Aball with mass of 0.050 kg is dropped from a height of h1 = 1 .5 m. it collides with the floor, then bounces up to a height of h2 = 1.0 m. the collison takes 0.015 s.
use part e to calculate impulse ( j) ? write the formula first .
use part f to calculate the average force acting ( f) on the ball by the floor during the collision?
*write the formula
calculate the change in the kinetic energy (∆k) in the collision. write formula first.
what happens to the lost kinetic energy?
what type of collision is this?

Explanation:

Impulse of reaction force of floor = change in momentum

Velocity of impact = √ 2gh₁

= √ 2 x 9.8 x 1.5 = 5.4 m /s.

velocity of rebound = √2gh₂

= √ 2x 9.8 x 1

= 4.427 m / s.

Initial momentum = .050 x 5.4 = .27 kg m/s

Final momentum = .05 x 4.427 = .22 kg.m/s

change in momentum = .27 - .22 = .05 kg m/s

Impulse = .05 kg m /s

Impulse = force x time

force = impulse / time

.05 / .015 = 3.33 N.

kinetic energy = 1/2 m v²

Initial kinetic energy = 1/2 x .05 x 5.4²

= 0.729 J

Final Kinetic Energy =1/2 x .05 x 4.427²

= 0.489 J

Change in Kinetic energy =0 .24 J

Lost kinetic energy is due to conversion of energy into sound light etc.

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