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

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form