A hammer can be modeled as a $750\mathrm{g}$ point mass on the end of a $42\mathrm{cm}$ long, $200\mathrm{g}$ uniform rod. How far from the head of the hammer is the center of mass?

• A. $2.3\mathrm{cm}$
• B. $3.1\mathrm{cm}$
• C. $5.6\mathrm{cm}$
• D. $6.3\mathrm{cm}$
• E. $4.4\mathrm{cm}$

Answer: (E)

Verified step by step guidance

1. First, understand the concept of the center of mass. The center of mass of a system is the point where the total mass of the system can be considered to be concentrated. For a system of particles, it is calculated using the weighted average of their positions.
2. Identify the masses and their positions. The hammer head is modeled as a point mass of 750 g, and the rod is a uniform mass of 200 g with a length of 42 cm. The center of mass of the rod is at its midpoint, which is 21 cm from the end of the rod.
3. Set up the equation for the center of mass. The center of mass \( x_{cm} \) can be calculated using the formula: \( x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} \), where \( m_1 \) and \( m_2 \) are the masses, and \( x_1 \) and \( x_2 \) are their respective positions.
4. Substitute the values into the equation. Let the position of the hammer head be at \( x_1 = 0 \) cm (since we are measuring from the head of the hammer), and the position of the center of mass of the rod be \( x_2 = 21 \) cm. The masses are \( m_1 = 750 \) g and \( m_2 = 200 \) g.
5. Calculate the center of mass using the formula: \( x_{cm} = \frac{750 \, \text{g} \times 0 \, \text{cm} + 200 \, \text{g} \times 21 \, \text{cm}}{750 \, \text{g} + 200 \, \text{g}} \). Simplify the expression to find the distance from the head of the hammer to the center of mass.