91Ó°ÊÓ

Mass of the ball is, m = 150 g = 0.150 kg .

The initial speed of the ball is,${\mathrm{v}}_{\mathrm{i}}=5.2\frac{\mathrm{m}}{\mathrm{s}}$.

Final energy of the ball = ½ Initial energy of the ball.

Time interval of the contact,$∆t=7.6\mathrm{ms}$ .

We can find the speed of the ball after rebounding using the given relation between the initial and final Kinetic Energy of the ball. Impulse and average force acting on the ball can be calculated using corresponding formulae.

Formula:

J = P

$F_{avg}=\frac{J}{t}$

Let’s consider the direction of the incidenceof the ballon the wall as negative.

Hence, the initial speed of the ball is,${\mathrm{v}}_{\mathrm{i}}=-5.2\frac{\mathrm{m}}{\mathrm{s}}$.

K.E of the ball after collision = ½ kinetic energy of the ball before the collision,

$\frac{1}{2}m{v}_f^2=\left(\frac{1}{2}\right)\frac{1}{2}m{v}_i^2$

$v_f^2=\frac{1}{2}{v}_i^2$

$v_f=\frac{v_i}{\sqrt{2}}$

Substitute the values in the above expressions, and we get the speed of the ball immediately after rebounding as,

$$\begin{gathered} v_f=\frac{5.2}{\sqrt{2}} \\ =3.7\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the ball immediately after rebounding is 3.7 m/s .

Impulse on the wall from the ball can be calculated as,

$$\begin{gathered} J=∆P \\ =m\left(v_f-v_i\right) \end{gathered}$$

Substitute the values in the above expressions, and we get,

$$\begin{gathered} J=\left(0.15\mathrm{kg}\right)\left[\left(3.7\mathrm{m}/\mathrm{s}\right)-\left(-5.2\mathrm{m}/\mathrm{s}\right)\right] \\ =1.3\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the magnitude of the impulse on the wall from the ball is 1.3 kg. m/s .

The average force on the ball from the wall is,

$F_{avg}=\frac{J}{∆t}$

Substitute the values in the above expressions, and we get,

$$\begin{gathered} F_{avg}=\frac{1.3}{0.0076} \\ =1.8×{10}^2\mathrm{N} \end{gathered}$$

Therefore, the magnitude of the average force on the ball from the wall during 7.6 ms is$1.8×{10}^2\mathrm{N}$ .