Question

A 544 g ball strikes a wall at 14.3 m/s and rebounds at 14.4 m/s. The ball is in contact with the wall for 0.042 s. What is the magnitude of the average force acting on the ball during the collision

Answer 1

-371.73 N

Step-by-step 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

Answer 2

`F = 371.738\,N`

Step-by-step 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:

`(0.544\,kg)\cdot (14.3\,(m)/(s))-F\cdot \Delta t = -(0.544\,kg)\cdot (14.4\,(m)/(s) )`

`F\cdot \Delta t = 15.613\,(kg)/(m\cdot s)`

The average force acting on the ball during the collision is:

`F = (15.613\,(kg)/(m\cdot s) )/(0.042\,s)`

`F = 371.738\,N`