Inelastic collision affecting periodic motion

[laurenm02]

Homework Statement

A block os mass m is attached to a horizontal spring, which is attached to a wall. The block is oscillating without friction with initiation amplitude A0 and maximum velocity v0. When the block is at its maximum amplitude (and therefore instantaneously at rest), is it struck by a second identical block, moving to the left at velocity v0, and the collision is completely inelastic.

What is the ratio of amplitudes after and before the collision A / A0

Homework Equations

E = (1/2)kA0^2 = (1/2)kx^2 + (1/2)mv^2
m1v1 + m2v2 = (m1 + m2)v'

The Attempt at a Solution

I first set up my conservation of momentum equation to determine v' after the collision:
0 + mv0 = 2mv'
v' = v0/2

I then tried to use my energy equation to solve for final amplitude, where the energy of the system after the collision is equal to the kinetic energy from the incoming block and the potential energy of the original system.

E2 = (1/2)(2m)(v0/2)^2 + (1/2)kA0^2 = (1/2)kA^2

Rearranging and simplifying, I've narrowed it down to:
kA^2 = (mv0^2)/2 + kA0^2

But I don't know how to get the ratio from this equation. Is there something else I should be plugging into substitute and simplify?

 

[Suraj M]

kA^2 = (mv0^2)/2 + kA0^2

Try expressing $\frac{1}{2}kA_0^2$ in terms of $V_0$
You know the max velocity of the initial oscillatory(SHM) motion.

 

[laurenm02]

I'm sorry, I'm not really sure how to do that? How do I figure out the max velocity?

 

[Suraj M]

not really sure how to do that? How do I figure out the max velocity?

You don't have to

The block is oscillating without friction with initiation amplitude A0 and maximum velocity v0.

 

[laurenm02]

So because the blocks collide at maximum potential energy, can I relate that direction to maximum kinetic energy?

(1/2)kA0^2 = (1/2)mv0^2
So A0 = √(mV0^2)/k ?

 

[Suraj M]

You can directly substitute this

(1/2)kA0^2 = (1/2)mv0^2

into

kA^2 = (mv0^2)/2 + kA0^2

so you'll get $A$ from this
and you have $A_0$ from this

A0 = √(mV0^2)/k

 

[laurenm02]

I see! So my final ratio A/A0 = 1/2 ?

 

[Suraj M]

I see! So my final ratio A/A0 = 1/2 ?

Could you check that?
Post you're calculations, i feel it's wrong.

 

[laurenm02]

So Energy after the collision = (1/2)(2m)(v0^2/4) + (1/2)kA0^2
Substituting A0 = √(mV0^2)/k
Then, E = mv0^2/4 + mv0^2 = (1/2)kA^2
Simplifying: E = (mv0^2)/2 + 2mv0^2 = kA^2
So A^2 = (5mv0^2)/2

And because the coefficient of A0 is just one, the ratio* is (5/2)?

 

[Suraj M]

Then, E = mv0^2/4 + mv0^2 = (1/2)kA^2

the second term should be $\frac{1}{2}mv_0^2$

 

[laurenm02]

Ahh, so instead of 5/2, it's 3/2?

 

[Suraj M]

Ahh, so instead of 5/2, it's 3/2?

You didn't take the root .