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How high must be the kinetic energy of a proton striking another, stationary, proton for their combined kinetic energy in the frame of the centre of inertia to be equal to the total kinetic energy of two protons moving toward each other with individual kinetic energies T=25.0 GeV? |
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Answer» Solution :Let `T^'=` KINETIC ENERGY of a proton striking another stationary particle of the same rest mass. Then, COMBINED kinetic energy in the CM FRAME `=2m_0c^2(sqrt(1+(T^')/(2m_0c^2)-1))=2T`, `((T)/(m_0c^2)+1)^2=(T^')/(2m_0c^2)` `(T^')/(2m_0c^2)=(T(2m_0c^2+T))/(m_0^2c^4)`, `T^'=(2T(T+2m_0c^2))/(m_0c^2)` |
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