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 Homework Statement:
 Two frictionless carts, with masses mA=43mB and mB= 0.55kg are on a level track. The carts have an ideal spring between them with spring constant κ= 250J/m2. The carts are compressing the spring a distance xi= 25cm from its equilibrium position.
 Relevant Equations:

Potential Energy of A Spring: U = 1/2 K x^2
Work Energy Theorem W = Change in K
a)What is the total energy in the system?
Only energy acting on the system assuming the track is level and there is no potential energy of the carts, is the potential energy of the spring.
Comes out to 7.8125 using the potential energy of a spring equation.
b) What are their velocities if the carts are simultaneously released?
This is what stumped me, initially I started by doing separate work energy theorems for cart A and cart B.
W = 1/2 ma Vf^2
7.8125 = 1/2(.733)vf^2
Came out to Vfa = 4.616
Using the same equation Vfb came out to Vfb = 5.33
That gave me velocities that made sense, the heavier cart(A) had a lesser velocity then b.
But then I realized, that potential energy of the spring is split up between the 2 carts, so setting that potential energy equal to work for both equations was probably incorrect. This is where I'm stuck.
Only energy acting on the system assuming the track is level and there is no potential energy of the carts, is the potential energy of the spring.
Comes out to 7.8125 using the potential energy of a spring equation.
b) What are their velocities if the carts are simultaneously released?
This is what stumped me, initially I started by doing separate work energy theorems for cart A and cart B.
W = 1/2 ma Vf^2
7.8125 = 1/2(.733)vf^2
Came out to Vfa = 4.616
Using the same equation Vfb came out to Vfb = 5.33
That gave me velocities that made sense, the heavier cart(A) had a lesser velocity then b.
But then I realized, that potential energy of the spring is split up between the 2 carts, so setting that potential energy equal to work for both equations was probably incorrect. This is where I'm stuck.