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???
Is there anything wrong with my thinking?
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!
To solve this problem, you need to use the principle of conservation of momentum. The total momentum before the man runs can be calculated by adding up the momentum of the man and the momentum of the flatcar.
The momentum of an object is given by the product of its mass and velocity. In this case, the mass of the man and the flatcar can be found by dividing their weights by the acceleration due to gravity (9.8 m/s^2).
So, the initial momentum is given by:
(915 N / 9.8 m/s^2 + 2805 N / 9.8 m/s^2) * 18.0 m/s
To find the final momentum, we need to consider the man's velocity relative to the flatcar. The man is running in the negative x direction at a velocity of 40.00 m/s relative to the flatcar. This means his velocity relative to the ground is the difference between the flatcar's velocity and his velocity relative to the flatcar:
Velocity of the man relative to the ground = velocity of the flatcar - velocity of the man relative to the flatcar
= 18.0 m/s - 40.00 m/s
= -22.00 m/s
So the final momentum is given by:
(915 N / 9.8 m/s^2) * (-22.00 m/s) + (2805 N / 9.8 m/s^2) * (V + 18.0 m/s)
Now, to solve for V, you equate the initial momentum to the final momentum:
(915 N / 9.8 m/s^2 + 2805 N / 9.8 m/s^2) * 18.0 m/s = (915 N / 9.8 m/s^2) * (-40.00 m/s) + (2805 N / 9.8 m/s^2) * (V + 18.0 m/s)
Solving this equation will give you the value of V, which represents the resulting increase in the speed of the flatcar.
If you got an answer of 13.04 m/s and it was marked as incorrect, it is possible that there was a calculation error. I recommend double-checking your calculations to ensure accuracy.