Duplicate Question
The question on this page has been marked as a duplicate question.
Original 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...Asked by COFFEE
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.
The original momentum is..
(915+2805)g18
That is equal to the final momentum
915g*(-40+18) + 2805g(V+18)
solve for V.
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.
The original momentum is..
(915+2805)g18
That is equal to the final momentum
915g*(-40+18) + 2805g(V+18)
solve for V.
Answers
Answered by
FAISAL
0.95
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.