A ball of mass 0.165 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.940 m. What impulse was given to the ball by the floor?

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wagonbelleville wagonbelleville · Answer 1 · 2020-01-24T08:05:02+01:00

Answer:

I = 1.525 kg.m/s

Explanation:

given,

mass of the ball = 0.165 Kg

height of drop, h = 1.25 m

ball rebound and reach to height, h' = 0.940 m

impulse = ?

using conservation of energy

Potential energy is converted into kinetic energy

[tex]mgh = \dfrac{1}{2}mv^2[/tex]

[tex]v=\sqrt{2gh}[/tex]

[tex]v=\sqrt{2\times 9.8 \times 1.25}[/tex]

v = 4.95 m/s

velocity of the ball after rebound

again using conservation of energy

[tex]mgh = \dfrac{1}{2}mv'^2[/tex]

[tex]v'=\sqrt{2gh}[/tex]

[tex]v'=\sqrt{2\times 9.8 \times 0.94}[/tex]

v' = 4.29 m/s

impulse is equal to change in momentum

I = m ( v' - v )

I = 0.165 x ( 4.29 - (-4.95))

I = 1.525 kg.m/s