Question

Three bodies A, B and C have equal kinetic energies and their masses are 400 g, 1.2 kg and 1.6 kg respectively. The ratio of their linear momenta is :

• A. $1 : \sqrt{3} : 2$
• B. $1 : \sqrt{3} : \sqrt{2}$
• C. $\sqrt{2} : \sqrt{3} : 1$
• D. $\sqrt{3} : \sqrt{2} : 1$

Answer 1

Answer: (A)

$1 : \sqrt{3} : 2$

Answer 2

The kinetic energy is given by:
$KE = \frac{P^2}{2m}, \quad \text{where } P \propto \sqrt{m}.$
For masses $m_A = 400 \, \text{g}, \, m_B = 1.2 \, \text{kg}, \, m_C = 1.6 \, \text{kg}$:
$P_A : P_B : P_C = \sqrt{400} : \sqrt{1200} : \sqrt{1600}.$
Simplify:
$P_A : P_B : P_C = 1 : \sqrt{3} : 2.$
Final Answer: 1 : $\sqrt{3}$ : 2.