# Variable mass system

1. Sep 15, 2016

### erisedk

1. The problem statement, all variables and given/known data
A pickup truck is driving at night through the rain. He drives onto a frozen lake. He immediately turns off his engine to save fuel (it wouldn't help anyway on frictionless ice) and lets the truck coast (move under no power). Unfortunately, the rain is accumulating in the back of his truck, at a constant rate of σ kg/sec. The given parameters of this problem are the truck's initial mass m0 and velocity v0 xˆ when it hits the lake at time t = 0, and the rate σ of rain accumulation. Also, the rain is falling straight downwards in the reference frame of the ground.

(a) Calculate the truck's speed v(t) as a function of time.

(b) New situation: As soon as he hits the frozen lake at t = 0, the trucker realizes that he can help himself by bailing the rain from the back of his truck. The trucker is very fit: he is able to get rid of every raindrop as it
falls by catching it in a bucket and hurling it backwards off the truck (–x direction) at speed u (relative to the truck). Calculate the truck's speed v(t) in this new situation.

(c) Final situation: This time, the trucker has some good luck → the rain is falling mostly downward but it also has a horizontal speed component of u (relative to the ground) in the +x direction. The trucker lets the rain accumulate as in part (a), hoping that the rain's forward speed component will help push him across the lake. Calculate the truck's speed v(t) in this last situation.

2. Relevant equations
Conservation of momentum

3. The attempt at a solution
I just feel like my answers are wrong, because they're not like the rocket problems that we get, and I didn't use any differentials or calculus of any sort.

(a) As there are no external forces in the x-direction, the momentum will be conserved in the x-direction.
Here, the system includes the truck and the rain.

$m_0v_0 = (m_0 + \sigma t)v$

$v = \dfrac{m_0v_0}{(m_0 + \sigma t)}$

(b) As there are no external forces in the x-direction, the momentum will be conserved in the x-direction.

$m_0v_0 = m_0v + (\sigma t)(v-u)$

$v = \dfrac{m_0v_0 + \sigma tu}{m_0 + \sigma t}$

(c) Again, conserving momentum in the x direction-
$m_0v_0 + \sigma tu = (m_0 + \sigma t)v$

$v = \dfrac{m_0v_0 + \sigma tu}{m_0 + \sigma t}$

2. Sep 15, 2016

### hmmm27

Are you sure ? I don't see any bouncing.
which bollixes up the picture of a raincloud hovering over the truck, matching speed.
You may have to rethink that.

Last edited: Sep 15, 2016
3. Sep 15, 2016

### haruspex

This is not valid.
It assumes all the rain was thrown off the back of the truck while it is doing the current speed, v. In fact, the speed of the truck has been changing. You will need a differential equation.

4. Sep 16, 2016

### erisedk

Would for (b) the answer then be something like this-
Let the truck's speed at a given instant be v. Let the trucker catch some raindrops of mass dm in his bucket. He throws the rain at a speed u relative to the car in the -x direction (v-u with respect to the ground)
So change in momentum of raindrops is dm(v-u)
⇒ change in momentum of car is -dm(v-u)
$dp = -dm (v-u)$
$\frac{dp}{dt} = -\frac{dm}{dt}(v-u)$
$\frac{dp}{dt} = -\sigma (v-u)$
$m_0 \int_{0}^{v} dv = \int_{0}^{t} -\sigma (v-u) dt$
$v = - \frac{\sigma (v-u) t}{m_0}$

5. Sep 16, 2016

### erisedk

And for (c)--
Let speed of the car at a given instant be v. Consider some rain falling into the car of mass dm. In the frame of the car, speed of the rain is u-v.
Change in momentum of rain = -dm (u-v)
Change in momentum of car = dm (u-v)
$\frac{dp}{dt} = \sigma ' (u-v)$
where
$\sigma ' = \frac{dm}{dt}$which is the rate at which rain hits the car.
$\sigma '$ is related to $\sigma$ by
$\sigma ' = \dfrac{\sigma (u-v)}{u}$
because although the rain is hitting the car with an x component of velocity (u) the relative speed of the balls and the car is only (u-v). We therefore have,
$m\dfrac{dm}{dt} = \dfrac{{(u-v)}^2\sigma}{u}$ --- 1
Also,
$\dfrac{dm}{dt} = \dfrac{(u-v)\sigma}{u}$

$m \dfrac{dv}{dm} = u - v$

$\int_{0}^{v} \dfrac{dv}{u-v} = \int_{m_0}^{m} \dfrac{dm}{m}$

$\ln{\dfrac{u}{u-v}} = \ln{\dfrac{m}{m_0}}$

$m = \dfrac{m_0u}{u-v}$

Substituting in 1
$\dfrac{m_0u}{u-v} \dfrac{dv}{dt} = \dfrac{{(u-v)}^2\sigma}{u}$

And then I solve for v from there. Is this right?

6. Sep 16, 2016

### haruspex

Fine until that last step. You integrated the right hand side as though v is a constant, but it is a function of time.
Go back to this point: $\frac{dp}{dt} = -\sigma (v-u)$ and treat it as a differential equation of v as a function of t. E.g. dp/dt is mdv/dt.

7. Sep 16, 2016

### erisedk

Oh, wow, sorry that was stupid and a result of sleep deprivation.

8. Sep 16, 2016

### erisedk

Is (c) correct?

