Momentum

(Note: If you see the symbol D, it represents delta, i.e., final minus initial)

We know that if you throw a basketball and a medicine ball with the same force, the basketball will move with a greater velocity. We can see this from Newton's Second Law,

(1) F = ma

If the force is the same, but the mass is greater, the acceleration must be smaller, thus giving lower velocities.

Since it is often times difficult to talk about forces that change such as those occurring during collisions, we define a new term, momentum, as an object's mass times its velocity, that is,

(2) p = mv

We can see momentum's relationship to force by considering the definition of acceleration: the change in velocity with time.

From equation (2), we have that v =p/m.
Thus the change in v = change in p/m.
Thus the acceleration = change in p/m.
Since m is constant, it doesn't change, so
a = (change in p) / m.
or m x a = change in p
that is F = change in p with time,

(3) F = D p / D t

.

So we see that force is just the change of momentum in time.

Momentum is also related to kinetic energy, as the mv term implies. From our definition of momentum, we have that velocity (v) is momentum / mass (p/m), so we can substitute back into the Kinetic Energy equation to find:

KE = 1/2 m v2
KE = 1/2 m (p / m) 2
KE = 1/2 m p2 / m 2
KE = 1/2 p2 / m
KE = p2 / 2m

So we find that kinetic energy increases as the square of momentum. That means that if we double an object's momentum (by doubling its velocity say), we quadruple its kinetic energy (double squared).

From Newton's First Law we know that an object in motion will stay in motion unless acted upon by an outside force. Usually, an object's motion is changed by a force acting for only a short period of time. We call a force that acts for a short period of time an impulse. An example of such a force is a baseball bat hitting a ball. It applies a force for a short period of time (< 1 second) and changes the direction of the ball, thus its velocity and momentum. Impulse is defined as the change in momentum,

Impulse = Dp = F D t

This definition of impulse can help us understand things such as why a jumping person isn't hurt as bad when they bend their knees or why a car crashing into a bush is not as dangerous as a car crashing into a brick wall.

Lets assume that the mass of the car, m, doesn't change, and that it starts at some initial velocity, v. We further assume that after the crash the car's velocity is zero.

Our change in momentum is

mvfinal - mvinitial

0 - mvi, which is a constant number.

When the car crashes into a wall, it takes a very short time to stop. When the car crashes into bushes, it takes a longer time to stop. Because Dp is a constant, the force acting on the car through the bushes must be smaller than that acting on the car crashing into the wall. Visually, and somewhat crudely, we can think of it like this:

Please don't try this at home.

 Wall: Dp = F Dt Bush: Dp = F Dt  We know that momentum and energy are related and that energy is always conserved. Thus it is reasonable to assume that momentum is conserved in some way, and this indeed is the case. The conservation of momentum states that the sum of final momentum is equal to the sum of initial momentum. For two objects, 1 and 2, that is,

m1fv1f + m2fv2f = m1iv1i + m2iv2i

Examples of the conservation of momentum are seen often in space flight, ice skating, and exploding projectiles. See college prep pages 121-122.

Rocket propulsion is made possible by conservation of momentum. Just as with two skaters pushing apart on an ice rink, so is it with the rocket and the gasses produced by its combusting rocket fuel. The force of the gas being shot out the rocket produces a force pushing the rocket skyward. The thrust of a rocket is the force that the gasses leave with. The rocket doesn't push against anything. It pushes its gasses out and its gasses push it up.

This thrust force is given by the velocity of the escaping exhaust gasses multiplied by the rate at which the fuel is burned

Thrust = F = D p / D t = D mv / D t = v (D m / D t) If we do a few calculations, we see that if the velocity of the escape gasses is 3 km/s an 75% of the mass is fuel, we find that the rocket attains a final velocity of vf = 4.2 km/s, which although is very fast, is nowhere near the approx. 11 km/s needed to leave the pull of earth's gravity. So how is it done?

Rockets use stages in order to accomplish this task.

After the fuel in each stage is used up, the stage is discarded thus making the rocket lighter and adding extra momentum.

Next Stop, collisions!