QJEEM15Ph3faf: JEE Main 2015 Physics - Class 11 JEEM15Ph3faf Solution: A particle of mass moving in the direction with speed is hit by …

Applying conservation of linear momentum in x direction

$2mv = \left( {2m + m} \right){V_x}$

$\Rightarrow {V_x} = {2 \over 3}v$

Applying conservation of linear momentum in y direction

$2mv = \left( {2m + m} \right){V_y}$

$\Rightarrow {V_y} = {2 \over 3}v$

Final speed of 3m mass, ${V_f}$ = $\sqrt {V_x^2 + V_y^2}$

= $\sqrt {{{4{v^2}} \over 9} + {{4{v^2}} \over 9}}$

= $\sqrt {{{8{v^2}} \over 9}}$

Initial kinetic energy

${E_i} = {1 \over 2}m{\left( {2v} \right)^2} + {1 \over 2}\left( {2m} \right){\left( v \right)^2}$

$= 2m{v^2} + m{v^2}$

= $3m{v^2}$

Final kinetic energy,

${E_f} = {1 \over 2}\left( {3m} \right)$ ${V_f^2}$

$= {{3m} \over 2}\left[ {{{8{v^2}} \over 9}} \right]$

$= {{4m{v^2}} \over 3}$

Energy loss = ${E_i} - {E_f}$

= $3m{v^2} - {{4m{v^2}} \over 3}$

= ${{5m{v^2}} \over 3}$

Percentage loss in the energy during the collision

= ${{{E_i} - {E_f}} \over {{E_i}}}$ $\times 100$

= ${{{{5m{v^2}} \over 3}} \over {3m{v^2}}}$ $\times 100$

= ${5 \over 9} \times 100$

$\simeq 56\%$