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3 years ago

6

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

1 answer:

bogdanovich [222]3 years ago

5 0

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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The height of the gate is $h = 3 \ m$

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