Question

A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball?

Answer 1

Answer:

800 N

Explanation:

Force: This can be defined as the product of mass and acceleration. The S.I unit of force is Newton(N).

From the question,

F = m(v-u)/t ....................... Equation 1

Where F = force exerted on the wall by the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time.

Note: Assuming the direction of the initial motion to be negative

Given: m = 0.80 kg, v = 25 m/s ( bounce back), u = -25 m/s, t = 0.05 s

Substitute into equation 1

F = 0.8[25-(25)]/0.05

F = 0.8(50)/0.05

F = 0.8(1000)

F = 800 N

Answer 2

Answer:

$F=800N$

Explanation:

The average force is defined as the mass of the body multiplied by its average velocity over the contact time. According to the third Newton's law,  the magnitude of the average force exerted on the wall by the ball is equal to  the magnitude of the average force exerted on the ball by the wall. Thus:

$F=m\frac{v_f-v_i}{t}\\F=0.80kg\frac{25\frac{m}{s}-(-25\frac{m}{s})}{0.05s}\\F=800N$