# Momentum

#### mentaaal

Joined Oct 17, 2005
451
Coming from a college student, this is going to sound ludicrous but what is momentum? I am fully aware of the formulae for momentum and kinetic energy but do not fully understand the difference. I am aware that momentum is the differential of kinetic energy but am just having a hard time trying to understand what it is?
Why is it that momentum is always conserved in a collision, even when some energy can be released from the system (inelastic collision)

#### studiot

Joined Nov 9, 2007
4,998
OK

Momentum is a vector quantity
= the product of mass (a scalar) and velocity (a vector)

Energy is a scalar quantity

Momentum is preserved because we need to preserve both all parts of the vectors involved in a collison, both magnitude and direction.

No scalar can achieve this.

Is this dummies explanation adequate or do you want the fancy maths?

#### mentaaal

Joined Oct 17, 2005
451
This dummies explanation is is what i am always told and made learn but it just isnt doing anything for me. Yes indeed, perhaps some fancy maths would be nice although if its very time consuming a nudge in the right direction will be more than sufficient. I dont spend my time on these forums asking questions in the interests of wasting other peoples' time ;-)

#### Mark44

Joined Nov 26, 2007
628
Greg,
I had to dig up my college Physics book to see what Halliday & Resnick had to say. By their definitions, kinetic energy is the work a body can do by virtue of its motion. Momentum is defined more technically, as the product of the mass of a particle and its velocity.

Given your original question, that's probably not very satisfying. H & R go on to talk about Newton's second law of motion, which they write as "The rate of change of momentum is proportional to the resultant force acting on the body and is in the direction of the force."

In symbols, F = d/dt(p)

So this formula gives a relationship (albeit indirect) between the force on a body, and its momentum. If we antidifferentiate both sides, we get
int[F dt] = p = mv (+ a constant, maybe)

From this, we see that momentum p can be thought of as the sum of forces over time on the body in question.
Mark

#### mentaaal

Joined Oct 17, 2005
451
Thanks for that mark, if anything that is one tasty chunk of food for thought. I also have to get my hands on that book you are mentioning. My physics lecturer also highly recommends it.

#### Mark44

Joined Nov 26, 2007
628
My edition is from 1967. I bought it used for \$13 in 1971, and have hung onto it, for times when I forget that F = ma.

My buddy that I shared a house with at the time and I used to joke that if you couldn't remember F = ma, then F = your grade.

#### thingmaker3

Joined May 16, 2005
5,084
Momentum is why the parked Ford Pinto bounces off the front of the speeding Cadillac.

#### Mark44

Joined Nov 26, 2007
628
Momentum is why the parked Ford Pinto bounces off the front of the speeding Cadillac.
But NOT why it bursts into flames.

#### veritas

Joined Feb 7, 2008
167
momentum is why a baseball hitting you at 30 mph hurts more than a paintball going 90 mph

It's why the 'm' is in the F=ma equation

#### Mark44

Joined Nov 26, 2007
628
momentum is why a baseball hitting you at 30 mph hurts more than a paintball going 90 mph
Another reason is the kinetic energy you are absorbing.

#### studiot

Joined Nov 9, 2007
4,998
'Dummies' was not meant as a criticism. I have learned some really useful stuff from the Dummies Book series and would recommend them to anyone.

Form your original post I though you knew the maths but were trying to gain a feel for the physics.

In physics there are several ideas that can be expressed (measured) in more than one way.

A simple example is the concept of 'quantity of matter' which may be measured by mass or by volume. Both are useful and of course are related by the simple relationship

mass = volume x density.

Neither measure are totally satisfactory by themselves.

In the world of mechanics the concept occurs of inertia or the resistance to change of its motion possessed by a particle. In coloquial terms, how hard it can hit back.

One measure of this is Kinetic energy, as has already been pointed out by others here.

But energy methods do not provide all the answers.

Energy methods can answer the question if I were to accelerate a particle how much energy would I have to supply or even, how hard would I have to push, thanks to Newton's laws.

However they cannot answer the question How hard would I have to push to change the direction of motion, without changing the velocity?

For this we need another model of inertia. We call this this property momentum, because it directly answers the question how much force would the particle exert if it hits a wall?

Consider this collision, the particle bounces back with reversal of direction but no change of actual velocity, the wall does not move.

Since the velocity remains constant there is no change of kinetic energy.

However the particle exerts a quantifiable force on the wall during the impact and ends up travelling the other way capable of exerting a similar force on a another wall in its new path.

So something has changed.

This something we call the momentum and is linked to the force exerted by the impulse equation

Fdt =mdv

Incidentally I note that you were looking at electron dynamics in you spectrum question.

