what is vilocity at which particle mass is double the rest mass?

Answer 1

M = gamma*m0

where M is the mass, and m0 is the rest mass, with gamma (the Lorentz factor) defined as follows

gamma = 1/SQRT(1 - (v/c)^2)

where c is the speed of light

You want to have gamma = 2

2 = 1/SQRT(1 - (v/c)^2)

SQRT(1 - (v/c)^2) = 1/2

1 - (v/c)^2 = 1/4

(v/c)^2 = 3/4

v/c = 1/2*SQRT3

v = 0.86 c

Answer 2

First, I assume you mean at what velocity is relativistic mass = 2 x rest mass....

equation is

M = 1 / (1 - (v^2/c^2))^.5 x m

where M = relativistic mass
m = rest mass
v = velocity
and c = speed of light

rearranging gives...

M/m = 1 / (1 - (v^2/c^2))^.5

and your question is when does M = 2m, ie when is M/m = 2

so let

2 = 1 / (1 - (v^2/c^2))^.5
4 = 1 / (1 - (v^2/c^2))
1/4 = 1 - (v^2/c^2)
v^2/c^2 = 3/4
v^2 = 3/4 c^2
v = 3^.5 / 2 x c = .866 x c = 2.6 x 10 ^8 m/s in a vacuum

Answer 3

The equation of the relationship between the mass of a particle in motion and that same particle at rest is:
Mm = Mr/sqrt [1 - (v/c)^2],
Where Mm is mass of particle in motion with velocity v,
Mr is mass of particle at rest,
c is the velocity of light.
For Mm = 2* Mr, we have (v/c)^2 = 3/4, or v = c * 1.7320/2 = c* 0.866 , i.e. when velocity of particle has attained about 86.6 % the speed of light.

Answer 4

At the point at which the object bounces. (?) I think check me though its been a long time since I took Physics.