Transcript II. Conservation of Momentum

Piri Reis University 2011-2012/ Physics -I
Physics for Scientists &
Engineers, with Modern
Physics, 4th edition
Giancoli
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Piri Reis University 2011-2012 Fall Semester
Physics -I
Chapter 9
Linear Momentum
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Piri Reis University 2011-2012
Lecture 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 Momentum
During a collision, measurements show that the total momentum does not
change:
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II. Conservation of Momentum
More formally, the law of conservation of momentum states:
The total momentum of an isolated system of objects remains constant.
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II. 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.
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Ex.1 (II) A 95-kg halfback moving at
4.1 m s
on an apparent breakaway for a touchdown is tackled from
behind. When he was tackled by an 85-kg cornerback
running at 5.5 m s
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.
pinitial  pfinal  mAvA  mB vB   mA  mB  v 
v 
mAvA  mB vB
mA  mB
 95 kg  4.1m s   85 kg  5.5m s 

 4.8m s
 95 kg   85 kg 
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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.
p  F t  mv 
v
 52.0 m s  39.0 m
F m
  0.145 kg  
t
3.00  103 s

s
3

4.40

10
N, towards the pitcher


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III. Collisions and Impulse
During a collision, objects are deformed
due to the large forces involved.
Since, we can write
The definition of impulse:
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III. Collisions and Impulse
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.
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III. Collisions and Impulse
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.
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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.


p  mv  4.5  10 kg  45 m s  0   2.0 kg m s
2
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Ex. 2(a)The average force is the impulse divided
by the interaction time.
F
p
t

2.0 kg m s
3
3.5  10 s
 5.8  10 N
2
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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 Dimension
Here 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 headon 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.
vA  vB    vA  vB   vB  vA  vA
Substitute this relationship into the momentum conservation equation
for the collision
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Ex.3
mA vA  mB vB  mA vA  mB vB  mA vA  mA vA  mB  vA  vA  
mA  mB 
0.220 kg

vA 
vA 
 3.30 m s   1.10 m s  east 
0.660 kg
 mA  mB 
vB  vA  vA  3.30 m s  1.10 m s  4.40 m s  east 
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VI. Inelastic Collisions
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.
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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 .
v
mM
m
2 gh
• Compare the two speeds with the same
masses.
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Ex.4
mM
v2
 m
v1 m  M
m
2 gh2
2 gh1

h2
h1

h2
h1

5.2
2.6
 2 
v2  2v1
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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
L
q
L-h
x
h
v
mM
m
2 gh 
1  mv 
1
  0.028 kg  230 m s  
h

 

2 
2 g  m  M  2 9.8 m s  0.028 kg  3.6kg 
2

2

 0.1607 m  0.16 m
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Ex.5
From the diagram, we
see that
q
L
L-h
x
h
L   L  h   x2
2
2
x
L   L  h 
2
2
 2.8 m    2.8 m  0.1607 m 
2
2
 0.94 m
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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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VII. Collisions in Two or Three Dimensions
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.
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VIII. Center of Mass
In (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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VIII. Center of Mass
For two particles, the center of mass lies closer to the one with the most
mass:
where M is the total mass.
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VIII. Center of Mass
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.
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VIII. Center of Mass
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.
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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 Motion
This 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, 46
References
o “Physics For Scientists &Engineers with Modern Physics” Giancoli 4th edition,
Pearson International Edition
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