A 12100 kg railroad car travels alone on a level frictionless track with a constant speed of 20.5 m/s. A 5850 kg load, initially at rest, is dropped onto the car. What will be the car's new speed? Please help!

Answer 1

Apply the law of conervation of linear momentum before and after the drop, along the direction of motion.

This does not seem like college level physics. In any case, you should be able to take it from there.

Answer 2

m₁v=(m₁+m₂)u
u= m₁v/(m₁+m₂)=