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a light body A and a heavy body B have same linear momentum. then kinetic energy of A with respect to B is???
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Linear momentum is given by: \[\bar p=m·\bar v\]then\[\bar p_A=m_A·\bar v_A\]and \[\bar p_B=m_B·\bar v_B\]On the other hand: \[\frac{ E_{K_A} }{ E_{K_B} }=\frac{ m_A·v_A }{ m_B·v_B }\frac{ v_A }{ v_B }=\frac{ p_A }{ p_B }\frac{ v_A }{ v_B }=\frac{ v_A }{ v_B }\]
Addition: E=1/2 mv^2 u left out the 1/2, but it can be cancelled anyway :)
\[E_k=\frac{ 1 }{ 2 }mv^2=\frac{ 1 }{ 2m }m^2v^2=\frac{p^2 }{ 2m }\] \[\frac{ E_{heavy} }{ E_{light} }=\frac{ \frac{ p^2 }{ 2m_{heavy} } }{ \frac{p^2 }{2m_{light} }}=\frac{ m_{light} }{ m_{heavy} }\]
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