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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You need to evaluate the momentum before and after collision such that:

`p_1 = 60*10^-3*32 = 1.92`

`p_2 = 60*10^-3*(-32) = -1.92`

You need to evaluate the change in momentum `Delta p`  such that:

`Delta p = p_1 - p_2 = 1.92 - (-1.92) = 3.84`

You need to remember that...

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You need to evaluate the momentum before and after collision such that:

`p_1 = 60*10^-3*32 = 1.92`

`p_2 = 60*10^-3*(-32) = -1.92`

You need to evaluate the change in momentum `Delta p`  such that:

`Delta p = p_1 - p_2 = 1.92 - (-1.92) = 3.84`

You need to remember that the change in momentum is also called impulse such that:

`Delta p = int_(t_1)^(t_2) F(t) dt `

You need to evaluate the area under the given curve, which is teh area of trapezoid, such that:

`J = (6+2)/2*10^(-3)*F_max`

Substituting `3.84`  for J yields:

`3.84 = 0.004*F_max => F_max = 3.84/0.004 = 960 N`

Hence, evaluating the maximum value of the contact force, during collision, yields`F_max = 960 N.`

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