A railroad hopper car has mass 50000 kg whenempty and contains 30000 kg of coal.
ID: 1750199 • Letter: A
Question
A railroad hopper car has mass 50000 kg whenempty and contains 30000 kg of coal. As itcoasts along the track at 7 m/s the hopper opensand steadily releases all the coal onto a platform below the railsover a period of 5 s.
a) How fast does the car travel after all the coal isdumped?
I was thinking that since Pf=Pi, you could set 50000*v= 80000*7.
and solve for v... that was not the right answer. I alsotried various different equations, but they all seemed to work outto the same answer. Could someone help me?
The other questions to this problem were....
What was the momentum absorbed by the platform?
If the coal comes to rest 1 second after thedumping is finished, what was the average horizontal force on theplatform?
If the coal fell 5 m, what was the total energyconverted to heat in the process of the coal coming to rest?
Explanation / Answer
assuming that force of engine on the hopper car isconstant which keeps the mometum of the car constant ifP1 and P2 be the mometum of the car before and after droppingthe coal from the car P1 =P2 P1 = (M + m) v P2 = M v ' v = 7 m/sec v' = ? v' = (M+m)v / M M = mass of the empty hopper car m' = mass of coal v =velocity of hoppercar before dropping the coal v ' =velocity of hoppercar after dropping the coal. pluggin in the knwon values in the above relation wecan solve for v' . if the coal losses all the energy in just a sec , then enrgyrate at which the energy dissapated can be given as P = E / t = [Mv'^2 / 2] / 1sec if F be the average force of the hopper car on therail which is moving with velocity v before comming torest F = P / vRelated Questions
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