Particle A and particle B are held togetherwith a compressed spring between them

Question

Particle A and particle B are held togetherwith a compressed spring between them. When they are released, thespring pushes them apart and they then fly off in oppositedirections, free of the spring. The mass of A is 2.2 timesthe mass of B, and the energy stored in the spring was 62J. Assume that the spring has negligible mass and that all itsstored energy is transferred to the particles. Once that transferis complete, what are the kinetic energies of (a)particle A and (b) particleB?

(a)

Number

Units

(b)

Number

Units

(a)

Number

Units

J

Explanation / Answer

Applying the law of conservation of momentum we have as, m1u1+m2u2=m1v1+m2v2 0=2.2mv1+mv2 v2=-2.2v1 Applying the law of conservation of energy we have as, PE=KEA+KEB 1/2*m1v12+1/2*m2v22=PE 1/2*2.2m(-v2/2.2)2+1/2*mv22=62 From this we get the KE of B. Plug in the values in the above to get of A. Hence we get by it.

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