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

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Physics

Goose

2 March, 14:13

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

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Kelley · 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