A block moving horizontally on a smooth surface with a speed of 40 m/s splits into two parts with masses in the ratio of 1 2: . If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is

Let a block of mass M splits into two masses m₁ and m₂ in ratio 1 : 2.

Then, mass of smaller part, ${\mathrm{m}}_1=\frac{\mathrm{M}}{1+2}=\frac{\mathrm{M}}{3}$

Mass of bigger part, m₂ = 2M/3

Given, speed of mass M, v = 40 m/s

Speed of mass m₁, v₁ = 60 m/s

Let speed of mass m₂, v₂ = v

Using conservation of linear momentum,

$\begin{array}{l}{\mathrm{p}}_{\mathrm{i}}={\mathrm{p}}_{\mathrm{f}}\\ \mathrm{Mv}={\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2\\ \mathrm{M}\times 40=\frac{\mathrm{M}}{3}\times 60+\frac{2\mathrm{M}}{3}\times \mathrm{v}\Rightarrow =30\mathrm{m}/\mathrm{s}\end{array}$

$\therefore$ The fractional change in kinetic energy,

$\begin{array}{l}\frac{\mathrm{\Delta K}}{\mathrm{K}}=\frac{{\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}}{{\mathrm{K}}_{\mathrm{i}}}=\frac{{\mathrm{K}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{i}}}-1\\ =\frac{\frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}_1^2+\frac{1}{2}{\mathrm{m}}_2{\mathrm{v}}_2^2}{\frac{1}{2}{\mathrm{Mv}}^2}-1\\ =\frac{\frac{1}{2}\left[\frac{\mathrm{M}}{3}\times (60{)}^2+\frac{2\mathrm{M}}{3}\times (30{)}^2\right]}{\frac{1}{2}\mathrm{M}\times (40{)}^2}-1=\frac{9}{8}-1=\frac{1}{8}\end{array}$