Question

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, then loss in energy$\left(\%\right)$ during the collision is close to

• A. $40\%$
• B. $50\%$
• C. $56\%$
• D. $62\%$

Answer 1

The correct option is C $56\%$

Since, there is no external force during the collision, momentum along $x\&y$ direction will remain conserved.
Both the particles will stick together as a mass $3m$, since it is a perfectly inelastic collision.


Considering the mass $3m$ to have velocity $v_x\&v_y$ along $x$ and $y$ directions after collision,
Using PCLM along $x-$ direction gives,
$P_i=P_f$
$m\left(2v\right)=\left(3m\right)v_x$
PCLM along $y-$ direction gives,
$2m\left(v\right)=3m\left(v_y\right)$
$\therefore v_y=\frac{2}{3}v$

$\%$ loss in K.E $=\frac{K.E_i-K.E_f}{K.E_i}\times 100$
$(K.E{)}_i=\frac{1}{2}(2m)v^2+\frac{1}{2}m(2v{)}^2$
$\Rightarrow (K.E{)}_i=3m{v}^2$
$(K.E{)}_f=\frac{1}{2}(3m\left)\right(v_x^2+v_y^2)$
$=\frac{1}{2}\left(3m\right)[{\left(\frac{2}{3}v\right)}^2+{\left(\frac{2}{3}v\right)}^2]$
$=\frac{1}{2}\left(3m\right)\left(\frac{8}{9}{v}^2\right)$
$\Rightarrow (K.E{)}_f=\frac{4}{3}m{v}^2$

$\%\text{loss in K.E}=\frac{3m{v}^2-\frac{4}{3}m{v}^2}{3m{v}^2}\times 100$
$=\frac{5}{9}\times 100\simeq 55.55\%$
Hence it is approximately $56\%$.