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

Hint: 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.