# Elastic Head-On Collision, Spring Compression.

## Homework Statement

A 0.60kg cart moving at 5.0m/s[W] collides with a 0.80kg cart moving at 2.0m/s[E]. The collision is cushioned by a spring(k=1200N/m).

[W] is Positive
PTo=PTf
ETo=ETf

## The Attempt at a Solution

PTo=PTf
m1v1o+m2v2o=(m1+m2)vf
(0.6kg)(5m/s)+(0.8kg)(-2m/s)=(0.6kg+0.8kg)Vf
3kgm/s-1.6kgm/s=(1.4kg)Vf
(1.4kgm/s)/(1.4kg)=Vf
Vf=1m/s[W]

ETo=ETf
1/2m1v1o2+1/2m1v2o2=1/2(m1+m2)vf2+1/2kx2
1/2(0.6kg)(5m/s)+1/2(0.8kg)(-2m/s)=1/2(0.6kg+0.8kg)(1m/s2)+1/2(1200N/m)x2
7.5kgm/s+1.6kgm/s=0.24kgm/s+600N/mx2
9.1kgm/s-0.24kgm/s=600Nmx2
0.0148=x2
x=0.12m

Im pretty sure this is correct. I think I actually figured out what my real question about this was while typing this out. Initially when I saw this problem I wanted to just take the combined initial energy of both carts 9.1J and use that equal to elastic potential energy, so 1/2kx2=9.1J. I assumed that since the max compression should happen in the moment before both carts change direction or the spring rebounds that the velocity of both carts should be 0. Now I think the kinetic energy in the ETf is actually the kinetic energy of the entire "system" of both carts and the spring moving since the cart moving West would push everything slightly west. And so the carts do have a velocity of 0 relative to the spring at max compression but the entire collection of stuff has a velocity relative to the earth or ground. Is that somewhat right? The only other thing that bothers me about this is why would the kinetic energy of all of them only be 1/2(m2+m2)Vf2 shouldn't this include the mass of the spring if the spring was also moving?

haruspex