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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 $\dfrac{E_1}{E_2}$ is :
A
$\dfrac{m_2}{m_1}$
B
$\dfrac{m_1}{m_2}$
C
1
D
$\dfrac{m_1v_2}{m_2v_1}$
Detailed Solution
The particle is initially at rest, so its momentum is zero. By conservation of momentum, the two fragments have equal and opposite momenta:
$m_1v_1 = m_2v_2$, i.e. $p_1 = p_2 = p$
Kinetic energy in terms of momentum: $E = \dfrac{p^2}{2m}$
$\dfrac{E_1}{E_2} = \dfrac{p^2/2m_1}{p^2/2m_2} = \dfrac{m_2}{m_1}$
So the lighter fragment carries more kinetic energy, and $\dfrac{E_1}{E_2} = \dfrac{m_2}{m_1}$.
$m_1v_1 = m_2v_2$, i.e. $p_1 = p_2 = p$
Kinetic energy in terms of momentum: $E = \dfrac{p^2}{2m}$
$\dfrac{E_1}{E_2} = \dfrac{p^2/2m_1}{p^2/2m_2} = \dfrac{m_2}{m_1}$
So the lighter fragment carries more kinetic energy, and $\dfrac{E_1}{E_2} = \dfrac{m_2}{m_1}$.