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 then apart, and they then fly off in opposite directions, free of the spring. The mass of A is 5.00 times the mass of and the energy stored in the spring was 88 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? J(particle A) J(particle B)

Explanation / Answer

mass of A = mA

mass of B = mB

mA = 5mB

Intial momentum of system = 0

Let the speeds of A and B be vA and vB after the srping is released

Final momentum of system = mA*vA + mB*vB

From conservation of linear momentum, mA*vA + mB*vB = 0

or, 5mB*vA + mB*vB = 0

or, vA = -0.2vB

Kinectic energy of the system after spring is relased = (1/2)mA*vA2 + (1/2)mB*vB2 = 68 J

Or, (1/2)mA*vA2 + (1/2)0.2mA*(5vA)2 = 68

Or, mA*vA2 = 68/(0.5+2.5) = 22.67

Therefore, (1/2)mAvA2 = 11.333 Joules

Therefore, (1/2)mBvB2 = 56.67 Joules

Kinetic energy of A = 11.33 Joules ; Kinetic energy of B = 56.67 Joules

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