g During a collision with a wall, the velocity of a 0.4 KgKg ball changes from 25 m/sm/s towards the vall to 12 m/sm/s away from the wall. If the time the ball was in contact with the call was 0.5 secsec, what was the magnitude of the avarage force applied to the ball

Question

Orlando Finley

Physics

8 October, 03:31

g During a collision with a wall, the velocity of a 0.4 KgKg ball changes from 25 m/sm/s towards the vall to 12 m/sm/s away from the wall. If the time the ball was in contact with the call was 0.5 secsec, what was the magnitude of the avarage force applied to the ball

+5

Answers (1)

Know the Answer?

Iliana Cisneros · Answer 1

F = 10.8N

Explanation:

Given the mass m = 0.4kg, v1 = 25m/s, v2 = 12m/s and t = 0.5s

From Newtown's second law of motion the average force can be found. This law states that the product of the force experienced by a body and the time t of the force acting on the body is equal to the change in momentum of the body. Mathematically it can be stated as follows

F*t = m (v2 - v1)

F = m (v2 - v1) / t = 0.4 (25 - 12) / 0.5 = 10.8N