Where will the masses lose contact?

[Rikudo]

Homework Statement

A thin plate A of mass m is affixed on the upper end of a spring, lower end of which is affixed on the ground. In equilibrium, the spring is compressed by an amount Xo. Another thin plate B of mass 2m is dropped from a height 3Xo above plate A hits plate A, moves downwards together with the plate A and after reaching a lowest position, both plates rebound upwards. What maximum height above the initial position of the plate A will the plate B rise? (see figure 1)

Relevant Equations

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In order to solve this problem, we can make use of energy and momentum conservation to solve this problem. But, I'm currently having a difficulty to find out where exactly plate B will lose contact with A.

Here is what I'm thinking.
First, B will collide inelastically with A, and then they will move downwards together. After stopping for the first time, they will move up. Eventually,
mass A and B will reach their new equilibrium position. In my opinion, this is where they will start to lose contact because after passing the equilibrium position, plate A will have more acceleration downwards than B, and hence make B lose contact with A.

Strangely, the solution video implies that this will happen when the spring is in its natural length.
Where are the flaws in my reasoning?

 

[haruspex]

after passing the equilibrium position, plate A will have more acceleration downwards than B

When they lose contact, what will B's acceleration be?
What is the state of the spring when A has that acceleration?

 

[Rikudo]

When they lose contact, what will B's acceleration be?

g downwards.

What is the state of the spring when B has that acceleration?

um... the spring must be at its equilibrium.

 

[jbriggs444]

g downwards.

Yes. If they lost contact at this point then B would be accelerating downward at one gee.

um... the spring must be at its equilibrium.

This spring is at the position where it would be in equilibrium if it were at rest and if both A and B were resting upon it. But B is no longer resting on it. Only A.

Given this, in what direction will A alone be accelerating under the net effect of gravity and the spring.

 

[haruspex]

the spring must be at its equilibrium.

The spring, as an isolated entity, is at equilibrium when at its natural length, so I assume you mean when the spring plus A is at equilibrium. In that state, what is A's acceleration?

 

[Rikudo]

In that state, what is A's acceleration?

0. Because it is an equilibrium point.

EDIT: Ah. I forgot that B is still in contact with it. So: $3ma = kx_o -3mg$

 

[haruspex]

Compare your three answers:

where they will start to lose contact… plate A will have more acceleration downwards than B

When they lose contact, what will B's acceleration be?

g downwards

what is A's acceleration?

0. Because it is an equilibrium point.

 

[Rikudo]

Ah! So it is not at the equilibrium point!

In order to lose contact, plate A should have acceleration g downwards or greater. The only possible location is when the spring is in its natural length or stretched. Am I right?

 

[haruspex]

Ah! So it is not at the equilibrium point!

In order to lose contact, plate A should have acceleration g downwards or greater. The only possible location is when the spring is in its natural length or stretched. Am I right?

Right.