Question

In a one-dimensional collision, a particle of mass $2m$ collides with a particle of mass $m$ at rest. If the particles stick together after the collision, what fraction of the initial kinetic energy is lost in the collision?

• A. $\frac{1}{4}$
  
• B. None
  
• C. $\frac{1}{2}$
  
• D. $\frac{1}{3}$
  

Answer 1

The correct option is D $\frac{1}{3}$


Let initial velocity of $2m$ be $v_0$
$\therefore$ total initial kinetic energy
$K_i=\frac{1}{2}\times 2m\times {\left(v_0\right)}^2+\frac{1}{2}\times m\times 0^2$
$=m{v}_0^2$
and initial momentum
$P_i=2m\times v_0=2m{v}_0$

Let, final velocity of combination be $v$

$\therefore$ final momentum

$P_f=(2m+m)v=3mv$

According to conservation of momentum
$P_i=P_f$

$2m{v}_o=3mv$

$\Rightarrow v=\frac{2}{3}{v}_0$
final kinetic energy
$K_f=\frac{1}{2}\left(3m\right)v^2=\frac{1}{2}\left(3m\right)\times \frac{4}{9}{v}_0^2$
$=\frac{2}{3}m{v}_0^2$

$\text{K.E}.$ lost $=K_i-K_f=\frac{1}{3}m{v}_0^2$

Fraction of lost $\text{K.E.}$ $=\frac{\text{Loss in K.E.}}{\text{Initial K.E.}}=\frac{\frac{1}{3}m{v}_0{}^2}{m{v}_0{}^2}=\frac{1}{3}$

Hence , option $\left(B\right)$ is correct.