Pool game, three balls, v0 = 5 m/s

Setup. All three balls have equal mass $m$, the table is frictionless and the impact is perfectly elastic, so linear momentum and kinetic energy are both conserved. Axes: $x$ horizontal, $y$ vertical. From the figure: $v_0=5$ m/s at $45^\circ$ above the horizontal, $V_A$ vertically upward, $V_B$ at $30^\circ$ below the horizontal and $V_C$ at $30^\circ$ above the horizontal (B and C on the same side).

Momentum in $x$:

$$m(5\cos45^\circ)=mV_B\cos30^\circ+mV_C\cos30^\circ$$ $$V_B+V_C=\frac{3.5355}{0.8660}=4.0825\qquad(1)$$

Momentum in $y$:

$$m(5\sin45^\circ)=mV_A-mV_B\sin30^\circ+mV_C\sin30^\circ$$ $$V_A=3.5355+0.5(V_B-V_C)\qquad(2)$$

Energy:

$$V_A^{2}+V_B^{2}+V_C^{2}=5^{2}=25\qquad(3)$$

Let $D=V_B-V_C$. From (1): $V_B^2+V_C^2=\dfrac{(4.0825)^2+D^2}{2}=8.333+0.5D^2$. From (2): $V_A^2=12.5+3.5355D+0.25D^2$. Substituting in (3):

$$0.75D^{2}+3.5355D-4.1667=0\ \Rightarrow\ D=0.9763$$

(The other root, $D=-5.69$, is rejected because it makes $V_B$ negative.)

Then

$$V_B=\frac{4.0825+0.9763}{2}=2.529\ \text{m/s},\qquad V_C=\frac{4.0825-0.9763}{2}=1.553\ \text{m/s}$$ $$V_A=3.5355+0.5(0.9763)=4.024\ \text{m/s}$$

Check: $4.024^2+2.529^2+1.553^2=16.19+6.40+2.41=25.0$.

Answer: $V_A=4.02$ m/s, $V_B=2.53$ m/s, $V_C=1.55$ m/s.