This is inelastic collision. So only the momentum is conserved, kinetic energy is NOT conserved.
Given Quantities & Unknowns:
#m_1=4kg; \quad vec v_(1i) = +2 ms^{-1}; \qquad vec v_{1f} = ?#
#m_2=2kg; \quad vec v_(2i) = -4 ms^{-1}; \qquad vec v_{2f} = ?#
Momentum Conservation : #vec P_f = vec P_i#
#m_1.vecv_{1f} + m_2vec v_{2f} = m_1vec v_{1i} + m_2vec v_{2i}#
#m_1.vec v_{1f} + m_2vec v_{2f} = [4kg\times2ms^{-1}][2kg\times(-4ms^{-1})] =0#
#vecv_{2f} = -(m_1/m_2)vecv_{1f} = - 2vecv_{1f}#
#v_{2f}^2 = vecv_{2f}.vecv_{2f} = +4v_{1f}^2# ...... (1)
Kinetic Energy Comparison: Given that there is a #20%# loss in kinetic energy, we conclude that the post collision kinetic energy is only #80%# of the kinetic energy before collision, #K_f = 0.8K_i#.
#1/2[m_1.v_{1f}^2 + m_2.v_{2f}^2] = 0.8\times1/2[m_1v_{1i}^2 + m_2.v_{2i}^2]# ...... (2)
#m_1.v_{1f}^2 + m_2.v_{2f}^2 = 0.8\times[m_1v_{1i}^2 + m_2.v_{2i}^2]=(8/10)\times48#
Eliminate #v_{2f}# in Eqn. (2) using Eqn. (1)
#m_1v_{1f}^2 + m_2(4v_{1f}^2)=(8/10)\times48#
#v_{1f}^2 = (8/10)(48)/(m_1+4m_2)=32/10; \qquad v_{1f}=4/\sqrt{5}ms^{-1};#
#vec v_{1f} = +4/\sqrt{5}ms^{-1}#
#vecv_{2f}=-2vecv_{1f}=-8/\sqrt{5}ms^{-1}#
