Momentum

Was this accurate? What really Happens? Does this make a difference:

We know that it is harder to get a more massive object moving from rest than a less massive object.◦ This is the concept of inertia previously

introduced. Building on the concept of inertia,

we ask the question, “How hard is it to stop an object?”◦ We call this property of an object that is related

to both its mass and its velocity, the momentummomentum of the object.

The linear momentum of an object is defined as the product of its mass and its velocity, and is measured in “kilogram-meters per second.”

The symbol for momentum is p, as in:p = mv

The bold-faced symbols above, momentum and velocity, are vectors with identical direction.

Look at the equation. Can you guess why the units are defined this way?

So an object may have a large momentum due to a large mass, a large velocity, or both. A slow battleship and a rocket-propelled race car (or kayak) have large momenta.

Conceptual Question: Which has the greater momentum, an 18-wheeler parked at the curb, or a Volkswagen rolling down a hill?

A. 19 wheelerB. VWC. Both the sameD. Can’t tell.

The word momentum is often used in our everyday language in a much looser sense, but it is still roughly consistent with its meaning in the physics world view; that is, something with a lot of momentum is hard to stop. You have probably heard someone say, “We don’t want to lose our momentum!” Coaches are particularly fond of this word. The catcher is protected

from the baseball’s momentum by body armor.

0

0

0

0

)(

)(

tt

mvmvF

tt

vvm

t

vmmaF

f

f

f

f

(mv)Impulse tF

Because impulse is a product of two things, there are many ways to produce a particular change in momentum.

For example, two ways of changing an object’s momentum by 10 kilogram-meters per second are:◦ exert a net force of 5 newtons on the object for 2 seconds, or◦ exert 100 newtons for 0.1 second.

They each produce an impulse of 10 newton-seconds (N·s) and therefore a momentum change of 10 kilogram-meters per second.

The units of impulse (newton-seconds) are equivalent to those of momentum (kilogram-meters per second).

A. a force of 2 newtons acting for 10 seconds, or

B. a force of 3 newtons acting for 6 seconds?

C. Both the same

Question: Which of the following will cause the larger change in the momentum of an

object:

Because our bones break when forces are large, the particular combination of force and time interval is important.◦ Sometimes, as in a gymnasium, a

wood floor may be enough of a cushion; in a car, the dashboards are made from foam rubber. Bumpers and air bags further increase the vehicle’s (and the passengers’) Δt.

A pole-vaulter lands on thick pads to increase the collision time and thus reduce the force. Otherwise, this could be a nasty fall.

Sometimes accidents can fill in for the lack of a parachute:

◦ During World War II, a Royal Air Force airman jumped in just his uniform from a flaming bomber flying at 18,000 ft. His terminal speed (Ch. 3, p. 46) was 120 mph, but tree branches and then 1½ feet of snow cushioned his fall.

The result: minor scratches and bruises.

◦ The record for surviving a fall without a parachute is held by Vesna Vulovic. She was serving as a hostess on a Yugoslavian DC-9 that blew up at 33,330 ft in 1972.

Play catch with a raw egg. After each successful toss, take one step away from your partner.

How do you catch the egg to keep from breaking it?

If you are falling, (or just moving along), you have a certain MOMENTUM, just before you come in contact with what is going to stop you.

Assume that your FINAL velocity (and momentum) will be zero.

Your momentum (final momentum=0) will be the impulse necessary to stop your fall.

constant)( collision thebefore justmvtF

ft 500floors 50 & floorsft 10 assume

Falls for 4 seconds

time speed

1 32 ft/s

2 64 ft/s

3 96 ft/s

4 128 ft/s catch

Speed=128 ft/sec Her weight = 100 lbs W=mg so her mass = 100/32 ~3 English

mass units (slugs) Momentum = mv = 3x128=384

(appropriate units) F x t=384 F=384/t

How Fast is the Stop??

