Question
A stationary particle explodes into two particles of masses $${m_1}$$ and $${m_2},$$ which move in opposite directions with velocities $${v_1}$$ and $${v_2}.$$ The ratio of their kinetic energies $$\frac{{{E_1}}}{{{E_2}}}$$ is
A.
$$1$$
B.
$$\frac{{{m_1}{v_2}}}{{{m_2}{v_1}}}$$
C.
$$\frac{{{m_2}}}{{{m_1}}}$$
D.
$$\frac{{{m_1}}}{{{m_2}}}$$
Answer :
$$\frac{{{m_2}}}{{{m_1}}}$$
Solution :
From conservation of linear momentum,
Initial momentum $${p_{{\text{initial}}}} =$$ Find momentum $${p_{{\text{final}}}}$$
$$\eqalign{
& 0 = {m_1}{v_1} - {m_2}{v_2} \cr
& {\text{or}}\,\,{m_1}{v_1} = {m_2}{v_2} \cr
& {\text{or}}\,\,\frac{{{v_1}}}{{{v_2}}} = \frac{{{m_1}}}{{{m_2}}}\,......\left( {\text{i}} \right) \cr} $$
Thus, ratio of kinetic energies,
$$\frac{{{K_1}}}{{{K_2}}} = \frac{{\frac{1}{2}{m_1}v_1^2}}{{\frac{1}{2}{m_2}v_2^2}} = \frac{{{m_1}}}{{{m_2}}} \times {\left( {\frac{{{m_2}}}{{{m_1}}}} \right)^2} = \frac{{{m_2}}}{{{m_1}}}$$
NOTE
In a collision of two bodies, whether it is perfectly elastic or inelastic, linear momentum is always conserved but kinetic energy need not be conserved.