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A proton moves with a momentum p=10.0GeV//c, where c is the velocity of light. How much (in per cent) does the proton velocity differ from the velocity of light? |
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Answer» <P> Solution :We have`(m_0v)/(SQRT(1-V^2/c^2))=p` or, `(m_0)/(sqrt(1-v^2/c^2))=sqrt(m_0^2+p^2/c^2)` or `1-v^2/c^2=(m_0^2c^2)/(m_0^2c^2+p^2)=1-(p^2)/(p^2+m_0^2c^2)` or `v=(c_p)/(sqrt(p^2+m_0^2c^2))=(c)/(sqrt(1+((m_0c)/(p))^2)` So `(c-v)/(c)=[1-(1+((m_0c)/(p))^2)^(-1//2)]xx100%~=1/2((m_0c)/(p))^2xx100%` |
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