Rainwater Accumulation on Moving Truck: Momentum Conservation

[CAF123]

Homework Statement

An open topped truck of mass $M$ is moving along a straight frictionless track at constant velocity $v_o$. At $t=0,$ the truck enters a rain shower and starts to collect rainwater. The rain falls vertically. Consequently, the trucks mass increases at a rate of $\lambda$ per second.

1) Show that its velocity $v(t)$ at time t after entering the shower is $$v(t) = \frac{Mv_o}{M + \lambda t}.$$

2)If the truck had been fitted with a drain hole to prevent rain from accumulating, find its velocity at time $t$, assuming that the water leaves the truck at zero relative velocity.

3) Compute the momentum of the system (the system being the truck plus water in 1) and truck in 2) as a function of time, and explain the qualitative difference between cases 1) and 2)).

Homework Equations

Equations will be derived from Momentum conservation

The Attempt at a Solution

1)Straightforward, but I have two questions. Momentum before entering rainstorm: $Mv_o.$ Momentum afterwards: $$M(t + δt) v(t + δt) = [M(t) + δ\dot{M}(t)][v(t) + δ\dot{v}(t)],$$ by Taylor expansion. Multiply out, ignore squared term, collect terms and simplify.

Two questions though: I took $\dot{v}(t) = 0$ here to attain the required 'show that'. Initially it was because I thought the truck is moving at constant speed, but now that I think about it , it is the velocity at time t we are interested in, so why does setting it to zero work? Also, I presume momentum is conserved in this case since if the mass increases, then the velocity of the truck must decrease.

2) and 3) I am struggling to start. I thought 2) might be similar to the decline in mass problem (ie. rocket eqn derivation), but then again the truck does not accumulate mass in the first place with the drain hole so I don't know if this is applicable.

Thanks for any advice

 

[TSny]

1)Straightforward, but I have two questions. Momentum before entering rainstorm: $Mv_o.$ Momentum afterwards: $$M(t + δt) v(t + δt) = [M(t) + δ\dot{M}(t)][v(t) + δ\dot{v}(t)],$$ by Taylor expansion. Multiply out, ignore squared term, collect terms and simplify.

Two questions though: I took $\dot{v}(t) = 0$ here to attain the required 'show that'. Initially it was because I thought the truck is moving at constant speed, but now that I think about it , it is the velocity at time t we are interested in, so why does setting it to zero work? Also, I presume momentum is conserved in this case since if the mass increases, then the velocity of the truck must decrease.

I'm not quite following your work here. For this problem, you don't have to derive a differential equation for v(t). You can just compare the horizontal momentum before the cart enters the rain with the horizontal momentum at any time t after the cart enters. There's no need to consider infinitesimal time increments, unless you want to as an exercise.

 

[tiny-tim]

Hi CAF123!

1)

Why are you using calculus?

This is a straightforward before-and-after equation. ​

2) and 3)

Hint: at time t, there is. loosely speaking, an amount of water λds moving along the track at speed v(s) for all times 0 < s < t

 

[mfb]

You do not need any derivatives, approximations or other stuff for (1). The truck starts at t=0, so M(0)*v(0) = M(t)*v(t). As you can calculate M(t), you get v(t) with a simple division.

In (2), the rain is accelerated to the current velocity of the truck with the rate given by the amount of water raining in. How does the influence the momentum of the truck?Edit: Oh, did not refresh the page. Where did the delete option go?

 

[CAF123]

In (2), the rain is accelerated to the current velocity of the truck with the rate given by the amount of water raining in. How does the influence the momentum of the truck?

The rain has become 'part' of the mass of the truck (or the truck system) so the momentum of the truck would decrease.

 

[CAF123]

Hi CAF123!


Why are you using calculus?

This is a straightforward before-and-after equation. ​

Hint: at time t, there is. loosely speaking, an amount of water λds moving along the track at speed v(s) for all times 0 < s < t

Hi tinytim,
So the mass of the truck is then $M(t = 0) + \lambda ds$ at some time s > 0. How is this taking into account the drain hole?

 

[TSny]

For this problem, you don't have to derive a differential equation for v(t). You can just compare the horizontal momentum before the cart enters the rain with the horizontal momentum at any time t after the cart enters. There's no need to consider infinitesimal time increments, unless you want to as an exercise.

I was only thinking about part 1 of the problem when I wrote this.

For part 2, calculus might be a good approach!

 

[CAF123]

I was only thinking about part 1 of the problem when I wrote this.

For part 2, calculus might be a good approach!

So I should integrate: $$v(t) = \int_0^t \frac{Mv_o}{M(t) + \lambda s} ds, $$ ie integrate over all s incremental time elements?

 

[TSny]

So I should integrate: $$v(t) = \int_0^t \frac{Mv_o}{M(t) + \lambda s} ds, $$ ie integrate over all s incremental time elements?

No, that doesn't look right. In fact, if you check, the right hand side does not have dimensions of velocity.

I'm not sure which part you are trying to solve here. No calculus needed in part 1. In part 2, the mass of the cart remains constant since the rain is not allowed to accumulate.

 

[mfb]

That equation is true for (1) only. And integrating a velocity over time would give a distance, not a velocity.

Edit: And too late again.

 

[CAF123]

If the mass is not allowed to accumulate (presumably due to the drain hole), then is this even simpler than part 1)? Do I assume that as the water hits the truck, it immediately leaves the truck? The bit about the zero relative velocity is a bit confusing and I am not sure how to incorporate that into my calculation.

