Understanding Momentum Conservation in Simple Harmonic Motion

[Prabs3257]

Homework Statement

A block A of mass m is in equilibrium after being suspended from the ceiling with the help of a spring of force constant k. The block B of mass m strikes the block A with a speed v and sticks to it.The value of v for which the spring just attains its natural length is

Relevant Equations

Momentum conservation

I first got the velocity of the combined mass with conservation of momentum and as it was in the mean position the velocity can be written as v = wA ( w= angular frequency , A = amplitude ) as we have to take it back to natural length i put A as the initial extension but i am getting a wrong ans can someone please tell me what i did wrong . Please refer to my work in the attachment below

 

[etotheipi]

As soon as the second mass coalesces with the first, the new mass starts to perform simple harmonic motion about a new equilibrium position. Determine the position of this new equilibrium and then use that $\frac{v}{2} = \omega_2 \sqrt{A_2^2 - x^2}$ for the necessary new amplitude $A_2$ so that it just reaches natural length at the top of the motion.

 

[Prabs3257]

As soon as the second mass coalesces with the first, the new mass starts to perform simple harmonic motion about a new equilibrium position. Determine the position of this new equilibrium and then use that $\frac{v}{2} = \omega_2 \sqrt{A_2^2 - x^2}$ for the necessary new amplitude $A_2$ so that it just reaches natural length at the top of the motion.

Thanks a lot man i got the answer