A0.5 kg ball is dropped from rest at a point 1.2m above the floor. the ball rebounds straight upward to a height of 0.7m. what are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor.

4.281 kgm/s upward

Explanation:

Impulse:This can be defined product of force and time. The S.I unit of impulse is Ns.

From Newton's second law of motion,

Impulse = Change in momentum.

I = mΔv Equation 1.

Where m = mass of the ball, Δv = change in velocity of the ball  

and Δv = v -u

Where u = velocity of the ball before it hit the floor, v = velocity of the ball after if hit the floor

I = m(v-u) Equation 2

But

the initial kinetic energy of the ball = potential energy at the initial height (1.2 m above)

1/2mu² = mgh₁

Where h₁ = initial height. or height of the ball before collision

making u the subject of the equation,

u = √(2gh₁) Equation 3

Where h₁ = 1.2 m g = 9.81 m/s²

Substitute into equation 3

u = √(2×1.2×9.81)

u =√(23.544)

u = -4.852 m/s.

Note: u is negative because the ball was moving downward at the first instance.

Similarly,

v = √(2gh₂) Equation 3

h₂ = height of the ball after collision

Given: h₂ = 0.7 m, g = 9.81 m/s²

Substitute into equation

v = √(2×9.81×0.7)

v = √13.734

v = 3.71 m/s.

Also given: m = 0.5 kg,

Substituting into equation 2

I = 0.5(3.71-(4.852)

I = 0.5(8.562)

I = 4.281 kgm/s. Upward.

Thus the impulse = 4.281 kgm/s upward

answer: b.

explanation:

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this