Three bodies A, B and C have equal kinetic energies and their masses are $400 \mathrm{~g}$. $1.2…

Three bodies A, B and C have equal kinetic energies and their masses are $400 \mathrm{~g}$. $1.2 \mathrm{~kg}$ and $1.6 \mathrm{~kg}$ respectively. The ratio of their linear momenta is :
  1. $\sqrt{2} ; \sqrt{3} ; 1$
  2. $1: \sqrt{3}: 2$
  3. $1: \sqrt{3}: \sqrt{2}$
  4. $\sqrt{3}: \sqrt{2}: 1$

Solution

$\begin{aligned} & \mathrm{KE}=\frac{\mathrm{P}^2}{2 \mathrm{~m}} \\ & \mathrm{P} \propto \sqrt{\mathrm{m}}\end{aligned}$ Hence, $\mathrm{P}_{\mathrm{A}}: \mathrm{P}_{\mathrm{B}}: \mathrm{P}_{\mathrm{C}}$ $=\sqrt{400}: \sqrt{1200}: \sqrt{1600}=1: \sqrt{3}: 2$

Asked in: JEE Main 2024 (08 Apr Shift 1)

Practice more Center of Mass Momentum and Collision questions on Aicharya