An automobile with a total mass of 1500 kg is placed on a moving flatbed railcar

Question

An automobile with a total mass of 1500 kg is placed on a moving flatbed railcar. (There are no additional horizontal forces acting on the system) The flatbed is 1.20 m above the ground. The railcar is moving eastward at a constant speed of 8.2 m/s and has a mass of 38000 kg. The car accelerates westward and when it leaves the flatcar it is traveling at 23.0 m/s relative to the ground. (a) What is the speed of the railcar when the car leaves? (b) What is the distance between the car and the west end of the flatcar, when the car touches the ground?

Explanation / Answer

The momentum is the combined mass of the railroad Mrr plus the mass of the car mc times the speed V o
initial momentum = (mc+Mrr)*Vo

When the car leaves, this is still the momentum.

Since the car and railroad are going in opposite directions, let make our unknown the postitve direction.

(mc+Mrr)*Vo = Mrr*Vf rr - mc Vc
solve this for velocity of the railroad
[(mc+Mrr)*Vrr + mc Vc]/Mrr

= vf rr = 9.43 m/s

(b) the car will fall 1.2m. In the time it takes to fall, it will be traveling horizontally away from the railroad at 23 m/s
speed = 0

distance = vo*t +(1/2)*a*t^2
t^2 = 2 distance/g = .245 s^2
t = 0.49s

Yes, if I drop something from a little higher than waist high, it takes about half a second to hit the ground

use distance equals rate times time for the horizontal. The combined rate is 23m/s + the new railroad speed = 32.43m/s
Distance = 32.43m/s*0.49s

= 16.05m or 16m

between the car and the west end of the flat car.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.