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
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.`