Two balls impact e = 0.84, energy lost

Assumptions (figure not shown): the x-axis is the line of impact. Ball A ($m_A=1.2$ kg, $20$ m/s) moves to the right, $30^\circ$ above the x-axis; ball B ($m_B=1.8$ kg, $25$ m/s) moves to the left, $45^\circ$ below the x-axis. The balls are smooth, $e=0.84$.

Components

The $y$ components are unchanged by the impact: $v'_{Ay}=10.00$ m/s, $v'_{By}=-17.68$ m/s.

Along the line of impact

Momentum:

$$1.2(17.32)+1.8(-17.68)=1.2v'_{Ax}+1.8v'_{Bx}\ \Rightarrow\ 1.2v'_{Ax}+1.8v'_{Bx}=-11.04$$

Restitution:

$$v'_{Bx}-v'_{Ax}=e(v_{Ax}-v_{Bx})=0.84(17.32+17.68)=29.40$$

Solving: $v'_{Ax}=-21.32$ m/s, $v'_{Bx}=8.08$ m/s.

Final velocities

$$v'_A=\sqrt{21.32^{2}+10^{2}}=23.55\ \text{m/s},\qquad \text{direction }25.1^\circ\text{ above the negative x-axis}$$ $$v'_B=\sqrt{8.08^{2}+17.68^{2}}=19.44\ \text{m/s},\qquad \text{direction }65.4^\circ\text{ below the positive x-axis}$$

Kinetic energy lost

$$\begin{aligned} T_1&=\tfrac12(1.2)(20)^{2}+\tfrac12(1.8)(25)^{2}=240+562.5=802.5\ \text{J}\\ T_2&=\tfrac12(1.2)(23.55)^{2}+\tfrac12(1.8)(19.44)^{2}=672.7\ \text{J}\\ \Delta T&=802.5-672.7=129.8\ \text{J} \end{aligned}$$

Answer: $v'_A=23.55$ m/s, $v'_B=19.44$ m/s; kinetic energy lost $\approx129.8$ J.