Question: Velocity of center of mass of two particles is $~v~$ and the sum of the masses of two particles is $~m~$. Kinetic energy of the system is:


B$)~~\ge \frac{1}{2}mv^2~$



Take the two blocks as $~m_1~$ and $~m_2~$ moving with velocities $~v_1~$ and $~v_2~$ then:

Total energy $~= \frac{1}{2}(m_1v_1^2 + m_2v_2^2)~$

Case $~2$

Velocity of center of mass $$V = \frac{m_1v_1+m_2v_2}{m_1+m_2}$$

Energy $~=\frac{1}{2}(m_1+m_2)~v^2~$

The energy of the block in case$~ 1~$ is not equal to the energy of block in case $~2~$.

How is this possible?

  • $\begingroup$ Your case 1 is correct. Case 2 is not. $\endgroup$
    – Bill Watts
    Aug 1, 2019 at 21:20

1 Answer 1


Take a very simple example of two equal masses receding from each other with equal velocities. In that case the center of mass velocity is zero. By your case 2 formula, you would get the total kinetic energy of the system as zero.

It is obvious that since each mass is in motion, the kinetic energy is not zero, your case 2 formula is wrong. You must use your case 1 formula to find the total kinetic energy.

Case 2 is essentially trying to add energies vectorially, as in the case of momentum, but since energy is a scalar, it just doesn't work that way.


The total kinetic energy is (also) the kinetic energy of the center of mass plus the kinetic energy of the particles with velocities measured in the frame of the center of mass.

$\frac{1}{2} \text{m1} (\text{v1}-v)^2+\frac{1}{2} \text{m2} (\text{v2}-v)^2+\frac{1}{2} (\text{m1}+\text{m2}) v^2$

Where $v$ is the center of mass velocity = $\frac{\text{m1} \text{v1}+\text{m2} \text{v2}}{\text{m1}+\text{m2}}$.

Expand all that and it will simplify to your case 1 formula. Of course in the general case, the $v$'s will be added and subtracted as vectors and the $v^2$'s will be dot products.

  • $\begingroup$ My book tells that the kinetic energy of a system of particles = kinetic energy of center of mass + kinetic energy of different particles in the frame of reference of center of mass. How can we add the energies of the centre of mass and the particles when they are two different cases? $\endgroup$
    – lamdeb
    Aug 2, 2019 at 8:17
  • $\begingroup$ Your book is correct, but your case 2 formula is using only one term of the total formula. Your com frame of reference velocities are $v1-v$ and $v2-v$. $\endgroup$
    – Bill Watts
    Aug 2, 2019 at 9:00
  • $\begingroup$ Could you please explain on how the formula is correct? $\endgroup$
    – lamdeb
    Aug 2, 2019 at 10:31
  • $\begingroup$ @lamdeb See my edited answer. $\endgroup$
    – Bill Watts
    Aug 2, 2019 at 17:56

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