A ball of mass 0.175 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.805 m. What impulse was given to the ball by the floor

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.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

the answer is 12 as i may remember

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