Question

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 1.50 × 10-3 s, and the force of the kick is given by F(t) = [(6.99 × 105)t - (4.66 × 108)t2] N for 0 less-than-or-equal-to t less-than-or-equal-to 1.50 times 10 Superscript negative 3 Baseline 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.

Answer 1

given

m = 0.45 kg
F(t) = 6.99*10^5*t - 4.66*10^8*t^2
0 < t < 1.50*10^-3 s

a) Impulse = integral F*dt

= integral (6.99*10^5*t - 4.66*10^8*t^2)*dt

= 6.99*10^5*t^2/2 - 4.66*10^8*t^3/3

= 6.99*10^5*((1.5*10^-3)^2 - 0^2 )/2 - 4.66*10^8*((1.5*10^-3)^3 - 0^3 )/3

= 0.262 N.s (or) kg.m/s <<<<<<<<<------------------Answer

b) now use, Impulse = F_avg*delta_t

F_avg = Impulse/delta_t

= 0.262/(1.5*10^-3)

= 175 N <<<<<<<<<------------------Answer

c) when F is maximum, dF/dt = 0

6.99*10^5 - 4.66*10^8*2*t = 0

==> t = 6.99*10^5/(2*4.66*10^8)

= 7.5*10^-4 s <<<<<<<<<------------------Answer

d) use, Impulse = change in momentum

0.262 = 0.45*v - 0

==> v = 0.262/0.45

= 0.582 m/s <<<<<<<---------------------Answer