Speed of freight car (momentum transfer)

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  • #1
MuIotaTau
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Homework Statement



This is problem 3.9 from Kleppner & Kolenkow's Intro to Mech.

A freight car of mass M contains a mass of sand of mass m. At t = 0, a force is applied in the direction of motion, and simultaneously sand is let out of the bottom of the car at a rate dm/dt. Assuming the car is at rest at t = 0, find the speed of the freight car when all the sand is gone.

Homework Equations



$$\vec{F} = \frac{\mathbb{d}\vec{p}}{dt}$$
$$\vec{p} = m\vec{v}$$

The Attempt at a Solution



I attempted to form the differential of the momentum of the system, as follows:

$$p(t) = \big(M + m - (\frac{dm}{dt})(t)\big)v$$
$$p(t + dt) = \big(M + m - (\frac{dm}{dt})(t + dt)\big)\big(v + dv\big)$$
$$dp = p(t + dt) - p(t) = -\frac{dm}{dt}(v dt + t dv + dv dt) + M dv + m dv$$

Dividing through by dt and neglecting a leftover dv, I get:

$$\frac{dp}{dt} = \big(M + m - t \frac{dm}{dt}\big)\big(\frac{dv}{dt}\big) - v \frac{dm}{dt}$$

Unfortunately, I have very little idea of what to do from here. Furthermore, thinking about the problem more makes me realize that I have no real idea of how to utilize the force F. It seems reasonable to place it into my last equation, but that seems a bit arbitrary to me and I can't imagine how I would even progress from there. I'm sure I'm making this more complicated than it should be, but still, some help would be greatly appreciated. Thanks!
 

Answers and Replies

  • #2
Princu
28
3
You should know that whenever mass is ejected from system,it applies a force whose magnitude is equal to=(dm/dt*v)
v=Relative velocity of ejected part will respect to system.Here v=0..So no force will apply on system when the sand is draining although mass of system will continuously decrease.
At any instant of time t
F(net)=(mass at time t)*a
F=(mg-ut)*a (u=dm/dt)
Put a=dv/dt and integrate the above equation under proper limits
 
  • #3
MuIotaTau
82
3
F=(mg-ut)*a (u=dm/dt)

I'm not sure I understand this step. Should it be F = mg - ut * a instead? And even then, where does this come from?
 
  • #4
gneill
Mentor
20,945
2,886
You should know that whenever mass is ejected from system,it applies a force whose magnitude is equal to=(dm/dt*v)
v=Relative velocity of ejected part will respect to system.Here v=0..So no force will apply on system when the sand is draining although mass of system will continuously decrease.
At any instant of time t
F(net)=(mass at time t)*a
F=(mg-ut)*a (u=dm/dt)
Put a=dv/dt and integrate the above equation under proper limits

Not in this case. The sand is just falling out the bottom and neither imparts nor absorbs an impulse to the train in the direction of motion. We can assume that the sand interacting with the ground when it hits the tracks does not alter the speed of the Earth :smile:
 
  • #5
Tanya Sharma
1,540
135
I attempted to form the differential of the momentum of the system, as follows:

$$p(t) = \big(M + m - (\frac{dm}{dt})(t)\big)v$$
$$p(t + dt) = \big(M + m - (\frac{dm}{dt})(t + dt)\big)\big(v + dv\big)$$
$$dp = p(t + dt) - p(t) = -\frac{dm}{dt}(v dt + t dv + dv dt) + M dv + m dv$$

Dividing through by dt and neglecting a leftover dv, I get:

$$\frac{dp}{dt} = \big(M + m - t \frac{dm}{dt}\big)\big(\frac{dv}{dt}\big) - v \frac{dm}{dt}$$

Unfortunately, I have very little idea of what to do from here. Furthermore, thinking about the problem more makes me realize that I have no real idea of how to utilize the force F. It seems reasonable to place it into my last equation, but that seems a bit arbitrary to me and I can't imagine how I would even progress from there. I'm sure I'm making this more complicated than it should be, but still, some help would be greatly appreciated. Thanks!

Your expression for P(t+dt) is incorrect,which leads to further errors.

The correct way of doing it is -

P(t) = [M+(m-(dm/dt)t]v

P(t+dt) = [M+(m-(dm/dt)t - dm](v+dv) + dm(v+dv)

dP = [M+(m-(dm/dt)t](dv)

or,dP/dt = [M+(m-(dm/dt)t](dv/dt)

Now dP/dt =Fext =F ,

So,F = [M+(m-(dm/dt)t](dv/dt)

Fdt/[M+(m-(dm/dt)t] = dv

Now,integrate with proper limits.
 
  • #6
MuIotaTau
82
3
Your expression for P(t+dt) is incorrect,which leads to further errors.

The correct way of doing it is -

P(t) = [M+(m-(dm/dt)t]v

P(t+dt) = [M+(m-(dm/dt)t - dm](v+dv) + dm(v+dv)

dP = [M+(m-(dm/dt)t](dv)

or,dP/dt = [M+(m-(dm/dt)t](dv/dt)

Now dP/dt =Fext =F ,

So,F = [M+(m-(dm/dt)t](dv/dt)

Fdt/[M+(m-(dm/dt)t] = dv

Now,integrate with proper limits.

Thank you, that makes much more sense. But now I have the problem of being unable to integrate the left hand side of the equation. I'm not entirely sure that integration is even possible, unless I'm just not seeing something crucial here.

##F## is a constant, right? So I can pull it out of the integral?
 
Last edited:
  • #7
MuIotaTau
82
3
Oh, no, I realize now how to integrate that! Thank you for all the help, everyone!
 

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