Posted by **D** on Wednesday, February 23, 2011 at 9:27pm.

Particle A and particle 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 3.00 times the mass of B, and the energy stored in the spring was 132 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 A and B in Joules?

- Physics Mechanics -
**bobpursley**, Wednesday, February 23, 2011 at 9:33pm
first conservation of momentum

MassA*Va+Massb*Vb=0

3Va+Vb=0

Vb=-3Va

then energy:

132=1/2 MassA*Va^2+1/2 MassB*Vb^2

264=massB (3Va^2+Vb^2) or

264=MassB (3Vb^2/9+Vb^2)) or

264=4/3 MassB*Vb^2

so you know then that KE massb is

KEmassB=132*3/4 so the remaining 1/4 must be KEmassA

check my work

- Physics Mechanics -
**D**, Wednesday, February 23, 2011 at 9:57pm
Thanks a lot. Your answer are both right. Thank you so very much.

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