Question

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mass of 0.425 kg and the player's foot is in contact with it for 5.30 ✕ 10−2 s, find the x- and y-components of the soccer ball's change in momentum and the magnitude of the average force exerted by the player's foot on the ball.

Answer 1

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

• $\Delta p_x= 6.6251 \,kg.m.s^{-1}$

• $\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

• $F_x=125.0018\,N$

• $F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$.....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$.........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

&

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$