A 500-kg sack of coal is dropped on a 2000-kg railroad flatcar which was initially moving at 3 m/s as shown. After the sack rests on the flatcar, the speed of the flatcar is:.

Before carrying the 500-kg sack, the flatcar has momentum

(2000 kg) (3 m/s) = 6000 kg•m/s

If momentum is conserved, and the sack has no initial momentum of its own, then the car-sack system has momentum

(2500 kg) v = 6000 kg•m/s

Solve for the new speed v :

v = (6000 kg•m/s) / (2500 kg) = 2.4 m/s

answer; i believe the correct answer is the college board;