Question

A 12404.7 kg railroad car travels alone on a level frictionless track with a constant speed of 11.6 m/s. a 5579.1 kg load, initially at rest, is dropped onto the car. what will be the car's new speed

Answer 1

The momentum must be conserved during both interactions

In the first case, the momentum was

P  = m*v

where m is the mass
v is the speed

P1 = 12404.7 kg *11.6 m/s = 143894.52 kg*m/s

P2 = (12404.7 +5579.1)kg* V2 = 143894.52 kg*m/s

Now, we find V2 (Car's new speed)

V2 = 143894.52 [kg*m/s]/17983.8[kg] = 8 m/s