A ball of mass 1 kg bounces vertically off the floor. Just before hitting the floor its velocity is 10 m/s downward. Just after hitting the floor its velocity is 8 m/s upward. What is the magnitude of the impulse experienced by the particle in this collision?

2kgm/s

Explanation:

Impulse is defined as the change in momentum of a body. It is expressed as;

Impulse = mass × change in velocity

Change in velocity = final velocity v - initial velocity u

Impulse = m(v-u)

Given mass of the ball = 1kg

Final velocity v = 8m/s (after hitting the floor)

Initial velocity = 10m/s (before hitting the floor)

Impulse = 1(8-10)

Impulse = -2kgm/s

Magnitude of the impulse experienced by the particle in the collision is 2kgm/s

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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