What is the solution to a box with a spring connected to it?

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The problem involves two blocks of mass M and 3M on a frictionless surface, with a spring between them. After a cord holding them together is burned, the 3M block moves at 2.00 m/s, prompting questions about the velocity of the M block, the system's original elastic potential energy, and momentum conservation. The solution reveals that total energy is conserved, equating the potential energy of the spring to the kinetic energy of both blocks. The original energy is attributed to the spring rather than the cord, and momentum conservation is explained despite initial stationary conditions. The discussion concludes with the realization that the spring's compression is not necessary for the calculations.
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[SOLVED] box with spring coneected

Homework Statement



2 Blocks of mass M and 3M are placed on a horizontal frictionless surface. A light spring is attatched to one of them and the blocks are pushed together with the spring between them. A cord initally holding the blocks together is burned; after that happens, the block of mass 3m moves to the right with a speed of 2.00m/s

a) What is the velocity of the block of mass M?

b)find system's original elastic potential E, taking the M= 0.350kg

c) Is the original energy in the spring or cord?

d) Is the momentum of the system conserved in the bursting-apart process?
How can it be with no motion beforehand and plenty of motion afterward?

http://img208.imageshack.us/img208/6215/boxwithstringsb5.th.jpg

Homework Equations



Ki= Kf

The Attempt at a Solution



SOLVED (figured it out)
 
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In this system the total energy is conserved. Therefore PE of the spring is equal to the total KE of M and 3M. Here spring constant or amount of compression is not needed. While calculating cancel M.
 
Oh..I just figured that out while waiting for a response to my post.

Thanks :smile:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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