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A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial kinetic energy. What is the magnitude of the change in momentum of the stone?

A) 1.63 kg middot m/s
B) 3.00 kg middot m/s
C) 0.000 kg middot m/s
D) 14.0 kg middot m/s
E) 18.4 kg middot m/s

Respuesta :

Answer:

E)  I = 18.4 N.s

Explanation:

For this exercise let's use momentum momentum

     I = Δp = [tex]p_{f}[/tex]- p₀

The energy of the stone is only kinetic

    K = ½ m v²

The initial energy is Ko and the final is 70% Ko

     [tex]K_{f}[/tex] = 0.70 K₀

energy equation

     [tex]K_{f}[/tex] = 0.7 ½ m v₀²

You can also write

     [tex]K_{f}[/tex] = ½ m vf²

   ½ m vf² = ½ m (0.7 v₀²)

   [tex]v_{f}[/tex] = v₀ √ 0.7

Now we can calculate and imposed

     I = m (-vo √0.7) - m vo

     I = m vo (1 +√0.7

     I = 0.5000 20.0 (1.8366)

     I = 18.4 N.s

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