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Physics for Scientists & Engineers, with Modern

Physics, 4th edition

Giancoli

Piri Reis University 2011-2012/ Physics -I

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Piri Reis University 2011-2012 Fall SemesterPhysics -I

Chapter 9

Linear Momentum

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Piri Reis University 2011-2012Lecture IX

I. Momentum and Its Relation to Force

II. Conservation of Momentum

III. Collisions and Impulse

IV. Conservation of Energy and Momentum in Collisions

V. Elastic Collisions in One Dimension

VI. Inelastic Collisions

VII. Collisions in Two or Three Dimensions

VIII. Center of Mass (CM)

IX. CM for the Human Body

X. Center of Mass and Translational Motion

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I. Momentum and Its Relation to Force

Momentum is a vector symbolized by the symbol p, and is defined as

The rate of change of momentum is equal to the net force:

This can be shown using Newton’s second law.

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II. Conservation of MomentumDuring a collision, measurements show that the total momentum does not change:

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More formally, the law of conservation of momentum states:

The total momentum of an isolated system of objects remains constant.

II. Conservation of Momentum

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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.

II. Conservation of Momentum

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sm 1.4

sm 5.5

Ex.1 (II) A 95-kg halfback moving at

on an apparent breakaway for a touchdown is tackled from behind. When he was tackled by an 85-kg cornerback running at

in the same direction, what was their mutual speed immediately after the tackle?

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Ex. 1The tackle will be analyzed as a one-dimensional momentum conserving situation. Let “A” represent the halfback, and “B” represent the tackling cornerback.

initial final

95 kg 4.1m s 85 kg 5.5m s4.8 m s

95 kg 85 kg

A A B B A B

A A B B

A B

p p m v m v m m v

m v m vv

m m

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Ex. 2 (II) A 0.145-kg baseball pitched at 39.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 52.0 m/s. If the contact time between bat and ball is 3.00 x 10-3 s, calculate the average force between the ball and bat during contact

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Answer to EX. 2 Choose the direction from the batter to the

pitcher to be the positive direction. Calculate the average force from the change in momentum of the ball.

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3

52.0 m s 39.0 m s0.145 kg 4.40 10 N, towards the pitcher

3.00 10 s

p F t m v

vF m

t

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III. Collisions and ImpulseDuring a collision, objects are deformed due to the large forces involved.

Since, we can write

The definition of impulse:

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Since the time of the collision is very short, we need not worry about the exact time dependence of the force, and can use the average force.

III. Collisions and Impulse

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The impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time.

This is why you should bend your knees when you land; why airbags work; and why landing on a pillow hurts less than landing on concrete.

III. Collisions and Impulse

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Ex. 2 (II) A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5 x 10-3 s. Find (a) the impulse imparted to the golf ball, and (b) the average force exerted on the ball by the golf club.

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Ex. 2(a) The impulse is the change in momentum. The direction of travel of the struck ball is the positive direction.

24.5 10 kg 45 m s 0 2.0 kg m sp m v

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Ex. 2(a)The average force is the impulse divided by the interaction time.

2

3

2.0 kg m s5.8 10 N

3.5 10 s

pF

t

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IV. Conservation of Energy and Momentum in Collisions

Momentum is conserved in all collisions.

Collisions in which kinetic energy is conserved as well are called elastic collisions, and those in which it is not are called inelastic.

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V. Elastic Collisions in One DimensionHere we have two objects colliding elastically. We know the masses and the initial speeds.

Since both momentum and kinetic energy are conserved, we can write two equations. This allows us to solve for the two unknown final speeds.

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Ex.3 (II) A ball of mass 0.440 kg moving east ( direction) with a speed of collides head-on with a 0.220-kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

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Ex.3 Let A represent the 0.440-kg ball, and B represent the 0.220-kg ball. We have and . Use Eq. 7-7 to obtain a relationship between the velocities.

