Chapter 5, Problem 37P

To determine

To Evaluate: The center of mass of the remaining of wreckage of an airplane of mass $1.0\times {10}^{4}Kg$ , that falls vertically downwards, crashing to the ground. The plane explodes just before impact. Investigators looking in to the cause of the crash search for the wreckage. Large fragments of mass $4.0\times {10}^{3}Kg$ , $3.0\times {10}^{3}Kg$ and $1.0\times {10}^{3}Kg$ are found respectively at the point of impact, $30m$ North and $50m$ East.

Answer to Problem 37P

The location of the remaining part of the wreckage is positioned at $25m$ West and $45m$ South

Explanation of Solution

Given info:

  Mass of airplane $=1.0\times {10}^{4}Kg$

  Mass of fragments $=4.0\times {10}^{3}Kg$

  Mass of fragments $=3.0\times {10}^{3}Kg$

  Mass of fragments $=1.0\times {10}^{3}Kg$

The fragments are found at a location $30m$ North and $50m$ East.

Formula used:

The $x$ - component of position of center of mass of a system is calculated as follows:

  $x_{cm}=\frac{{\displaystyle \sum x\times m}}{{\displaystyle \sum m}}$

Where, $m$ is the mass of individual units of the system and $x$ its respective distances. ${\displaystyle \sum m}$ gives the sum total of masses of all the units in the system.

Similarly the $y$ -component of location for center of mass of a system is calculated as follow:

  $y_{cm}=\frac{{\displaystyle \sum y\times m}}{{\displaystyle \sum m}}$

Where, $m$ is the mass of the individual units of the system and $y$ its respective distances. The total mass of all the units in the system, when added up is given by ${\displaystyle \sum m}$ .

Calculation:

The sum total of the masses of wreckage pieces will be equal to the mass of the airplane. Assuming mass of airplane as $M$ and $m_1,m_2,m_3$ are the masses of fragments, then

  $M=m_1+m_2+m_3+m_r$

Substituting $M$ $=1.0\times {10}^{4}Kg$ , $m_1$ $=4.0\times {10}^{3}Kg$ , $m_2$ $=3.0\times {10}^{3}Kg$ , $m_3$ $=1.0\times {10}^{3}Kg$

  $\begin{array}{l}1.0\times {10}^{4}Kg=4.0\times {10}^{3}Kg+3.0\times {10}^{3}Kg+1.0\times {10}^{3}Kg+m_r\\ m_r=\text{ 2}\text{.0}\times {\text{10}}^{3}Kg\end{array}$

Considering the center of mass to be at the origin itself:

  $0=(0\times m_1)+(30m\times m_2)+(50m\times m_3)+m_{r}x+m_{r}y$

Therefore the x-component is derived as:

  $\begin{array}{l}{m}_{1}x=-m_3\times 50m\\ x=\frac{m_3\times 50m}{m_1}\end{array}$

Substituting $m_3=1.0\times {10}^{3}Kg\text{ and }{m}_1=2.0\times {10}^{3}Kg$

  $\begin{array}{l}x=-\frac{(50m)\times (1.0\times {10}^{3}Kg)}{(2.0\times {10}^{3}Kg)}\\ \text{ = }-25m\end{array}$

The negative sign is an indication that the center of mass is positioned towards West.

Similarly the $y$ -component is derived as follows:

  $\begin{array}{l}{m}_{1}y=-(30m)\times m_2\\ \text{ y =}-\frac{(30m)\times m_2}{m_1}\end{array}$

Substituting $m_2=3.0\times {10}^{3}Kg\text{ and }{m}_1=2.0\times {10}^{3}Kg$

  $\begin{array}{l}y=-\frac{(30m)\times (3.0\times {10}^{3}Kg)}{(2.0\times {10}^{3}Kg)}\\ \text{ = }-45m\end{array}$

Hence the location of the center of mass is towards the South direction as indicated by the negative sign of the result.

Conclusion:

Thus, the location of the remaining portion of the wreckage is positioned at $25m$ in West, and $45m$ in south direction.