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

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

Answer 2

Answer:

$F = 371.738\,N$

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:

$(0.544\,kg)\cdot (14.3\,\frac{m}{s})-F\cdot \Delta t = -(0.544\,kg)\cdot (14.4\,\frac{m}{s} )$

$F\cdot \Delta t = 15.613\,\frac{kg}{m\cdot s}$

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

$F = \frac{15.613\,\frac{kg}{m\cdot s} }{0.042\,s}$

$F = 371.738\,N$