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
