Particle $A$ has a mass of $5.00 \mathrm{~g}$ and particle $B$ has a mass of $1.40 \mathrm{~g}$. Particle $A$ is located at the origin and particle $B$ is at the point $(x, y)=(25.0 \mathrm{~cm}, 4.30 \mathrm{~cm})$.
What is the $y$-component of the $C M$ ?
$\mathrm{cm}$

Step 1 ：Given that particle $A$ has a mass of $5.00 \, g$ and is located at the origin, and particle $B$ has a mass of $1.40 \, g$ and is located at the point $(25.0 \, cm, 4.30 \, cm)$.

Step 2 ：The center of mass (CM) of a system of particles is given by the formula: \[CM = \frac{\sum_{i=1}^{n} m_i r_i}{\sum_{i=1}^{n} m_i}\] where $m_i$ is the mass of the $i$th particle and $r_i$ is the position vector of the $i$th particle.

Step 3 ：We are asked to find the $y$-component of the CM, which is given by: \[CM_y = \frac{m_A y_A + m_B y_B}{m_A + m_B}\] where $y_A$ and $y_B$ are the $y$-coordinates of particles $A$ and $B$, respectively.

Step 4 ：In this case, $y_A = 0$ and $y_B = 4.30 \, cm$.

Step 5 ：Substituting the given values into the formula, we get: \[CM_y = \frac{5.0 \times 0.0 + 1.4 \times 4.3}{5.0 + 1.4} = 0.94 \, cm\]

Step 6 ：Final Answer: The $y$-component of the center of mass is \(\boxed{0.94 \, cm}\)