Time to stop Force in pounds

.1 seconds 3840 (ouch!)

.5 seconds 768

1 second 384

Imagine standing on a giant skateboard, at rest.◦ The total momentum of you and the

skateboard must be zero, because everything is at rest.

Now suppose that you walk on the skateboard. What happens to the skateboard?

When you walk in one direction, the skateboard moves in the other direction, as shown in the figure.

An analogous thing happens when you fire a rifle: the bullet goes in one direction, and the rifle recoils in the opposite direction.

The force you exert on the skateboard is, by Newton’s third law, equal and opposite to the force the skateboard exerts on you.◦ Because you and the skateboard each experience the same

force for the same time interval, you must each experience the same-size impulse and, therefore, the same-size change in momentum.

Even though you and the skateboard are moving and, individually, have nonzero

momenta, the total momentum remains zero.

Because the impulses are in opposite directions (red arrows, foot/skateboard & you), the changes in the momenta are also in opposite directions.

Thus, your momentum and that of the Thus, your momentum and that of the skateboard still add to zero.skateboard still add to zero.

Conservation of Momentum

http://www.youtube.com/watch?v=KL8-PbdRYY0

Because the changes in the momenta of the two objects are equal in size and opposite in direction, the value of the total momentum does not change. We say that the total momentum is conserved.

We can generalize these findings. Whenever any object is acted on by a force, there must be at least one other object involved. ◦ This other object might be in actual contact with the first, ◦ or it might be interacting at a distance of 150 million km, but it is

there.

Consider the objects as a system. Whenever there is no net force acting on the system from the outside (that is, the system is isolated, or closed), the forces that are involved act only between the objects within the system. As a consequence of Newton’s third law, the total momentum of the system remains constant.

This generalization is known as the law of conservation of linear momentum.

Consider a “system” of “interacting” particles.

At one instant of time add up all the momenta of all of the particles in the system.◦ Vector addition, of course, but let’s not worry much

about that. Wait some amount of time. After that time, the sum of the momenta is

the same as it was when initially measured.

MOMENTUM IS CONSERVED

The Law of Conservation of Linear Momentum: ◦ The total linear momentum of a system does not change if

there is no net external force.

A GAS??

Air ???

You experience conservation of momentum firsthand when you try to step from a small boat onto a dock. ◦ As you step toward the dock, the boat moves

away from the dock, and you may fall into the water.

Although the same effect occurs when we disembark from an ocean liner, the large mass of the ocean liner reduces the speed given it by our stepping off. ◦ A large mass requires a small change in velocity

to undergo the same change in momentum.

If a rifle of mass 4.5 kg fires 10-gram bullets at a speed of 900 m/s (2000 mph), what will be the speed of its recoil?◦ The magnitude of the momentum p of the bullet

is:p = mv = (0.01 kg)(900 m/s)

= 9 kg·m/s◦ Because the total momentum of the bullet and

rifle was initially zero, conservation of momentum requires that the rifle recoil with an equal momentum in the opposite direction. If the mass of the rifle is M, the speed V of its recoil is given by

=2 m/s

If you do not hold the rifle snugly against your shoulder, the rifle will hit your shoulder at this speed (2 m/s = 4.5 mph) and hurt you.

Question: Why does holding the rifle snugly reduce the recoil effects?

Answer: Holding the rifle snugly increases the recoiling mass (your mass is now added to that of the rifle) and therefore reduces the recoil speed.

Interacting objects don’t need to be initially at rest for conservation of momentum to be valid. Suppose a ball moving to the left with a certain momentum crashes head-on with an identical ball moving to the right with the same-size momentum.

◦ Before the collision, the two momenta are equal in size but opposite in direction, and because they are vectors, they add to zero.

◦ After the collision the balls move apart with equal momenta in opposite directions.

Because the balls are identical and their masses are the same, the speeds after the collision are also the same. These speeds depend on the ball’s physical properties.

