91Ó°ÊÓ

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 $F\mathrm{is}\left(6.0×{10}^6\right)\mathrm{t}-\left(2.0×{10}^9\right){\mathrm{t}}^2$ .

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

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.

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

(a)

The impulse - linear momentum theorem can be stated as:

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

Hence, the impulse can be calculated as

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

(b)

The average force can be calculated as

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

(c)

The force is maximum when

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

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

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

(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.