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
Expert's answer
2020-03-10T11:12:32-0400

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

1010210\cdot10^2 m/h is 10102/3600=0.28 m/s.10\cdot10^2/3600=0.28\text{ m/s}.


Now 2010920\cdot10^9 mm/yr. 1 meter has 1000 millimeters, 1 year is 365 days long each 24 hours long each 3600 seconds long:


20109[365243600]1000=0.634 m/s.\frac{20\cdot10^9}{[365\cdot24\cdot3600]\cdot1000}=0.634 \text{ m/s}.

The kinetic energy of a body is


KE=mv2/2.KE=mv^2/2.

The kinetic energy of our system is


KE=12(0.090.282+0.0140.6342)==6.34103 J.KE=\frac{1}{2}(0.09\cdot0.28^2+0.014\cdot0.634^2)=\\=6.34\cdot10^{-3}\text{ J}.


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