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 should remember that the impulse of collision is equal to the change in momentum.
Since the problem provides the mass of the ball and the speed of the ball before and after the collision, you may evaluate the momentum before collision `M_1` and momentum after collision,`M_2` , such that:
`M_1 = 60*10^-3*32 = 0.06*32 = 1.92 (Kg*m)/s`
`M_2 = 60*10^-3*(-32) = -1.92 (Kg*m)/s`
`Delta M = M_1 - M_2 = 1.92 - (-1.92) = 3.84 (Kg*m)/s`
Since the problem provides the graph of force, you may evaluate the impulse of collision as area under the graph, such that:
Impulse = `(6+2)/2*10^(-3)*F_max`
Notice that 6 and 2 represents the lengths of the parallel sides of trapezoid formed by the graph of force with x axis.
`Impulse = Delta M => (6+2)/2*10^(-3)*F_max = 3.84`
`F_max = 3.84/(4*10^(-3)) = 960 N`
Hence, evaluating the magnitude of the contact force during collision yields `F_max = 960 N` .