Question

A man (weighing 915 N) stands on a long railroad flatcar (weighing 2805 N) as it rolls at 18.0 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 40.00 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?

For Further Reading

* Physics - bobpursley, Sunday, February 25, 2007 at 5:25pm

Use conservation of momentum. I will be happy to critique your thinking.

* Re: Physics - COFFEE, Sunday, February 25, 2007 at 11:19pm

Ok, so I use mv = m1v1 + m2v2??? And I would get m1 and m2 by dividing the weights by 9.8 m/s^2? Please explain.

For Further Reading

* RE: PHYSICS - bobpursley, Monday, February 26, 2007 at 6:10pm

The original momentum is..
(915+2805)g18
That is equal to the final momentum
915g*(-40+18) + 2805g(V+18)
solve for V.

I did:
(915+2805)g(18)=(915)g(-40+18)+(2805)g(v+18)
g cancels out on both sides.
I ended up with 13.04 m/s and this is the wrong answer. Why???

For Further Reading

* Re: PHYSICS - bobpursley, Tuesday, February 27, 2007 at 7:26am

Is there anything wrong with my thinking?

* Re: PHYSICS - COFFEE, Tuesday, February 27, 2007 at 9:58pm

Not that I can see but it's telling me that it's the wrong answer :( I don't know how that can be, your reasoning makes sense to me!

Answer 1

for momentum after, the man has a momentum of his own that is -40m/s. but there is also the combined mass of the man and cart which has its own momentum.