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Naddik [55]
3 years ago
6

The coefficient of the restitution of an object is defined as the ratio of its outgoing to incoming speed when the object collid

es with a rigid surface. For an object with a coefficient of 0.72, what fraction of the object's kinetic energy is lost during a single collision?
Physics
1 answer:
IgorLugansk [536]3 years ago
5 0

Answer:

48.16 %

Explanation:

coefficient of restitution = 0.72

let the incoming speed be = u

let the outgoing speed be = v

kinetic energy = 0.5 x mass x x velocity^{2}

  • incoming kinetic energy = 0.5 x m x x u^{2}

     

  •  coefficient of restitution =\frac{v}{u}

       0.72 =\frac{v}{u}

       v = 0.72u

        therefore the outgoing kinetic energy = 0.5 x m x (0.72u)^{2}

        outgoing kinetic energy = 0.5 x m x 0.5184 x u^{2}

        outgoing kinetic energy = 0.5184 (0.5 x m x x u^{2})

recall that 0.5 x m x x u^{2} is our incoming kinetic energy, therefore

outgoing kinetic energy = 0.5184 x (incoming kinetic energy)

from the above we can see that the outgoing kinetic energy is 51.84 % of the incoming kinetic energy.

The energy lost would be 100 - 51.84 = 48.16 %

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A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w
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Answer:

Explanation:

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Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

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Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

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Using the same principle above

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v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

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