Author was seeing a foot ball match. Ball & players are moving all over the ground. Author struck something different. In matter electrons & other particles are also moving randomly & vibrating all over the matter similar to players in ground. (Similar to gas molecules in box) This random velocity increases individual mass of particle; cumulative effect will be increase in total mass of matter. When this matter moves with relative to other observer, mass of matter again increases due to relativity. Author consider that at the place of total matter if we consider each individual moving particles in matter & if we consider relative increase in mass of each particle individually & sum the all relative masses with relative to observer, we will get relative mass of matter A with relative to observer. Means, relative cumulative mass of each constituent of matter A is equal to the relative mass of matter A with relative to the observer. To calculate such a individual mass of particles is impossible in real world. So, Author created similar situation.

Consider cabin A of train moving with relative to the man on platform with velocity Vx. Consider at the centre of mass of cabin horizontal metal plank is fixed in such a way that this plank is perpendicular to velocity of train. On this plank two balls of same mass are moving in opposite directions with velocity Vy from same distance in such a way that centre of zero moment of cabin is same as centre of mass of cabin.

Part 1 :-

Let, consider mass of each ball (at rest) = mb

Consider rest mass of cabin A excluding two ball = mc

Then, Total rest mass of cabin including two balls, Mr = mc + 2 x mb x {1/(1-Vy^{2}/C^{2})^{0.5}}----------(1)

This mass will act at the centre of mass of train cabin which is also a point where summation of total moment of substances in cabin is zero.

When this cabin A moves with velocity Vx with relative to man on platform.

Mass of cabin with relative to man on platform = Mr. {1/(1-Vx^{2}/C^{2})^{ 0.5}}

By equation (1) put value of Mr

Mass of cabin with relative to man on platform with two balls= [mc + 2 x mb x {1/(1-Vy^{2}/C^{2})^{ 0.5}}]. [1/(1-Vx^{2}/C^{2})^{ 0.5}]

= mc x [1/(1-Vx^{2}/C^{2})^{ 0.5}] + 2 x mb x [1/(1-Vy^{2}/C^{2})^{ 0.5}]. [1/(1-Vx^{2}/C^{2})^{ 0.5}]

----------------------------------------------------------equa (2)Part II :-Each individual moving matter has taken separately

Now, consider relative mass of balls & cabin separately with relative to man on platform.

Relative mass of only cabin= mc x [1/(1-Vx^{2}/C^{2})^{ 0.5}] ---------------------(a)

Velocity of each ball with relative to man on platform is sum of velocity Vx & Vy in perpendicular direction.

So, Relative velocity of ball with relative to observer on platform = (Vx^{2}+Vy^{2})^{ 0.5}

Relative mass of each ball = mb x [1/{1-(Vx^{2 }+Vy^{2})/C^{2}}^{ 0.5}] ---------------------(a)

Total relative mass of cabin with balls =mass of cabin+2 x mass of individual ball

=eq(1) + 2 eq(2)

= mc x [1/(1-Vx^{2}/C^{2})^{ 0.5}]+2 x mb x [1/{1-(Vx^{2 }+Vy^{2})/C^{2}}^{ 0.5}]

--------------eq(3)

As eq(2) & eq(3) gives same mass of cabin with balls, so R.H.S. results can be equated

i.e. mc x [1/(1-Vx^{2}/C^{2})^{ 0.5}] + 2 x mb x [1/(1-Vy^{2}/C^{2})^{ 0.5}]. [1/(1-Vx^{2}/C^{2})^{ 0.5}] = mc x [1/(1-Vx^{2}/C^{2})^{ 0.5}]+2 x mb x

[1/{1-(Vx^{2 }+Vy^{2})/C^{2}}^{ 0.5}]

[1/(1-Vy^{2}/C^{2})^{ 0.5}]. [1/(1-Vx^{2}/C^{2})^{ 0.5}] = [1/{1-(Vx^{2 }+Vy^{2})/C^{2}}^{ 0.5}]

(1-Vy^{2}/C^{2})^{ 0.5}. (1-Vx^{2}/C^{2})^{ 0.5}= {1-(Vx^{2 }+Vy^{2})/C^{2}}^{ 0.5}

(1-Vy^{2}/C^{2}). (1-Vx^{2}/C^{2}) = {1-(Vx^{2 }+Vy^{2})/C^{2}}

1- Vy^{2}/C^{2}-Vx^{2}/C^{2}+(Vy^{2}/C^{2}).(Vx^{2}/C^{2)}) =1-Vx^{2 }/C^{2}-Vy^{2}/C^{2}

(Vy^{2}/C^{2}).(Vx^{2}/C^{2)}) =0

Vy^{2}.Vx^{2}=0

Vy.Vx =0

But as Vy & Vx is not zero. Vx. Vy is not equal to zero.

So, L.H.S. is not equal to R.H.S.

This will create inconsistency.

Is anything wrong in above calculation?