Evidence For The Relativistic Mass Formula

Consider identical particles that approach each other along a line, each with speed v relative to the Earth.  Assume the particles collide elastically and continue along a perpendicular line with the same speed relative to the Earth.  The following are the velocity vectors before the collision relative to the Earth:

(v, 0) and (-v, 0).

The following are the velocity vectors after the collision relative to the Earth:

(0, v) and (0, -v).

Consider a rigid body moving with velocity v in the initial direction of one of the particles.  Let c be the speed of light in a vacuum.  The following are the velocity vectors before the collision relative to this rigid body:

(0, 0) and (-2v / (1 + v² / c²), 0).

The following are the velocity vectors after the collision relative to this rigid body:

(-v, v√1 - v² / c²) and (-v, -v√1 - v² / c²).

The law of conservation of momentum appears to not be true relative to the rigid body!  Assuming the law is still true, that suggests that mass varies with speed.  Consider the following particle mass formula where m_o is the mass relative to the particle, and, u is the speed relative to the rigid body:

m_o / √1 - u² / c².

Let γ(u) equal 1 / √1 - u² / c². The following is the new initial total momentum relative to the rigid body:

γ(2v / (1 + v² / c²))(-2m_ov / (1 + v² / c²), 0).

The following is the new final total momentum relative to the rigid body:

γ(v√2 - v² / c²)(-2m_ov, 0).

This implies:

γ(2v / (1 + v² / c²)) / (1 + v² / c²) = γ(v√2 - v² / c²).

Because both sides of the equation are always positive, it is sufficient to confirm whether the difference of the squares of both sides is zero for all v.  Here is Sage code that confirms this:

var("v c")

def gamma(v, c):
        return 1 / sqrt(1 - v^2 / c^2)

lhs  = gamma(2 * v / (1 + v^2 / c^2), c) / (1 + v^2 / c^2)
rhs  = gamma(v * sqrt(2 - v^2 / c^2), c)
diff = lhs^2 - rhs^2
print(diff)
print(diff.full_simplify())

This relativistic mass formula has been confirmed by experiment.