A B A B B A A v v v v v v v Substitute this relationship into the momentum conservation equation for the collision

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Ex.3

A A B B A A B B A A A A B A A

A BA A

A B

B A A

0.220 kg3.30 m s 1.10 m s east

0.660 kg

3.30 m s 1.10 m s 4.40 m s east

m v m v m v m v m v m v m v v

m mv v

m m

v v v

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VI. Inelastic CollisionsWith 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.

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Ex.4 (I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second projectile causes the the pendulum to swing twice as high, h2 = 5.2 cm. The second projectile was how many times faster than the first?

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The initial projectile speed is given by .

• Compare the two speeds with the same masses.

2m M

v ghm

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Ex.4

222 2

2 1

1 111

25.2

2 22.62

m Mgh hv hm v v

m Mv hhghm

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Ex.5 (II) A 28-g rifle bullet traveling 230 m/s buries itself in a 3.6-kg pendulum hanging on a 2.8-m-long string, which makes the pendulum swing upward in an arc. Determine the vertical and horizontal components of the pendulum’s displacement.

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Ex.5 From Example 4,

we know that

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2

2

0.028 kg 230 m s1 1

2 0.028 kg 3.6kg2 9.8 m s

0.1607 m 0.16 m

m Mv gh

m

mvh

g m M

L

x

L - h

h

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Ex.5

From the diagram, we

see that

22 2

2 2 22 2.8 m 2.8 m 0.1607 m 0.94 m

L L h x

x L L h

L

x

L - h

h

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VII. Collisions in Two or Three Dimensions

Conservation of energy and momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy.

Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities.

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Problem solving:

1. Choose the system. If it is complex, subsystems may be chosen where one or more conservation laws apply.

2. Is there an external force? If so, is the collision time short enough that you can ignore it?

3. Draw diagrams of the initial and final situations, with momentum vectors labeled.

4. Choose a coordinate system.

5. Apply momentum conservation; there will be one equation for each dimension.

6. If the collision is elastic, apply conservation of kinetic energy as well.

7. Solve.

8. Check units and magnitudes of result.

VII. Collisions in Two or Three Dimensions

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VIII. Center of MassIn (a), the diver’s motion is pure translation; in (b) it is translation plus rotation.

There is one point that moves in the same path a particle would take if subjected to the same force as the diver. This point is called the center of mass (CM).

The general motion of an object can be considered as the sum of the translational motion of the CM, plus rotational, vibrational, or other forms of motion about the CM.

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For two particles, the center of mass lies closer to the one with the most mass:

where M is the total mass.

VIII. Center of Mass

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The center of gravity is the point where the gravitational force can be considered to act. It is the same as the center of mass as long as the gravitational force does not vary among different parts of the object.

VIII. Center of Mass

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The center of gravity can be found experimentally by suspending an object from different points. The CM need not be within the actual object – a doughnut’s CM is in the center of the hole.

VIII. Center of Mass

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IX.CM for the Human Body

High jumpers have developed a technique where their CM actually passes under the bar as they go over it. This allows them to clear higher bars.

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X. Center of Mass and Translational Motion

The total momentum of a system of particles is equal to the product of the total mass and the velocity of the center of mass.

The sum of all the forces acting on a system is equal to the total mass of the system multiplied by the acceleration of the center of mass:

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X. Center of Mass and Translational MotionThis is particularly useful in the analysis of separations and explosions; the center of mass (which may not correspond to the position of any particle) continues to move according to the net force.

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Summary of Chapter 9

• Momentum of an object:

• Newton’s second law:

•Total momentum of an isolated system of objects is conserved.

• During a collision, the colliding objects can be considered to be an isolated system even if external forces exist, as long as they are not too large.

• Momentum will therefore be conserved during collisions.

• Impulse:

• In an elastic collision, total kinetic energy is also conserved.

• In an inelastic collision, some kinetic energy is lost.

• In a completely inelastic collision, the two objects stick together after the collision.

• The center of mass of a system is the point at which external forces can be considered to act.

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HOMEWORK

Giancoli, Chapter 9

4, 15, 17, 19, 25, 28, 34, 40, 41, 46Referenceso “Physics For Scientists &Engineers with Modern Physics” Giancoli 4th edition, Pearson International Edition