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5.Work, Energy, Power and Collision
hard
A stationary particle breaks into two parts of masses $m_A$ and $m_B$ which move with velocities $v_A$ and $V_B$ respectively. The ratio of their kinetic energies $\left(\mathrm{K}_B: \mathrm{K}_{\mathrm{A}}\right)$ is :
A
$v_B: v_A$
B
$m_B: m_A$
C
$m_B v_B: m_A v_A$
D
$1: 1$
(JEE MAIN-2024)
Solution
Initial momentum is zero.
$\text { Hence }\left|P_A\right|=\left|P_B\right|$
$\Rightarrow m_A v_B=m_B v_B$
$\frac{(\mathrm{KE})_A}{(\mathrm{KE})_B}=\frac{\frac{1}{2} m_A v_A^2}{\frac{1}{2} m_B v_B^2}=\frac{v_A}{v_B}$
$\frac{(\mathrm{KE})_B}{(\mathrm{KE})_A}=\frac{v_B}{v_A}$
Standard 11
Physics
