*A 15000 kg railroad car travels on a horizontal track with a constant speed of 12 m/s. A 6000 kg load is dropped onto the car. What will be the car's speed?

8.57 m/s

Explanation:

From the question,

The total momentum of the car before the load was dropped = Total momentum of the call after the load was dropped.

Mu =  (M+m)V Equation 1

Where M = mass of the car, u = initial speed of the car, m = mass of the load, V = final speed of the load.

make V the aubject of the equation

V = mu/(M+m) Equation 2

Given: M = 15000 kg, m = 6000 kg, u = 12 m/s,

Substitute these values into equation 2

V = (15000×12)/(15000+6000)

V = 180000/21000

V = 8.57 m/s

c because the action was the roller coaster reaches bottom of hill (action) reaction would be blood tends to drain away from head

hope this !