A ball of mass 0.05 kg is strike a smooth wall normally for 4time in2 second with velocity 10 metre per second is time the ball rebounds with the same speed of 10 metre per second calculate the average force of the wall​

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ANSWER:

[tex]\huge\boxed{0.5newton}[/tex]_____________________________________

DATA:

m = 0.05 kg

Vf = 10 m/s

Vi = -10 m/s

t = 4 times hits in 2 second

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SOLUTION:

The momentum  ρ is given by,

                              ρ = mV

where,

           m = mass

           V = velocity

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The Equation for average force is given by,

                               F = Δρ/Δt

where,

           Δt = time taken 

           Δρ = change in momentum. Which is given by,

                               Δρ=[tex]MVf-Mvi[/tex]

                       Vf = final velocity which is the rebound velocity here

                       Vi = Initial velocity which is the velocity before striking the wall

So Substitute the equation of momentum in the equation of force

                        F = [tex]\frac{MVf - MVi}{t}[/tex]

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SUBSTITUTING THE VARIABLES IN THE EQUATION

                               F = [tex]\frac{MVf - MVi}{t}[/tex]

                               F = [tex]\frac{(0.05x10) - (0.05x-10)}{2}[/tex]

                               F = [tex]\frac{(0.5 + 0.5)}{2}[/tex]

                               F = [tex]\frac{1}{2}[/tex]

                               F = 0.5N

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0.5 NEWTON is the average force on the wall

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