Solved Problem on Center of Mass
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Two particles, A and B, have masses of 4 kg and 6 kg, respectively. Both move with
constant speeds vA = 5 m/s e vB = 3 m/s such that their directions
make an angle of 60°. Determine:
a) The velocity of the center of mass;
b) The momentum of the system.
a) The velocity of the center of mass;
b) The momentum of the system.
Problem data:
- Mass of particle A: mA = 4 kg;
- Mass of particle B: mB = 6 kg;
- Particle A velocity: vA = 5 m/s;
- Particle B velocity: vB = 3 m/s.
a) The velocity of the center of mass will be given by the following equation in vector form
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\vec{v}=\frac{m_{A}{\vec{v}}_{A}+m_{B}{\vec{v}}_{B}}{m_{A}+m_{B}}}
\end{gather}
\]
this equation can be decomposed in the x and y directions
\[
\begin{gather}
v_{x}=\frac{m_{A}v_{Ax}+m_{B}v_{Bx}}{m_{A}+m_{B}} \tag{I-a}
\end{gather}
\]
\[
\begin{gather}
v_{y}=\frac{m_{A}v_{Ay}+m_{B}v_{By}}{m_{A}+m_{B}} \tag{I-b}
\end{gather}
\]
We draw the
\( {\vec{v}}_{A} \)
and
\( {\vec{v}}_{B} \)
velocities on a cartesian coordinate system and get their components, the
\( {\vec{v}}_{B} \)
velocity coincides with the x-axis, so the angle between them will be 0° (Figure 1)
- Horizontal direction:
\[
\begin{gather}
v_{Ax}=v_{A}\cos 60°
\end{gather}
\]
from Trigonometry
\( \cos 60°=\dfrac{1}{2} \)
\[
\begin{gather}
v_{Ax}=5\times\frac{1}{2}\\[5pt]
v_{Ax}=2.5\;\text{m/s} \tag{II}
\end{gather}
\]
\[
\begin{gather}
v_{Bx}=v_{B}\cos 0°
\end{gather}
\]
from Trigonometry
\( \cos 0°=1 \)
\[
\begin{gather}
v_{Bx}=3\times 1\\[5pt]
v_{Bx}=3\;\text{m/s} \tag{III}
\end{gather}
\]
- Vertical direction:
\[
\begin{gather}
v_{Ay}=v_{A}\sin 60°
\end{gather}
\]
from Trigonometry
\( \sin 60°=\dfrac{\sqrt{3\;}}{2} \)
\[
\begin{gather}
v_{Ay}=5\times\frac{\sqrt{3\;}}{2}\\[5pt]
v_{Ay}=4.3\;\text{m/s} \tag{IV}
\end{gather}
\]
\[
\begin{gather}
v_{By}=v_{B}\sin 0°
\end{gather}
\]
from Trigonometry
\( \sin 0°=0 \)
\[
\begin{gather}
v_{By}=3\times 0\\[5pt]
v_{By}=0 \tag{V}
\end{gather}
\]
Substituting the values of the masses and the velocities (II) and (III) in the x direction in the
expression (I-a)
\[
\begin{gather}
v_{x}=\frac{4\times 2.5+6\times 3}{4+6}\\[5pt]
v_{x}=\frac{10+18}{10}\\[5pt]
v_{x}=2.8\;\text{m/s} \tag{VI}
\end{gather}
\]
Substituting the values of the masses and the velocities (IV) and (V) in the y direction into the
expression (I-b)
\[
\begin{gather}
v_{y}=\frac{4\times 4.3+6\times 0}{4+6}\\[5pt]
v_{y}=\frac{17.2+0}{10}\\[5pt]
v_{y}=1.7\;\text{m/s} \tag{VII}
\end{gather}
\]
Vectors
\( {\vec{v}}_{x} \)
and
\( {\vec{v}}_{y} \)
are represented in Figure 2-A, and their vector sum will give us the velocity of the center of mass of
the system. The magnitude of this vector can be found by applying the Pythagorean Theorem to the
right triangle of Figure 2-B, where the legs represent the speeds in the x and y
directions, and the hypotenuse is the speed of the center of mass.
\[
\begin{gather}
v^{2}=v_{x}^{2}+v_{y}^{2}
\end{gather}
\]
substituting the results (VI) and (VII) for the velocities
\[
\begin{gather}
v^{2}=(2.8)^{2}+(1.7)^{2}\\[5pt]
v^{2}=7.84+2.89\\[5pt]
v^{2}=10.73\\[5pt]
v=\sqrt{10.73\;}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v=3.3\;\text{m/s}}
\end{gather}
\]
b) The momentum of the system will be
\[
\begin{gather}
\bbox[#99CCFF,10px]
{Q=mv}
\end{gather}
\]
where m is the total mass of the system.
\[
\begin{gather}
Q=(m_{A}+m_{B})v\\[5pt]
Q=(4+6)\times 3.3
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{Q=33\;\text{kg.m/s}}
\end{gather}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .