Let #u_1=1 m/s# & #u_2=-5 m/s# be the initial velocities of two balls having masses #m_1=7\ kg\ # & #\m_2=2\ 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#
#7(1)+2(-5)=7v_1+2v_2#
#7v_1+2v_2=-3\ .......(1)#
Now, loss of kinetic energy is #75%# hence
#(1-\frac{75}{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)#
#1/4(\frac{1}{2}7(1)^2+\frac{1}{2}2(5)^2)=\frac{1}{2}7v_1^2+\frac{1}{2}2v_2^2#
#28v_1^2+8v_2^2=57 \ ......(2)#
substituting the value of #v_2=\frac{-7v_1-3}{2}# from (1) into (2) as follows
#28v_1^2+8(\frac{-7v_1-3}{2})^2=57#
#42v_1^2+28v_1-13=0#
solving above quadratic equation, we get #v_1=0.315, -0.982# & corresponding value of #v_2=-2.602, 1.937#
Hence, the final velocities of both the balls are either #v_1=0.315\ m/s# & #v_2=-2.602\m/s#
or
#v_1=-0.982\ m/s# & #v_2=1.937\m/s#
