Answer:
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
