2 Carts on a Track Compressed by a Spring -- What are their Velocities?

  • #1
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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.
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
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Hi,

You already understand that the energy that is stored in the spring will be converted to kinetic energy of the carts and that the sum of the kinetic energies of the carts will be 7.8125 J. One relevant equation (energy conservation) used up. But there are two unknowns (the speeds) so another relationship (equation) will be needed. Know of any other variable (other than energy) that might be conserved ?

Oh, and, eh:
:welcome: !​


##\ ##
 
  • #3
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OH I see now that I might be able to use conservation of momentum to solve this equation.
 
  • #4
BvU
Science Advisor
Homework Helper
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Bingo !
 

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