Question

Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is:
$
  {\text{A}}{\text{. }}\dfrac{1}{9} \\
  {\text{B}}{\text{. }}\dfrac{8}{9} \\
  {\text{C}}{\text{. }}\dfrac{4}{9} \\
  {\text{D}}{\text{. }}\dfrac{5}{9} \\
 $

Answer 1

Hint: During the collision, body A transfers it energy to B so, first we need to find out the final velocity of the body B after collision. Then the fraction of energy lost by body A is equal to the final kinetic energy of body B divided by the initial kinetic energy of body A. Using this expression, we can get the required answer.

Complete step by step answer:
We are given two bodies A and B whose masses are given as
$
  {m_1} = 4m \\
  {m_2} = 2m \\
 $
The body A moving with velocities u collides with B at rest. Therefore, their initial velocities are given as
$
  {u_1} = u \\
  {u_2} = 0 \\
 $
Let ${v_1}$ be the final velocity of body A and ${v_2}$ be the final velocity of body B after collision.
In an elastic collision, the final velocity of the body B is given as
$
  {v_2} = \left( {\dfrac{{{m_2} - {m_1}}}{{{m_1} + {m_2}}}} \right){u_2} + \dfrac{{2{m_1}{u_1}}}{{{m_1} + {m_2}}} \\
   = 0 + \dfrac{{2 \times 4mu}}{{4m + 2m}} = \dfrac{{8mu}}{{6m}} = \dfrac{{4u}}{3} \\
 $
The body A will transfer some of its kinetic energy to body B. So, the fractional energy lost by body A is given as
$\dfrac{{{\text{Final kinetic energy of B}}}}{{{\text{Initial kinetic energy of A}}}} = \dfrac{{\dfrac{1}{2}{m_2}v_2^2}}{{\dfrac{1}{2}{m_1}v_1^2}} = \dfrac{{2m \times v_2^2}}{{4m \times u_1^2}} = \dfrac{{v_2^2}}{{2{u^2}}}$
Inserting the expression for final velocity of body B here, we get
$\dfrac{{{\text{Final kinetic energy of B}}}}{{{\text{Initial kinetic energy of A}}}} = \dfrac{{{{\left( {\dfrac{4}{3}u} \right)}^2}}}{{2{u^2}}} = \dfrac{8}{9}$
This is the required answer. Hence, the correct answer is option B.

Additional information:
Applying the law of conservation of linear momentum to the above head on collision, we get
${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$
Inserting the known values, we get
$
  4mu + 0 = 4m{v_1} + 2m{v_2} \\
  2u = 2{v_1} + {v_2} \\
 $
Applying the law of conservation of kinetic energy to the above head on collision, we get
$\dfrac{1}{2}{m_1}u_1^2 + \dfrac{1}{2}{m_2}u_2^2 = \dfrac{1}{2}{m_1}v_1^2 + \dfrac{1}{2}{m_2}v_2^2$
Inserting the known values, we get
$
  \dfrac{1}{2} \times 4m{u^2} = \dfrac{1}{2} \times 4mv_1^2 + \dfrac{1}{2} \times 2mv_2^2 \\
  2{u^2} = 2v_1^2 + v_2^2 \\
 $

Note:
It should be noted that in an elastic collision, the initial momentum is equal to the final momentum and also the initial kinetic energy is equal to the final kinetic energy. In an inelastic collision, the momentum is conserved but the kinetic energy of the reaction is not conserved.