Calculating velocity of 2 carts after spring between them released?

AI Thread Summary
The discussion focuses on calculating the velocities of two air track carts after a spring between them is released. The carts, with masses of 0.1 kg and 0.3 kg, are initially moving at 1.0 m/s, and the spring is compressed by 0.042 m with a constant of 150 N/m. The conservation of energy and momentum principles are applied to derive the final velocities after the string is burned. The user expresses confusion about their approach and seeks clarification on incorporating momentum conservation into their calculations. Understanding these principles is crucial for accurately determining the final velocities of the carts.
m84uily
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Homework Statement



Two air track carts are sliding to the right tied together by a string with a spring between them at 1.0m/s. The cart on the left has a mas of 0.1kg and the cart on the right has a mass of 0.3kg. The spring between them has a constant of 150N/m and is compressed at 0.042m . The carts pass over a flame that burns the string holding them together.

What is the velocity of the cart on the left after the string is destroyed?

m1 = 0.1kg
m2 = 0.3kg

vi(1 + 2) = 1 m/s
vf1 = ?
vf2 = ?

k = 150 N/m
xi = 0.042 m
xf = 0
Uf = 0



Homework Equations


K = (1/2)mv^2
U = (1/2)k(xi)^2

The Attempt at a Solution



Ki + Ui = Kf + Uf
Ki(1+2) + Ui = Kf1 + Kf2 + Uf

Ki(1+2) + Ui = Kf1 + Kf2

(1/2)m(1+2)(vi(1+2))^2 + (1/2)k(xi)^2 = (1/2)m1(v1f)^2 + (1/2)m2(v2f)^2


(1/2)(0.4)(1) + (1/2)(150)(0.042)^2 = (1/2)(0.1)(v1f^2) + (1/2)(0.3)(v2f^2)
(0.2) + 0.1323 = (0.05)(v1f^2) + (0.15)(v2f^2)
0.3323 = (0.05)(v1f^2) + (0.15)(v2f^2)


I think I may be taking a very wrong approach? I'm quite dumbfounded and would certainly appreciate some help.
 
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You need to consider conservation of momentum also.
 
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