Q 11-05-079JEE MainJEE Main 2024 (8 Apr, Shift 1)Easy
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 $(K_B : K_A)$ is:
Answer: (A) $v_B : v_A$
By momentum conservation, $m_Av_A = m_Bv_B = p$. Then $K = \frac12pv$:
$$\frac{K_B}{K_A} = \frac{\frac12pv_B}{\frac12pv_A} = \frac{v_B}{v_A}$$
Solution by Sreeraj P, M.Sc Physics