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. I

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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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Helga 5 years 2021-08-05T05:28:18+00:00 1 Answers 17 views 0

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  1. Euphemia
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    2021-08-05T05:29:27+00:00
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    Answer:

    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

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