In all cases the two momenta are the same size and in opposite directions, and the total momentum remains zero.

A question on the final exam asks, “What do we mean when we claim that the total momentum is conserved during a collision?” The following two answers are given:◦ Answer 1: Total momentum of the system stays the same

before and after the collision.◦ Answer 2: Total momentum of the system is zero before and

after the collision.

Which answer (if either) do you agree with? Answer: While we have considered several examples in

which the total momentum of the system is zero, this is not the most general case. The momentum of a system can have any magnitude and any direction before the collision. If momentum is conserved, the momentum of the system always has the same magnitude and direction after the collision. Therefore, answer 1 is correct. This is a very powerful principle because of the word always.

A boxcar traveling at 10 meters per second approaches a string of four identical boxcars sitting stationary on the track. The moving boxcar collides and links with the stationary cars, and the five boxcars move off together along the track.

What is the final speed of the five cars immediately after the collision?

Conservation of momentum tells us that the total momentum must be the same before and after the collision. ◦ Before the collision, one car is moving at 10 meters per second. ◦ After the collision, five identical cars are moving with a common

final speed. ◦ Because the amount of mass that is moving has increased by a

factor of five, the speed must decrease by a factor of five. The cars will have a final speed of 2 meters per second.

Notice that we did not have to know the mass of each boxcar, only that they all had the same mass.

We can use the conservation of momentum to measure the speed of fast-moving objects. For example, consider determining the speed of an arrow shot from a bow.

We first choose a movable, massive target—a wooden block suspended by strings. Before the arrow hits the block (a), the total momentum of the system is equal to that of the arrow (the block is at rest).

After the arrow is embedded in the block (b), the two move with a smaller, more measurable speed.

The final momentum of the block and arrow just after the collision is equal to the initial momentum of the arrow. Knowing the masses, the arrow’s initial speed can be determined.

Accident investigators use conservation of momentum to reconstruct automobile accidents. Newton’s laws can’t be used to

analyze the collision itself because we do not know the detailed forces involved.

Conservation of linear momentum, however, tells us that regardless of the details of the crash, the total momentum of the two cars remains the same.

The total momentum immediately before the crash must be equal to that immediately after the crash.

As an example, consider a rear-end collision. Assume that the front car was stopped and the two cars locked bumpers on impact.

From an analysis of the length of the skid marks made after the collision and the type of surface, the total momentum of the two cars just after the collision can be calculated. ◦ (We will see how to do this in Chapter 7; for now, assume we

know their total momentum.) Because one car was stationary, the total momentum

before the crash must have been due to the moving car. Knowing that the momentum is the product of mass and velocity (mv), we can compute the speed of the car just before the collision. We can thus determine whether the driver was speeding.

Let’s use conservation of momentum to analyze such a collision. ◦ For simplicity assume that each car has a mass of 1000 kg (a

metric ton) and that the cars traveled along a straight line. ◦ Further assume that we have determined that the speed of the

two cars locked together was 10 m/s (about 22 mph) after the crash.

The total momentum after the crash was equal to the total mass of the two cars multiplied by their combined speed:

But because momentum is conserved, this was also the value before the crash. Before the crash, however, only one car was moving. So if we divide this total momentum by the mass of the moving car, we obtain its speed:

The car was therefore traveling at 20 m/s (~45 mph) at the time of the accident.

How fast would you have to throw a baseball (m = 145 g) to give it the same momentum as a 10-g

bullet traveling at 800 m/s? a. 0.55 m/sb. 1.8 m/sc. 55 m/s d. 800 m/s

What change in momentum occurs when a force of 20 N acts for 4 s?

a. 5 kg·m/sb. 16 kg·m/sc. 24 kg·m/s

* d. 80 kg·m/s

A. a. 500 kg·m/s to the rightB. b. 500 kg·m/s to the leftC. c. 500 kg·m/sD. d. zero