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

Question #104835
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?
1
2020-03-10T11:12:32-0400

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}."

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