Chapter 7- Linear Momentum

Assignment 6 Textbook (Giancoli, 6th edition), Chapter 6:

Due on Thursday, November 12 1. On page 163 of Giancoli, problem 38. 2. On page 165 of Giancoli, problem 69. 3. On page 188 of Giancoli, problem 12. 4. A gymnast of mass 52 kg is jumping on a trampoline. She jumps so that her feet reach a maximum height of 2:5 m above the trampoline and, when she lands, her feet stretch the trampoline down 75 cm. How far does the trampoline stretch when she stands on it at rest? [Hint: Assume the trampoline obeys Hook's law when it is stretched.]

Chapter 7

• Momentum and Its Relation to Force • Conservation of Momentum • Collisions and Impulse • Conservation of Energy and Momentum in Collisions • Elastic Collisions in One Dimension • Inelastic Collisions • Collisions in Two or Three Dimensions

Recalling Last Lecture

Conservative and Nonconservative Forces When the work done by a force does NOT depend on the path taken, this FORCE is said to be CONSERVATIVE.

When the work done by a force DOES depend on the path taken, this FORCE is said to be NONCONSERVATIVE.

Nonconservative forces do NOT have a potential energy associated to them.

Thus, potential energy can be defined only for conservative forces.

(6-14) Eq. 6-14 tells us that the work done by nonconservative forces is equal to the total change in kinetic and potential energies.

Mechanical Energy and its Conservation

“ If only conservative forces are acting, the TOTAL MECHANICAL ENERGY of a system neither increases nor decreases in any process. It stays constant
it is CONSERVED”

POWER: Power is the rate at which work is done, or the rate at which energy is transformed. (6-18)

In the SI system the unit of power is: In general, if an object moves with average velocity

1 hp = 746 W , the power can be written as (6-19)

Kinetic Energy, and the Work Energy Principle Problem 6-63 (textbook): A driver notices that her 1150-kg car slows down from 85 Km/h to 65 Km/h in about 6.0 s on the level when it is in neutral. Approximately what power (watts and hp) is needed to keep the car traveling at a constant 75 km/h?

Kinetic Energy, and the Work Energy Principle Problem 6-63 The energy transfer from the engine must replace the lost kinetic energy. From the two speeds, calculate the average rate of loss in kinetic energy while in neutral.

 1m s

  1m s  = 23.61 m s v = 65 km h = 18.06 m s 2     3.6 km h   3.6 km h  2 2 ∆ KE = 12 mv 22 − 12 mv12 = 12 (1150 kg )  (18.06 m s ) − ( 23.61 m s )  = − 1.330 × 10 5 J

v1 = 85 km h 

P=

W t

=

1.330 × 10 5 J 6.0 s

(

= 2.216 × 10 W , or 2.216 × 10 W 4

4

) 746 W = 29.71 hp 1 hp

Note now that 75 Km/h is the average between the car’s initial and final speeds: ⇒ We can use the

2.2 × 10 4 W or 3.0 × 101 hp is needed from the engine.

to conclude

Today

Momentum and Its Relation to Force The momentum of an object tells how hard (or easy) is to change its state of motion.

Example: It is easier to stop a car when it is moving at 10 km/h than when it is moving at 100 Km/h.


But, note that it will also depend on the mass of the car: a heavy truck moving at 10 Km/h is more difficult to be brought to rest than a small Mercedes SMART (which is much lighter than a truck) moving at the same speed.

Momentum is a vector symbolized by the symbol p, and is defined as (6-20)

Momentum and Its Relation to Force Recalling that acceleration is the rate at which velocity changes, we can then write:

But,

Defining:

and

, we can write (6-21)

The right side of the equation 6.21 is the net force acting on the object. Then:

(6-22)

Momentum and Its Relation to Force (6-22) Eq. 6-22 is another way of expressing Newton’s second law. However, it is a more general definition because it introduces the situation where the mass may change. Example: A rocket burns fuel when lifting off or maneuvering in space:
As it burns fuel, the rocket becomes lighter
So, its total mass changes and so does its momentum manage to keep its velocity constant)

(even if you

Collision and Impulse Collisions are common event in everyday life. During a collision, objects are deformed due to the large forces involved. These forces are generally very strong, much stronger than other interactions between the colliding objects and their surrounding environment, and act for a very short period of time ∆t. We can use eq. 6-22 and define the impulse on an object as:

(6-23) In general, the forces involved in the process are not constant during ∆t, but we can approximate the resulting force by the average force acting during this interval of time: (6-24)

Conservation of Momentum Let’s consider a collision between two billiard balls as shown in the figure. Assume that the two balls form a system isolated from the external world:
in other words
there is NO net force acting on the billiard balls other than the interaction between them.

The momentum of each ball before and after the collision is: Object A: before


,

after


Object B: before


,

after


Conservation of Momentum Recalling previous slide: The momentum of each ball before and after the collision is: Object A:

,

Object B:

,

Now, ball A exerts a force

on ball B.

According to the general form of Newton’s second law:

But, On the other hand, according to Newton’s third law, B exerts a force

Conservation of Momentum We have then:

Combining the two equations, we find:

(6-25) Equation 6-25 tells that the total momentum of the system (the sum of the momentum of the two balls) before the collision is equal to the total momentum of the system after the collision IF the external net force acting on the system is zero
isolated system. This is known as Conservation of Total Momentum. The above equation can be extended to include any number of objects such that the only forces are the interaction between the objects in the system.

Conservation of Momentum Momentum conservation works for a rocket as long as we consider the rocket and its fuel to be one system, and account for the mass loss of the rocket.

