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 than fly off in opposite directions, free of the spring. The mass of A is 5.00 times the mass of B, and the energy stored in the spring was 55 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 (a) particle A and (b) particle B? (a) Number Unit (b) Number Unit

Explanation / Answer

Let the mass of the particle B be m units

so the mass of the particle A is

mA = 5mB

mA = 5m

Let vA and vB be the final velocities of the particles A and B respectively. We also know that particle A moves in the opposite direction to the particle B

Conservation of momentum

The momentum before release = the momentum after the release

since both the particles are at rest before the release their initial momentum is zero

so

0 = mAvA – mBvB

0 = 5mvA -mvB

VB = 5vA

Conservation of energy

The change is kinetic energy of the particle = the elastic potential energy of the spring

½ mAvA2 + ½ mBvB2 = 55 J

½ x 5m x vA2 + ½ m x (5vA)2 = 55

5 x ½ mvA2 + 25 x ½ mvA2 = 55

½ mvA2 (5 + 25) = 55

½ mvA2 = 55/30

The KE of particle A = 5 x (½ mvA2)

EA = 5 x 55/30

EA = = 9.17 J

KE of particle B = 55 – 18.33

EB = 45.83 J

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