In all collisions momentum is always conserved. In elastic collisions kinetic energy is also conserved.
Let momentum before collision be
#p_i = m_8v_(8i) + m_2v_(2i)#
As the second mass is stationary, we have
#p_i = m_8v_(8i)# ........(1)
Using the expression for kinetic energy we have
#E=1/2m_8 v_(8i)^2# ......(2)
Assuming that initial velocity is along #x#-direction, final velocities are also in the same direction. From Law of conservation of Momentum we have
#p_i= p_f#
#=>m_8v_(8i) = m_8v_(8f)+ m_2v_(2f)#
We know that the #m_2=1/4m_8#. Hence, above equation becomes
#m_8v_(8i) = m_8v_(8f)+ 1/4m_8v_(2f)#
#v_(8i) = v_(8f)+ 1/4v_(2f)# ......(3)
From Law of conservation of energy we have
#E_i = E_f#
#1/2m_8 v_(8i)^2 = 1/2m_8v_(8f)^2+ 1/2m_2v_(2f)^2#
Using (2) and (3) we get
#1/2m_8 v_(8i)^2 = 1/2m_8v_(8f)^2+ 1/2m_8/4(4(v_(8i) - v_(8f)))^2#
#=>1/2m_8 v_(8i)^2 = 1/2m_8v_(8f)^2+ 1/2m_8 [4(v_(8i)^2 + v_(8f)^2-2v_(8i)v_(8f))]#
#=> 5 v_(8f)^2-8v_(8i)v_(8f)+3v_(8i)^2 =0#
Solving the quadratic of unknown #v_(8f)# by splitting the middle term
# 5 v_(8f)^2-5v_(8i)v_(8f)-3v_(8i)v_(8f)+3v_(8i)^2 =0#
#=> 5 v_(8f)[v_(8f)-v_(8i)]-3v_(8i)[v_(8f)-v_(8i)] =0#
#=>[5 v_(8f)-3v_(8i)][v_(8f)-v_(8i)] =0#
The two roots are
#5 v_(8f)=3v_(8i) and v_(8f)=v_(8i) #
Ignoring second root as it implies #2kg# mass remains stationary after the collision, we have
#5 v_(8f)=3v_(8i) #
Squaring both sides we get
#25 v_(8f)^2=9v_(8i)^2#
Converting in terms of kinetic energy
#E_f=9/25E#
#E_f=0.36E#
