- #1

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

[itex] x_{icm}=x_{fcm} [/itex]

## 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 [itex] v_{icm}=v_{fcm} [/itex]

[itex] \frac{m_cv_{ci}+m_m+v_{ci}}{m_c+m+m} + \frac{m_cv_{cf}+m_mv_{mf}}{m_c+m_m} [/itex]

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:

[itex] v_{mc} = v_{mg} + v_{gc} [/itex] those should be vectors but idk how to do latex vectors

[itex] v_{mc} = v_{mg} - v_{cg} [/itex] i swapped the subscripts

we need Vmg... so

[itex] v_{mg} = v_{mc}+v_{cg} [/itex]

[itex] v_{cg} [/itex] is the same as [itex] v_{cfinal} [/itex] or [itex] v_{cf} [/itex] 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

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

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