_cf) is correct, and the user is advised to substitute the appropriate values to find the change in velocity.

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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 the 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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