Understanding Momentum in Continuous Mass Flow Problems

AI Thread Summary
The discussion focuses on the momentum calculations for a freight car losing mass as sand flows out while a force is applied. There is confusion regarding the correct expression for momentum at time t, with some participants arguing that it should simply be (mc(t))v without adding the change in mass. The debate continues with differing opinions on the momentum at time t + Δt, with suggestions that it should incorporate both the change in mass and velocity. Ultimately, the consensus leans towards maintaining the fundamental equation F = m(dv/dt) as the correct representation of the system's dynamics. Clarifying these momentum expressions is essential for accurately solving continuous mass flow problems.
unscientific
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Homework Statement



From 2.2 Worked Examplehttp://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/continuous-mass-flow/MIT8_01SC_coursenotes19.pdf"

Emptying a Freight Car - A freight car of mass mc
contains a mass of sand ms
At t = 0 a constant horizontal force of magnitude F is applied in the direction of rolling and at the same time a port in the bottom is opened to let the sand flow out at the constant rate b = dms/ dt . Find the speed of the freight car when all the sand is gone. Assume that the freight car is at rest at t = 0 .

Homework Equations



It is written that the momentum of vehicle at time t is ( \Deltam + m ) v

The Attempt at a Solution



Shouldn't the momentum of the vehicle at time t be just (mc(t))v ? by adding \Deltam you're doing it twice!

I don't understand how they got the momentum for time t+ \Deltat too! shouldn't it be (mc - \Deltam)(v + \Deltav)?
 
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unscientific said:

Homework Statement



The Attempt at a Solution



Shouldn't the momentum of the vehicle at time t be just (mc(t))v ? by adding \Deltam you're doing it twice!

I don't understand how they got the momentum for time t+ \Deltat too! shouldn't it be (mc - \Deltam)(v + \Deltav)?

I would agree with you that the momentum at time t is (mc(t))v. But disagree with the momentum at t+ \Deltat. It should be
P(t+\delta t)=(m-\delta m)(v+\delta v)+\delta mv.

So the final result should remain as F=m\frac{dv}{dt}.
 
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