Hukum Kekekalan Energi Mekanik

Bola jatuh di titik B → N_B = 0

\begin{equation*}
\begin{split}
\sum F_{\text{sentripetal}} & = m \: \frac{v^2}{R} \\\\
N_B + mg \cos \theta & = m \: \frac{v_B^2}{R} \\\\
0 + \cancel {m} g \cos \theta & = \cancel {m} \: \frac{v_B^2}{R} \\\\
g \cos \theta & = \frac{v_B^2}{R} \\\\
v_B^2 & = gR \cos \theta \\\\
v_B & = \sqrt {gR \cos \theta} \\\\
v_B & = \sqrt {10 \:.\: 2 \:.\: \cos 37} \\\\
v_B & = 4 \text{ m/s}
\end{split}
\end{equation*}

Kecepatan di A

\begin{equation*}
\begin{split}
EM_1 & = EM_2 \\\\
EP_1 + EK_1 & = EP_2 + EK_2 \\\\
0 + \tfrac{1}{2} \:.\: \cancel {m} \:.\: v_A^2 & = \cancel {m} \:.\: g \:.\: h_B + \tfrac{1}{2} \:.\: \cancel {m} \:.\: v_B^2 \\\\
\tfrac{1}{2} \:.\: v_A^2 & = g (R + R \cos \theta) + \tfrac{1}{2} \:.\: v_B^2 \\\\
\tfrac{1}{2} \:.\: v_A^2 & = 10 (2 + 2 \:.\: 0,8) + \tfrac{1}{2} \:.\: 4^2 \\\\
v_A^2 & = 88 \\\\
v_A & = 2 \sqrt{22} \text{ m/s}
\end{split}
\end{equation*}