Question

A body of mass m moving with velocity v makes a head on collision with another body of mass 2m which is initially at rest. The ratio of kinetic energies of colliding body before and after collision will be:
A. 9:1
B. 1:1
C. 4:1
D. 2:1

Answer 1

Hint: When the front of two vehicles hit each other, then the collision is called a head on collision. This happens when they are travelling in opposite directions.

Complete step by step solution:
Let the body mass ’m’ be donated by letter ‘A’ and the body mass ‘2m’ be denoted by letter ‘B’
Using law of conservation of energy,

$\dfrac{1}{2}{m_A}u_A^2 + \dfrac{1}{2}{m_B}u_B^2 = \dfrac{1}{2}{m_A}v_A^2 + \dfrac{1}{2}{m_B}v_B^2$

For perfectly elastic collision
The final velocity of first body is

${v_A} = \dfrac{{\left( {{m_A} - {m_B}} \right)}}{{\left( {{m_A} + {m_B}} \right)}}{u_A} + \left( {\dfrac{{2{m_B}{u_B}}}{{{m_A} + {m_B}}}} \right)$

But ${u_B} = 0$

$\therefore {v_A} = \left( {\dfrac{{{m_A} - {m_B}}}{{{m_A} + {m_B}}}} \right){u_A}$
${v_A} = \left( {\dfrac{{m - 2m}}{{m + 2m}}} \right){u_A}{\text{ }}\left[ {\because \begin{array}{*{20}{c}}
  {{m_A} = m} \\
  {{m_B} = 2m}
\end{array}} \right]$
${v_A} = \dfrac{{ - {u_A}}}{3}$

Hence ratio of kinetic energies $ = \dfrac{{\dfrac{1}{2}{m_A}u_A^2}}{{\dfrac{1}{2}{m_A}v_A^2}}$

Note: We can also find out the ratio of loss of kinetic energy and initial kinetic energy from this result. This is equal to 8/9.