A 350-g ball strikes a wall with horizontal velocity 7.0 m/s and rebounds with horizontal velocity −5.5 m/s. The ball was in contact with the wall for 0.075 s. What was the average horizontal force on the ball while in contact with the wall?

-58.33 N

Explanation:

From Newton's second law of motion,

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

Where F = Average horizontal force, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t = time.

Given: m = 350 g = 0.35 kg, u = 7.0 m/s, v = -5.5 m/s, t = 0.075 s

Substitute into equation 1

F = 0.35(-5.5-7)/0.075

F = 0.35(-12.5)/0.075

F = -58.33 N.

Hence the average horizontal force on the ball while in contact with the wall = -58.33 N

answer: b.

explanation:

momentum of soft ball three times less than baskey ball