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Instantaneous acceleration of blocks

  • #1
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Homework Statement



The blocks are of mass 2 kg shown is in equilibrium. At t = 0 right spring in fig (i) and right string in
fig (ii) breaks. Find the ratio of instantaneous acceleration of blocks?


Homework Equations





The Attempt at a Solution



In fig (i) before the string is cut,(2kx)sin37° = 2g , kx=g/sin37°

In fig (ii) before the string is cut,(2T)sin37° = 2g , T=g/sin37°

Is the instantaneous acceleration in the two cases same i.e ratio 1:1


I would be grateful if somebody could help me with the problem .
 

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Answers and Replies

  • #2
BvU
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I know physics isn't democracy, but I'm inclined to vote for 1:1 too!
Net force after cutting will be minus the missing force, and these are equal.
I'll be listening in to see if someone can prove me (us) wrong!
 
  • #3
PhysicoRaj
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What is the state of the spring at t=0? Stretched? Equilibrium? Compressed?
 
  • #4
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I know physics isn't democracy, but I'm inclined to vote for 1:1 too!
Net force after cutting will be minus the missing force, and these are equal.
I'll be listening in to see if someone can prove me (us) wrong!
I don't think so. When the string is cut, the acceleration of the mass is purely tangential. When the spring is cut, the acceleration is not purely tangential.

Chet
 
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  • #5
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I don't think so. When the string is cut, the acceleration of the mass is purely tangential. When the spring is cut, the acceleration is not purely tangential.

Chet
Excellent thinking!!!

Are you getting 25/24 ?
 
  • #9
BvU
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Impressed ! :redface:
 
  • #10
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Excellent thinking!!!

Are you getting 25/24 ?
Tanya, can you please show how you solved this problem?

Thanks!
 
  • #11
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Tanya, can you please show how you solved this problem?

Thanks!
In the first case i.e having springs

From equilibrium condition kx = 5g/3

After string is cut , the acceleration of the block will have both radial and tangential components .

at = 2gcos37° = 8g/5

ar = kx - 2gsin37° = 5g/3 - 6g/5 = 7g/15

∴ net acceleration a1 = 5g/3

In the second case i.e having strings

From equilibrium condition T = 5g/3

After string is cut , the acceleration of the block will be purely tangential i.e ar = 0

at = 2gcos37° = 8g/5

∴ net acceleration a2 = 8g/5


So a1/a2 = 25/24
 
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  • #12
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In the first case i.e having springs

From equilibrium condition kx = 5g/3

After string is cut , the acceleration of the block will have both radial and tangential components .

at = 2gcos37° = 8g/5

ar = kx - 2gsin37° = 5g/3 - 6g/5 = 7g/15

∴ net acceleration a1 = 5g/3

In the second case i.e having strings

From equilibrium condition T = 5g/3

After string is cut , the acceleration of the block will be purely tangential i.e ar = 0

at = 2gcos37° = 8g/5

∴ net acceleration a2 = 8g/5


So a1/a2 = 25/24
Excellent, thanks a lot Tanya! :smile:
 
  • #13
BvU
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@Balasruthi: post in the thread if you have a question, like 'How is it purely tangential?' -- or (since this is a very old thread) ask your question in a new thread you create.

Acceleration cannot be radial since the wire does not stretch: the tension in the wire cancels any radial component from an ecternal forcce such as gravity.
 
  • #14
haruspex
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Yes. I confirm that result.

Chet
An old thread, but more interesting than it looks.

We forever encounter questions involving unrealistic entities such as inelastic or inextensible strings, perfectly elastic collisions, etc. These are defensible if the result matches the limit as that ideal is approached; but it doesn’t in this question.

If we allow the string just a bit of extensibility (finite k) then the force exerted by the remaining string cannot instantly drop. The immediate acceleration of the mass is exactly the same in both cases. Taking the limit as the string's k value tends to infinity does not alter this.

Well done @BvU (post #2).
 
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