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 the energy during the collision is close to:
(A) 44%
(B) 50%
(C) 56%
(D) 62%

Answer 1

Hint : The collision of the two particles is elastic so the linear momentum for both the x-direction and y-direction is calculated. We need to apply the concepts of law of conservation of linear momentum.

Complete step by step answer
Conservation of linear momentum can be applied but energy is not conserved.
Conserving linear momentum in x-direction:
${\left( {{p_i}} \right)_x} = {\left( {{p_f}} \right)_x}$
It is given that, 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 $v$.
Initial linear momentum in the x-direction = ${\left( {{p_i}} \right)_x} = 2mv$
After collision, linear momentum in the x-direction = ${\left( {{p_f}} \right)_x} = \left( {2m + m} \right){v_x}$ where ${v_x}$ is the final velocity.
$\therefore$ By conservation of linear momentum, ${\left( {{p_i}} \right)_x} = {\left( {{p_f}} \right)_x}$
$\Rightarrow 2mv = \left( {2m + m} \right){v_f}$
Thus the final velocity ${v_x}$ is given by, ${v_x} = \dfrac{2}{3}v$.
Conserving linear momentum in y-direction:
${\left( {{p_i}} \right)_y} = {\left( {{p_f}} \right)_y}$
Initial linear momentum in the y-direction = ${\left( {{p_i}} \right)_x} = 2mv$
After collision, linear momentum in the x-direction = ${\left( {{p_f}} \right)_x} = \left( {2m + m} \right){v_y}$ where ${v_y}$ is the final velocity.
$\Rightarrow 2mv = \left( {2m + m} \right){v_y}$
Thus the final velocity ${v_y}$is given by, ${v_y} = \dfrac{2}{3}v$.
Let us now calculate the initial kinetic energy of the system. Given that,a particle moving in the x-direction has speed $2v$ and that moving in the y-direction has speed $v$.
$K.E{._{initial}} = \dfrac{1}{2}m{\left( {2v} \right)^2} + \dfrac{1}{2}\left( {2m} \right){v^2}$
$= \dfrac{1}{2}4m{v^2} + \dfrac{1}{2}2m{v^2}$
Simplifying the equation further, we get,
$= 2m{v^2} + m{v^2} = 3m{v^2}$.
Final energy of the system after collision is-
$K.E{._{final}} = \dfrac{1}{2} \cdot 3m \cdot {\left( {{v_x} + {v_y}} \right)^2}$
$\Rightarrow K.E{._{final}} = \dfrac{1}{2}\left( {3m} \right)\left[ {\dfrac{{4{v^2}}}{9} + \dfrac{{4{v^2}}}{9}} \right]$
We simplify the equation further,
$= \dfrac{{3m}}{2}\left[ {\dfrac{{8{v^2}}}{9}} \right] = \dfrac{{4m{v^2}}}{3}$
The loss in energy is given the difference between the initial and final energies of the system.
$\Delta K.E. = K.E{._{initial}} - K.E{._{final}}$
$\Rightarrow \Delta K.E. = m{v^2}\left[ {3 - \dfrac{4}{3}} \right] = \dfrac{5}{3}m{v^2}$
Percentage loss in the energy during the collision is given by the formula: $\dfrac{{\Delta K.E.}}{{K.E{._{initial}}}} \times 100$.
\dfrac{{{\raise0.7ex\hbox{5} \\!\mathord{\left/ {\vphantom {5 3}}\right.} \\!\lower0.7ex\hbox{3}}m{v^2}}}{{3m{v^2}}} \times 100 = \dfrac{5}{9} \times 100 \simeq 56\%
So the correct answer is option C.

Note
An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same. In an ideal, perfectly elastic collision, there is no net conversion of kinetic energy into other forms such as heat, noise, or potential energy.
During the collision of small objects, kinetic energy is first converted to potential energy associated with a repulsive force between the particles (when the particles move against this force, i.e. the angle between the force and the relative velocity is obtuse), then this potential energy is converted back to kinetic energy (when the particles move with this force, i.e. the angle between the force and the relative velocity is acute).