# 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 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...

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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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