A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?

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Solution

$p_{1} = m_{1} v_{1}$ $p_{2} = m_{2} v_{2}$
But $p_{1} = p_{2}$ or $m_{1} v_{1} = m_{2} v_{2}$

If $m_{1} < m_{2}$ then $v_{1} > v_{2}$
$\begin{matrix} (K . E)_{1} = \frac{1}{2} m_{1} v_{1}^{2} & (K . E)_{2} = \frac{1}{2} m_{2} v_{2}^{2} (K . E)_{1} = \frac{1}{2} (m_{1} v_{1}) v_{1} = \frac{1}{2} p_{1} v_{1} & (K . E)_{2} = \frac{1}{2} (m_{2} v_{2}) v_{2} = \frac{1}{2} p_{2} v_{2} \end{matrix}$

considering $p_{1} = p_{2}$
$\frac{(K . E)_{1}}{(K . E)_{2}} = \frac{\frac{1}{2} p_{2} v_{1}}{\frac{1}{2} p_{2} v_{2}} = \frac{v_{1}}{v_{2}}$

But $v_{1} > v_{2}$
Therefore, $(K . E)_{1} > (K . E)_{2}$

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A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?

p1=m1v1 p2=m2v2 But p1=p2 or m1v1=m2v2 If m1<m2 then v1>v2 (K.E)1=12m1v21(K.E)2=12m2v22(K.E)1=12(m1v1)v1=12p1v1(K.E)2=12(m2v2)v2=12p2v2 considering p1=p2 (K.E)1(K.E)2=12p2v112p2v2=v1v2 But v1>v2 Therefore, (K.E)1>(K.E)2