Question

The reference frame, in which the centre of inertia of a given system of particles is at rest, translates with a velocity $\overrightarrow V$ relative to an inertial reference frame $K$. The mass of the system of particles equals $m$, and the total energy of the system in the frame of the centre of inertia is equal to $\overrightarrow E$. The total energy $\overrightarrow E$ of this system of particles in the reference frame $K$ is given as $E = \overrightarrow E + \dfrac{1}{x}m{V^2}$. Find x.

Answer 1

We will first form a link between the values of a system's mechanical energy in the K and C reference frames and the using the formula of velocity for any ${i^{th}}$particle we will derive at an equation for total energy. We can then compare this derived equation to find the value of x.
Formula used:
$P = \sum {\dfrac{1}{2}{m_i}v_i^2}$
Where P= kinetic energy.

Complete answer:
To discover the link between the values of a system's mechanical energy in the K and C reference frames, start with the system's kinetic energy P.
In the K frame, the velocity of the ${i^{th}}$ particle may be written as $\overrightarrow {{v_i}} = \overrightarrow {{v_i}} + \overrightarrow {{v_c}}$ ( where $\overrightarrow {{v_c}}$ is the velocity with respect to frame C and $\overrightarrow {{v_i}}$ is velocity of first reference ).
We also know that kinetic energy P:
$P = \sum {\dfrac{1}{2}{m_i}v_i^2}$
Or, $P = \sum {\dfrac{1}{2}} {m_i}\left( {\overrightarrow {{v_1}} + \overrightarrow {{v_c}} } \right)\left( {\overrightarrow {{v_i}} + \overrightarrow {{v_c}} } \right)$
Or, $P = \sum {\dfrac{1}{2}{m_i}v_i^2} + \overrightarrow {{v_c}} \sum {{m_i}} \overrightarrow {{v_1}} + \sum {\dfrac{1}{2}} {m_i}v_c^2......(1)$
Also in the frame C, $\sum {{m_i}{v_i} = 0}$
Hence equation now becomes:
$P = \overrightarrow P + \dfrac{1}{2}mv_c^2$
Or, $P = \overrightarrow P + \dfrac{1}{2}m{V^2}$. (because here in this problem ${v_c} = V$)
Hence the equation will not change and be given by :
$P = \overrightarrow P + \dfrac{1}{2}m{V^2}$
But according to the given question, in frame K, they have considered the equation as $E = \overrightarrow E + \dfrac{1}{x}m{V^2}$ here they have considered kinetic energy as E. thus by comparing $E = \overrightarrow E + \dfrac{1}{x}m{V^2}$
And $P = \overrightarrow P + \dfrac{1}{2}m{V^2}$ we can see that $x = 2$.
Hence the correct value of x will be 2.

Note:
Here keep in mind that the magnitude of a system's internal potential energy U is the same in all reference frames since the internal potential enemy U of a system depends solely on its configuration. We get the same solution by adding U to the left and right sides of Eq. 1, hence there’s no need to add it.