Question

A 0.5 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.

Answer 1

4.281 kgm/s upward

Step-by-step 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