Solution to Question 30
Center of Mass Principle
Since there is no external horizontal force on the system (wedge + 2 balls), the horizontal position of the Center of Mass (CM) remains constant.
Let $X$ be the displacement of the wedge. Let $h$ be the height of the wedge. Let angles be $\alpha$ and $\beta$. Horizontal displacement of left ball relative to wedge: $-h \cot \alpha$. Horizontal displacement of right ball relative to wedge: $+h \cot \beta$.
Equation
$$ M(\Delta x_{wedge}) + m(\Delta x_{ball1}) + m(\Delta x_{ball2}) = 0 $$ $$ M(X) + m(X – h \cot \alpha) + m(X + h \cot \beta) = 0 $$ Note: Signs depend on geometry. If balls roll down opposite sides: One moves left (relative), one moves right (relative). $$ (M + 2m)X + mh(\cot \beta – \cot \alpha) = 0 $$ $$ X = \frac{mh(\cot \alpha – \cot \beta)}{M + 2m} $$