# Homework Help: Space Station,Exploding Object

1. May 17, 2012

### ZxcvbnM2000

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I solved questions a) and b) no problem but the thing is that in c)

i said that B will hit the end first so at first the tube will move with a speed Vtube1 ( Vb=x/t).So t=23.25s . After some time which i found to be 47.17 s , object A will hit the other end so the tube will now move by δV = Va-Vb .

But in the solutions it says that the tube's velocity will be zero!!!

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2. May 17, 2012

### tiny-tim

Hi ZxcvbnM2000!
How can it take 23 s to go only 1 m ?

A will hit first.

(It would help in future if you show all your calculations at the start.)

3. May 17, 2012

### ZxcvbnM2000

Explosion (system of objects A&B) : 0 = 0.8*1 + 3* ub so ub = -0.26 m/s .

Object A : Ma*ua = (Ma+Mtube)*Vtube <=> Vtube = 0.0615 m/s

Object B : Mb*ub=(Mb+Mtube)*V'tube so V'tube=-0.052 m/s

And because this is space we're talking about the movement is linear so V=x/t for each object and we can find the times until they hit the ends.

What do you mean conservation of momentum ?
This is really not intuitive for me!

So you are saying that if we consider the tube as system how can we prove that it won't move ?

4. May 17, 2012

### tiny-tim

A yes; B no, because you need to use the new velocity and mass after A collides with the tube
if we consider the whole thing as the system, yes

5. May 17, 2012

### ZxcvbnM2000

Momentum Conservation between ( A&Tube) & B

so mb*ub + (ma + mtube)*V = (ma +mb +mtube)*V'

but we know that ma*ua + 0 =(ma +mtube)*V , also -maua=mbub

so V' = 0 right ?? So it won't move.

But i have a question as The tube is going to the left after colliding with A and B is going to the right doesnt this mean that B will cover less distance ??

6. May 18, 2012

### tiny-tim

quicker would be to compare before and after for the whole thing
yes, but so what? it won't affect the velocities, and the question doesn't ask for it!
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