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?

Question

bobpursley · Answer

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

D · Answer

Thanks a lot. Your answer are both right. Thank you so very much.

Omar · Answer

Thanks for making it straighforward and simple, not even chegg could help me with this one this time