Suppose I have two particles with different masses and velocities. The momentums are as follows: $$\mathbf{p}_1 = m_1 \mathbf{v_1} $$ $$\mathbf{p}_2 = m_2 \mathbf{v_2}$$ The total momentum is: $$\mathbf{P} = m_1 \mathbf{v_1}+ m_2 \mathbf{v_2}$$ Now, for the individual particles, the kinetic energies are given by: $$k_1 = \frac{\mathbf{p}_1^2 }{2m_1}$$ $$k_2 = \frac{\mathbf{p}_2^2}{2m_2}$$ The total mass is: $$M=m_1 + m_2$$ And the total kinetic energy is: $$K = k_1 + k_2 = \frac{\mathbf{p}_1^2 }{2m_1} + \frac{\mathbf{p}_2^2}{2m_2}$$ This perhaps suggests a natural question though; is there any meaning that can be assigned to the quantity: $$\frac{\mathbf{P}^2}{2M}?$$ Kinetic energy can be interepted by the work-kinetic-energy-theorem as that things which changes commensurately with the work done on a particle. Maybe there is another type of similar quantity which can be related to the above?
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$\begingroup$ What’s the point of using momentum? Mass is constant $\endgroup$– Bob DCommented Apr 12 at 21:04
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$\begingroup$ @BobD I do not understand your question. Momentum is a very important physical quantity $\endgroup$– Jbag1212Commented Apr 12 at 21:13
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$\begingroup$ I mean for this purpose why not just $$\frac{mv^2}{2}$$ $\endgroup$– Bob DCommented Apr 12 at 21:30
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$\begingroup$ @BobD The formula $k=\frac{1}{2}mv^2 = \frac{p^2}{2m}$ is used quite ubiquitously. The point is that using this formula is perhaps suggestive when you have the momentum of an entire system. $\endgroup$– Jbag1212Commented Apr 12 at 22:13
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$\begingroup$ The kinetic energy of the entire system is the sum of the kinetic energies of the particles of the system which only depends on each particles speed, not their velocities or momenta, vector quantities. I guess I just don't understand the whole point of your post. But that's OK. I will be interested in others comments. $\endgroup$– Bob DCommented Apr 12 at 22:51
2 Answers
Introducing the total momentum $$\mathbf{P}=\mathbf{p}_1+\mathbf{p}_2$$ of the two particles, the momentum of their relative motion, $$\mathbf{p}= \frac{m_2 \mathbf{p}_1-m_1 \mathbf{p}_2}{m_1+m_2}, $$ the total mass $$M=m_1+m_2$$ and the reduced mass $\mu$ defined by $$ \frac{1}{\mu}= \frac{1}{m_1}+\frac{1}{m_2},$$ the total kinetic energy can be rewritten in the form $$\frac{\mathbf{p}_1^2}{2m_1}+\frac{\mathbf{p}_2^2}{2m_2}=\frac{\mathbf{P}^2}{2M}+\frac{\mathbf{p}^2}{2\mu},$$ where $\mathbf{P}^2/2M$ is the contribution of the center-of-mass motion and $\mathbf{p}^2/2\mu$ the one of the relative motion.
We can define the net work done on the center-of-mass (known as pseudo-work or center-of-mass work) using the sum of the external forces acting on an extended object and the displacement of the center of mass. Explicitly, $$ W_{\textrm{com}} = \int \vec{F}_{\textrm{net}}\cdot d\vec{R}\,, $$ where $\vec{R}$ is the position of the center of mass, and $$ \vec{F}_{\textrm{net}} = \sum_i \vec{F}_{\textrm{net on particle }i} = \sum_i \vec{F}_{\textrm{net external on particle }i}\,, $$ where the internal forces, i.e, forces exerted by particles in the system on other particles in the system all cancel each other out, by Newton's third law. (Note that $\vec{F}_{\textrm{net}}$ is truly the net force acting on the system of particles, but by computing the work using this net force rather than summing the works done on individual particles, we are losing information.)
It is possible to show that $$ W_{\textrm{com}} = \Delta K_{\textrm{tr}}\,, $$ where $K_{\textrm{tr}}$ is the translational kinetic energy of the center of mass, i.e., it is exactly $$ K_{\textrm{tr}} = \frac{P^2}{2m}\,, $$ where $\vec{P} = \sum_i \vec{p}_i$ is the total momentum of the system.
This establishes a kinematic relationship between motion of the center of mass of a system and the forces exerted by external agents on the system of particles. One must be careful to interpret this relationship as an energetic one, because it doesn't include any of the energies associated with the motion of the particles relative to each other. For that, we have to do a more careful accounting that includes the forces that the particles interior to the system exert on each other, and thereby the works that the particles do on each other. In addition, if there are things like friction, we have to be even more careful about thinking about this relation as an energy relation.
