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A 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds at 20 m/s. The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is _______.

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Answer:

10 kgm/s

Explanation:

Change in momentum: This can be defined as the product of mass and change in velocity. The S.I unit of change in momentum is Kgm/s.

From the question,

ΔM = m(v-u)...................... Equation 1

Where ΔM = change in momentum, u = initial velocity, v = final velocity.

Note: Let upward direction be negative, and downward direction be positive.

Given: m = 0.2 kg, v = -20 m/s, u = 30 m/s

Substitute into equation 1

ΔM = 0.2(-20-30)

ΔM = 0.2(-50)

ΔM = -10 kgm/s.

The negative sign shows that the change in momentum is Upward

The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.

Given data:

The mass of rubber ball is, m = 0.2 kg.

The initial speed of ball is, u = 30 m/s.

The final rebounding speed of ball is, v = - 20 m/s ( Negative sign shows that during the rebounding, the ball changes its direction)

The momentum of any object is defined as the product of mass and change in velocity. The S.I unit of momentum is Kg-m/s. And the expression for the change in momentum is given as,

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

Solving as,

[tex]p= 0.2 \times ( -20-30)\\\\p=-10 \;\rm kg.m/s[/tex]

Thus, we can conclude that the magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.

Learn more about the change in momentum here:

https://brainly.com/question/904448