A stationary bomb explodes into two fragments of masses $m_{1}$ and $m_{2}$ . IF the momentum of these fragments are $p_{1}$ and $p_{2}$ then what will be the ratio of their kinetic energies?

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Solution

Given : Initial condition stationary
$∴ P_{1} + P_{2} = 0 m_{1} v_{1} = - m_{2} v_{2} (m_{1} v_{1})^{2} = - (- m_{2} v_{2})^{2} \Rightarrow m_{1}^{2} v_{1}^{2} = m_{2}^{2} v_{2}^{2} \Rightarrow \frac{1}{2} m_{1}^{2} v_{1}^{2} = \frac{1}{2} m_{2}^{2} v_{2}^{2} \Rightarrow \frac{\frac{1}{2} m_{1} v_{1}^{2}}{\frac{1}{2} m_{2} v_{2}^{2}} = \frac{m_{2}}{m_{1}} \Rightarrow \frac{K . E_{1}}{K . E_{2}} = \frac{m_{2}}{m_{1}}$

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A stationary bomb explodes into two fragments of masses m1 and m2. IF the momentum of these fragments are p1 and p2 then what will be the ratio of their kinetic energies?

Given : Initial condition stationary∴P1+P2=0m1v1=−m2v2(m1v1)2=−(−m2v2)2⇒m21v21=m22v22⇒12m21v21=12m22v22⇒12m1v2112m2v22=m2m1⇒K.E1K.E2=m2m1

A stationary bomb explodes into two fragments of masses $m_{1}$ and $m_{2}$ . IF the momentum of these fragments are $p_{1}$ and $p_{2}$ then what will be the ratio of their kinetic energies?