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A m = 500 kg sack of coal falls vertically onto a m = 2000 kg railroad flatcar which was initially moving horizontally at 3 m/s. no external forces, including friction, act on the flatcar. after the sack rests on the flatcar, the speed of the flatcar is:

User Lenny D
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2 Answers

4 votes
This can be pointed out as the inelastic collision between a 2000 kg object moving at 3 m/s, and a stationary 500 kg object.
The total momentum of the objects before the collision would be equal to the momentum after collision.
So before collision: flatcar = 2000*3 = 6000 kg*m/s. coal = 0*500 = 0.
After collision: flatcar & coal = 2500*V = 6000, V = 2.4 m/s.
User Dima Mamchur
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3 votes
You can view this as an inelastic collision between a 2000 kg object moving at 3 m/s, and a stationary 500 kg object.
The total momentum of the objects before the collision equals their momentum after collision.
Before:
flatcar = 2000*3 = 6000 kg*m/s.
coal = 0*500 = 0.
After:
flatcar & coal = 2500*V
then
2500*V = 6000,
V = 2.4 m/s.
Additional:
the definition of an inelastic collision is that the objects "stick together" after colliding.
User Tom McQuarrie
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