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?

Question

Kolton Hickman

Physics

1 June, 11:47

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?

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Answers (1)

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Jax Andrews · Answer 1

Given:

m = 0.5 kg, the mass of the ball

v₁ = 20 m/s, the impact velocity

The KE of the ball at impact is

KE = (1/2) * (0.5 kg) * (20 m/s) ² = 100 J

Because the ball rebounds with 70% of its KE, the rebound velocity, v₂, is

(1/2) * (0.5 kg) * (v₂ m/s) ² = 0.7 * (100 J)

0.25v₂² = 70

v₂ = 16.733 m/s (in the opposite direction)

The change in momentum of the ball is

ΔP = (0.25 kg) * ) 20 - (-16.733) m/s)

= 9.1833 (kg-m) / s

Because momentum is conserved, this change in momentum applies to the stone as well.

Answer: 9.2 (kg-m) / s