Question

chap09-P37 Please write the answer neatly for a thumps up, thank you!

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 1.30 x 10-3 s, and the force of the kick is given by F(t) = [(9.69 x 105) - (7.45 x 103)+2] N for osts 1.30 X 10's, where t is in seconds. Find the magnitudes of the following: (a) the impulse on the ball due to the kick, (b) the average force on the ball from the player's foot during the period of contact, (c) the maximum force on the ball from the player's foot during the period of contact, and (d) the ball's speed immediately after it loses contact with the player's foot. (a) Number Units Units (b) Number (c) Number (d) Number Units Units The number of significant digits is set to 3; the tolerance is +/-3%

Answer 1

Force as a function of time F(t) = a t - b t²   , 0 < t < 1.3 $\times$ 10^-3 s

where a = 9.69 $\times$ 10⁵ N/s and b = 7.45 $\times$ 10⁸ N/s²

Impulse if force is acting for a period $\Delta$ t = 1.3 $\times$ 10^-3 s is given by

$F \Delta t = \int_{0}^{\Delta t}\left ( a t - b t^{2} \right ) dt = \frac{a}{2}\left ( \Delta t\right )^{2} - \frac{b}{3}\left ( \Delta t \right )^{3}$

Hence impulse = { (9.69 $\times$ 10⁵$\times$ 1.3 $\times$ 1.3 $\times$ 10^-6) / 2 } - { ( 7.45 $\times$ 10⁸$\times$ 1.3 $\times$ 1.3 $\times$ 1.3 $\times$ 10^-9 )/ 3 }

impulse = 0.2732 Ns

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average force

$\bar{F} = \frac{1}{\Delta t}\int_{0}^{\Delta t}F(t) dt = \frac{1}{\Delta t}\int_{0}^{\Delta t}\left ( a t - b t^{2} \right ) dt = \frac{a}{2} \left ( \Delta t\right ) - \frac{b}{3}\left ( \Delta t \right )^{2}$

average force = { (9.69 $\times$ 10⁵$\times$1.3 $\times$ 10^-3 ) /2 - (7.45 $\times$ 10⁸$\times$ 1.3 $\times$ 1.3 $\times$ 10^-6)/3 } = 210 N

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maximum force is obtained by differentiating force function and equating to zero

dF(t)/dt = a - (2b)t = 0

hence maximum force occurs at t = a / (2b) = 9.69 $\times$ 10⁵ / ( 2 $\times$ 7.45 $\times$ 10⁸ ) = 0.65 $\times$ 10^-3 s

maximum force = a (0.65 $\times$ 10^-3) - b $\times$ (0.65 $\times$ 10^-3 )² = { 9.65 $\times$ 0.65 - 7.45 $\times$ 0.65$\times$0.65 } $\times$ 10²

maximum force = 312.5 N

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Ball's speed is obtained from , F $\Delta$ t = change in momentum = m v

( Initially ball is at rest, hence change in momentum is mv )

v = ( F $\Delta$ t ) / m = 0.2732 / 0.45 = 0.607 m/s