A girl of mass 60 kg throws a ball of mass 0.8 kg against a wall. The ball strikes the wall horizontally with a speed of 11 m/s, and it bounces back with this same speed. The ball is in contact with the wall for 0.05 s. What is the average force exerted on the wall by the ball?
A. 6600 N
B. 26,400 N
C. 13,200 N
D. 180 N
E. 350 N

352 N

Explanation:

From Newton's second law of motion,

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

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

From the question,

Assuming: The direction of the initial velocity to be negative

Given: m = 0.8 kg, u = -11 m/s, v = 11 m/s (bounce back with the same speed), t = 0.05 s

Substitute into equation 1

F = 0.8[11-(-11)]/0.05

F = 0.8(11+11)/0.05

F = 0.8(22)/0.05

F = 352 N

Hence the average force = 352 N

the answer is a. at the equator