Static equilibruim in light spring

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In the scenario of a light spring with a 0.20 kg mass and a 0.10 kg mass suspended from it, the total weight acting on the spring is calculated to be 3 N. When the thread is burned, the tension from the cut string is removed, allowing the spring to exert its force without this additional load. At this moment, the upward acceleration of the 0.20 kg mass is determined to be 5.0 m/s², based on the forces acting on it. The analysis focuses on the forces just before the string is severed, as the spring's extension remains unchanged at that instant. Understanding these dynamics is crucial for solving static equilibrium problems involving springs and suspended masses.
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A light spring has a mass of 0.20 kg suspended from its lower end. A second mass of 0.10 kg is suspended from the first by a thread. The arrangement is allowed to come into static equilibrium and then is burned through. At this instant, what is the upward acceleration of the 0.20 kg mass. ( take g= 10 m s^-2)

Answer according to the booklet is 5.0 m s^-2

total weight acting on the first spring is 3 N ( i think so..)

Then i don't really know how to proceed...help..!
 
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You can do this : look at the force on the upper mass just before the string is burnt through since the spring continues to exert the same force as long as the extension does not change (which does not at that instant as the problem says), all you have to do is take out the tension of the now cut string and you will have the force and acc.
 
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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