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 .

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

answer///terminator or twilight zone;

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y