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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.

Respuesta :

Answer:

change in momentum, [tex]\Delta p=7.65 \,kg.m.s^{-1}[/tex]

  • [tex]\Delta p_x= 6.6251 \,kg.m.s^{-1}[/tex]
  • [tex]\Delta p_y= 3.825 \,kg.m.s^{-1}[/tex]

Average Force, [tex]F=144.3396\,N[/tex]

  • [tex]F_x=125.0018\,N[/tex]
  • [tex]F_y=72.1698\,N[/tex]

Explanation:

Given:

angle of kicking from the horizon, [tex]\theta= 30^{\circ}[/tex]

velocity of the ball after being kicked, [tex]v=18 m.s^{-1}[/tex]

mass of the ball, [tex]m=0.425\, kg[/tex]

time of application of force, [tex]t=5.3\times 10^{-2}\,s[/tex]

We know, since body is starting from the rest

[tex]\Delta p=m.v[/tex].....................(1)

[tex]\Delta p=0.425\times 18[/tex]

[tex]\Delta p=7.65 \,kg.m.s^{-1}[/tex]

Now the components:

[tex]\Delta p_x= 7.65\times cos 30^{\circ}[/tex]

[tex]\Delta p_x= 6.6251 \,kg.m.s^{-1}[/tex]

similarly

[tex]\Delta p_y= 7.65\times sin 30^{\circ}[/tex]

[tex]\Delta p_y= 3.825 \,kg.m.s^{-1}[/tex]

also, impulse

[tex]I=F\times t[/tex].........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

[tex]F\times t=m.v[/tex]

[tex]F\times 5.3\times 10^{-2}= 7.65[/tex]

[tex]F=144.3396\,N[/tex]

Now, the components

[tex]F_x=144.3396\times cos 30^{\circ}[/tex]

[tex]F_x=125.0018\,N[/tex]

&

[tex]F_y=144.3396\times sin 30^{\circ}[/tex]

[tex]F_y=72.1698\,N[/tex]

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