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A 2.2 kg ball strikes a wall with a velocity of 7.4 m/s to the left. The ball bounces off with a velocity of 6.2 m/s to the right. If the ball is in contact with the wall for 0.25 s, what is the constant force exerted on the ball by the wall?

User Beek
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1 Answer

2 votes

Answer:

The constant force exerted on the ball by the wall is 119.68 N.

Explanation:

Consider the provided information.

It is given that the mass of the ball is m = 2.2 kg

The initial velocity of the ball towards left is 7.4 m/s

So the momentum of the ball when it strikes is =
2.2* 7.4=16.28

The final velocity of the ball is -6.2 m/s

So the momentum of the ball when it strikes back is =
2.2* -6.2=-13.64

Thus change in moment is:
16.28-(-13.64)=29.92

The duration of force exerted on the ball t = 0.25 s

Therefore, the constant force exerted on the ball by the wall is:


(29.92)/(0.25)=119.68

Hence, the constant force exerted on the ball by the wall is 119.68 N.

User Pyko
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