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Gnesinka [82]
3 years ago
14

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

m apart, and they then fly off in opposite directions, free from the spring. The mass of A is 2.00 times the mass of B, and the energy stored in the spring was 60 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?
Physics
1 answer:
Alinara [238K]3 years ago
4 0

Answer:

Kinetic energy of mass A = 20 J

Kinetic energy of mass B = 40 J

Explanation:

Lets take final speed of mass A  = v

The final speed of mass B = v'

mass of A= m

mass of B = m'

m = 2 m'

There is no any external force so the linear momentum will be conserve.

Pi = pf

0 = m v + m ' v'

0 =  2 m ' v +m ' v'

v ' = - 2 v                          

Now from energy conservation

1/2 k x²=1/2 m v² + 1/2 m' v'²

60 = 1/2 m v² + 1/2 m' v'²

m v² +  m' v'² = 120

2 m ' (-v'/2)² + m' v'² = 120

m ' v'²/2 + m' v'² = 120

m' v '² = 80

m =2 m'  and v ' = - 2 v

So

m/2 ( 4 v²) = 80

m v ²=40

So the kinetic energy of mass A= 1/2 m v² = 20 J

Kinetic energy of mass B = 1/2 m' v'² = 40 J

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Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

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           Em₀ = U = mg h

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             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

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            h = ½ v_{cm }^{2}/g    5/3

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           h = 5/6 6.1 2 / 9.8

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b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

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             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

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