# Find V(t) of train car with changing mass.

oddjobmj

## 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.

Data: M = 1900 kg; F0 = 47000 N; b = 190 kg/s; t = 6.8 s.

F=m*a
V(t)=v0+a*t

## The Attempt at a Solution

M(t)=M+bt

F=ma=M$\frac{dv}{dt}$

F dt=M dv

dv=$\frac{F}{M}$dt=$\frac{F}{M=bt}$dt

Integrating both sides I get:

v=$\frac{F*ln((bt/M)+1)}{b}$

Plugging in my known values I find that v(6.8)=128.33 m/s which is incorrect. Any suggestions?

Thank you!

lep11
Newtons 2nd law says Ftot=dp/dt= dmv/dt=(dm/dt)*v+(dv/dt)m

1 person
oddjobmj
Thank you, lep11. I'm actually not sure how to solve that. I am very rusty with regard to differential equations.

Can I multiply through by dt to get Ftotaldt=vdm+mdv ?

Then I can integrate both sides to get Ft=2mv where m=M+bt

Ft=2v(M+bt)

v=$\frac{Ft}{2(M+bt)}$

I tried this and came up with 50.06 m/s which is also not correct.

Tanya Sharma

1 person
oddjobmj
I don't have the correct answer, unfortunately. It is an online based program where I have a certain number of attempts.

Homework Helper
Gold Member
Newtons 2nd law says Ftot=dp/dt= dmv/dt=(dm/dt)*v+(dv/dt)m
You have to be careful with that formulation. E.g. how would you represent the case where the coal is falling from a hopper which is itself moving?
Oddjobmj, the problem with your formulation is that it treats the added coal as already having the same forward velocity as the cart. Instead, just think about the total momentum provided by the force in time t and the total momentum obtained by cart plus its load of coal at that time.

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2 people
Tanya Sharma
oddjobmj...You should approach the variable mass system problems using momentum approach.

P(t) = M(t)v(t)

P(t+dt) = ( M(t)+dm )( v(t)+dv )

dP = M(t)dv+v(t)dm

or,dP/dt = M(t)dv/dt+v(t)dm/dt

Now dP/dt =F , dm/dt = b and M(t)=M+bt

F = M(t)dv/dt+bv(t)

F-bv(t) = M(t)dv/dt

dt/M(t) = dv/(F-bv(t))

Integrate with proper limits and you should end up with v=Ft/(M+bt) .

You have to be careful with that formulation. E.g. how would you represent the case where the coal is falling from a hopper which is itself moving?
Oddjobmj, the problem with your formulation is that it treats the added coal as already having the same forward velocity as the cart.

Does that mean we can apply dP/dt approach directly in variable mass problems only when the velocity of the outgoing/incoming mass(added coal) is same as that of the system(car+coal at time t) ?

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1 person
Homework Helper
Gold Member
Integrate with proper limits and you should end up with v=Ft/(M+bt) .
Yes, but it's much easier than that. No calculus is needed. Just think about the total momentum that has been gained by the cart and the coal in it at time t.
Does that mean we can apply dP/dt approach directly in variable mass problems only when the velocity of the outgoing/incoming mass(added coal) is same as that of the system(car+coal at time t) ?
Think about what the v dm/dt term means. It says some amount of mass has been added / created at speed v. Since in reality mass does not get created out of nothing, it must have already existed with some momentum. Taking the gain in momentum to be v dm/dt assumes it had no prior momentum (within the reference frame). In the present problem that works because we are taking the added coal to be initially at rest, but it does not work in general.

2 people
voko
Does that mean we can apply dP/dt approach directly in variable mass problems only when the velocity of the outgoing/incoming mass(added coal) is same as that of the system(car+coal at time t) ?

In theory, this approach is always valid. In practice, total momentum ## p(t) = m(t)v(t) ## only in special cases, in general ## p(t) ## is the sum (or integral) of various masses moving at various speeds. For example, if sand is ejected at relative velocity ##v_e##, then ## p(t) = \int_0^t (-m'(\tau)) (v(\tau) + v_e(\tau)) d\tau + m(t) v(t) ## and ## p'(t) = - m'(t) (v(t) + v_e(t)) + m'(t) v(t) + m(t) v'(t) = m(t) v'(t) - m'(t) v_e(t) ##.

2 people
Tanya Sharma
Yes, but it's much easier than that. No calculus is needed. Just think about the total momentum that has been gained by the cart and the coal in it at time t.

I know what you have been hinting at :) .This is an excellent way of thinking in the context of the problem.Let oddjobmj figure it out.

Think about what the v dm/dt term means. It says some amount of mass has been added / created at speed v. Since in reality mass does not get created out of nothing, it must have already existed with some momentum. Taking the gain in momentum to be v dm/dt assumes it had no prior momentum (within the reference frame). In the present problem that works because we are taking the added coal to be initially at rest, but it does not work in general.

Okay…So vdm/dt is the change in the momentum of the added/ejected mass ‘dm’ in inertial reference frame. For this reason ,in rocket equation we have (v-u)dm/dt instead of vdm/dt .

Suppose the cart and hopper both would have been moving at same velocity ‘v’.In that case the term representing change in momentum of the increased mass dm, vdm/dt would have been zero.

Is my understanding correct ?

1 person
Homework Helper
Gold Member
Okay…So vdm/dt is the change in the momentum of the added/ejected mass ‘dm’ in inertial reference frame. For this reason ,in rocket equation we have (v-u)dm/dt instead of vdm/dt .

Suppose the cart and hopper both would have been moving at same velocity ‘v’.In that case the term representing change in momentum of the increased mass dm, vdm/dt would have been zero.

Is my understanding correct ?
Yes. In the context of the OP, but in a reference frame where the coal in the hopper has a constant nonzero velocity u and mass is transferred to the cart at rate b, you could reason that the change in the momentum for hopper+coal is u dm/dt + m du/dt = -ub. Combining that with vb+m(t)dv/dt for the cart+coal we have F = (v-u)b+m(t)dv/dt.

1 person
oddjobmj
Sorry, I was away for a few days!

oddjobmj...You should approach the variable mass system problems using momentum approach.

P(t) = M(t)v(t)

P(t+dt) = ( M(t)+dm )( v(t)+dv )

dP = M(t)dv+v(t)dm

or,dP/dt = M(t)dv/dt+v(t)dm/dt

Now dP/dt =F , dm/dt = b and M(t)=M+bt

F = M(t)dv/dt+bv(t)

F-bv(t) = M(t)dv/dt

dt/M(t) = dv/(F-bv(t))

Integrate with proper limits and you should end up with v=Ft/(M+bt) .

Does that mean we can apply dP/dt approach directly in variable mass problems only when the velocity of the outgoing/incoming mass(added coal) is same as that of the system(car+coal at time t) ?

This worked, thank you! I was pretty darn close.