Energy of a system-two colliding balls

[haha1234]

Homework Statement

Two identical steel balls,each of diameter 25.4mm and moving in opposite directions at 5m/s,run into each other head-on and bounce apart.Prior to the collision,one of the balls is squeezed in a vise while precise measurements are made of the resulting amount of compression.The results show that Hooke's law is a fair model of the ball's elastic behaviour.For one datum,a force of 16kN exerted by each jaw of the vise results in a 0.2mm reduction in the diameter.The diameter returns to its original value when the force is removed.
Compute an estimate for the kinetic energy of each of the balls before they collide.

Homework Equations

The Attempt at a Solution

The mass of the two balls are not mentioned in the question,so how to find the kinetic energy?

 

[collinsmark]

Homework Statement

Two identical steel balls,each of diameter 25.4mm and moving in opposite directions at 5m/s,run into each other head-on and bounce apart.Prior to the collision,one of the balls is squeezed in a vise while precise measurements are made of the resulting amount of compression.The results show that Hooke's law is a fair model of the ball's elastic behaviour.For one datum,a force of 16kN exerted by each jaw of the vise results in a 0.2mm reduction in the diameter.The diameter returns to its original value when the force is removed.
Compute an estimate for the kinetic energy of each of the balls before they collide.

Homework Equations

The Attempt at a Solution

The mass of the two balls are not mentioned in the question,so how to find the kinetic energy?

I think the key is the problem statement mentioned that the balls are made of steel. You can look up the density of steel. (The density of steel varies depending on the alloying constituents, but I'm guessing the problem is only asking for a rough estimate. Maybe your textbook has a table in it for metal densities that you can use?) Multiply that times the volume of each ball to get the mass of each ball. (Note: when calculating the volume of each ball, don't forget that 25.4 mm is the diameter, not the radius. )

Don't worry about the Hooke's law and the vise experiment for the this part. I'm guessing that information is to be used in a latter part of the problem (not mentioned here).

That's my guess anyway.