Solve Extension of Spring After Collision: 1kg & 50g Mass Merge

[roovs]

Spring with k=100N/m attached to the ceiling. Someone attaches mass of 1kg to the spring. After the spring reaches equilibrium another mass of 50g is dropped from a height of 2m above the 1st mass. After the collision of m1 and m2 the masses merge and continue to move.

a) Find extension of the spring 2s after the collision.
b) Find the maximal extension. Find amplitude and frequency.

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Ok, we can find the initial extension (when it reaches equilibrium point):

mg=k*$$\Delta$$l
$$\Delta$$l = 0.098 m

and the speed of the new mass right after the collision:

m1v1+m2v2=(m1+m2)v
v = 0.298 m/s

Now I am lost. How is it possible to solve the first question without finding the amplitude first?

Since the new mass (after the collision) starts to move from equilibrium starting phase is T/4 and we can replace cos in x(t)=Acos(wt+$$\phi$$) with sin. -> x(t)=Asin(wt)
Therefore extension of the spring after 2s should be: $$\Delta$$l + x(t) => $$\Delta$$l + Asin(wt)
Is this correct?

 

[Doc Al]

Hint: Consider conservation of energy.

 

[roovs]

So mechanical energy will stay the same after the collision? I thought it will change because it depends on angular frequency which in turn depends on mass.

 

[Doc Al]

So mechanical energy will stay the same after the collision?

Sure. After the collision, all forces are conservative: no mechanical energy is dissipated.

I thought it will change because it depends on angular frequency which in turn depends on mass.

The energy does not depend on angular frequency.

 

[roovs]

Oh, I had wrong understanding of involvement of energies in harmonic motion, apparently...
Thanks for you help Doc Al.