Here you can consider a frame at rest relative to the rocket before it lifts off
Its initial momentum is then zero. As it takes off, the fuel burns expelling gas in one direction
this gives momentum to the gas
in order to obey momentum conservation, the rocket has to move in opposite direction such that the total momentum (rocket + gas) remains zero.

Conservation of Energy and Momentum in Collisions In general, we can identify two different types of collisions: 1. Elastic collision 2. Inelastic collision In elastic collisions the total kinetic energy of a system is conserved.
No energy dissipates in form of heat or other form of energy. An example is the collision between the two billiard balls discussed in the previous slides: (6-26)

In inelastic collisions, there is NO conservation of kinetic energy.
Some of the total initial kinetic energy is transformed into some other form of energy.

Conservation of Energy and Momentum in Collisions Example of inelastic collision: With inelastic collisions, some of the initial kinetic energy is lost to thermal or potential energy. It may also be gained during explosions, as there is the addition of chemical or nuclear energy. A completely inelastic collision is one where the objects stick together afterwards, so there is only one final velocity.

Conservation of Energy and Momentum in Collisions Note: Regardless whether we have inelastic or elastic collisions, the total momentum is always conserved if the system is isolated. For instance, the collision between two trains as depicted below is inelastic. Part of the total initial kinetic energy might have been transformed into thermal or some other form of energy. However, the total momentum of the closed (isolated) system (two trains) should be conserved. (The textbook has a good example of this problem).

Conservation of Energy and Momentum in Collisions Note: Also…

The total energy (the sum of all energies) in a closed (isolated system) is ALWAYS conserved.

Conservation of Energy and Momentum in Collisions Elastic collision in one dimension: Here we have two objects colliding elastically. We know the masses and the initial speeds. Since both total momentum and kinetic energy are conserved, we can write two equations. This allows us to solve for the two unknown final speeds. From conservation of momentum:

We can rearrange this equation and write:

Conservation of Energy and Momentum in Collisions Elastic collision in one dimension: Since it is an elastic collision, the total kinetic energy is also conserved:

using that

, then

Conservation of Energy and Momentum in Collisions Elastic collision in one dimension: We then have two equations from total momentum and kinetic energy conservation:

Replacing the last equation into the first one:

(6-27) Eq. 6-27 tell us that in ONE DIMENSION elastic head-on collision, the relative velocity between the objects have the same magnitude but in opposite direction before and after the collision.

Conservation of Energy and Momentum in Collisions Collision in two or more dimensions

That is just a generalization of what we have discussed in the previous slides. We have to make use of the concept of vector components. In this particular case, the reference system is taken such B is initially (before the collision) at rest and A moves in the x direction before the collision. In the figure, considering the system isolated, the total momentum has to be conserved in both x and y directions.

Conservation of Energy and Momentum in Collisions Collision in two or more dimensions We then have: (i) (ii)

It follows then, using the above expressions into (i) and (ii), that: ⇒ This gives a system of two equations and three variables. If you can measure any of these variables, the other two can be calculated from system of equations.

Linear Momentum Problem 7-1 (textbook) A constant friction force of 25 N acts on a 65-kg skier for 20 s. What is the skier’s change in velocity?

Linear Momentum Problem 7-1 : From Newton’s second law,

r r ∆p = F∆t For a constant mass object,

r r ∆p = m ∆v r Equate the two expressions for ∆ p r r F∆t = m∆v

→

r ∆v =

r F∆t m

If the skier moves to the right, then the speed will decrease, because the friction force is to the left. F ∆t ( 2 5 N )( 2 0 s ) ∆v = − = − = − 7 .7 m s m 65 kg The skier loses 7.7 m/s of speed.

Linear Momentum Problem 7-4 (textbook) A child in a boat throws a 6.40-kg package out horizontally with a speed of 10.0 m/s. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 26.0 kg, and that of the boat is 45.0 kg. Ignore water resistance.

Linear Momentum Problem 7-4 : The throwing of the package is a momentum-conserving action, if the water resistance is ignored. Let “A” represent the boat and child together, and let “B” represent the package. Choose the direction that the package is thrown as the positive direction. Apply conservation of momentum, with the initial velocity of both objects being 0.

p in itia l = p fin a l v ′A = −

m B v B′ m

A

→ = −

(m A

+ m

B

)v

= m A v ′A + m B v B′

( 6 . 4 0 k g ) (1 0 . 0 m s ) ( 2 6 .0 k g + 4 5 .0 k g )

→

= − 0 .9 0 1 m s

The boat and child move in the opposite direction as the thrown package.

Linear Momentum Problem 7-34 (textbook) An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.5 times the mass of the other. If 7500 J were released in the explosion, how much kinetic energy did each piece acquire? (Assume that all available released energy is carried away by the two particles). .

Linear Momentum Problem 7-34 Use conservation of momentum in one dimension, since the particles will separate and travel in opposite directions. Call the direction of the heavier particle’s motion the positive direction. Let A represent the heavier particle, and B represent the lighter particle. We have

m A = 1.5 m B

vA = vB = 0

p initial = p final

→

0 = m A v A′ + m B v B′

→ v A′ = −

m B v B′ mA

= − 23 v B′

The negative sign indicates direction. Since there was no mechanical energy before the explosion, the kinetic energy of the particles after the explosion must equal the energy added.

E added = KE A′ + KE B′ = 12 m A v ′A2 + 12 mB v B′ 2 = KE B′ = 53 E added = Thus:

3 5

( 7500 J ) = 4500 J

1 2

(1.5mB ) ( 23 vB′ )

2

+ 12 m B v B′ 2 =

5 3

(

1 2

)

m B v B′ 2 = 53 KE B′

KE A′ = Eadded − KE B′ = 7500 J − 4500 J = 3000 J

K E A′ = 3.0 × 10 3 J K E B′ = 4.5 × 10 3 J