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

AI Thread Summary
The discussion revolves around calculating the velocity of the center of inertia after bar 1 breaks off from the wall. It is established that bar 1 separates when bar 2 passes its original position, and the conservation of energy equation is used to relate the velocities of both bars. Participants clarify that linear momentum is not conserved while bar 1 is in contact with the wall, leading to the conclusion that assuming v_1 equals zero at the moment of separation is valid. This assumption simplifies the problem, allowing for the correct calculation of the system's dynamics. The key takeaway is that bar 1's velocity is zero when it breaks off, which is crucial for determining the center of inertia's velocity.
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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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What can you tell us about the momentum of the system at various times?
 
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,
m_1v_1^2+m_2v_2^2=kx^2
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,
m_1v_1^2+m_2v_2^2=kx^2
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.
 
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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:
 
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