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?
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.
A complete relativistic relation is E^{2} = (m c^{2})^{2} = (m_{0}c^{2})^{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 m_{0} = 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 m_{0}c^{2} = sqrt[1100^{2} - 953.94^{2}] = 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 10^{8} meters per sec = 2.599 x 10^{8} meters per sec
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 doesnt change which makes my question a bit whacko obsolete.. and no my body doesnt travel to the speed of light so I guess I got my answer .. Thanks
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.
Hey you refered 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
Here is the exact equation for relating particle relativistic total energy E, momentum p, and rest mass m_{0}c^{2}. I show here that for nonrelativistic particles, it reduces to the standard non-relativistic relation. 1) E^{2}= (m_{0}c^{2})^{2} + (pc)^{2} where E is total energy (rest mass + kinetic energy), m_{0}c^{2} is rest mass, and pc is momentum in energy units (momentum x speed of light) Using K as kinetic energy 2) E = m_{0}c^{2} + K 3) (m_{0}c^{2} + K)^{2}= (m_{0}c^{2})^{2} + (pc)^{2} 4) (m_{0}c^{2})^{2} + 2m_{0}c^{2} K + K^{2} = (m_{0}c^{2})^{2} + (pc)^{2} 5) K (1+ K/2m_{0}c^{2} )= (pc)^{2}/2m_{0}c^{2} = p^{2}/2m_{0} For non relativistic cases, we drop the small term K/m_{0}c^{2} and get the familiar equation: 6) K = p^{2}/2m_{0} = m_{0}^{2} v^{2}/2m_{0} = (1/2)m_{0}v^{2 }which is the non relativistic equation.