Why Does the Tube's Velocity Become Zero After Collisions?

[ZxcvbnM2000]

Homework Statement

Homework Equations

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!

can you please explain ?!

 

[tiny-tim]

Hi ZxcvbnM2000!

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 .

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.)

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

what about conservation of momentum?

 

[ZxcvbnM2000]

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

Object A : M_a*u_a = (M_a+M_tube)*V_tube <=> V_tube = 0.0615 m/s

Object B : M_b*u_b=(M_b+M_tube)*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 ?

 

[tiny-tim]

Object A : M_a*u_a = (M_a+M_tube)*V_tube <=> V_tube = 0.0615 m/s

Object B : M_b*u_b=(M_b+M_tube)*V'_tube so V'_tube=-0.052 m/s

A yes; B no, because you need to use the new velocity and mass after A collides with the tube

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 ?

if we consider the whole thing as the system, yes

 

[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 doesn't this mean that B will cover less distance ??

 

[tiny-tim]

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.

quicker would be to compare before and after for the whole thing

The tube is going to the left after colliding with A and B is going to the right doesn't this mean that B will cover less distance ??

yes, but so what? it won't affect the velocities, and the question doesn't ask for it!