If the momentum question is related to this you need to consider relativistic momentum equations rather than Newtonian ones.

#### mentaaal

Joined Oct 17, 2005
451
Thanks for that reply studiot, I think thats about as good an answer as I could hope to get. Yes i see the point your making and as you have expressed it very well! The rest is up to me to think on it.

Cheers and kudos!

#### mentaaal

Joined Oct 17, 2005
451
Energy methods can answer the question if I were to accelerate a particle how much energy would I have to supply or even, how hard would I have to push, thanks to Newton's laws.

However they cannot answer the question How hard would I have to push to change the direction of motion, without changing the velocity?

For this we need another model of inertia. We call this this property momentum, because it directly answers the question how much force would the particle exert if it hits a wall?

Consider this collision, the particle bounces back with reversal of direction but no change of actual velocity, the wall does not move.

Since the velocity remains constant there is no change of kinetic energy.

However the particle exerts a quantifiable force on the wall during the impact and ends up travelling the other way capable of exerting a similar force on a another wall in its new path.

So something has changed.

This something we call the momentum and is linked to the force exerted by the impulse equation

Fdt =mdv

Incidentally I note that you were looking at electron dynamics in you spectrum question.

If the momentum question is related to this you need to consider relativistic momentum equations rather than Newtonian ones.
I have been pondering your words and am afraid i am still at a loss...

You said " Since the velocity remains constant there is no change of kinetic energy." For the ball to have an elastic collision with the wall such that kinetic energy is conserved then momentum has to also be conserved. When kinetic energy is conserved momentum has to be conserved but not vice versa. ALso when you say that momentum changed by exerting a force on the wall i agree, but so too did the kinetic energy. It had to have as at some point it wasnt moving at all, the energy was being stored in the elastic structure of the partical. Your explanation of momentum is however very good and i am starting to see things a little clearer. Thanks for going to the trouble of writing such a comprehensive response.

#### Caveman

Joined Apr 15, 2008
471
An interesting note about momentum involves Newton's second law. People often simplify it to F = m*a. However, they often forget to state that the mass must be constant for this to be true (or that it is actually a simplification at all).

Newton's second law actually states that force is the time rate of change of momentum, so F = dp/dt as already stated in this thread. To get to the simplification, you need to do a little math...

Since momentum is mass * velocity, you can differentiate it using the product rule to get
F = dp/dt = d(m*v)/dt = dm/dt * v + m * dv/dt = dm/dt * v + m*a.
If the mass is constant, dm/dt = 0, so the first term goes away.

If you want to know a good use of the full equation, think about rockets. As a rocket travels, it burns its fuel. This reduces it's mass, and so dm/dt != 0, and so that term plays a role. If you ignore it, you're calculated acceleration will be smaller than the real acceleration.

#### mentaaal

Joined Oct 17, 2005
451
A point indeed worth noting! So thats why my laser guided rockets kept overshooting....

#### Caveman

Joined Apr 15, 2008
471
I have been pondering your words and am afraid i am still at a loss...

You said " Since the velocity remains constant there is no change of kinetic energy." For the ball to have an elastic collision with the wall such that kinetic energy is conserved then momentum has to also be conserved. When kinetic energy is conserved momentum has to be conserved but not vice versa. ALso when you say that momentum changed by exerting a force on the wall i agree, but so too did the kinetic energy. It had to have as at some point it wasnt moving at all, the energy was being stored in the elastic structure of the partical. Your explanation of momentum is however very good and i am starting to see things a little clearer. Thanks for going to the trouble of writing such a comprehensive response.
Technically, the velocity does change. Velocity is a vector, so a change in direction is a change in velocity. You are right, though, about the storage of energy. Note that momentum, acceleration, and force are all vector quantities as well. K.E. is a scalar (ie, not a vector).

1. At first the particle has a specific momentum and K.E., but no potential energy due to compression.
2. Then as the particle contacts the wall, the wall exerts a force on the particle. F = dp/dt, but the force is opposite of the direction of the velocity, so the velocity will decrease. The K.E. is decreasing. The energy is being transferred into a form of potential energy in the elasticity. Also, in the theoretical case of a perfectly elastic particle, incompressible wall, and homogenous particle, the source of the force will be due less and less to the velocity of the ball and more and more due to the particle attempting to decompress.
3. Eventually, the velocity will reach zero. At this point, the K.E. is 0 as well. The force remains the same, but now the particle's attempt at decompression is the sole source.
4. Now the magnitude of the velocity will begin to increase, but in the opposite direction from original. The reverse occurs until all of the potential energy is transferred back to kinetic energy.