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

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`

Step-by-step 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* 10^(-2)\,s`

We know, since body is starting from the rest

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

`\Delta p=0.425* 18`

`\Delta p=7.65 \,kg.m.s^(-1)`

Now the components:

`\Delta p_x= 7.65* cos 30^(\circ)`

`\Delta p_x= 6.6251 \,kg.m.s^(-1)`

similarly

`\Delta p_y= 7.65* sin 30^(\circ)`

`\Delta p_y= 3.825 \,kg.m.s^(-1)`

also, impulse

`I=F* t`.........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

`F* t=m.v`

`F* 5.3* 10^(-2)= 7.65`

`F=144.3396\,N`

Now, the components

`F_x=144.3396* cos 30^(\circ)`

`F_x=125.0018\,N`

&

`F_y=144.3396* sin 30^(\circ)`

`F_y=72.1698\,N`