A 17000 kg railroad car travels along on a level frictionless track with a constant speed of 23.0 m/s. A 5100 kg additional load is dropped onto the car. What then will be its speed in meters/second?

Answer 1

m₁v=(m₁+m₂)u
u= m₁v/(m₁+m₂)=
=17000•23/(17000+5100)=17.7 m/s