Answer:
[tex]F = 371.738\,N[/tex]
Explanation:
Let assume that ball strikes a vertical wall in horizontal direction. The situation can be modelled by the appropriate use of the definition of Moment and Impulse Theorem, that is:
[tex](0.544\,kg)\cdot (14.3\,\frac{m}{s})-F\cdot \Delta t = -(0.544\,kg)\cdot (14.4\,\frac{m}{s} )[/tex]
[tex]F\cdot \Delta t = 15.613\,\frac{kg}{m\cdot s}[/tex]
The average force acting on the ball during the collision is:
[tex]F = \frac{15.613\,\frac{kg}{m\cdot s} }{0.042\,s}[/tex]
[tex]F = 371.738\,N[/tex]
Answer:
-371.73 N
Explanation:
From Newton's Second Law of motion,
F = m(v-u)/t..................... Equation 1
Where F = Force acting on the ball, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t= time.
Note: Let the direction of the initial velocity be positive
Given: m = 544 g = (544/1000) kg = 0.544 kg, u = 14.3 m/s, v = -14.4 m/s (rebounds), t = 0.042 s.
Substitute into equation 1
F = 0.544(-14.4-14.3)/0.042
F = 0.544(-28.7)/0.042
F = -371.73 N
Note: The force is negative because it act against the direction of the initial motion of the ball
