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, the percentage loss in energy during the collision is close to

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

Answer 1

The correct option is C $56\%$

For inelastic collision, conservation of linear momentum can be applied, but energy is not conserved.
Consider the movement of the two particles as shown below.


Conserving linear momentum in $x-$ direction,
$(p_i{)}_x=(p_f{)}_x$
or $m\left(2v\right)+0=(2m+m)v_x$
or $v_x=\frac{2}{3}v$

Conserving linear momentum in $y-$ direction,
$(p_i{)}_y=(p_f{)}_y$
or $0+\left(2m\right)v=(2m+m)v_y$
or $v_y=\frac{2}{3}v$

Initial kinetic energy of the two particle system is
$E_i=\frac{1}{2}m(2v{)}^2+\frac{1}{2}(2m\left)\right(v{)}^2$
$=\frac{1}{2}\times 4m{v}^2+\frac{1}{2}\times 2m{v}^2$
$=2m{v}^2+m{v}^2=3m{v}^2$

Final energy of the combined two particle system is
$E_f=\frac{1}{2}\left(3m\right)(v_x^2+v_y^2)$
$=\frac{1}{2}\left(3m\right)[\frac{4v^2}{9}+\frac{4v^2}{9}]=\frac{3m}{2}\left[\frac{8v^2}{9}\right]=\frac{4m{v}^2}{3}$

Loss in the energy $\Delta E=E_i-E_f$
$=m{v}^2[3-\frac{4}{3}]=\frac{5}{3}m{v}^2$

$\therefore$ Percentage loss in energy during the collision
$\frac{\Delta E}{E_i}\times 100=\frac{\left(5/3\right)m{v}^2}{3m{v}^2}\times 100=\frac{5}{9}\times 100=55.55\approx 56\%$