That latest value for the Angle is in Grads, not in Kilograms.
Apply law of conservation of momentum along vertical direction.
[tex]m_1v_1sin\theta_1-m_2v_2sin\theta_2=0[/tex]
[tex]v_2=\frac{v_1sin\theta_1}{sin\theta_2}=\frac{sin54}{sin36}v_1 = 1.376v_1[/tex]
Apply law of conservation of momentum along the horizontal direction
[tex]m_1u_1=m_1v_1cos\theta_1+m_2v_2cos\theta_2[/tex]
[tex]u_1=v_1(cos\theta_1+1.376cos\theta_2)[/tex]
[tex]u_1=v_1(cos(54)+1.376cos(36))[/tex]
[tex]u_1=(1.7) v_1[/tex]
[tex]v_1= \frac{3.14}{1.7}=1.84m/s[/tex]
The second ball velocity is [tex]v_2 = (1.376)(1.84)=2.531m/s[/tex]
The magnitud of final total momentum is
[tex]m(v_1+v_2)=(1.37)(2.531+1.84)=5.98kgm/s[/tex]
The magnitude of final energy is
[tex]\frac{1}{2}m(v^2_1+v^2_2)=\frac{1}{2}(1.37)(2.531^2+1.84^2)=6.07J[/tex]
