Answer:
The impulse received by the ball is - 1.561 kg.m/s
Explanation:
Given;
mass of the ball, m = 0.175 kg
initial displacement of the ball, h₁ = 1.25 m
final displacement of the ball, h₂ = 0.805 m
Assumptions:
let the downward direction of the ball be positive
let the upward direction of the ball be negative
The following equation of motion will be used to determine the final velocity of the ball at each displacement.
v² = u² ± 2gh
The final velocity of the ball when it is dropped downwards to 1.25 m;
v² = u² + 2gh
v² = 0 + 2gh
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 1.25)
v = 4.95 m/s
The final velocity of the ball when it rebounds from the floor to 0.805 m;
vf² = u² - 2gh
vf² = 0² - 2gh
vf² = -2gh
vf = -√2gh
vf = - √(2 x 9.8 x 0.805)
vf = -3.97 m/s
The impulse received by the ball is calculated as;
J = ΔP = mΔv = m(vf - v)
= 0.175(-3.97 - 4.95)
= - 1.561 kg.m/s
The negative sign indicates upward direction of the impulse.
