Particle A and particle B are held together with a compressed spring between the

Question

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 8.00 times the mass of B, and the energy stored in the spring was 74 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles.

(a) Once that transfer is complete, what is the kinetic energy of particle A?


(b) Once that transfer is complete, what is the kinetic energy of particle B?

Explanation / Answer

we know, K.E= (1/2)mv^2= (1/2)(mv)^2/m= .5*p^2/m where p= linear momentum

so K.E is proportional to (1/m)
by conservation of momentum, initial momentum = final momentum
so, initial momentum of the system =0 so final momentum must also be 0
so, by magnitude the momenta of the two particles must be same and must be directed in opposite directions

so let K.Eof particle B be x
so K.E of particle A=(1/8)x .... as K.E is proportional to (1/m)
so total energy = x+(1/8)x=(9/8)x
by conservation of energy spring energy stored in the spring must be transferred to the particles.
so 74= (9/8)x
so x=65.8J
so K.E of particle A=(1/8)x= 8.2J
K.E of particle A=x=65.8J

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