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lana [24]
3 years ago
11

A 0.150-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa

ll and rebounds with 70.0% of its initial kinetic energy. What is the magnitude of the change in momentum of the stone?
Physics
1 answer:
Hitman42 [59]3 years ago
3 0

Answer:

The magnitude of the change in momentum of the stone is 5.51kg*m/s.

Explanation:

the final kinetic energy = 1/2(0.15)v^2

                1/2(0.15)v^2  = 70%*1/2(0.15)(20)^2

                              v^2 = 21/0.075

                              v^2 = 280

                                 v = 16.73 m.s

if u is the initial speed and v is the final speed, then:

u = 20 m/s and v = - 16.73m/s

change in momentum = m(v-u)

                                     = 0.15(- 16.73-20)

                                    = -5.51 kg*m.s

Therefore, The magnitude of the change in momentum of the stone is 5.51kg*m/s.

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Answer:

<h2>7.5 N</h2>

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3 0
3 years ago
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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23
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