Work, Energy and Power — Question 13

We know that,

$\text{K.E.} = \dfrac{1}{2} \times m \times v^2 \\[0.5em]$

Let the two bodies be A and B.

Let mass of body A be m_A and body B be m_B. Let velocity of body A be v_A and body B be v_B.

Given,

Ratio of masses:

$\dfrac {m_A}{m_B} = \dfrac {5}{1} \\[0.5em]$

Ratio of K.E.:

$\dfrac {KE_A}{KE_B} = \dfrac {125}{9} \\[1em] \dfrac {KE_A}{KE_B} = \dfrac{\dfrac{1}{2} \times m_A \times v_A^2}{\dfrac{1}{2} \times m_B \times v_B^2} \\[0.5em] \Rightarrow \dfrac {125}{9} = \dfrac{\dfrac{1}{2} \times m_A \times v_A^2}{\dfrac{1}{2} \times m_B \times v_B^2} \\[0.5em] \Rightarrow \dfrac {125}{9} = \dfrac{m_A}{m_B} \times \Big(\dfrac{v_A}{v_B}\Big)^2 \\[0.5em] \Rightarrow \dfrac {125}{9} = \dfrac{5}{1} \times \Big(\dfrac{v_A}{v_B}\Big)^2 \\[0.5em] \Rightarrow \Big(\dfrac{v_A}{v_B}\Big)^2 = \dfrac{125}{9} \times \dfrac{1}{5} \\[0.5em] \Rightarrow \Big(\dfrac{v_A}{v_B}\Big)^2 = \dfrac{25}{9} \\[0.5em] \Rightarrow \dfrac{v_A}{v_B} = \sqrt{\dfrac{25}{9}} \\[0.5em] \Rightarrow \dfrac{v_A}{v_B} = \dfrac{5}{3} \\[0.5em]$

∴ Ratio of their velocities = 5:3