Relation between mass and velocity

[sundar0206]

Hi fellas..

I am not so good with physics .. Clearly defined because I am a programmer... I came across this question during one of my simulation experiments .

I was using RK4 integrator to solve equations . Which means i feed in momentum to calculate the velocity.

Momentum = mass x velocity

I hope i did get that equation correctly. This meant that to find the velocity i multiple momentum with inverse mass. .
So for increasing the velocity of the particle u decrease the mass .. but is nt it a practical observation that a heavier object falls down faster...? Am i wrong in saying this? can someone tell me the physics behind it?

 

[madmike159]

P = mv (momentum = mass x velocity). Mass doesn't change (so long a v is less than 10% of the speed of light). If v decreases then so will p. So momentum is proportional to velocity, with mass being the constant.
This should be fine unless your modling massless objects (light), or very fast objects (over 10% of the speed of light), in which case it gets more complex.

 

[Bob S]

A complete relativistic relation is

E² = (m c²)² = (m₀c²)² + (pc)²
where pc = momentum in energy units (divide by c to get momentum) (note: the mass used in the momentum is the relativistic mass)
E = total energy
m = relativistic mass
m₀ = rest mass

Example problem: If a moving neutral particle decays into two photons, a 500-MeV photon and a 600-MeV photon, with an opening angle of 60 degrees, what is the rest mass of the neutral perticle?
After balancing transverse momentum, the longitudinal momentum (pc units) is
pc = 600 cos(27 degrees) + 500 cos(33 degrees) = 953.94 MeV
E = 500 + 600 = 1100 MeV
Therefore
m₀c² = sqrt[1100² - 953.94²] = 547.72 MeV
The closest particle is the eta meson, with a rest mass of 547.30 MeV.

The momentum beta of the neutral particle = 953.94/1100 = 0.8672
So velocity v = .8672 x 3 x 10⁸ meters per sec = 2.599 x 10⁸ meters per sec

 

[sundar0206]

Thanks guys..
Think I understood... Thought mass to vary and wondered how the velocity would change relative to the varying mass. But ya.. silly of me.. mass doesn't change which makes my question a bit whacko obsolete.. and no my body doesn't travel to the speed of light so I guess I got my answer .. Thanks

 

[Cantab Morgan]

but is nt it a practical observation that a heavier object falls down faster...? Am i wrong in saying this? can someone tell me the physics behind it?

Actually, that is the opposite of an observation. It's a myth. If you drop two objects of different weights from the same height, you'll observe that they hit the ground at the same time. Heavier objects do not fall faster.

 

[Amrit N]

Momentum = mass x velocity

I hope i did get that equation correctly. This meant that to find the velocity i multiple momentum with inverse mass. .
So for increasing the velocity of the particle u decrease the mass .. but is nt it a practical observation that a heavier object falls down faster...? Am i wrong in saying this? can someone tell me the physics behind it?

Hey you referred that for increasing velocity , there is decreasing mass.This should be when momentum is suppose to be constant quantity, but actually it increases with increase in mass and velocity. i like to tell that its total quantity involved in motion.here mass and motion are part of quantities , if among two one increases, momentum increases
And other you told the heavier object fall faster,its wrong and note that all the boidies fall towards Earth at 9.8m/s^2, they fall together

 

[Bob S]

Here is the exact equation for relating particle relativistic total energy E, momentum p, and rest mass m₀c². I show here that for nonrelativistic particles, it reduces to the standard non-relativistic relation.

1) E²= (m₀c²)² + (pc)²
where E is total energy (rest mass + kinetic energy), m₀c² is rest mass, and pc is momentum in energy units (momentum x speed of light)
Using K as kinetic energy
2) E = m₀c² + K
3) (m₀c² + K)²= (m₀c²)² + (pc)²
4) (m₀c²)² + 2m₀c² K + K² = (m₀c²)² + (pc)²
5) K (1+ K/2m₀c² )= (pc)²/2m₀c² = p²/2m₀

For non relativistic cases, we drop the small term K/m₀c² and get the familiar equation:

6) K = p²/2m₀ = m₀² v²/2m₀ = (1/2)m₀v² which is the non relativistic equation.