Two objects attached with a spring

[giokrutoi]

hey I guess I found solution I realized that I had to choose reference frame on center of mass
are my conclusion write

 

[haruspex]

hey I guess I found solution I realized that I had to choose reference frame on center of mass
are my conclusion write

Those images are simply unreadable. Please make the effort to type your equations in. Use parentheses, superscript (X²) and subscript (X₂) as appropriate.
Diagrams would be good, but draw them in black and try to get the lighting even on the page.

 

[giokrutoi]

I rewrote it
is it readable?

 

[haruspex]

I rewrote it
is it readable?

Yes, this readable.
Check the signs in the first equation on the right hand page.
Some places you write "due to centre of mass" where I think you mean "relative to centre of mass".
Other than that, you seem to be on the right track now.

 

[giokrutoi]

I think all signs are write
in relation to center of mass I think i have to subtract velocity of center of mass

 

[haruspex]

I think all signs are write
in relation to center of mass I think i have to subtract velocity of center of mass

Actually, it's not a sign issue, but something is wrong.
You defined v₁ as the velocity of the mass centre after releasing m₂, and v as the velocity of m₁ at the moment of release. Thus you should have m₁v=(m₁+m₂)v₁.
v and v₁ are fixed, but v₂ varies, so the equation cannot be right. Do you use that equation?

 

[ehild]

do you know Russian

I understand Russian. Show the original problem, please.

 

[giokrutoi]

if both v and v1 is fixed v2 should also be fixed but it varies so it's magnitude derived from formula of conversation of momentum it may be maximal

 

[giokrutoi]

here is Russian version

 

[haruspex]

if both v and v1 is fixed v2 should also be fixed but it varies so it's magnitude derived from formula of conversation of momentum it may be maximal

Wrong conclusion. Since v₂ varies the equation is wrong. If you want to change the equation by replacing the reference to v₂ to a reference to v_{2, max} then that could be OK, but you should show some justification. At the moment I cannot see where the equation comes from.

 

[giokrutoi]

as the initial momentum is zero it may stay constant so m1(V-v1)-m2v2=0 v2 is vmax and minus in formula is because they are moving in opposite way from each other

 

[giokrutoi]

are my conclusion write

 

[ehild]

here is Russian version

Thank you.

Two blocks connected by a spring are placed on a horizontal table. m1= 0.9 kg, m2= 1.6 kg, the spring constant is k=20 N/m.
Initially the spring is held compressed by 10 cm. Then the first block is released and when the spring gets relaxed, the second one is also released. What is the maximum magnitude of acceleration of the second block during the further motion of the system?

 

[giokrutoi]

I have stated that before
what can you say about solution

 

[ehild]

maybe m1a1=m2a2

That is right, for the magnitude of accelerations. And ma=F. What is the force acting on the blocks?

 

[giokrutoi]

kx

 

[giokrutoi]

That is right, for the magnitude of accelerations. And ma=F. What is the force acting on the blocks?

aren't solutions by momentum and center of mass write?

 

[ehild]

Do you mean "right" when you write "write"?
I can not follow your argument. The momentum of the whole system is constant after both blocks has been released, but it is zero in the center of mass frame of reference only. The spring force and the accelerations are the same in all inertial frames of reference. You have to give the maximum acceleration of block 2.
The center of mass moves with constant velocity and the blocks perform simple harmonic motion with respect to the center. During that motion, the spring gets compressed, gets relaxed and gets stretched. What is the maximum compression, and what is the spring force then? What are the accelerations of the blocks?

 

[giokrutoi]

maximal compression may be 10 centimeters and if that's true force should be 10 centimeter multiplied by k and acceleration of second mass should be this magnitude divided by m2
and yes I mean right when I write write

 

[ehild]

maximal compression may be 10 centimeters and if that's true force should be 10 centimeter multiplied by k and acceleration of second mass should be this magnitude divided by m2

Yes, but you need to calculate the compression in meters, and that should be multiplied by k=20 N/m.

