Question

A particle of mass $m$ moving in the $x$ distance 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 :

• A. $44\%$
• B. $50\%$
• C. $56\%$
• D. $62\%$

Answer 1

The correct option is C $56\%$

energy of the system before collision,

$E_1=\frac{1}{2}m(2v{)}^2+\frac{1}{2}2m{v}^2$

$\therefore E_1=3m{v}^2$

from given, after the collision, the mass becomes, $2m+m=3m$

$3mV=\sqrt{(2mV{)}^2+(2mV{)}^2}$

$V=\frac{2\sqrt{2}v}{3}$

$E_2=\frac{1}{2}3m(V{)}^2$

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

$E_2=\frac{4}{3}{v}^2$

$\frac{E_1-E_2}{E_1}=\frac{3m{v}^2-\frac{4}{3}{v}^2}{3m{v}^2}=\frac{\frac{5}{3}}{3}$

$\frac{E_1-E_2}{E_1}\times 100=\frac{5}{9}\times 100=56$%