1
$\begingroup$

Please help me understand this equation :

Total kinetic energy of two particle system = K.E. of system w.r.t. centre of mass + K.E. of centre of mass

Here's the derivation from my book :

enter image description here

enter image description here

The author starts off by deriving the individual momenta of the two particles w.r.t. the centre of mass, and later uses the formula given below to find their kinetic energies in the centre of mass frame. (Is that the correct way to put it?)

$$K = {p^2}/2m$$

However, when they add the two energies (please see the pictures), they get a value which seems to be half of the correct value (1/2 + 1/2 = 1 and not 1/2). This is one of my doubts.

Next, they find the kinetic energy of the centre of mass in the ground frame. This is pretty straightforward :

$$K_{com (ground frame)} = (1/2)({m_1} + {m_2}){(V_{com (ground frame)})}^2$$

Now comes the confusing part (my second doubt) :

The kinetic energy of a system of particles ($K$) is equal to the kinetic energy of the system relative to centre of mass ($K'$) plus kinetic energy of the centre of mass.

$K_{Total}$ = K.E. of system w.r.t. centre of mass + KE of centre of mass

(Again, please refer the pictures.)

They are almost treating kinetic energies like vectors and adding them up! How is that correct?

...

I found a related question, but didn't understand the answer provided. :(

...

A few solved examples related to this are provided in my textbook. Here's one of them :

enter image description here

Why have they used such a complicated formula to calculate the change in kinetic energy? Couldn't they have used a much simpler formula :

$$K = {1/2}mv^2 + {1/2}MV^2$$

...

Any help is appreciated. 👍

|cite|improve this question
$\endgroup$
1
$\begingroup$

In the second photo you have (with m1+m2 = M)

$K_{system,cm} = \frac{1}{2}m_1 m_2^2 / M^2 + \frac{1}{2} m_1^2m_2 / M^2$

= $\frac{1}{2M^2} (m_1m_1m_2 + m_1m_2m_2)$

= $\frac{1}{2M^2} (m_1m_2(m_1+m_2))$

= $\frac{1}{2M^2} (m_1m_2M)$

= $\frac{1}{2} m_1m_2/M$

So the authors are correct, apart from the m2->m1 type that you corrected. Maybe you missed that the numerators squared different mi?

In the "Important point" box in your second photo, the book makes the useful point that we can decompose the energy a multi-particle system (whether 2 particles or a gas in a box) as a sum of two terms: the internal motions (measured in a center-of-mass frame) + the average motion. In this case, once you know the internal energy, $K_{system,cm}$ you can then easily figure it out for a moving system: just add $\frac{1}{2}MV^2$ -- no need to recompute the velocities of the particles, which may be easy for 2 particles, but a pain for 10 or 100 particles.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks a lot! I got it now. You have my upvote for that. I still don't understand the last paragraph of your answer, though. How do we know that the internal motion and the average motion are exclusive (don't have anything in common)? After all, energy is not a vector, which can be very easily decomposed into two parts without any confusion/ambiguity. :( $\endgroup$ – Anurag B. Jul 16 '18 at 13:40
  • $\begingroup$ I edited my last paragraph to make it less confusing. Still, you are right that the photographed pages do not derive the result in the "Important point" box. Maybe it is nearby in your book? However, for this 2-particle case you can, as a "textbook exercise" simply compute the total energy of two particles when the center-of-mass moves, then subtract the $K_{system,cm}$ and verify that you do get the $\frac{1}{2}MV^2$ term. $\endgroup$ – bernander Jul 16 '18 at 15:07
  • $\begingroup$ Thanks, your answer looks better now. And nope, my book doesn't provide any more proof. :( ... Maybe the author thought that it's supposed to be an intuitive thing. But I did check the formula by plugging in the values of mu (the reduced mass), M and V ... and indeed, it simplifies to (1/2)(m1)((v1)^2) + (1/2)(m2)((V2)^2) !! It's only a verification, and not a proper derivation, but I'm happy that the formula checks out. Besides, your answer clears things up, a bit. Maybe tomorrow I'll check if this logic works for frames other than that of the centre of mass. :) $\endgroup$ – Anurag B. Jul 16 '18 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.