A Rubber Ball Rebound
A rubber ball of mass 2.0kg falls to the floor. The ball hits with a speed of 8m/s and rebounds with approximately the same speed. High speed photographs show that the ball is in contact with the floor for 10^-3s. What can we say about the force exerted on the ball by the floor?

Explanation

The momentum of the ball just before it hits the floor is `P_a=1.6hatk kg-m//s` and its momentum `10^-3` later is `P_b=+1.6hatk kg-m//s`
Since `underset(t_(a))overset(t_(b))int Fdt = P_(b) - P_(a)`
`underset(t_(a))overset(t_(b))int Fdt = 1.6 hat(k) - (- 1.6 hat(k)) = 3.2 hat(k) kg-m//s`

Although the exact variation of F with time is not known, it is easy to find the average force exerted by the floor on the ball. IF the collision time is `Deltat=t_b-t_a`, the average force `F_av` acting during the collision is
`F_(av) Deltat int_(ta)^(ta+Deltat)Fdt`
Since `Deltat=10^-3s`,
`F_(av)=(3.2hatkkg-m/s)/(10^-2s)=3200hatkN`
The average force is directed upward, as we expect. In more familiar units, `3200N=720lb --a` sizable force. The instantanwous force on the ball is even larger at the peak, as the sketch shows. If the ball hits a resillient surface, the collision time is longer and the peak force is less.
Actually, there is a weakness in our treatment of the rubber ball rebound. In calculating the impulse `intFdt, F` is the total force. this includes the gravitational force, which we have neglected. Proceeding more carefully, we write.
`F=F_(floor)+F_(grav)=F_(floor)-Mghatk`.
The impulse equation then becomes
`int_0^(10^-3)F_("floor")dt- int_0^(10^-3)Mghatkdt=3.2hatkkg-m//s` The impulse due to the gravitational force is `-int_0^(10^-3) Mghatkdt=-Mghatkint_0^(10-3)dt=(0.2)(9.8)(10^-3)hatk=-1.96xx10^-3hatkkg-m//s`