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A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for 0.1 , what force did the wall impart onto the ball?

Sagot :

Answer:

5N

Explanation:

We have a simple problem of momentum here.

ΔMomentum= mΔv= FΔt

Solve for F

mΔv/Δt=F

Plug in givens

1*(2-1.5)/0.1=F

F=5N

abidemiokin

The amount of force that the wall imparts on the ball is 5.0N

According to Newton's second law, the formula for calculating the force applied is expressed as:

[tex]F=ma[/tex]

m is the mass of the object

a is the acceleration of the object

Since acceleration is the change in velocity of an object, hence [tex]a=\frac{\triangle v}{t}[/tex]

The applied force formula becomes [tex]F=\frac{m\triangle v}{t}[/tex]

Given the following parameters

m = 1.0kg

[tex]\triangle v=2.0-1.5\\\triangle v=0.5m/s[/tex]

t = 0.1sec

Substitute the given parameter into the formula

[tex]F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N[/tex]

Hence the amount of force that the wall imparts on the ball is 5.0N

Learn more here: https://brainly.com/question/17811936

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