Body A of mass 4m moving with speed u collides with another body B of mass 2m at test. The collision is head on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is  [NEET 2019]

The given situation of collision can be drawn as

Applying law of conservation of linear momentum, Initial momentum of system = Final momentum of system

$\Rightarrow \left(4m\right)u+\left(2m\right)u\text{'}=\left(4m\right)v_1+\left(2m\right)v_2$

                 $4\mathrm{mu} + \left(2\mathrm{m}\right)\times 0 = 4{\mathrm{mv}}_1 + 2{\mathrm{mv}}_2$

or                              $2\mathrm{u}=2{\mathrm{v}}_1+{\mathrm{v}}_2$                                                   …(i)

The kinetic energy of A before collision,

                                ${\left(\mathrm{KE}\right)}_A=\frac{1}{2}\left(4m\right)u^2=2m{u}^2$

Kinetic energy of B before collision,
                            (KE)_B = 0
The kinetic energy of A after collision,

            ${\left(K{E}^{\text{'}}\right)}_A=\frac{1}{2}\left(4m\right)v_1^2=2m{v}_1^2$

Kinetic energy of B after collision,

               ${\left({\mathrm{KE}}^{\text{'}}\right)}_B=\frac{1}{2}\left(2m\right)v_2^2=m{v}_2^2$

As, initial kinetic energy of the system = final kinetic energy of the system

$\Rightarrow (\mathrm{KE}{)}_A+(\mathrm{KE}{)}_B={\left(\mathrm{KE}\text{'}\right)}_A+{\left(\mathrm{KE}\text{'}\right)}_B$

                                 $2m{u}^2 + 0 = 2m{v}_1^2 + m{v}_2^2$

                                        $2m{u}^2=2m{v}_1^2 + m{v}_2^2$

or                                        $2u^2=2v_1^2 + v_2^2$                                   …(ii)

Solving Eqs. (j) and (ii), we get

                             $v_1=\frac{1}{3}u$  and $v_2=\frac{4}{3}u$

or the final velocity of A can be directly calculated by using the formula,

                  $v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\frac{2m_2{u}_2}{m_1+m_2}$

                        $=\left(\frac{4m-2m}{4m+2m}\right)u+\frac{2\left(2m\right)\times 0}{(4m+2m)}$                  $\left(\because u_2=u\text{'}=0\right)$

                        $=\frac{2m}{6m}u=\frac{1}{3}u$

$\therefore$  Net decrease in kinetic energy of A,

            $\Delta \mathrm{KE}=(\mathrm{KE}{)}_A-{\left(\mathrm{KE}\text{'}\right)}_A=2m{u}^2-2m{v}_1^2$

                     $= 2m (u^2-v_1^2)$

Substituting the value of v₁, we get

                   $\Delta \mathrm{KE}=2m\left(u^2-\frac{u^2}{9}\right)=\frac{16m{u}^2}{9}$

$\therefore$  The fractional decrease in kinetic energy,

             $\frac{\Delta \mathrm{KE}}{(\mathrm{KE}{)}_A}=\frac{16m{u}^2}{9}\times \frac{1}{2m{u}^2}=\frac{8}{9}$