Calculating velocity of 2 carts after spring between them released?

  • Thread starter m84uily
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
jtbell
Mentor
15,755
3,963
You need to consider conservation of momentum also.
 

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