8. A soccer player kicks a soccer ball of mass 0.45 kg that is initially at rest

Question

8. A soccer player kicks a soccer ball of mass 0.45 kg that is initially at rest. The player’s foot is in contact with the ball for 3.0x10-3 s. The force exerted on the ball by the foot is given by for . a. Find the magnitude of the impulse delivered to the ball. b. Find the magnitude of the average force exerted on the ball by the player’s foot. c. Find the magnitude of the maximum force exerted on the ball by the player’s foot. d. Find the magnitude of the ball’s velocity immediately after it loses contact with the player’s foot.

Explanation / Answer

(A) The impulse on the ball due to the kick is the integral of the force of the time. Let a = 6x10^6 and b = 2x10^9, then

Impulse = integral(F(t))dt = integral(a*t - b*t^2)dt
= (0.5*a*t^2 - b*t^3/3) | (3x10-3,0)
= 9.0 kg x m/s


(B) The average force is the same integral divided by the time interval, or the impulse divided by the time over which the impulse is applied.

Impulse/dT = Impulse/(3x10^-3 - 0)

=3.0 kN

(C) The time of maximum force can be found by taking the derivative of the force and looking at which time the derivative equals zero.

dF/dt = a - 2*b*tmax = 0

This leads to

tmax = a/(2*b)

=4.5 kN

(D) The average force can be expressed as

<F> = m*dV/dT

where dV is the change in velocity and dT is the time over which the force is applied. Solving for dV

dV = <F>*dT/m
= 20 m/s

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