What is the velocity of the center of inertia after bar 1 breaks off the wall?

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Saitama
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Homework Statement


Two bars of masses ##m_1## and ##m_2## connected by a weightless spring of stiffness ##k## rest on a smooth horizontal plane. Bar 2 is shifted a small distance ##x## to the left and then released. Find the velocity of the centre of inertia of the system after bar 1 breaks off the wall.
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Homework Equations


The Attempt at a Solution


The bar 1 breaks off from the wall when the bar 2 just passes its original position but I don't have any idea about how to begin making the equations here. :confused:
 

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What will be the velocity of 1 when it breaks off the wall?
What will be the velocity of 2 when 1 breaks off the wall?(conserve energy)
 
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consciousness said:
What will be the velocity of 1 when it breaks off the wall?
What will be the velocity of 2 when 1 breaks off the wall?(conserve energy)

Conserving energy,
[tex]m_1v_1^2+m_2v_2^2=kx^2[/tex]
where ##v_1## and ##v_2## are the velocities of 1 and 2 when 1 breaks off the wall.

verty said:
What can you tell us about the momentum of the system at various times?
Do you ask me to conserve linear momentum? :confused:
 
Pranav-Arora said:
Conserving energy,
[tex]m_1v_1^2+m_2v_2^2=kx^2[/tex]
where ##v_1## and ##v_2## are the velocities of 1 and 2 when 1 breaks off the wall.

What will be the velocity of block 1 when it breaks off the wall?

Pranav-Arora said:
Do you ask me to conserve linear momentum? :confused:

No need to conserve momentum.Linear momentum will not be conserved till block 1 remains in contact with the wall.
 
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Tanya Sharma said:
What will be the velocity of block 1 when it breaks off the wall?

If I assume ##v_1=0##, I get the right answer. Thanks! :smile:
 
Pranav-Arora said:
If I assume ##v_1=0##, I get the right answer. Thanks! :smile:

Great ...:approve:

But do you understand why should you do so ?
 
Tanya Sharma said:
But do you understand why should you do so ?

As the question asks ##v_{CM}## at the instant 1 breaks off the wall, so we can assume ##v_1=0## and I don't see how will ##v_1## reach a finite value in an instant, right?
 
When block 2 is released,it will move towards right,but block 1 will remain pushed against the wall.Block 2 will reach a point A such that the spring is in its natural uncompressed length.The force on block 1 at this point will be zero .

Now due to the momentum gained,Block 2 will continue moving towards right.Just as block 2 crosses point A,there will be a stretching in the spring which will pull the block 1 off the wall with an initial speed zero.
 
Tanya Sharma said:
When block 2 is released,it will move towards right,but block 1 will remain pushed against the wall.Block 2 will reach a point A such that the spring is in its natural uncompressed length.The force on block 1 at this point will be zero .

Now due to the momentum gained,Block 2 will continue moving towards right.Just as block 2 crosses point A,there will be a stretching in the spring which will pull the block 1 off the wall with an initial speed zero.

Thanks!

I had somewhat similar scenario in my mind when I began with this question. I figured out that 1 will separate from wall when 2 just passes A. I did not think of ##v_1=0##. Should have been more careful. :rolleyes: