Question

Two bodies of masses m1 and m2 have the same linear momentum. What is the ratio of their kinetic energies?

Answer 1

m1v1 = m2v2 = p
⇒v1 = p/m1     and     v2 =p/m2

Ratio of KE = $\frac{0.5*m1*(v1)^{2} }{0.5*m2* (v2)^{2} }$
                    
                    =$\frac{m1*(v1)^{2} }{m2* (v2)^{2} }$

                    =$\frac{m1*(p/m1)^{2} }{m2* (p/m2)^{2} }$
                 
                     = $\frac{ \frac{ p^{2} }{m1} }{ \frac{ p^{2} }{m2} }$

                     =$\frac{m2}{m1}$

Answer 2

$p_1 = momentum\ of\ m_1 = m_1\ v_1\\ p_2 = momentum\ of\ v_2=m_1\ v_2 = p_1 = m_1\ v_1\\ \\Kinetic\ energy=\frac{p^2}{2m}\\ \\Ratio\ of\ Kinetic\ Energies = KE_1 : KE_2 = \frac{p_1^2}{2m_1} : \frac{p_2^2}{2m_2} \\ \\= 1/m_1 : 1/m_2\ as\ p_1=p_2 \\ \\Multiply\ with\ m_1m_2,\\ \\m_2 : m_1\ \ \ or\ \ m_2/m_1$