1. Analysis of Motion and Constraints

Let the velocity of prism A (mass $m_A$) be $v_A$ to the left, and the velocity of prism B (mass $m_B$) be $v_B$ to the right. Let the cylinder (mass $m$) move with velocity components $v_x$ (horizontal, to the right) and $v_y$ (vertical, downwards).

Since the surfaces are frictionless and the system starts from rest with no external horizontal forces, the horizontal momentum of the entire system is conserved.

Constraint Equations (Contact Condition)

For the cylinder to maintain contact with the prisms, the velocity components perpendicular to the contact surfaces must match.

Contact with Prism A (Angle $\alpha$):
The relative velocity perpendicular to the face must be zero. $$v_y = (v_A + v_x) \tan \alpha \quad \dots(1)$$

Contact with Prism B (Angle $\beta$):
Similarly, for prism B: $$v_y = (v_B – v_x) \tan \beta \quad \dots(2)$$

From (1) and (2), we equate $v_y$ to find $v_x$:

$$(v_A + v_x) \tan \alpha = (v_B – v_x) \tan \beta$$ $$v_x (\tan \alpha + \tan \beta) = v_B \tan \beta – v_A \tan \alpha$$ $$v_x = \frac{v_B \tan \beta – v_A \tan \alpha}{\tan \alpha + \tan \beta} \quad \dots(3)$$

2. Conservation of Momentum

Since the initial horizontal momentum is zero:

$$P_{initial} = 0$$ $$P_{final} = m_B v_B – m_A v_A + m v_x = 0$$ Substituting $v_x$ from eq (3): $$m_B v_B – m_A v_A + m \left[ \frac{v_B \tan \beta – v_A \tan \alpha}{\tan \alpha + \tan \beta} \right] = 0$$ Multiply by $(\tan \alpha + \tan \beta)$ to clear the denominator: $$m_B v_B (\tan \alpha + \tan \beta) – m_A v_A (\tan \alpha + \tan \beta) + m v_B \tan \beta – m v_A \tan \alpha = 0$$ Group terms by $v_A$ and $v_B$: $$v_B [ m_B (\tan \alpha + \tan \beta) + m \tan \beta ] = v_A [ m_A (\tan \alpha + \tan \beta) + m \tan \alpha ]$$ $$\frac{v_A}{v_B} = \frac{m_B (\tan \alpha + \tan \beta) + m \tan \beta}{m_A (\tan \alpha + \tan \beta) + m \tan \alpha}$$

3. Calculation

Given values:

• $m_A = 9.0 \text{ kg}, \quad \alpha = 53^\circ \Rightarrow \tan 53^\circ = 4/3$
• $m_B = 16.0 \text{ kg}, \quad \beta = 37^\circ \Rightarrow \tan 37^\circ = 3/4$
• $m = 25 \text{ kg}$

Calculate the term $(\tan \alpha + \tan \beta)$:

$$\tan \alpha + \tan \beta = \frac{4}{3} + \frac{3}{4} = \frac{16+9}{12} = \frac{25}{12}$$

Numerator (coefficient of $v_B$):

$$\text{Num} = 16\left(\frac{25}{12}\right) + 25\left(\frac{3}{4}\right) = \frac{400}{12} + \frac{75}{4} = \frac{625}{12}$$

Denominator (coefficient of $v_A$):

$$\text{Den} = 9\left(\frac{25}{12}\right) + 25\left(\frac{4}{3}\right) = \frac{225}{12} + \frac{100}{3} = \frac{625}{12}$$ $$\text{Ratio } \frac{v_A}{v_B} = \frac{625/12}{625/12} = 1$$