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?
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
Thanks a lot. Your answer are both right. Thank you so very much.
Thanks for making it straighforward and simple, not even chegg could help me with this one this time
To find the kinetic energies of particles A and B, we need to use the principle of conservation of energy. The total energy of the system, which includes both the spring and the particles, remains constant. This means that the energy stored in the spring is transferred entirely to the kinetic energies of particles A and B.
Given that the energy stored in the spring is 132 J, we can assume that this energy is equally divided between particles A and B since they fly off in opposite directions.
Let's denote the kinetic energy of particle A as KA and the kinetic energy of particle B as KB.
Since particle A has a mass 3.00 times that of particle B, we can say:
KA = 3KB
The total energy of the system is the sum of the energies of particles A and B:
132 J = KA + KB
Substituting the value of KA in terms of KB, we get:
132 J = 3KB + KB
Combining like terms:
132 J = 4KB
Now, we can solve for KB:
KB = 132 J / 4
KB = 33 J
Since KA = 3KB, we can calculate KA:
KA = 3 * 33 J
KA = 99 J
Therefore, the kinetic energy of particle A is 99 J, and the kinetic energy of particle B is 33 J.