Momentum and Impulse

Center of Mass

The point located at the object’s average position of mass.

The Center of Mass and the

Center of Gravity

Are located at the same point AS LONG AS

gravity is consistent throughout the object.

The center of mass does not always lie within the object itself….

When an object is in motion, its center of mass will follow a smooth line.

Locating the Center of Gravity, CG

Method One:

It’s a balance point along the object.

Method Two:

If you suspend any object, its cg will be located along a vertical line drawn from the suspension point.

Center of Gravity, cg, vs Center of Mass, cm

In this course, we virtually ALWAYS use cm rather than cg, and for almost all situations, they are located at the same place. The cg is the average location of the weight- the cm is the average location of the mass. As long as the gravitational force is the same throughout the body, these two will be at the same place.

(Near the event horizon of a black hole, the gravitational force could be very different in an extended body depending on which portion of the object is closer to the black hole! In this case the cg would NOT be at the cm.)

The center of mass for multiple objects of total mass M is given by

rM

m rcm i ii

1

Finding the x-coordinate for the center of mass for 2 separate

masses.

21

2211

mm

xmxmxcm

And the same for ycm!

Newton’s First Law: When there’s no net external force acting on a system of masses (or a continuous mass), if the center of mass was at rest, it will remain at rest, if it was in motion, it continues that motion, regardless of what the individual masses are doing!

Newton’s Second Law:If there is a net external force acting upon a

system of masses, the center of mass will accelerate.

When we draw free-body diagrams, we generally draw forces acting at the cm to determine the acceleration of the center of mass (which may NOT be the acceleration of every point of the body or every particle in the system)

F = mtacm

Newton’s Third Law- we will revisit later when we look at recoil and collisions from a center of mass reference frame.

The center of mass of a system moves like a particle of mass M = mi under the influence of the net external force acting on the system.

Let’s look at an example….

A projectile is fired into the air over level ground with an initial velocity of 24.5m/s at 36.9° to the horizontal. At its highest point, it explodes into two fragments of equal mass. One fragment falls straight down to the ground. Where does the other fragment land?

Since the only external force acting on the system is gravity, the center of mass continues on its parabolic path as if there had been no explosion. The cm lands at R, where R is the range. The first fragment lands at 0.5R. The other fragment of equal mass must land at 1.5R.

What if the fragment were 0.3M? At what location X will the other fragment land?

The Range (for cm) = (vo sing x 2 x vo cos = 58.8m (the cm will fall at ½ that range)

58 8

0 31

258 8 0 7

.( . )( . ) ( . )

M x M X

M

X = 71.4m

cm

m x m x

M

( )( ) ( )1 1 2 2

The Calculus ConnectionTo find the center of mass of a continuous object, we replace the

sum with an integral

Where dm is a differential element of mass located at position r.MEMORIZE this equation, but you probably won’t have to

evaluate the integral!The AP objective states that the student should “use integration to

find the center of mass of a thin rod of non-uniform density”. I have seen questions where the student was asked to write the integration expression, but not to evaluate it, however… we’ll look at one or two…but not today…

rM

r dmcm 1

rM

m rcm i ii

1

Momentum:

A measure of how difficult it is to stop a moving object.

Momentum = mass x velocity

p = mv

Unit for momentum:sm

kg

Can a VW bug have the same momentum as an ocean liner?

Can a bullet have the same momentum as an Army tank?

If the boulder and the boyhave the same momentum,will the boulder crush the boy?

Hint:  Which would have thelarger speed?

Momentum of a system of particles

Which also yields the formula to calculate the velocity of the center of mass:

P m v Mvi ii

cm

vm v

Mcm

i i

Calculus Connection!

dt

d

dt

mvd

dt

dvmmaF

p

Kinetic Energy in terms of momentum– MEMORIZE!!

Kmv mv mv

m

2

2 2

( )( )

Kp

m

2

2Honestly, this could save you a lot of time on occasions!

How do you change the momentum of an object?

PUSH on it for a period of TIME.

Impulse: the product of the force exerted on an object and the time interval during which it acts.

Impulse = Force x time

J = FtSome texts symbolize impulse with “I”, but AP uses “J”

Example:  Wall exertsa force of 100,000 N.

The contact time is0.2 s.------------------------------Impulse = F t              = 20000 N-s

The impulse given to an object is equal to the change in momentum of the object.

Ft = mv = mvf – mvo

Change in Momentum = Impulse(mv) = Ft     mv = Ft

The same change in momentum may be the result of a SMALL force exerted for a LONG time, or a LARGE force exerted for a SHORT time.

Varying Forces

If the force exerted on an object is NOT a constant force, finding the impulse/change in momentum is a little more difficult.

As the saying goes,

“If the forces varies…

You must integrate!”

In most cases dealing with collisions, the varying force is shown to be a function of time.

