Question

A particle of mass $m$ moving in 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 of the energy retained by the colliding particles after the collision is close to:

• A. $56$%
• B. $80$%
• C. $35$%
• D. $44$%

Answer 1

The correct option is A $56$%

Energy before collision

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

$E_1=3m{v}^2$

After collision mass become $=(2m+m)=3m$

Hence,

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

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

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

$E_2=\frac{1}{2}3m{\left(\frac{2\sqrt{2}v}{3}\right)}^2$

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

$\therefore \frac{E_1-E_2}{E_1}=\frac{3m{v}^2-4/3m{v}^2}{3m{v}^2}=\frac{5}{9}\times 100=$56%