What force does the safety stop apply to the carts

Question

A full mining car gets lose and rolls down a hill. The cart was 50 m up the hill the cart hits a second empty cart at the bottom of the hill. The two carts lock together and travel and hit the safety stop at the end of the track which stops the carts in 0.250 s. The full cart has a mass of 985 kg and an empty cart has a mass of 150.0 kg. What force does the safety stop apply to the carts?

Answer 1

M = 985 kg

m = 150.0 kg

h = 50 meters

Δt = 0.250 s

 

At bottom of hill, M has initial speed

 

v = √(2gh)

 

and initial momentum

 

Mv = M√(2gh)

 

Momentum is conserved in the collision with m at the bottom of the hill.

 

M√(2gh) = (M + m)u

 

u = M√(2gh)/(M + m)

 

Stopping acceleration is

 

a = Δv/Δt = (0 - u)/Δt = -u/Δt

 

Stopping force is

 

F = (M + m)a

 

Plug in the numbers step by step and get F in newtons.  F will be negative because the acceleration is negative.  You should be able to finish from here.