# Momentum of spring between two objects

moonbase

## Homework Statement

A massless spring of spring constant 20 N/m is placed between two carts on a frictionless surface. Cart 1 has a mass M1 = 5 kg and Cart 2 has a mass M2 = 2.5 kg. The carts are pushed toward one another until the spring is compressed a distance 1.2 m. The carts are then released and the spring pushes them apart. After the carts are free of the spring, what are their speeds?

PEspring=0.5kx2
KE=0.5mv2=p2/2m
p=mv

## The Attempt at a Solution

I plugged in the given values to find that the potential energy of the spring is 14.4 J. Since there is no friction, the energy is conserved and the resulting kinetic energy should be the same value, and plugging it into the equation in terms of momentum results in p=14.69 since momentum is also conserved. So when I plug that into p=mv, I get v1=2.94 and v2=5.88 but apparently these are incorrect. Am I skipping a step?

issacnewton
Your calculations are not correct. check it

moonbase
Well I'm trying to figure out where exactly I went wrong. What I did so far:

0.5*20*1.22=14.4 J

p2/2(7.5)=14.4 -> p2=216 -> p=14.69

v=p/m -> 14.69/5=2.94 m/s and 14.69/2.5=5.88 m/s

Any ideas?

issacnewton
moon,

$$K=\frac{p^2}{2m_1}+\frac{p^2}{2m_2}$$

moonbase
Oh okay, I was combining them but I see now. Thank you!