9. Sep 16, 2016

### haruspex

I don't see a reason for using a different σ. The vertical rate of the rain (mass per unit horizontal area per unit time) is the same.

10. Sep 16, 2016

### erisedk

That's what I did initially. Then someone pointed out a very similar problem from a textbook (Morin) and said that what I did was wrong.
Here's the problem and the solution as published in Morin:

5.15.Propelling a car **

For some odd reason, you decide to throw baseballs at a car of mass Mthat is free to move frictionlessly on the ground. You throw the balls at the back of the car at speed u, and they leave your hand at a mass rate of σ kg/s (assume the rate is continuous, for simplicity). If the car starts at rest, find its speed and position as a function of time, assuming that the balls bounce elastically directly backward off the back window.

5.16.Propelling a car again **

Do the previous problem, except now assume that the back window is open, so that the balls collect inside the car.

5.15.Propelling a car

Let the speed of the car be v(t). Consider the collision of a ball of mass dm with the car. In the instantaneous rest frame of the car, the speed of the ball is uv. In this frame, the ball reverses velocity when it bounces (because the car is so much more massive), so its change in momentum is −2(uv)dm. This is also the change in momentum in the lab frame, because the two frames are related by a given speed at any instant. Therefore, in the lab frame the car gains a momentum of 2(uv)dm from each ball that hits it. The rate of change in momentum of the car (that is, the force) is thus

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where σ′ ≡ dm/dt is the rate at which mass hits the car. σ′ is related to the given σ by σ′ = σ(uv)/u, because although you throw the balls at speed u, the relative speed of the balls and the car is only (uv). We therefore have

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Note that vu as t → ∞, as it should. Writing this speed as u(1 − 1/(1 + 2σt/M)), we can integrate it to obtain the position,

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where the constant of integration is zero because x = 0 at t = 0. We see that even though the speed approaches u, the car will eventually be an arbitrarily large distance behind an object that moves with constant speed u (for example, pretend that your first ball misses the car and continues forward at speed u).

5.16.Propelling a car again

We can carry over some of the results from the previous problem. The only change in the calculation of the force on the car is that since the balls don’t bounce backward, we don’t pick up the factor of 2.
The force on the car is therefore

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where m(t) is the mass of the car-plus-contents, as a function of time. The main difference between this problem and the previous one is that this mass mchanges because the balls are collecting inside the car. From the previous problem, the rate at which mass enters the car is σ′ = σ(uv)/u. Therefore,

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We must now solve the two preceding differential equations. Dividing Eq. (5.128) by Eq. (5.129), and separating variables, gives33
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Note that m → ∞ as vu, as it should. Substituting this value of m into either Eq. (5.128) or Eq. (5.129) gives

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11. Sep 16, 2016

### erisedk

12. Sep 16, 2016

### haruspex

This is a different situation. Because the arrival of the balls is in the same direction as the movement of the car, and the throwing rate is constant, the arrival rate (relative to the car) is affected by the car's speed.
In the present problem, the vertical rate of the rain does not change; it says in c) that it also has a horizontal speed. The truck has no vertical component of speed, so the mass arrival rate does not change.

13. Sep 16, 2016

### erisedk

Oh, ok, I get it. So here's what it should be then:
Let speed of the car at a given instant be v. Consider some rain falling into the car of mass dm. In the frame of the car, speed of the rain is u-v.
Change in momentum of rain = -dm (u-v)
Change in momentum of car = dm (u-v)
$\frac{dp}{dt} = \sigma (u-v)$

$m\dfrac{dv}{dt} = \sigma (u-v)$ --- 1
Also,
$\dfrac{dm}{dt} = \sigma$

$m-m_0 = \sigma t$

$m = \sigma t + m_0$

$(\sigma t + m_0)\dfrac{dv}{dt} = \sigma (u-v)$

And then I integrate and solve for v.
$v = \dfrac{u\sigma t}{m_0 + \sigma t}$

Is this correct?

14. Sep 16, 2016

### haruspex

I agree with your equation (1), but the way you arrived at it is open to question. The mass is changing, so dp/dt is no longer just mdv/dt. As against that, you only considered the change in momentum of the added rain as equal and opposite to the change in momentum of truck+rain, i.e. you ignored the fact that the arriving rain becomes part of the total. These two cancel out, but not in a way that is clear to the reader.
More convincing would be:
dp=mdv+vdm=udm
You do not show the integration steps. The final answer cannot be right since the truck had speed at t=0. Maybe you left out a constant of integration.
Anyway, (c) is much easier than (b), not requiring any calculus.

By the way, you did not post your final answer for (b).

15. Sep 16, 2016

### erisedk

$v = u(1-e^{-\frac{\sigma t}{m_0}})$

I'm trying to correct (c)

16. Sep 16, 2016

### erisedk

Wait, I think I forgot $v_0$ in part (b) as well, and took all my limits of integration from 0 to $v$

17. Sep 16, 2016

### erisedk

Ignore

Last edited: Sep 16, 2016
18. Sep 16, 2016

### erisedk

Actually, for (c) I used the equation for rocket propulsion that's given in my homework. That's why I wrote equation 1.

19. Sep 16, 2016

### erisedk

After correcting part (c) for the constant of integration, I get
$v = \dfrac{m_0v_0 + u\sigma t}{m_0 + \sigma t}$

20. Sep 16, 2016

### erisedk

Part (b):

$v = u + (v_0 - u)e^{-\frac{\sigma t}{m_0}}$