A 2.9 kg ball strikes a wall with a velocity of 8.6 m/s to the left. The ball bounces off with a velocity of 7.3 m/s to the right. If the ball is in contact with the wall for 0.33 s, what is the constant force exerted on the ball by the wall? Answer in units of N.

The constant force exerted by the wall is F=11.4N

Explanation:

The problem bothers on the impulse of a force

Which is given as

Ft=mv-mu

Ft=m(v-u)

Given data

mass of ball m =2.9kg

Final speed v=8.6m/s

Initial speed u=7.3m/s

Time t= 0.33s

Substituting to find F

F*0.33=2.9(8.6-7.3)

F*0.33=2.9*1.3

F=3.77/0.33

F=11.4N

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

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