Conceptual Physics: III

NOTE TO CHALLENGE MASTERS/MODERATORS:

This solution applies to any combination of throws as long as the forces involved are all equal, as stated in the Details section. Even if the second and the third astronauts swap places and the third astronaut pushes off of the second one and flings the second one into space and him/herself toward the first, same kinematics will apply as described below (the only difference being which astronaut went where).

If there is anything else that you'd like to dispute, go ahead, but don't just change the wording of the problem without my consent, please.

Thank you.

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SOLUTION:

To fully understand this, first see Conceptual Physics: I

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From Conceptual Physics I, we should already know that answer C (can't throw in space) is wrong; we don't need to "push off" of anything in order to push other objects. As Newton's Third Law states:

"Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first."

In other words,

To every action there is always an opposed equal reaction

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But, this is is really a Momentum Conservation problem rather than a Laws of Motion problem. So let's get into it:

The astronaut on the left pushes the astronaut in the middle, accelerating the middle astronaut towards the astronaut on the right with a certain speed (let's call it $s$ )while also accelerating oneself to the left with an equal amount ( $a=\frac{F_{net}}{m}$ - since $F$ and $m$ is the same for all the astronauts, $a$ is the same also.)

The score is 1 .

Now, when the middle astronaut collides with the astronaut on the right, they both continue in their line of motion toward right with half the velocity the middle astronaut was traveling before the collision (this is due to momentum distribution for both astronauts in an inelastic collision. See Conceptual Physics: II .

Then, when the right astronaut throws the middle astronaut, the middle astronaut travels at $\frac{1}{2}s$ to the left. This is because the right astronaut, along with the middle astronaut, was traveling at $\frac{1}{2}s$ to the right, so half of this acceleration went into canceling the rightward momentum and the other half generating the leftward momentum.

The score is 2 .

And so, the left astronaut is traveling left with speed $s$ , while the middle astronaut is hurling in the same direction, with half the speed, $\frac{1}{2}s$ , and never catching up with the astronaut. And thus, the gave only lasts two throws (or one successful one).

Extra credit: Which astronaut is traveling at the highest speed?

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Law of Conservation of Momentum:

$mv_{BeforeEvent}=mv_{AfterEvent}$

Since they're all stationary (with respect to each other), the net momentum of the three astronaut system is zero.

$mv_{Before}=0$

When the first push occurs, the net momentum also must be zero.

$mv_{Left}+mv_{Middle}+mv_{Right}=0$ .

Since the right astronaut is initially stationary,

$mv_{Right}=0 \Rightarrow mv_{Left}=-mv_{Middle}$ .

(Since masses are the same, velocities must be the opposite, as expected. (Newton's Third Law))

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We can look at the Middle-Right astronaut system for now. When the right astronaut catches the middle astronaut, they share the momentum:

(net $mv$ ) $_{before}$ = (net $mv$ ) $_{after}$

( $m\times v$ ) $_{before}$ = ( $2m\times V$ ) $_{after}$

By simple algebra, $V=\frac{1}{2}v$ . This makes sense because, since twice as much mass is moving after the collision, the velocity mst be half as much as the velocity before collision.

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Briefly for the second throw:

$mv_{MiddleBefore}+mv_{RightBefore}=mv_{MiddleAfter}+mv_{RightAfter}$ .

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$v$ is going to change by the same about in accord with Newton's Third and Second laws; that is, the astronaut pushes with a force and is pushed by a force equal and opposite to the push, and thus accelerates both oneself and the other astronaut by the same amount. Or, when we use $v$ as the initial speed of the push as reference,

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$(2m)(\frac{1}{2}v)=m((\frac{1}{2}v)+v)+m(Av)$

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Solving for A reveals $A=-\frac{1}{2}$ . So, the Middle Astronaut travels in opposite direction with half the initial velocity.

Since the velocity of the Left astronaut is still $v$ and the velocity of the middle astronaut is now $\frac{1}{2}v$ , the Middle astronaut is left forever floating in space.

Also note the net momentum of the system is always conserved: $mv_{Left}+mv_{Middle}+mv_{Right}=m(-v)+m(-\frac{1}{2}v)+m(v+\frac{1}{2}v)=0$ .

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Poor astronauts...

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But really, poor right astronaut...

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