Mechanical Energy Transfer in a Spring System

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SUMMARY

The discussion focuses on the mechanical energy transfer in a spring system involving two particles, A and B, where particle A has three times the mass of particle B. The spring releases 76 J of energy, which is divided between the two particles upon release. The kinetic energies calculated reveal that particle B has 57 J while particle A has 19 J, demonstrating the conservation of both energy and momentum in the system. The solution emphasizes the importance of recognizing momentum conservation alongside energy conservation in such problems.

PREREQUISITES
  • Understanding of mechanical energy concepts
  • Knowledge of kinetic and potential energy equations
  • Familiarity with conservation laws in physics
  • Basic algebra for solving equations
NEXT STEPS
  • Study the principles of conservation of momentum in elastic collisions
  • Explore the equations of motion for systems involving springs
  • Learn about energy transfer in mechanical systems
  • Investigate the role of mass ratios in energy distribution
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This discussion is beneficial for physics students, educators, and anyone interested in understanding energy transfer mechanisms in mechanical systems, particularly in relation to springs and particle dynamics.

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Homework Statement



Particle A and partidcle B are held together with a compressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of A is 3 times the mass of B, and the energy stored in the sring was 76 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of each particle?


Homework Equations



E=const.

For the system: total U_i + K_i=total U_f + K_f

The Attempt at a Solution



[tex]U_i + K_i = U_f + K_f[/tex]

K_i=0, U_f=0

[tex]U_i=K_f[/tex]

[tex]U_i=K_{Af}+K_{Bf}[/tex]

[tex]U_i=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2[/tex]

Let m_B=m, so m_A=3m

[tex]76 J=\frac{1}{2}3m+\frac{1}{2}mv_B^2[/tex]

At this point, I noticed that if the velocities are the same, K_A will be 3 times as large as K_B. Sure enough, the answer is that the U_i has been divided into 4 parts--but paticle B has 57 J and paticle A has 19 J!

How was I supposed to get there?
 
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Hint: Energy is not the only thing conserved in this problem.
 
Dude. I thought of momentum and said, "Nope. Outside force going on, baby." Must remember that WHENEVER energy is conserved, so is momentum. And sometimes when energy isn't conserved, also...

OK. so [tex]4mv_i=0=3mv_{Af}+mv_{Bf}[/tex]

Rearrange and get [tex]v_{Af}=\frac{v_{Bf}}{3}[/tex] and shove it into my potential energy equation... and have oodles of fun and find out all sorts of stuff I'm not asked for, and get the answer!

Bless you, Doc Al! :!)
 

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