momentum

The classical equation for momentum (p=mv), frustratingly enough, becomes inaccurate when velocity approaches the speed of light—that is, momentum increases exponentially with increased velocity. The formula for relativistic momentum is:

p=mv/sqrt(1-(v²/c²))

where c is the speed of light. So, a physics textbook with a mass of 5 kg moving at .95c would, under Newtonian physics, have a momentum of 1,425,000 kg·m/s (the unit for momentum), but with this newfangled theory of relativity, its momentum is 4,563,652 kg·m/s-- more than three times as much! As velocity approaches c, the denominator of the fraction approaches 0, so the momentum approaches infinity—further proof that going faster than light is impossible, as it would require infinite force to produce an infinite change in momentum.

Note also that at everyday velocities, like 30 m/s, the denominator is so close to 1 (in this case, it's something like 1 minus 6.6x10^-15) that it's virtually identical to the Newtonian model. Only at relativistic velocities is this model required.

This is very much like the effect from Newton's Cradle. Things get interesting when one car is heavier than the other. If the moving car is twice the mass of the stationary car, then you can say that the mass of the moving car is "2M" (2*M) and the stationary car is "M". If the moving car is moving at "V" ms^-1, then the momentum is 2*1 = 2 kg ms^-1. When the moving car hits the stationary car in an elastic collision all momentum will be transferred, so car number 2 also has momentum of "2M". Because its mass is M, its velocity will be 2V, ie. twice as fast as car number 1.

This effect can also be seen in golf when the ball leaves the surface of the club at a much faster velocity than the club was moving, because the club has a much higher mass than the ball. An explosion can be said to have momentum. An unexploded bomb has no momentum because it is not moving. A second after it explodes shrapnel of varying sizes will be travelling in all directions. Because velocity, and momentum, are vectors, something travelling in one direction will cancel out the thing travelling in the opposite direction. This is because the thing travelling in one direction has a mass of M and a velocity of V, so a momentum of MV, while its opposite counterpart will also have a mass of M but a velocity of -V, so a momentum of -MV. If you add MV and -MV you get 0, so the momentum before the explosion, and the momentum 1 second after is equal. Shrapnel from an explosion will always be balanced in all directions due to this law of physics (or should i say that this law of physics is so because shrapnel is always balanced?).

This is not to say an explosion is perfectly symmetrical, just that all the momentums will add up to 0 in every direction (there may be a few slow big things going in one direction, and the same amount of small things going in the opposite direction but much faster).

p = mv

Since velocity depends on reference frames, all magnitudes must include the reference frame. Looking at the equation, we see that a larger massed object will have a greater momentum than a smaller massed one with the same velocity. What this basically means is that a big truck going at 70 km/h can smash your house, but a 2 year old baby going at 1 m/s will not (at least by walking into it).

Since velocity can only be changed by applying a force, it follows that momentum also must be changed in the same manner. As a matter of fact, Newton's second law was originally stated in a more general form, dealing with momentum (which he referred to as quantity of motion). The original second law, paraphrased, is:

The rate of change of momentum of a body is proportional to the net force applied to it.

Or...

ΣF = Δp/Δt

Where ΣF stands for its normal value, the net force applied to an object and Δt is the change in time. Some simple arithmetic (which is left to the reader) is all that is required to translate the above equality to the familiar ΣF = ma.

Conservation of Momentum

A simple concept -- the total momentum of an isolated system is the same at all times (i.e., no momentum is gained or lost).

The total momentum of an isolated system of bodies remains constant.

But 'Wait!', you say. 'What if we drop a ball from 5 metres? It will accelerate, thus increasing its momentum, and then reach a velocity of zero, reducing its momentum to zero. This means there was a loss of momentum.' Not so. The solution in all such cases is to expand the system. In this case, we would include the Earth. In the case of the ball, all of the ball's momentum was transferred to the Earth. However, since the mass of the Earth is so large, very little change in velocity is needed to impart to the Earth the same momentum as the ball. This idea is extremely important in the field of Mechanics.

*Note: Bold letters are designated as vectors.

p = mv / √[ 1 - (v/c)² ] = γmv

where γ = 1 / √[ 1 - (v/c)² ] and γ > 1. Some prefer to associate the γ with the mass and introduce a relativistic mass m_R = γm.. That allows you to write the momentum as p = m_Rv, but m_R is speed dependent. Usually one mass m is used, which is independent of speed, just like the two other fundamental properties of particles of mattter, charge and spin.

Mo*men"tum (?), n.; pl. L. Momenta (#), F. Momentums (#). [L. See Moment.]

1. Mech.

The quantity of motion in a moving body, being always proportioned to the quantity of matter multiplied into the velocity; impetus.

2.

Essential element, or constituent element.

I shall state the several momenta of the distinction in separate propositions. Sir W. Hamilton.

© Webster 1913.