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Ksju [112]
4 years ago
7

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:
Physics
2 answers:
wolverine [178]4 years ago
7 0
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.
vladimir2022 [97]4 years ago
4 0
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. 
<span> coal = 0*500 = 0.
 After:
 flatcar & coal = 2500*V
 then
</span> 2500*V <span>= 6000,
 V = 2.4 m/s. 
</span> Additional:
 the definition of an inelastic collision is that the objects "stick together" after colliding.
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