Respuesta :
Answer:
The correct answer is a
Explanation:
The concept of center of mass strongly simplifies the analysis of problems, since all the external bridging forces = denote the point of the center ce plus.
xcm = 1 / M ∑xi mi
v cm = 1 / M ∑ xi vi
When we do this, the momentum and the kinetic energy are conserved, that is, it is equivalent to the movement of the particle system
The correct answer is a
Answer:
d. The momentum, but not the kinetic energy, of the center of mass is equivalent to that of the system of particles.
Explanation:
The center of mass is equal to:
[tex]x_{cm} =\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2} }[/tex] (eq. 1)
The total mass is:
mtotal = m₁ + m₂
Replacing:
[tex]x_{cm} =\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{total} }\\x_{cm}*{m_{total} ={m_{1}x_{1}+m_{2}x_{2}} }[/tex] (eq. 2)
Differentiate respect to time is:
[tex]\frac{d}{dt} x_{cm} m_{total} =\frac{d}{dt} m_{1} x_{1} +m_{2} x_{2}\\m_{total}v_{cm} =m_{1} v_{1}+m_{2} v_{2}[/tex]
The momentum is:
[tex]P_{cm} =m_{total} v_{cm} =m_{1} v_{1}+m_{2} v_{2}[/tex]
Differentiate the equation 1 respect to time is:
[tex]\frac{d}{dt} x_{cm} =\frac{d}{dt} \frac{m_{1}x_{1}+m_{2}x_{2}}{m_{total} }\\v_{cm} =\frac{1}{m_{total} } {m_{1}v_{1}+m_{2}v_{2}} }[/tex]
The kinetic energy is:
[tex]E_{k} =\frac{1}{2} m_{total} v_{cm} ^{2} =\frac{1}{2m_{total} } (m_{1}v_{1} -m_{2}v_{2})^{2}[/tex]
Observing equations 1 and 2, it can be seen that the kinetic energy of the center of mass is not equal to that of the particle system.
