Question

A ball impinges directly on a similar ball at rest. The first ball is brought to rest by the impact. If half of the kinetic energy is lost by impact, the value of coefficient of restitution is
A) \[\dfrac{1}{{2\sqrt 2 }}\]
B) \[\dfrac{1}{{\sqrt 3 }}\]
C) \[\dfrac{1}{{\sqrt 2 }}\]
D) \[\dfrac{{\sqrt 3 }}{2}\]

Answer 1

Hint:Here we will use a kinetic energy formula. Also, net change in the kinetic energies will also be calculated.

Complete answer:Kinetic Energy: When an object is given some motion then it possesses kinetic energy. Its standard formula is given as, \[{\text{k}} = \dfrac{1}{2}{\text{m}}{{\text{u}}^2}\]where, m is the mass of that object and u is the velocity at which it is moving.

Let the first ball is moving with initial velocity \[{{\text{u}}_1}\], final velocity be \[{{\text{v}}_1}\]. Similarly, the initial velocity of the second ball be \[{{\text{u}}_2}\]and final velocity be \[{{\text{v}}_2}\]. Let the mass of the ball is m.
Now, we will compute the initial kinetic energy and the final kinetic energy of the balls. The formula for total initial kinetic energy is given as,

\[{{\text{K}}_{\text{i}}} = \dfrac{1}{2}{\text{m}}{{\text{u}}_1}^2 + \dfrac{1}{2}{\text{m}}{{\text{u}}_2}^2\]

Total final kinetic energy is given as,

\[{{\text{K}}_{\text{f}}} = \dfrac{1}{2}{\text{m}}{{\text{v}}_1}^2 + \dfrac{1}{2}{\text{m}}{{\text{v}}_2}^2\]

Velocities \[{{\text{u}}_2}\]and\[{{\text{v}}_1}\]are zero. Net change or loss of the kinetic energy is given as,

\[\Delta = \]Total initial kinetic energy –Total final kinetic energy
\[
  \Delta = \dfrac{1}{2}{\text{m}}{{\text{u}}_1}^2 + \dfrac{1}{2}{\text{m}}{{\text{u}}_2}^2 - \left( {\dfrac{1}{2}{\text{m}}{{\text{v}}_1}^2 + \dfrac{1}{2}{\text{m}}{{\text{v}}_2}^2} \right) \\
   = \dfrac{1}{2}{\text{m}}{{\text{u}}_1}^2 + 0 - 0 - \dfrac{1}{2}{\text{m}}{{\text{v}}_2}^2 \\
  \Delta = \dfrac{1}{2}{\text{m}}{{\text{u}}_1}^2 - \dfrac{1}{2}{\text{m}}{{\text{v}}_2}^2 \\
\]

Considering the given question we observe that half of the kinetic energy is lost by impact. Therefore,
Following expression is obtained,

\[ \dfrac{1}{2}\left( {\dfrac{1}{2}{\text{m}}{{\text{v}}_1}^2} \right) = \dfrac{1}{2}{\text{m}}{{\text{u}}_1}^2 - \dfrac{1}{2}{\text{m}}{{\text{v}}_2}^2 \\
  {\text{m}}{{\text{v}}_1}^2 = 2{\text{m}}{{\text{u}}_1}^2 - 2{\text{m}}{{\text{v}}_2}^2 \\
  {{\text{u}}_1}^2 = 2{{\text{v}}_2}^2 \\
  \dfrac{{{{\text{u}}_1}}}{{\sqrt 2 }} = {{\text{v}}_2} \\
\]

Formula of coefficient if restitution is given as,

\[{\text{e}} = \left| {\dfrac{{{{\text{v}}_2} - {{\text{v}}_1}}}{{{{\text{u}}_1} - {{\text{u}}_2}}}} \right|\]Since, the velocities \[{{\text{u}}_2}\] and \[{{\text{v}}_1}\] are zero therefore,

\[\
  {\text{e}} = \dfrac{{{{\text{v}}_2}}}{{{{\text{u}}_1}}} \\
  {\text{e}} = \dfrac{1}{{\sqrt 2 }} \\
\]

Therefore, the value of the coefficient of restitution is \[\dfrac{1}{{\sqrt 2 }}\].

Hence, option c is correct.

NOTE:In such types of problems, we must try to apply conservation law of momentum or energy as per the question’s requirement. Also, it is important to learn all the standard formulas of energies like kinetic energy and potential energy.