A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:

${\text{v}}_1=2\text{v,}{\text{v}}_{\text{2}}=\text{v}$

E before collision $={\text{E}}_1$

$=\frac{\text{1}}{\text{2}}\text{m}{\left(\text{2v}\right)}^{\text{2}}+\frac{1}{2}\text{2m}{\left(\text{v}\right)}^{\text{2}}$

$=3{\text{mv}}^{\text{2}}$

After collision $\text{m}=3\text{m}$

$\text{3mv}=\sqrt{{\left(\text{2mv}\right)}^{\text{2}}+{\left(\text{2mv}\right)}^{\text{2}}}$

${\text{v}}_{\text{1}}=\frac{\text{2}\sqrt{\text{2}}\text{v}}{\text{3}}$                                       

${\text{E}}_{\text{2}}=\frac{1}{2}\text{3m}{\left({\text{v}}_{\text{1}}\right)}^{\text{2}}$

$=\frac{1}{2}3\text{m}{\left(\frac{2\sqrt{2}}{3}\right)}^2$

$=\frac{4}{3}{\text{v}}^{\text{2}}\text{m}$

Percentage loss in energy

$\text{=}\frac{{\text{E}}_{\text{1}}-{\text{E}}_{\text{2}}}{{\text{E}}_{\text{1}}}\times 100$

$=\frac{{\text{3mv}}^2-\frac{4}{3}{\text{mv}}^{\text{2}}}{{\text{3mv}}^{\text{2}}}\times 100$

$=\frac{5}{9}\times 100=56\%$