Question

Two identical steel balls, each of mass 71.4 g, are moving in opposite directions at 4.20 m/s.They collide head-on and bounce apart elastically. By squeezing one of the balls in a vise while precise measurements are made of the resulting amount of compression, you find that Hooke's law is a good model of the ball's elastic behavior. A force of 15.1 kN exerted by each jaw of the vise reduces the diameter by 0.110 mm. Model the motion of each ball, while the balls are in contact, as one-half of a cycle of simple harmonic motion. Compute the time interval for which the balls are in contact.

Answer 1

Since the collision is perfectly elastic, the ball will rebound for some height and then repeat the motion over and over again. Thus, the motion we found will be periodic.

To determine the period, use

$x=\frac{1}{2}gt^2\Rightarrow t=\sqrt{\frac{2x}{g}}$

where x= rebound height

This t will equal to the half cycle so time period will be

T=2t

To calculate x: use

$\frac{1}{2}k\Delta x^2=Fx, \textup{ Since k is elastic constant of ball}$

$\textup{Where }\Delta x=0.110\times 10^{-3}m \textup{ and }F=15.1kN$