Solve for Spring Compression: Physics 11 Momentum and Energy Problem

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The problem involves a 1.1 kg cart and a 2.2 kg cart connected by a compressed spring with a force constant of 600 N/m. When released, the carts separate, and the goal is to determine the spring's compression needed for the 2.2 kg cart to reach a speed of 1.5 m/s. The total kinetic energy of the system is calculated using the formula KE = 0.5mv², yielding a combined kinetic energy of 7.425 J. This energy must equal the initial potential energy stored in the spring, leading to the equation KE = 0.5kx² to solve for the compression distance. The final answer for the spring compression is determined to be 0.16 m.
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Homework Statement



A 1.1 kg cart and a 2.2 kg cart are at rest with a compressed string between them. the force constant of the spring is 600 N/m. when the spring is released, the carts rapidly separate. how far must the spring have been compressed for the 2.2 kg cart to end up moving at 1.5 m/s?
(involves both momentum and energy)

Homework Equations



p = mv
Ek = 0.5mv2
Ep = 0.5kx2

The Attempt at a Solution



i don't even know where to start !
 
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Figure out the total KE of the carts. Hint: What's the momentum of the system?
 
Doc Al said:
Figure out the total KE of the carts. Hint: What's the momentum of the system?

ok so i got:
0.5(1.1)(3^2) = 4.95 J for Cart 1
0.5(2.2)(1.5^2) = 2.475 J for Cart 2
 
Using what you know about the system, apply the law of conservation of energy now.
 
jgens said:
Using what you know about the system, apply the law of conservation of energy now.

ok so
0.5mv2
= 0.5(7.425)(1.5)2
=8.35

but the answer should be 0.16 m.. did i do something wrong or is there another setp now?
 
Well, your total kinetic energy is the sum of the kinetic energy of each kart, so KE = 7.425. You know that the resultant KE must equal the initial potential energy so KE = .5(k)(x^2).
 
great thanks !
 
You're welcome.
 
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