Center of Mass

Given:The final velocity of the first sphere, $v_1 = \frac{7}{9} u_1$$u_2 = 0$ (the second sphere is at rest)The mass of a sphere is proportional to the cube of its radius: $m \propto R^3$Let $m_1$ and $m_2$ be the masses, and $R_1$ and $R_2$ be the radii of the first and second spheres.Step 1: Use the formula for an elastic collisionAfter the collision, the velocity of the first sphere is:$v_1 = \frac{(m_1 - m_2) u_1 + 2 m_2 u_2}{m_1 + m_2}$Step 2: Substitute $u_2 = 0$$v_1 = \frac{(m_1 - m_2) u_1}{m_1 + m_2}$We know $v_1 = \frac{7}{9} u_1$ :$\frac{7}{9} u_1 = \frac{(m_1 - m_2) u_1}{m_1 + m_2}$Divide both sides by $u_1$ (since $u_1 \neq 0$ ):$\frac{7}{9} = \frac{m_1 - m_2}{m_1 + m_2}$ Step 3: Solve for $m_1$ in terms of $m_2$Multiply both sides by ( $m_1 + m_2$ ):$7 (m_1 + m_2) = 9 (m_1 - m_2)$$7 m_1 + 7 m_2 = 9 m_1 - 9 m_2$$7 m_2 + 9 m_2 = 9 m_1 - 7 m_1$$16 m_2 = 2 m_1$$m_1 = 8 m_2$Step 4: Relate masses to radiiSince mass of a sphere, $m \propto R^3$ :$\frac{m_1}{m_2} = \left( \frac{R_1}{R_2} \right)^3$$8 = \left( \frac{R_1}{R_2} \right)^3$Step 5: Solve for $R_2$Take the cube root of both sides:$\frac{R_1}{R_2} = 2$$R_1 = 2 R_2$ $R_2 = \frac{R_1}{2}$Plug in $R_1 = 1.2 \text{ cm}$ :$R_2 = \frac{1.2}{2} = 0.6 \text{ cm}$