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Question: A ball impinges directly on a similar ball at rest. The first ball is brought to rest by the impact....

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) 122\dfrac{1}{{2\sqrt 2 }}
B) 13\dfrac{1}{{\sqrt 3 }}
C) 12\dfrac{1}{{\sqrt 2 }}
D) 32\dfrac{{\sqrt 3 }}{2}

Explanation

Solution

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, k=12mu2{\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 u1{{\text{u}}_1}, final velocity be v1{{\text{v}}_1}. Similarly, the initial velocity of the second ball be u2{{\text{u}}_2}and final velocity be v2{{\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,

Ki=12mu12+12mu22{{\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,

Kf=12mv12+12mv22{{\text{K}}_{\text{f}}} = \dfrac{1}{2}{\text{m}}{{\text{v}}_1}^2 + \dfrac{1}{2}{\text{m}}{{\text{v}}_2}^2

Velocities u2{{\text{u}}_2}andv1{{\text{v}}_1}are zero. Net change or loss of the kinetic energy is given as,

Δ=\Delta = Total initial kinetic energy –Total final kinetic energy

Δ=12mu12+12mu22(12mv12+12mv22) =12mu12+0012mv22 Δ=12mu1212mv22  \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,

mv12=2mu122mv22 u12=2v22 u12=v2  {\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,

e=v2v1u1u2{\text{e}} = \left| {\dfrac{{{{\text{v}}_2} - {{\text{v}}_1}}}{{{{\text{u}}_1} - {{\text{u}}_2}}}} \right|Since, the velocities u2{{\text{u}}_2} and v1{{\text{v}}_1} are zero therefore,

e=v2u1 e=12  {\text{e}} = \dfrac{{{{\text{v}}_2}}}{{{{\text{u}}_1}}} \\\ {\text{e}} = \dfrac{1}{{\sqrt 2 }} \\\

Therefore, the value of the coefficient of restitution is 12\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.