# Speed of freight car (momentum transfer)

• MuIotaTau
In summary, the conversation discusses problem 3.9 from Kleppner & Kolenkow's book on Intro to Mechanics, which involves finding the speed of a freight car after a force is applied and sand is released from the car. The conversation includes equations for momentum and force, as well as an attempt at a solution. Ultimately, it is determined that the equation F = [M+(m-(dm/dt)t](dv/dt) must be integrated to solve the problem.

## 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!

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

Princu said:
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?

Princu said:
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

MuIotaTau said:
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!

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.

1 person
Tanya Sharma said:

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:
Oh, no, I realize now how to integrate that! Thank you for all the help, everyone!

## What is the speed of a freight car?

The speed of a freight car varies depending on factors such as weight, incline, and external forces. It can range from 0 mph when stationary to over 60 mph when traveling downhill.

## How is the speed of a freight car calculated?

The speed of a freight car is calculated by dividing the distance traveled by the time it took to travel that distance. This formula is known as speed = distance / time.

## What factors affect the speed of a freight car?

The speed of a freight car is affected by factors such as weight, incline, external forces like wind or friction, and the power of the locomotive pulling the car. Inertia and momentum also play a role in determining the speed of a freight car.

## What is momentum transfer in a freight car?

Momentum transfer in a freight car refers to the transfer of kinetic energy from the locomotive to the car. When the locomotive starts or stops, it transfers its momentum to the car, causing it to accelerate or decelerate accordingly.

## How does the speed of a freight car impact its momentum transfer?

The higher the speed of a freight car, the greater the momentum transfer. This is because momentum is directly proportional to both mass and velocity. Therefore, a faster-moving freight car will have more momentum transfer than a slower-moving one.