Conservation of Momentum Set-Up

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SUMMARY

The discussion focuses on solving a conservation of momentum problem involving a freight car and coal. The parameters include a mass (M) of 1400 kg, an external force (F0) of 42000 N, and a coal inflow rate (b) of 150 kg/s over a time period (t) of 6.5 seconds. The initial approach incorrectly treated the problem as a rocket ship scenario, leading to an erroneous velocity calculation of 140.45 m/s. The correct approach involves recognizing the changing mass of the system as m(t) = M + bt, which adjusts the differential equation and leads to the accurate solution.

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Homework Statement



An empty freight car of mass M starts moving from rest with a constant applied force F0. At the same time, coal runs into the car at steady rate b from a hopper above the car at rest along the track. Calculate the velocity of the car as a function of time.

GIVEN: M = 1400 kg; F0 = 42000 N; b = 150 kg/s; t = 6.5 s.

Homework Equations



I have a problem setting up conservation of momentum problems; I was wondering if there are some basic steps or factors to always look at, and help me understand the anatomy of them.

The Attempt at a Solution



I tried treating this like a rocket ship problem-

Fexternal = F0 = d(mv)/dt

F0*dt = d(mv) -->

F0 = v*dM/dt + M*dv/dt = v*b + M*dv/dt

to differentiate--> F0 - v*b = M*dv/dt
--> (1/m)dt = (1/b)(1/F0/b - v)*dv

V(t) = [F0/b]*[1-e-b*t/m]

And plugging in I get 140.45 m/s which is not correct.

will someone help me understand the anatomy of the conservation of momentum, any helpful links are also appreciated. Thanks!
 
Last edited:
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Your problem is that you forgot that m(t) = M + bt, so when you separated variables, your LHS should have been (1/(M+bt))dt

Think that should fix it.

I got: << solution edited out by berkeman >>
 
Last edited by a moderator:
YEAH! Thank you for your help!
 

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