# Inelastic collision affecting periodic motion

1. Mar 28, 2015

### laurenm02

1. The problem statement, all variables and given/known data
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

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

3. 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 in to substitute and simplify?

2. Mar 28, 2015

### Suraj M

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

3. Mar 28, 2015

### laurenm02

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

4. Mar 28, 2015

### Suraj M

You dont have to

5. Mar 28, 2015

### 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 ?

6. Mar 28, 2015

### Suraj M

You can directly substitute this
into
so you'll get $A$ from this
and you have $A_0$ from this

7. Mar 28, 2015

### laurenm02

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

8. Mar 28, 2015

### Suraj M

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

9. Mar 28, 2015

### 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)?

10. Mar 28, 2015

### Suraj M

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

11. Mar 28, 2015

### laurenm02

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

12. Mar 29, 2015

### Suraj M

You didn't take the root .