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 $$E=\overrightarrow{E}+{\displaystyle \frac{1}{x}}m{V}^2$$. 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:
$$P=\sum {\displaystyle \frac{1}{2}}{m}_i{v}_i^2$$
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:
$$P=\sum {\displaystyle \frac{1}{2}}{m}_i{v}_i^2$$
Or, $$P=\sum {\displaystyle \frac{1}{2}}{m}_i\left(\overrightarrow{v_1}+\overrightarrow{v_c}\right)\left(\overrightarrow{v_i}+\overrightarrow{v_c}\right)$$
Or, $$P=\sum {\displaystyle \frac{1}{2}}{m}_i{v}_i^2+\overrightarrow{v_c}\sum m_i\overrightarrow{v_1}+\sum {\displaystyle \frac{1}{2}}{m}_i{v}_c^2......(1)$$
Also in the frame C, $$\sum m_i{v}_i=0$$
Hence equation now becomes:
$$P=\overrightarrow{P}+{\displaystyle \frac{1}{2}}m{v}_c^2$$
Or, $$P=\overrightarrow{P}+{\displaystyle \frac{1}{2}}m{V}^2$$. (because here in this problem $$v_c=V$$)
Hence the equation will not change and be given by :
$$P=\overrightarrow{P}+{\displaystyle \frac{1}{2}}m{V}^2$$
But according to the given question, in frame K, they have considered the equation as $$E=\overrightarrow{E}+{\displaystyle \frac{1}{x}}m{V}^2$$ here they have considered kinetic energy as E. thus by comparing $$E=\overrightarrow{E}+{\displaystyle \frac{1}{x}}m{V}^2$$
And $$P=\overrightarrow{P}+{\displaystyle \frac{1}{2}}m{V}^2$$ 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.