Question

Two identical steel balls, each of mass 67.8 g, are moving in opposite directions at 4.80 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 15.9 kN exerted by each jaw of the vise reduces the diameter by 0.130 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

Step-by-step explanation:

To detrmine the time interval at which the balls are in contact.

Given information

The mass of each steal ball 67.8g. The speed of ball towards each other is 4.80 m/s. The exerted force by each jaw 15.9 kN and the forece reduce the diameter by 0.130 mm.

Expression for the effective spring constant ball is shown below.

K = |F|/|x|

Here,

k is a spring constant

F is the force exerted on the ball

x is dispalcement due force

substitute 15.9 kN for F and 0.130 mm in above equation

K = (15.9 kN)(1X10³N) / (0.130 mm)(1x10⁻³m/1mm)

122 x 10⁶ N/m

The spring constant is 122 x 10⁶ N/m