Answer:
Contact maintained by the soccer player, with the ball is of 0.01 seconds.
Explanation:
Given:
Change in momentum of the ball, Δp = [tex]11\ kg.m.s^-1[/tex]
Net applied force on the ball, F = [tex]910\ N[/tex]
We have to find for how long the soccer player maintains contact with the ball.
Formula to be used:
Acceleration = Ratio of change in velocity and time.
⇒ [tex]a=\frac{\triangle v}{\triangle t}[/tex]
⇒ [tex]\triangle v=a \triangle t[/tex]
Momentum = Product of mass of an object and applied time.
⇒ [tex]\triangle p=m \triangle v[/tex]
⇒ [tex]\triangle p=ma \triangle t[/tex] ..plugging [tex]\triangle v=a \triangle t[/tex]
⇒ [tex]\triangle p=F \triangle t[/tex] ..Newtons seconds law,[tex]F=ma[/tex]
⇒ [tex]\triangle t=\frac{\triangle p}{F}[/tex] ..re-arranging the equation.
Plugging the value of force and momentum we have.
⇒ [tex]\triangle t=\frac{\triangle p}{F}[/tex]
⇒ [tex]\triangle t=\frac{11}{910}[/tex]
⇒ [tex]\triangle t =0.01[/tex] sec
So,
The soccer player maintain contact with the ball for 0.01 seconds.
