Question

A 12000 kg railroad car is traveling at a 2 m/s when it strikes another 10000 kg railroad is at rest. If the car lock together what is the final speed of the two

Answer 1

The final speed is
`1.09m(m)/(s)`

Why?

To calculate the final speed, we need to know that we are working with an inelastic collision, it means that both bodies stick together after the colission. We can calculate the final speed of the bodies (if exists) using the following equation:

`m_(1)v{1}+m_(2)v_(2)=(m_(1)+m_(2))v_(f)`

We are given the following information:

`m_(1)=12000kg\\v_(1)=2(m)/(s)\\m_(2)=10000kg\\v_(2)=0(m)/(s)(at rest)`

So, substituting and calculating, we have:

`m_(1)v{1}+m_(2)v_(2)=(m_(1)+m_(2))v_(f)\\\\12000kg*2(m)/(s)+10000kg*0(m)/(s)=(12000kg+10000kg)*v_(f)\\\\v_(f)=(24000(kg.m)/(s) )/(22000kg)=1.09m(m)/(s)`

Hence, the final speed is 1.09 m/s.

Have a nice day!