Sure! Let's break down the solution step-by-step for each part of the question. We'll find the impulse, average force, maximum force, and the ball's speed:
### Given:
- Mass of the soccer ball, [tex]\( m = 0.45 \, \text{kg} \)[/tex]
- Contact time, [tex]\( t_{\text{contact}} = 1.70 \times 10^{-3} \, \text{s} \)[/tex]
- Force function, [tex]\( F(t) = (7.93 \times 10^5) t - (4.66 \times 10^8) t^2 \)[/tex]
### (a) Impulse on the ball
The impulse [tex]\( J \)[/tex] on the ball is given by the integral of the force over the time interval. This can be mathematically expressed as:
[tex]\[ J = \int_{0}^{t_{\text{contact}}} F(t) \, dt \][/tex]
Using the force function [tex]\( F(t) = (7.93 \times 10^5) t - (4.66 \times 10^8) t^2 \)[/tex] and evaluating the integral from [tex]\( 0 \)[/tex] to [tex]\( 1.70 \times 10^{-3} \, \text{s} \)[/tex], we find:
[tex]\[ J = 0.3827323333333333 \, \text{N} \cdot \text{s} \][/tex]
### (b) Average force on the ball
The average force [tex]\( F_{\text{avg}} \)[/tex] can be found using the impulse and the contact time. It is given by:
[tex]\[ F_{\text{avg}} = \frac{J}{t_{\text{contact}}} \][/tex]
Substituting the values for impulse and contact time:
[tex]\[ F_{\text{avg}} = \frac{0.3827323333333333 \, \text{N} \cdot \text{s}}{1.70 \times 10^{-3} \, \text{s}} \approx 225.13666666666666 \, \text{N} \][/tex]
### (c) Maximum force on the ball
The maximum force [tex]\( F_{\text{max}} \)[/tex] is simply the highest value of the force function [tex]\( F(t) \)[/tex] during the period [tex]\( 0 \leq t \leq 1.70 \times 10^{-3} \, \text{s} \)[/tex].
Evaluating [tex]\( F(t) \)[/tex] over this period:
[tex]\[ F_{\text{max}} = 337.3653433212993 \, \text{N} \][/tex]
### (d) Ball's speed immediately after it loses contact with the player's foot
The speed of the ball [tex]\( v \)[/tex] after it loses contact with the player's foot can be found using the relationship between impulse and change in momentum. Since the initial velocity is zero (ball is at rest), the final momentum is equal to the impulse:
[tex]\[ J = m \cdot v \][/tex]
Solving for [tex]\( v \)[/tex]:
[tex]\[ v = \frac{J}{m} \][/tex]
Substituting the values for impulse and mass:
[tex]\[ v = \frac{0.3827323333333333 \, \text{N} \cdot \text{s}}{0.45 \, \text{kg}} \approx 0.8505162962962962 \, \text{m/s} \][/tex]
### Summary:
(a) Impulse on the ball: [tex]\( 0.3827323333333333 \, \text{N} \cdot \text{s} \)[/tex]
(b) Average force on the ball: [tex]\( 225.13666666666666 \, \text{N} \)[/tex]
(c) Maximum force on the ball: [tex]\( 337.3653433212993 \, \text{N} \)[/tex]
(d) Speed of the ball after contact: [tex]\( 0.8505162962962962 \, \text{m/s} \)[/tex]
