Question

A soccer player kicks a soccer ball of mass 0.45 kg that is initially at rest. The foot of the player is in contact with the ball for$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{s}$, and the force of the kick is given by

role="math" localid="1661489189931" $\mathit{F}\left(t\right)\mathbf{=}\left[\left(6.0\times {10}^6\right)t-\left(2.0\times {10}^9\right)t2\right]\mathbf{N}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathbf{0}\mathbf{\le }\mathbf{t}\mathbf{\le }\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$

Where t is in seconds. Find the magnitudes of the (a) 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 velocity immediately after it loses contact with the player’s foot.

Short answer

a) The impulse on the ball due to the kick J is 9.0 N.s.

b) The average force on the ball during the period of contact $F_{\mathrm{avg}}\mathrm{is}3.0\times {10}^3\mathrm{N}$.

c) The maximum force on the ball during the period of contact $F_{max}\mathrm{is}4.5\times {10}^3\mathrm{N}$.

d) The ball’s velocity immediately after losing contact with the player’s foot v =20 m/s.

Step-by-step solution

Understanding the given information

i) The mass of the ball m is 0.45 kg.

ii) The initial speed of the ball is 0 m/s.

iii) The period of contact with the ball t is 3.0 x 10-3 s .

iv) The force acting on the ball $F\mathrm{is}\left(6.0\times {10}^6\right)\mathrm{t}-\left(2.0\times {10}^9\right){\mathrm{t}}^2$ .

v) The time for which the force acts is $0\le t\le 3.0\times {10}^{-3}\mathrm{s}$.

Concept and formula used in the given question

The player kicks the ball. It imparts an impulse on the ball during the contact with the ball. We can use the impulse – linear momentum theorem to determine the impulse, average force and maximum force on the ball and given as.

$$\begin{gathered} \mathbf{J}\mathbf{=}\mathbf{∆}\mathbf{p} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{m}\mathbf{}\mathbf{∆}\mathbf{}\mathbf{v} \\ \mathbf{J}\mathbf{=}{\mathbf{\int }}_{{\mathbf{t}}_{\mathbf{1}}}^{{\mathbf{t}}_{\mathbf{2}}}\mathbf{Fdt} \end{gathered}$$

Calculation for the magnitudes of the impulse on the ball due to the kick

(a)

The impulse - linear momentum theorem can be stated as:

$\mathrm{J}={\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\mathrm{Fdt}$

Hence, the impulse can be calculated as

$$\begin{gathered} \mathrm{J}={\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\mathrm{Fdt} \\ ={\int }_0^{3\times {10}^{-3}}\left(\left(6.0\times {10}^6\right)t-\left(2.0\times {10}^9\right)t^2\right)dt \\ ={\left[\left(6.0\times {10}^6\right)\frac{t^2}{2}-\left(2.0\times {10}^9\right)\frac{t^3}{3}\right]}_0^{3\times {10}^{-3}} \\ =\left[\left(3.0\times {10}^6\times 9\times {10}^{-6}\right)-\left(0.66\times {10}^9\times 27\times {10}^{-9}\right)\right]-\left(0\right) \\ =27-18 \\ =9.0\mathrm{N}.\mathrm{s} \end{gathered}$$

 Calculation for the average force on the ball from the player’s foot during the period of contact

(b)

The average force can be calculated as

$$\begin{gathered} J=F_{\mathrm{avg}}∆t \\ =\frac{J}{∆t} \\ =\frac{9.0}{3\times {10}^{-3}} \\ =3\times {10}^3\mathrm{N} \\ =3.0\mathrm{kN} \end{gathered}$$

Calculation for the maximum force on the ball from the player’s foot during the period of contact

(c)

The force is maximum when

$$\begin{gathered} \frac{dF}{dt}=0 \\ \frac{dF}{dt}=\left(6.0\times {10}^6\right)-\left(2\times 2\times {10}^9\right)t \\ \frac{dF}{dt}=0 \\ =\left(6.0\times {10}^6\right)-\left(2\times 2\times {10}^9\right)t=0 \\ \left(2\times 2\times {10}^9\right)t=6.0\times {10}^6 \\ t=\frac{6.0\times {10}^6}{4.0\times {10}^9}=1.5\times {10}^{-3}\mathrm{s} \end{gathered}$$

Thus, force is zero at $1.5\times {10}^{-3}\mathrm{s}$

$$\begin{gathered} F_{max}=F\left(t=1.5\times {10}^{-3}\right)=\left(6.0\times {10}^6\right)\left(1.5\times {10}^{-3}\right)-\left(2\times {10}^9\right){\left(1.5\times {10}^{-3}\right)}^2 \\ {F}_{max}=9.0\times {10}^3-4.5\times {10}^3 \\ =4.5\times {10}^3\mathrm{N} \\ \mathrm{Thus},F_{max}=4.5\mathrm{kN} \end{gathered}$$

Calculation for the ball’s velocity immediately after it loses contact with the player’s foot. 

(d)

The ball’s velocity immediately after it loses contact with the foot of the player can be determined using the definition of impulse as follows

$$\begin{gathered} J=∆\mathrm{p} \\ =m∆v \\ ∆v=\frac{J}{m} \\ =\frac{9.0}{0.45} \\ =20\mathrm{m}/\mathrm{s} \end{gathered}$$

But the initial velocity is zero. Hence change in velocity is equal to final velocity of the ball

Thus, the velocity of the ball after it loses contact with the foot is 20 m/s.