Posted by **Physics/Math** on Friday, February 23, 2007 at 2:47am.

A soccer player kicks a soccer ball of mass 0.45 kg that is initially at rest. The force of the kick is given by F(t) = [(5×106)t - (2.5×109)t2] N for 0 ≤ t ≤ 2.0×10-3 s, where t is in seconds. (The player's foot is in contact with the ball for 2.0×10-3 s.) Find the magnitude of the impulse on the ball due to the kick. Find the average force on the ball from the player's foot during the period of contact. Find the maximum force on the ball from the player's foot during the period of contact. Find the ball's velocity immediately after it loses contact with the player's foot.

Integrate the force vs time function during the period of contact. This will tell you the momentum delivered to the ball.

Impulse = Integral of a t- b t^2

from t = 0 to t = 2*10^-3 s,

where

a = 5*10^6 and b = 2.5*10^9

The integral is at^2/2 - bt^3/3.

Evaluate it at the limits of integration and take the difference.

The average force is the impulse divided by the time of contact.

The maximum force occurs when f'(t) is zero, which is when

a = 2 b t

t = a/(2 b) = 1.0*10^-3 s.

The ball's velocity when it leaves the foot = (momentum change)/mass = (impulse)/mass

(since the initial momentum is zero)

Great...I got it. Thanks :)

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