How Does the Order of People Jumping Affect the Final Velocity of a Flatcar?

[Leo Liu]

Homework Statement

This is a statement

Relevant Equations

This is an equation

In the question above, the car's final velocity in the first case which is a group of people together jumping off the car, is less than that in the second case which is the same group of people jumping off the car one by one.

My work (subquestion b only):
$$M\Delta v=um \text{(from rocket equation)}$$
$$v_{i+t}-v_i=\frac{um}{M}$$
$$v_{j+1}=v_{j}+\frac{mu}{M_{j}}=v_j+\frac{um}{M+(N-j)m},$$
$$\text{j is the number of persons already jumped}$$
$$v_b=\sum_{j=0}^{N-1}\frac{um}{M+m(N-j)}$$

My questions are as follows:
1. For b, why is $N-1$ the upper bound of the series rather than $N$?
2. Could you provide an intuition for why the final velocity of the car in b is greater than that in a although both cases expel the same amount of mass?

Answer provided by my answer book:

Thanks in advance!

 

[haruspex]

why is N-1 the upper bound of the series rather than N?

j is defined as the number that have already jumped, so it ranges from 0 to N-1.

intuition for why the final velocity of the car in b is greater than that in a although both cases expel the same amount of mass?

Think about how it feels for the jumpers in the two cases. What difference would they notice? It may help to consider the jump taking a little time and analyse forces and accelerations.

 

[Leo Liu]

Think about how it feels for the jumpers in the two cases. What difference would they notice?

I believe that the j th jumper feels jumping off the car harder than the j+1 th jumper does, as both of them need to reach the same velocity in the car's frame and the mass on the car decreases.

Due to my ignorance, I can't really tell how they feel differently in the two cases.

 

[haruspex]

I believe that the j th jumper feels jumping off the car harder than the j+1 th jumper does, as both of them need to reach the same velocity in the car's frame and the mass on the car decreases.

Due to my ignorance, I can't really tell how they feel differently in the two cases.

On second thoughts, there is an easier way to think about it. Consider the momentum that each jumper retains in the lab frame. If they land on a frictionless surface, how would they be moving at the end?

 

[etotheipi]

It's quite interesting to note that in the discrete case jumping one at a time does result in a different change in velocity; an effect which disappears when you take the continuum limit. For instance, take the variable mass equation (here I use $\vec{v}_{\mathrm{rel}} = v_{\mathrm{rel}} \hat{x}$, so remember that my $v_{\mathrm{rel}}$ is negative here!)$$\begin{align*}\frac{dm}{dt} v_{\mathrm{rel}} &= m\frac{dv}{dt} \\

\int_{m_1}^{m_2} \frac{dm}{m} &= \int_{v_1}^{v_2} \frac{dv}{v_{\mathrm{rel}}} \\

\ln{\frac{m_2}{m_1}} &= \frac{\Delta v}{v_{\mathrm{rel}}} \implies \Delta v = v_{\mathrm{rel}} \ln{\frac{m_2}{m_1}}

\end{align*}$$That's just the Tsiolkovsky equation, and if $v_{\mathrm{rel}}$ is constant then all that matters are the final and end masses.

Here's another question for you to think about... imagine you have an array of $N$ identical cells connected in series, and a capacitor that you wish to charge. You can either connect the entire battery across the capacitor in one go, or alternatively you could connect one cell to the capacitor, and then the next cell (so now two cells across the capacitor), and then the next (so now three cells across the capacitor), one by one... which option results in the highest proportion of supplied energy being transferred to the capacitor? A hint is that $dU = V dQ$.

In both cases, the difference between the discrete and continuous cases can be visualised as the difference between a jagged, zig-zag curve on a graph and a smooth curve on a graph...

An interesting effect not limited to mechanics, worth thinking about!

 

[Leo Liu]

On second thoughts, there is an easier way to think about it. Consider the momentum that each jumper retains in the lab frame. If they land on a frictionless surface, how would they be moving at the end?

I am afraid that my intuition has failed to help me figure out the answer. I could write out several math equations to answer your question, yet it's not the method I have been seeking. Anyway, I would view this phenomenon as an interesting result of the equations to describe the motion. Thank you.

 

[Leo Liu]

It's quite interesting to note that in the discrete case jumping one at a time does result in a different change in velocity; an effect which disappears when you take the continuum limit.

Thanks for your reply. But I think the first case is also discrete because the entire combination of masses leaves the flatcar at the same instant :).

 

[etotheipi]

Both cases in your example are discrete, but I was comparing this to what would happen if you replaced everyone with an equal mass of, say, grasshoppers, each of which jumped off at the same relative velocity. And really what happens when you take the continuum limit $\Delta m \rightarrow 0$.

And you find that if the mass is leaving continuously, then there is actually no difference whether you dump it all at the start or eject it at some constant rate. As can be seen from the Tsiolkovsky equation, in this case the change of speed only depends on the initial and final masses (assuming constant ejection velocity w.r.t. the rocket).

And the question about which way to charge a capacitor in order to transfer the highest proportion of useful energy is completely analogous, in that you can charge it either all in one go, in several discrete steps (i.e. the voltage across the capacitor is a step function), or by continuously increasing the voltage. And it's interesting to see how much energy is transferred in each case, and then look at how it relates to the mechanics question!

 

[haruspex]

I am afraid that my intuition has failed to help me figure out the answer. I could write out several math equations to answer your question, yet it's not the method I have been seeking. Anyway, I would view this phenomenon as an interesting result of the equations to describe the motion. Thank you.

Then try this...
As each man jumps he gives impulse mu to the flatcar and whichever men remain in it. Although those other men gained a share of that, each only gives back mu when they jump. But if they all jump together, the whole of each mu goes to the flatcar.

 

[Leo Liu]

Then try this...
As each man jumps he gives impulse mu to the flatcar and whichever men remain in it. Although those other men gained a share of that, each only gives back mu when they jump. But if they all jump together, the whole of each mu goes to the flatcar.

I find this answer quite helpful.
If I understand it correctly, the following two equations are the impulses for the two cases.
$$\Delta P_1=\int_i^f F_{men}dt=M\Delta v$$
$$\Delta P_2=\int_i^f F_{men}dt=(M+nm)\Delta v$$
Where n denotes the number of persons remaining on the car.
It is obvious that the second one has the greater impulse than the first one does. I consider the excess impulse comes from the term $nm\Delta v$. When a men jumping off the flatcar in the second case, the momentum of that term along with the momentum the same as the first one, are released. In contrast, in the first case, the men jump off the car together, which does not result in transporting some momentum to the next guy. So I think this is why the final velocity of the car in the 2nd case is greater.
Is my interpretation valid?