# Mechanical Energy Transfer

## 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

$$U_i + K_i = U_f + K_f$$

K_i=0, U_f=0

$$U_i=K_f$$

$$U_i=K_{Af}+K_{Bf}$$

$$U_i=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2$$

Let m_B=m, so m_A=3m

$$76 J=\frac{1}{2}3m+\frac{1}{2}mv_B^2$$

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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Doc Al
Mentor
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 $$4mv_i=0=3mv_{Af}+mv_{Bf}$$

Rearrange and get $$v_{Af}=\frac{v_{Bf}}{3}$$ 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! :!!)