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Anton [14]
3 years ago
8

In space (no gravity or friction), you throw a ball with mass 0.1 kg at a target with mass 1 kg. You throw the ball at a speed o

f 4 m/s. When the ball impacts the target, it sticks to it and they drift off together.
1) How much energy is in the translational energy of the block+ball system after the collision?
2) How much of the initial translational kinetic energy of the ball was converted to internal energy in this collision? KEinitial-KEfinal= ___ J
3) The initial temperatures of the ball and block are 300K. By how much does the entropy of the entire system increase in the collision? Assume all the energy stays in the system, which reaches thermal equilibrium. You can also assume that the temperature is constant in the calculation.
Sfinal-Sinitial = ____ J/K
4) How much did the dimensionless entropy, S/k, change?
Physics
1 answer:
ioda3 years ago
6 0

Answer:

Explanation:

We shall apply law of conservation of  momentum in space to know the velocity of combination after the impact

m₁v₁ = m₂v₂

.1 x 4 = ( 1 + .1 ) v₂

v₂ = .3636 m /s

1  )  

Kinetic energy of the combination

= 1/2 x 1.1 x ( .3636)²

= 7.3 x 10⁻² J

2 )

Initial kinetic energy of the system

= 1/2 x 0.1 x 4²

= 0.8 J

Final  kinetic energy of the system = 7.3 x 10⁻²

Loss of energy = .8 - .073

= .727 J

This energy was converted into internal energy of the system .

3 )

increase in entropy = dQ / T

Here dQ = .727 J

T  = 300 ( Constant )

dQ / T = 2.42 X 10⁻³ J/K

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(a) -39.4^{\circ}

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u is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

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