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A 0.1-kg steel ball is dropped straight down onto a hard horizontal floor and bounces straight up. Its speed just before and just after impact with the floor is 10 m/s. Determine the magnitude of the impulse delivered to the floor by the steel ball. A) zero N • s B) 1 N • s C) 2 N • s D) 10 N • s E) 100 N • s

Respuesta :

Answer:

Impulse, [tex]J=2\ N.s[/tex]

Explanation:

Given that,

Mass of the steel ball, m = 0.1 kg

Initial speed of the ball, u = -10 m/s (it straight down)

Final speed of the ball, v = +10 m/s (it straight up)

To find,

The magnitude of the impulse delivered to the floor by the steel ball.

Solution,

Let J is the magnitude of impulse. The change in momentum of an object to another is equal to the impulse delivered. Mathematically, it i given by :

[tex]J=m(v-u)[/tex]

[tex]J=0.1\ kg\times (10\ m/s-(-10\ m/s))[/tex]

[tex]J=2\ kg.m/s[/tex]

J = 2 N.s

Therefore, the magnitude of the impulse delivered to the floor by the steel ball is 2 N.s. Hence, this is the required solution.

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