 

[Saitama]

You can also solve the first question using $$F=\frac{dp}{dt}$$

 

[tiny-tim]

Hi CAF123!

Do I assume that as the water hits the truck, it immediately leaves the truck? The bit about the zero relative velocity is a bit confusing …

No, the water entering at time t has zero horizontal speed, but the water leaving has speed v(t).

(and, as TSny says, you'll probably need to use calculus)

 

[CAF123]

Hi CAF123!


No, the water entering at time t has zero horizontal speed, but the water leaving has speed v(t).

(and, as TSny says, you'll probably need to use calculus)

Ok, so the water that enters at some time t will contribute some mass to the truck while at the same time the water that was already in the truck is leaving. I don't know how to incorporate this into something that I can easily solve.

Also, when you said the water is leaving at speed v(t), why is this the case? Does the question not state the water leaves with zero relative velocity (I presume 0 velocity relative to the truck that is). Many thanks.

 

[tiny-tim]

Also, when you said the water is leaving at speed v(t), why is this the case? Does the question not state the water leaves with zero relative velocity (I presume 0 velocity relative to the truck that is). Many thanks.

yes, the speed of the truck is v(t) at time t, so the water leaves at the same speed, and presumably continues to move along the track (ahead of the truck) forever at that speed, v(t)

(so there's a trail of water in front of the truck, moving at all speeds greater than v(t) and less than v(0))

Ok, so the water that enters at some time t will contribute some mass to the truck while at the same time the water that was already in the truck is leaving. I don't know how to incorporate this into something that I can easily solve.

find the total momentum of everything at time t, including of that trail of water

as i said …

Hint: at time t, there is. loosely speaking, an amount of water λds moving along the track at speed v(s) for all times 0 < s < t

 

[haruspex]

Hint: at time t, there is. loosely speaking, an amount of water λds moving along the track at speed v(s) for all times 0 < s < t

I don't think that works. As the truck slows, the quantity of water draining in distance ds will increase. λdt = λds/v(s).

 

[tiny-tim]

hi haruspex!

i said "loosely speaking" so that he would know to check it for himself, and correct it if necessary

(oh, and my s is time, not distance)

 

[haruspex]

my s is time, not distance

Ah, quite so. Apologies.

 

[CAF123]

So initial momentum of system: $Mv_o$ (before t = 0)
Final momentum of system:$(M + \lambda t)v(t) + (-dm)(v(t))$, the first term deals with the momentum of the truck at time t and the second term the momentum of the rain. I am not quite sure what dm (the amount of water leaving the truck)would be.

I presume the rate of water leaving is less than that entering otherwise we wouldn't have any change in mass.

I am trying to work with your hints but I am not quite sure how to piece it together.

 

[haruspex]

So initial momentum of system: $Mv_o$ (before t = 0)
Final momentum of system:$(M + \lambda t)v(t) + (-dm)(v(t))$, the first term deals with the momentum of the truck at time t and the second term the momentum of the rain. I am not quite sure what dm (the amount of water leaving the truck)would be.

I presume the rate of water leaving is less than that entering otherwise we wouldn't have any change in mass.

I am trying to work with your hints but I am not quite sure how to piece it together.

No, in this part of the question you assume that the water leaves at pretty much the same rate as it enters, so the mass of the truck does not vary enough to worry about. But along the way, it is giving some of its momentum to the rain that passes through. Assume that in time dt a mass of rain λdt is given velocity v=v(t) then drained.

 

[CAF123]

So what I have now is:

Momentum before: $Mv_o$
Momentum after: $Mv(t) - \lambda t v(t)$.
So $v(t) = Mv_o/(M - \lambda t)$

Since it says the mass does not accumulate, I assumed that the water left the truck as soon as it entered the truck system. So the mass of the truck stays constant. However, since the rain now moves through the truck system, when it leaves the momentum of the truck decreases by the term (momentum of rain leaving) $\lambda t v(t)$. (The rain has the speed of the truck).
Since no external forces act in x direction, I can equate before = after and solve to get above v(t). Is it correct?

For the third question, $p = mv = (M + λt)((Mv_o)/(M + \lambda t)) = Mv_o$ for system A. For system B, similarly, $p = M((Mv_o)/(M-\lambda t)) = (M^2 v_o)/(M - \lambda t)$

Do these seem fine?

 

[tiny-tim]

So initial momentum of system: $Mv_o$ (before t = 0)
Final momentum of system:$(M + \lambda t)v(t) + (-dm)(v(t))$

no, if you use dm, you must have Mv(t) on the LHS, and Mv(t + dt) at the start of the RHS

(because the mass M of water in the truck stays the same)

then solve the differential equation

(if you don't use dm, you'll need an integral for the momentum of all the water that left at all speeds between v(t) and v(0))

 

[haruspex]

So what I have now is:

Momentum before: $Mv_o$
Momentum after: $Mv(t) - \lambda t v(t)$.

Momentum after what? After time t?
At time t, the speed is v(t) and the momentum Mv(t). In time dt, what mass of water passes through? What momentum does that water acquire? What is the speed of the truck at t+dt?

 

[CAF123]

Yes, momentum after time t. After time t, an amount of water $\lambda t$ of water has Left the truck with velocity v(t). So the momentum is $\lambda t v(t)$ of the rain. Since the rain leaves the truck system, put a negative in front and I get v(t) as in my last post. No?

 

[haruspex]

Yes, momentum after time t. After time t, an amount of water $\lambda t$ of water has Left the truck with velocity v(t).

No, only the most recent left with velocity v(t). Earlier water, at time s < t, left with velocity v(s) > v(t).