Chapter 7, Problem 60P

(a)

To determine

The location of $m_{II}$ landing

Answer to Problem 60P

If $m_{II}=3m_I$ then $m_{II}$ would land at $\frac{7}{3}d$

Explanation of Solution

Given:

The center of mass of the system will still be $2d$ from the launch point and $m_{II}=3m_I$

The given system is shown below.

  

Formula used:

The center of mass is calculated as

  $x_{cm}=\frac{m_I{x}_I+m_{II}{x}_{II}}{m_I+m_{II}}$

Calculation:

Using the center of mass formula,

  $\begin{array}{l}{x}_{cm}=\frac{m_I{x}_I+m_{II}{x}_{II}}{m_I+m_{II}}\\ 2d=\frac{m_{I}d+3m_I{x}_{II}}{4m_I}=\frac{d+3x_{II}}{4}\\ x_{II}=\frac{7}{3}d\end{array}$

Conclusion:

If $m_{II}=3m_I$ then $m_{II}$ would land at $\frac{7}{3}d$

(b)

To determine

The location of $m_{II}$ landing

Answer to Problem 60P

If $m_I=3m_{II}$ then $m_{II}$ would land at $5d$

Explanation of Solution

Given:

The center of mass of the system will still be $2d$ from the launch point and $m_I=3m_{II}$

The given system is shown below.

  

Formula used:

The center of mass is calculated as

  $x_{cm}=\frac{m_I{x}_I+m_{II}{x}_{II}}{m_I+m_{II}}$

Calculation:

Using the center of mass formula,

  $\begin{array}{l}{x}_{cm}=\frac{m_I{x}_I+m_{II}{x}_{II}}{m_I+m_{II}}\\ 2d=\frac{3m_{II}d+m_{II}{x}_{II}}{4m_{II}}=\frac{3d+x_{II}}{4}\\ x_{II}=5d\end{array}$

Conclusion:

If $m_I=3m_{II}$ then $m_{II}$ would land at $5d$