How Does Releasing Coal Affect the Speed and Dynamics of a Railroad Car?

  • Thread starter Thread starter pleasehelpme6
  • Start date Start date
  • Tags Tags
    Car Coal Momentum
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a railroad hopper car releasing coal. The problem explores concepts related to momentum, impulse, forces, and energy transformations as the car interacts with the coal being dumped.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the implications of impulse and momentum changes as coal is released from the hopper car. There are varying interpretations of the velocity of the car after the coal is dumped, with some asserting it remains at 7 m/s while others suggest it should increase. Questions arise regarding the calculation of average force and energy conversions during the process.

Discussion Status

Some participants have provided calculations for impulse and average force, while others express uncertainty about the correct approach to certain parts of the problem. There is an ongoing exploration of the relationships between momentum, force, and energy, with no clear consensus reached on all points.

Contextual Notes

Participants are navigating the constraints of an online homework assignment, which may influence their reasoning and the interpretations of the problem setup. There is mention of specific time intervals and conditions affecting the calculations, such as the coal coming to rest and the method of release.

pleasehelpme6
Messages
62
Reaction score
0

Homework Statement



A railroad hopper car has a mass of 20000 kg when empty and contains 30000 kg of coal. As it coasts along the track at 7 m/s the hopper opens and steadily releases all the coal onto a platform below the rails over a period of 5 s.

a) How fast does the car travel after all the coal is dumped?

b) What is the horizontal impulse delivered to coal by the platform?

c) If the coal comes to rest 1 second after the dumping is finished, what was the average horizontal force on the platform?

d) If the coal fell 4 m, what is the total energy converted into heat in the process of the coal coming to rest?

e) Let's say that instead of the coal falling out of the bottom of the coal car, some very strong physics student threw it backwards from the car, so that the velocity of each piece of coal was zero just after it left her shovel. How fast does the car travel after all the coal is thrown instead of dumped?

Homework Equations



p = mv
pi =pf

The Attempt at a Solution



The answer for part a was 7 m/s, so apparently the platform was attached to the hopper, but I have no idea where to begin with solving any of the others. Please help.
 
Last edited:
Physics news on Phys.org
impulse is the change in momentum of a body,so if u calculate the velocity with which coal goes out and hit platform you know initial momentum of the coal before hitting platform after hitting its zero as it rests there... so you get impulse...
finding average force is change of momentum /time taken to change,
actually time derivative of momentum is force.. the average force came from there...
 
impulse is not zero, which is why its so difficult.

7 m/s is the velocity, so there should be NO impulse, but that's not the right answer.
 
well i solved for impulse, it was 210000, but I am still not getting part c
 
solved for part c...

i was only using 1 s as the interval, but i had to add that to the time over which the coal was dumped.

so average force = mass * average acceleration.
average acceleration = change in velocity / time

time =6, change in velocity = 210000
average force = 35000
 
finding average force is change of momentum /time taken to change,

so if coal comes to rest in 1 sec then averge force would be the change in momentum itself...
 
7 m/s is n't the velocity of car after dumping coal it should definitely increase
 
7 is correct, since it is not propelled off the back, there is no external force acting on the cart.

its an online assignment, trust me, that's the right answer.

im still unsure how to find the energies for part d though.
 
for d its the potential energy mgh that may convert to heat..for e apply conservation of momentum you'll get it
 
Last edited by a moderator:
  • #10
yup, just got it.

i had to use potential energy + kinetic energy, since it also had a vi = 7.
so mgh + 1/2mv^2 = 1912200.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 13 ·
Replies
13
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
14K
Replies
4
Views
2K