Momentum Calculations from 'ThePhysicsClassroom'

[AmirTimur]

---

Members don't see this ad.

This Problem is straight from the website of "The Physics Classroom": http://www.physicsclassroom.com/Class/momentum/u4l2a.cfm

Chubby, Tubby and Flubby are astronauts on a spaceship. They each have the same mass and the same strength. Chubby and Tubby decide to play catch with Flubby, intending to throw her back and forth between them. Chubby throws Flubby to Tubby and the game begins. Describe the motion of Chubby, Tubby and Flubby as the game continues. If we assume that each throw involves the same amount of push, then how many throws will the game last?

Answer (highlight): The game will last two throws and one catch. When Chubby throws Flubby, the two will travel in opposite directions at the same speed. When Tubby catches Flubby, Flubby will slow down to half her original speed and move together with Tubby at that same speed. When Tubby throws Flubby towards Chubby, the greatest speed which Flubby can have is one-half the original speed. The game is now over since Flubby will never catch up to Chubby.

Now their explanation makes sense, but I would like to do it mathematically as well:

First, Chubby throws Flubby: We have m1=m2, Vinitial chubby=0, Vinitial flubby=0, so 0 = -m(Vfinal chubby)+m(Vfinal flubby). Minus because they move opposite directions.

Second, Tubby catches Flubby: We have m(Vfinal flubby) + m(Vinitial tubby) = 2m(Vfinal chubby and tubby). Since m(Vinitial tubby) = 0, we have V final (chubby and tubby) = 1/2 V final flubby (makes sense, right? because of conservation of momentum).

Third, Tubby throws Flubby: We have (2m)(V final flubby/2)=-m(V flubby)+m(V tubby). How can we find V flubby now? And how do we know, that an answer is: 'the greatest speed which Flubby can have is one-half the original speed.'?

Sorry for the lengthy explanation. If you made it through the end, you're the man (or the woman) and thanks so much. BTW, the physics classroom is an awesome resource, it has done an amazing job in crushing my misconceptions about physics.

 

[milski]

The total momentum of Flubby and Tubby has to be mV, no matter what they do to each other. That's because Chubby already has a momentum of -mV and does not interact with them.

One of the two possible things that Flubby and Tubby can do is move together, after Tubby catches Flubby and in that case their new speed obeys 2mVn = mV or Vn=V/2.

If Tubby throws Flubby again, they will have the same change of velocity but in different direction - that's the only way to keep their momentum constant. In general, their new speeds after the second depend only on how hard Tubby pushes Flubby. But we are told that all pushes are the same - that means that the change of moment (aka impulse) will be the same for all pushes.

We've already established that this change is mV. If Tubby pushes Flubby directly towards Chubby, Flubby's momentum will become mV/2-mV=-mV/2 and his velocity will be -V/2 or half of her initial velocity.

PS: Keeping track of who Tubby, Flubby and Chubby is is way too painful. 😀