The impulse is defined as the integral of the force over the time interval during which the force acts. It equals the total change in momentum of the particle.

pimpulsedt)t(F

For example:If F(t) = 4t2 + 2,J = 4/3 t3 + 2t (+ C)Which will equal the CHANGE in momentum

And, of course, on a Force vs time graph,The impulse/change in momentum, Is the “area under the curve”

F t dt p( )

Conservation of Momentum

System- a collection of objects that is being observed.

Closed system- no objects enter or leave the systemIsolated system- no external forces act on any

object in the system. The objects can exert forces on one another.

In a closed, isolated system, the total momentum will remain the same- it will be CONSERVED.

IF THE NET EXTERNAL FORCE ACTING ON A SYSTEM IS ZERO, THE TOTAL MOMENTUM OF THE SYSTEM IS CONSERVED.

(this can sometimes be true in one direction, but not in the other direction! For example, in projectiles with no air resistance, momentum is conserved in the horizontal direction, but NOT in the vertical direction)

Recoil

Momentum Before = 0-------------MomentumAfter = 0-------------After firing, the oppositemomentacancel.

Rocket Thrust

Fthrust = Ru

R- the rate fuel is consumed, kg/s

u – the velocity of the exhaust relative to the rocket, m/s

Do the fundamental units for Ru correspond to the unit for Force??

Two rocket examples:

F Ru mdv

dt

F Ru mg mdv

dt

A railroad car moves at a constant speed of 3.2 m/s under a container of sand. The sand drops onto it at the rate of 540 kg/min. What force must be applied to the railroad car, in the absence of friction, to keep it moving at a constant speed?For an acceleration of zero, there must be a net force of zero acting on the car. Therefore, the force applied to the railroad car must balance the “thrust” force of the sand, Ru, where R is the rate the sand drops and u is the relative horizontal velocity between the sand and car- which is actually the velocity of the car itself.Force = 540 kg/min x 1 min/60sec x 3.2 m/sForce = 28.8 N

A railroad car moves at a constant speed of 3.2 m/s under a container of sand. The sand drops onto it at the rate of 540 kg/min. If the mass of the railroad car was 500 kg, what is its initial acceleration?F = ma-Ru = maa = -Ru/ma = (-540 kg/min x 1 min/60 sec x 3.2 m/s ) / 500 kga = 0.0576 m/s2

Two long barges are moving in the same direction in still water, one with a speed of 10 km/h and the other with a speed of 20 km/h. While they are passing each other, coal is shoveled from the slower to the faster one at a rate of 1000 kg/min. How much force must be provided by the driving engines of each barge if neither is to change speed? Assume that the shoveling is always perfectly sideways and that the frictional forces between the barges and the water do not depend on the weight of the barges.

First the faster barge:For a net force of zero, the additional force provided by the engines must balance the “thrust” force from the coal so thatFengines = (-) RuWhat is u, the relative velocity between the coal and the faster barge?u = 10 km/h – 20 km/h u = -10 km/h = -2.78 m/sF = 1000 kg/min (1 min/60 s) x -2.78 m/sFengines = 46.3 N

Now, the slower barge: In a similar way…Fengines = RuWhat is u, the relative velocity between the coal and the slower barge?It’s ZERO, since the coal is moving at exactly the same velocity as the barge it is on.So…NO force is required on the slower barge!!

Conservation of Momentum in Collisions

Conservation of Linear Momentum is a direct consequence of Newton’s Third Law, “for every force, there is an equal but opposite force”.

Linear Momentum is conserved in a closed, isolated system.

Mechanical Energy, U + K, is conserved when no non-conservative do work

Conservation of Momentum in Collisions

• Elastic collision- A collision in which objects collide and bounce off each other such that the total kinetic energy of the SYSTEM is the same before and after the collision. (Total K will most likely change during the actual encounter)

• Inelastic collision- A collision in which objects collide and some of the kinetic energy is lost to the system. (almost all collisions)

• “Completely”, “perfectly”, “totally” inelastic collision- the objects stick together after the collision (maximum loss of K) and will move together with the velocity of the center of mass (since there are no external forces, the velocity of the cm is constant- the same before and after the collision!)

• Super elastic collision- the system’s kinetic energy INCREASES after the encounter.

Conservation of Momentum is really a consequence of Newton’s Third Law… for every force, there is an equal, but opposite force…

Think of the recoil example:

The gun pushes on the bullet for an amount of time, imparting an impulse. The bullet imparts an equal but opposite impulse on the gun. The impulse results in an equal but opposite change in momentum, resulting in the recoil relationship of

mv = mv.

m1v1o = m2v2fm1v1o - m2v2o = - m1v1f + m2v2f

m1v1o + m2v2o = m1v1f + m2v2f

Three Examples of conservation of momentum in collisions with equations:

Center of Mass Reference FrameVERY VERY USEFUL!!!

The center-of-mass reference frame is one that moves with the velocity of the center of mass. In this frame, the total momentum of a system is zero.

In perfectly inelastic collisions, the particles remain at rest within the center-of-mass reference frame after the collision.

In elastic collisions in one dimension, the velocity of each particle is reversed.