Let #u_1=1 m/s# & #u_2=-8 m/s# be the initial velocities of two balls having masses #m_1=4\ kg\ # & #\m_2=6\ kg# moving in opposite directions i.e. first one is moving in +ve x-direction & other in -ve x-direction, After collision let #v_1# & #v_2# be the velocities of balls in +ve x-direction
By law of conservation of momentum in +ve x-direction, we have
#m_1u_1+m_2u_2=m_1v_1+m_2v_2#
#4(1)+6(-8)=4v_1+6v_2#
#2v_1+3v_2=-22\ .......(1)#
Now, loss of kinetic energy is #10%# hence
#(1-\frac{10}{100})(\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2)=(\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2)#
#9/10(\frac{1}{2}4(1)^2+\frac{1}{2}6(8)^2)=\frac{1}{2}4v_1^2+\frac{1}{2}6v_2^2#
#2v_1^2+3v_2^2=174.6 \ ......(2)#
substituting the value of #v_2=\frac{-2v_1-22}{3}# from (1) into (2) as follows
#2v_1^2+3(\frac{-2v_1-22}{3})^2=174.6#
#5v_1^2+44v_1-19.9=0#
solving above quadratic equation, we get #v_1=0.431, -9.231# & corresponding values of #v_2=-7.621, -1.179#
Hence, the final velocities of both the balls after collision are either #v_1=0.431\ m/s# & #v_2=-7.621\m/s#
or
#v_1=-9.231\ m/s# & #v_2=-1.179\m/s#
