CM and Translational Motion problem

[The_Fritz]

A 280-kg flatcar 25 m long is moving with a speed of 6.0m/s along horizontal frictionless rails. A 95-kg worker starts walking from one end of the car to the other in the direction of motion, with speed 2.0m/s with respect to the car. In the time it takes for him to reach the other end, how far has the flatcar moved?


This problem falls into the category of center of mass and translational motion
M$$\rightarrowa$$_CM = $$\Sigma$$F_ext


Considering the rails are frictionless I figured both the mass of the flatcar and mass of the person were both irrelevant.
Time it takes for person to reach end of flatcar= (25m)(s/2.0m)=12.5 s
Distance traveled by flatcar = (6.0m/s)(12.5 s)= 75m
Is 75m the correct answer or am I missing a large part of this problem?

 

[Nate810]

Yes, 75 m is the correct answer. The mass of the flatcar and the worker are both irrelevant since there is no friction. The only relevant factor is the speed of the flatcar and the time it takes for the worker to walk from one end to the other.

 

[kerimek]

Your calculations for the distance traveled by the flatcar are correct. However, it is important to consider the concept of center of mass and how it relates to translational motion in this problem. The center of mass of the system (flatcar and person) remains at a constant velocity, as there are no external forces acting on the system. This means that the distance traveled by the flatcar is equal to the distance traveled by the center of mass of the system, which is also 75m.

Additionally, the masses of the flatcar and person are not irrelevant in this problem. They are necessary in calculating the center of mass of the system, which is a crucial concept in understanding the motion of the system. In this case, the center of mass of the system can be calculated using the formula:

x_CM = (m1x1 + m2x2) / (m1 + m2)

Where m1 and x1 are the mass and position of the flatcar, and m2 and x2 are the mass and position of the person on the flatcar. Plugging in the values given in the problem, we get:

x_CM = (280kg)(0m) + (95kg)(25m) / (280kg + 95kg) = 75m

This shows that the center of mass of the system is also located at 75m, which further supports the idea that the distance traveled by the flatcar is equal to the distance traveled by the center of mass. Therefore, your answer of 75m is correct.