A 500 kg sack of coal is dropped vertically onto a 2000 kg railroad flatcar which was initially moving at 3 m/s toward the right. Just after the sack comes to rest on the flatcar what is the speed of the flatcar ?
2006-11-21 13:19:01 · 4 answers · asked by Maria M 1 in Science & Mathematics ➔ Physics
Please show how you get the answer. Thanks
2006-11-21 13:25:24 · update #1
merv has the right answer.. just dont know how she got it
2006-11-21 13:35:23 · update #2
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 = 6000, V = 2.4 m/s.
Additional: the definition of an inelastic collision is that the objects "stick together" after colliding.
2006-11-21 13:38:15 · answer #1 · answered by pack_rat2 3 · 1⤊ 0⤋
It's either you have insufficient data or this is a trick question. By the way you have phrased the problem, the flatcar does not seem to be accelerating/decelerating. Therefore, s = 3 m/s.
If you were asking the speed of the flatcar WITH RESPECT TO THE FALLING SACK, you have to indicate the height from which the sack was dropped and the distance of the flatcar from the point of drop.
2006-11-21 13:29:54 · answer #2 · answered by Askhole Ninja 3 · 0⤊ 2⤋
2 2/5 m/s
Okay, okay. I kind of assumed that the problem was frictionless. By doing this, I took 2000(starting mass)/2500(finishing mass)*3m/s. It wouldn't matter how high the bag was dropped. As long as it's frictionless, the bag can be hurtling towards the flatcar.
2006-11-21 13:24:16 · answer #3 · answered by merviedz trespassers 3 · 0⤊ 2⤋
yes, this is inelastic collision in the x-direction
linear momentum is conserved, but the mechanical energy of the system not (energy difference goes into heat)
2006-11-21 14:02:36 · answer #4 · answered by oracle 5 · 0⤊ 0⤋