$(Given)\ m_A=1\ kg, \ m_B= 2 kg,\\ Let\ u\ and\ v\ be\ the\ initial\ and\ final\ velocities.\\ u_A= 3\ m/s,\ u_B= 1 m/s;\\ Let's\ take\ X-axis\ be\ the\ dotted\ line\ as\ the\ line\ of\ impact\ as\ shown\ in\ fig.$

• According to law of conservation of momentum,

$m_A.u_{Ax} + m_B.u_{Bx} = m_A.v_{Ax} + m_B.v_{Ax}\\ 1 \times 3\cos 30+2\times (-1\cos 45)=1\ v_{Ax} + 2\ v_{Bx}\\ 1.1839= v_{Ax}+ 2\ v_{Bx} \cdots(1)$

$Coefficient\ of\ restitution\ (e)\ = \dfrac {v_{Bx} - {v_{Ax}}} {u_{Ax}-{u_{Bx}}}\\ 0.75= \dfrac {v_{Bx} - {v_{Ax}}} {3\cos30 + 1\cos45}\\ 0.75= \dfrac {v_{Bx} - {v_{Ax}}} {3.3052}\\ 0.75 \times 3.3052 = {v_{Bx} - {v_{Ax}}}\\ 2.4789 = {v_{Bx} - {v_{Ax}}} \cdots (2)\\ Equating\ (1)\ and\ (2),\ we\ get,\\ V_{Ax}= -1.2580\ and\ V_{Bx}= 1.2209; $

• Velocities of both spheres will be same along y-axis as impact is along x-axis.

$v_{Ay}=u_{Ay}= 3sin30 = 1.5\ and\ v_{By}=u_{By}= -1sin45 = -0.7071$

$After\ impact,\ Velocity\ of\ ball\ A\ = \sqrt {(V_Ax)^2 + (V_Ay)^2}= \sqrt {(-1.2580)^2 + (1.5)^2}\\ =1.9577\ m/s$

$\alpha_A= tan^{-1} \dfrac {V_{Ay}} {V_{Ax}}\\ =tan^{-1} \dfrac {1.5} {1.2580}\\ =50.0146 ^\circ$

• $After\ impact,\ Velocity\ of\ ball\ B\ = \sqrt {(V_Bx)^2 + (V_By)^2}= \sqrt {(1.2209)^2 + (-0.7071)^2}\\ =1.4109\ m/s$

$\alpha_B= tan^{-1} \dfrac {V_{By}} {V_{Bx}}\\ =tan^{-1} \dfrac {0.7071} {1.2209}\\ =30.0778 ^\circ$

• $The\ velocities\ after\ impact\ of\ ball\ A\ = 1.9577\ m/s,\ \angle50.0146 ^\circ and\ of\ ball\ B\ = 1.4109\ m/s, \angle30.0778 ^\circ$