Question 8.13: Quarreling pets Now let’s apply Equations 8.18 and 8.20 to a...
Quarreling pets
Now let’s apply Equations 8.18 and 8.20 to a simple one-dimensional system. Suppose a 2.0 kg cat and a 3.0 kg dog are moving toward each other along the x axis, heading for a fight. At a particular instant, the cat is 1.0 m to the right of the origin and is moving in the +x direction with speed 3.0 m/s, and the dog is 2.0 m to the right of the origin, moving in the -x direction with speed 1.0 m/s. Find the position and velocity of the center of mass of the two-pet system, and also find the total momentum of the system.
\begin{matrix} x_{cm}=\frac{m_Ax_A + m_Bx_B + m_Cx_C + \cdot \cdot \cdot }{m_A + m_B + m_C + \cdot \cdot \cdot }, \\ \\ y_{cm}=\frac{m_Ay_A + m_By_B + m_Cy_C + \cdot \cdot \cdot }{m_A + m_B + m_C + \cdot \cdot \cdot}.\end{matrix} (8.18)
\begin{matrix} \upsilon_{cm,x} =\frac{m_A\upsilon_{A,x} + m_B\upsilon_{B,x} + m_C\upsilon_{C,x} + \cdot \cdot \cdot }{m_A + m_B + m_C + \cdot \cdot \cdot }, \\ \\ \upsilon_{cm, y}=\frac{m_A\upsilon_{A,y} + m_B\upsilon_{B,y} + m_C\upsilon_{C,y} + \cdot \cdot \cdot }{m_A + m_B + m_C + \cdot \cdot \cdot } .\end{matrix} (8.20)
SET UP Figure 8.25 shows our sketch. We represent the pets’ motion as x components of velocity, being careful with signs.
SOLVE To find the center of mass of the two-pet system, we use Equations 8.18:
x_{cm}=\frac{m_{cat}(x_{cat})+m_{dog}(x_{dog})}{m_{cat}+m_{dog}}
=\frac{(2.0 kg)(1.0 m)+(3.0 kg)(2.0 m)}{2.0 kg +3.0 kg}=1.6 m.
To find the velocity of the center of mass, we use Equations 8.20:
\upsilon _{cm,x}=\frac{m_{cat}\upsilon _{cat,x} + m_{dog}\upsilon _{dog,x}}{m_{cat} + m_{dog}}
=\frac{(2.0 kg)(3.0 m/s) + (3.0 kg)(-1.0 m/s)}{2.0 kg + 3.0 kg}
= 0.60 m/s.
The total x component of momentum is the sum of the x components of momenta of the two animals:
P_{x} = m_{cat}(\upsilon _{cat,x}) + m_{dog}(\upsilon _{dog,x})
=(2.0 kg)(3.0 m/s) + (3.0 kg)(-1.0 m/s)
= 3.0 kg · m/s.
Alternatively, the total momentum is the total mass M times the velocity of the center of mass:
P_x = M \upsilon _{cm,x} = (5.0 kg)(0.60 m/s) = 3.0 kg \cdot m/s
REFLECT As always, remember that an object moving in the -x direction has a negative x component of velocity. The total momentum of the system is equal to the momentum of a single particle with mass equal to the total mass of the system and with velocity equal to the velocity of the center of mass of the system.
Practice Problem: At the instant described, the cat decides to avoid combat. It quickly turns around and runs in the -x direction with speed 2.0 m/s. Find the position and velocity of the center of mass, and the total momentum of the system, at this instant. Answers: x_{cm} = 1.6 m, \upsilon_{cm }= -1.4 m/s, P_x = -7.0 kg \cdot m/s.
