Solution to Question 10

Physics Solution: Elastic Collision Deformation

1. Understanding the System

We have a lighter particle of mass $ m $ moving with initial velocity $ v $ and a stationary heavier particle of mass $ 2m $. We need to find the kinetic energy lost by the lighter particle at the point of maximum deformation.

Mass of projectile: $ m_1 = m $
Velocity of projectile: $ u_1 = v $
Mass of target: $ m_2 = 2m $
Velocity of target: $ u_2 = 0 $

2. Velocity at Maximum Deformation

During a head-on collision, maximum deformation occurs when both particles move with the same common velocity. This common velocity is the velocity of the Center of Mass ($ v_{cm} $).

$$ v_{cm} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} $$

Substituting the values:

$$ v_{cm} = \frac{m(v) + 2m(0)}{m + 2m} = \frac{mv}{3m} = \frac{v}{3} $$

So, at the moment of maximum deformation, the lighter particle $ m $ is moving with velocity $ \frac{v}{3} $.

3. Calculating Kinetic Energy Loss

We compare the initial kinetic energy of the lighter particle with its kinetic energy at the moment of maximum deformation.

Initial Kinetic Energy ($ K_i $):

$$ K_i = \frac{1}{2} m v^2 $$

Kinetic Energy at Deformation ($ K_f $):

$$ K_f = \frac{1}{2} m (v_{cm})^2 = \frac{1}{2} m \left( \frac{v}{3} \right)^2 = \frac{1}{2} m \left( \frac{v^2}{9} \right) = \frac{1}{18} m v^2 $$

Loss in Kinetic Energy ($ \Delta K $):

$$ \Delta K = K_i – K_f $$ $$ \Delta K = \frac{1}{2} m v^2 – \frac{1}{18} m v^2 $$ $$ \Delta K = \left( \frac{9}{18} – \frac{1}{18} \right) m v^2 = \frac{8}{18} m v^2 = \frac{4}{9} m v^2 $$

Final Answer

The kinetic energy lost by the lighter particle during the period of deformation is (c) $ \frac{4}{9}mv^2 $.