Answer:
[tex]I=20\ kg.m.s^{-1}[/tex]
Explanation:
Given:
mass of the ball, [tex]m=2\ kg[/tex]
initial speed of the ball before collision, [tex]u=30\ m.s^{-1}[/tex]
final speed of hte ball after collision, [tex]u=20\ m.s^{-1}[/tex]
We have mathematical relation for impulse as:
[tex]I=F.dt[/tex] ......................(1)
From the Newton's second law of motion:
[tex]F=\frac{d}{dt} (p)[/tex]
[tex]F=m.\frac{dv}{dt}[/tex] ................................(2)
where:
[tex]dt=[/tex] time duration
[tex]dv=[/tex] change in velocity
[tex]dp=[/tex] change in momentum
Now from (1) & (2)
[tex]I=m.\frac{dv}{dt} .dt[/tex] (since this impulse is due to the impact force of the ball falling on the ground)
[tex]I=m.dv[/tex]
putting the values
[tex]I=2\times (30-20)[/tex]
[tex]I=20\ kg.m.s^{-1}[/tex] is the impulse transfer to the bowling by the sidewalk.