 

[giokrutoi]

yes I forgot that but I don't think that it's the write solution because that 10 centimeters are maximal extension for both but not for m2

 

[jbriggs444]

maximal compression may be 10 centimeters and if that's true force should be 10 centimeter multiplied by k and acceleration of second mass should be this magnitude divided by m2
and yes I mean right when I write write

That simple analysis is attractive, but incorrect. After m1 is released and before m2 is released, m1 is undergoing simple harmonic motion. It is easy to see that the maximal compression is 10 cm shorter than relaxed and that the maximal extension is 10 cm longer than relaxed. This motion can optionally be maintained for many cycles before m2 is finally released.

When m2 is released, the spring is at its relaxed length. If m2 were still held stationary, the spring would eventually extend 10 cm longer than relaxed. But because m2 has been released, we are no longer assured that this will be the case.

One line of analysis that indicates that it cannot be the case is based on energy. Think initially about the state with mass m2 pinned in place. In its compressed state, the spring has a certain amount of potential energy. That amount of energy can be calculated based on the 10 cm number and the spring constant. In its extended state, the spring has that same amount of potential energy. In both cases the system has zero kinetic energy. All of the energy is in the spring.

The pin holding m2 in place is doing no work. It applies a force, but it is motionless. Following release the pin holding m2 in place still does no work. Accordingly, energy is conserved within the m1, m2, spring system both before and after the release of m2.

If m1 and m2 were both motionless then the spring would have to extend or compress to the full 10 cm in order for energy to be conserved. But as long as either are in motion, the spring cannot ever extend that far without violating energy conservation. But momentum is conserved following release of m2, so it can never be the case that both m1 and m2 are motionless following the release of m2.

 

[giokrutoi]

That simple analysis is attractive, but incorrect. After m1 is released and before m2 is released, m1 is undergoing simple harmonic motion. It is easy to see that the maximal compression is 10 cm shorter than relaxed and that the maximal extension is 10 cm longer than relaxed. This motion can optionally be maintained for many cycles before m2 is finally released.

When m2 is released, the spring is at its relaxed length. If m2 were still held stationary, the spring would eventually extend 10 cm longer than relaxed. But because m2 has been released, we are no longer assured that this will be the case.

One line of analysis that indicates that it cannot be the case is based on energy. Think initially about the state with mass m2 pinned in place. In its compressed state, the spring has a certain amount of potential energy. That amount of energy can be calculated based on the 10 cm number and the spring constant. In its extended state, the spring has that same amount of potential energy. In both cases the system has zero kinetic energy. All of the energy is in the spring.

The pin holding m2 in place is doing no work. It applies a force, but it is motionless. Following release the pin holding m2 in place still does no work. Accordingly, energy is conserved within the m1, m2, spring system both before and after the release of m2.

If m1 and m2 were both motionless then the spring would have to extend or compress to the full 10 cm in order for energy to be conserved. But as long as either are in motion, the spring cannot ever extend that far without violating energy conservation. But momentum is conserved following release of m2, so it can never be the case that both m1 and m2 are motionless following the release of m2.

I understood what you have said and may guess was very silly but how can I solve this problem i wrote conservation of momentum up in commentary's and some of them said it was wrong
so please if you can read my upper comments and tell me were my conclusions wrong or not

 

[SammyS]

yes I forgot that but I don't think that it's the write solution because that 10 centimeters are maximal extension for both but not for m2

Do you mean "right" when you write "write"?

I was going to point this out, but ehild has done so previously.

"Right" = Correct ≠ "Write"

Right ?​

 

[giokrutoi]

I was going to point this out, but ehild has done so previously.

"Right" = Correct ≠ "Write"

Right ?​

ok I made mistake sorry

 

[giokrutoi]

I was going to point this out, but ehild has done so previously.

"Right" = Correct ≠ "Write"

Right ?​

but what can you say about solute

 

[SammyS]

but what can you say about solute

Too many helpers at one time is not good.

ehild and briggs are excellent.

 

[giokrutoi]

Too many helpers at one time is not good.

ehild and briggs are excellent.

but whom should I trust they are saying opposite things

 

[jbriggs444]

but whom should I trust they are saying opposite things

We are saying identical things.

 

[giokrutoi]

We are saying identical things.

but you say that it is impossible 10cm to be maximal extention