Answer to Question #104835 in Mechanics | Relativity for Rethabile force

Question

An elastic collision takes place between a 0.090-kg
toy car moving at 10 ×10^2 m/h and a 0.014-kg toy car moving at 20 *10^9 mm/yr

Part A
What is the kinetic energy of the system?

Answer 1

Just add their kinetic energies! But don't forget to convert the units:

$10\cdot10^2$ m/h is $10\cdot10^2/3600=0.28\text{ m/s}.$

Now $20\cdot10^9$ mm/yr. 1 meter has 1000 millimeters, 1 year is 365 days long each 24 hours long each 3600 seconds long:

$$\frac{20\cdot10^9}{[365\cdot24\cdot3600]\cdot1000}=0.634 \text{ m/s}.$$

The kinetic energy of a body is

$$KE=mv^2/2.$$

The kinetic energy of our system is

$$KE=\frac{1}{2}(0.09\cdot0.28^2+0.014\cdot0.634^2)=\\=6.34\cdot10^{-3}\text{ J}.$$