A 60 g tennis ball with an initial speed of 32 m/s hits a wall and rebounds with the same speed. The figure shows the force of the wall on the ball during the collision. What is the value of F_max, the maximum value of the contact force during the collision? See link for graph: http://i652.photobucket.com/albums/uu245/estara89/graphFigureP0908.jpg
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The figure representing the force of the wall as a function of time is below. We have three regions for the force:
`F(t) = (F_max/2)*t ` for `t in [0,2] ms`
`F(t) =F_max ` for `t in [2,4] ms`
`F(t) = -(F_max/2)*t +3*Fmax` for `t in [4,6] ms`
The total impulse change delivered by the wall to the ball is
`Delta(P) = int_0^0.006F(t)dt =int_0^0.002(F_max/2)*t*dt +int_0.002^0.004F_max*dt +int_0.004^0.006(3*F_max-F_max/2)*dt=`
`=F_max*t^2/4 (0->0.002) +F_max*t (0.002->0.004) +3*F_max*t (0.004->0.006) -F_max*t^2/4 (0.004->0.006) =`
`=F_max*(10^-6 +0.004-0.002 +3*0.006-3*0.004-9*10^-6 +4*10^-6)=`
`=7.996*10^-3*F_max "(kg*m/s)"`
The total change in impulse for the ball is simply
`Delta(P) = m*V_f -m*V_i =2*0.06*32 =3.84 kg*m/s`
Therefore
`7.996*10^-3*F_max =3.84`
`F_max =3.84/(7.996*10^-3) =480.24 N`
The maximum value of the contact force during the collision is F=480.24 N
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