Download Gravitation and Momentum

Journal #
• Think about your experience with the CM
project so far. What advice would you give
Ms. Doyle about making the project directions
clearer for next year?
• Are there any rules you think should be
changed? Why?
Chapter 9 Terms Due Wed. 12/1
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Impulse (include formula and unit)
Momentum (include formula and unit)
Impulse-momentum theorem
Law of Conservation of Momentum
Closed System
Isolated System
Elastic Collision (P. 298)
Inelastic Collision (P. 298)
Center of Mass
Law of Universal Gravitation
Momentum & Impulse
Excerpts from
Chapters 9 and 11
Newton’s Law of Universal Gravitation
• The gravitational force is equal to the
universal gravitational constant, times the
mass of object 1, times the mass of object 2,
divided by the distance (between the centers
of the objects) squared.
Newton’s Law of Universal Gravitation
• Formula:
m1m2
F G 2
d
The Universal Gravitational Constant
• How large is G?
– We know from experience that the attractive force
between two normal sized objects is not
noticeable.
– In 1798, Henry Cavendish created an experiment
that finally allowed this number to be measured.
Cavendish’s Experiment
Cavendish’s Experiment
• The main idea was that by using manageable but
massive spheres of lead, the gravitational force
between the large and small spheres would be
enough to cause motion. The force was measured
for multiple sets of spheres and the constant was
found.
• G = 6.67x10-11Nm2/kg2
Important!!!
• Cavendish’s experiment paved the way for us
to know the mass of the Earth, Sun, and other
planetary objects.
• Numbers for the force of gravity will be small
(x10-? N) unless working with REALLY massive
objects.
Example 1
• What is the gravitational force between two
15-kg packages that are 0.35m apart?
Example 1 Answer
m1  15kg
m2  15kg
d  0.35m


m1m2
F G 2
d
15 15
F  6.67 10
2
0.35
7
F  1.2 10 N
11
Homework Problems
• P. 191: #53, 55-57
Journal # 30
• What do we mean when we ask people to
“conserve water”?
• What do you think it will mean if we say that
momentum is conserved?
• Get HW and Vocab out when finished!
Answers to HW
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53:
55:
56:
57:
F=6.1x10-9 N
F=5.84x10-10 N
F=8.0x10-10 N
F=6.5x10-8 N
Journal # 31
• Explain how an airbag protects you by making
you come to a stop differently than hitting
steering wheel.
Try to use the words impulse, force, and time in
your answer.
Journal # 31
• Impulse is a force applied over an interval of
time. In this question, the impulse of hitting the
steering wheel and hitting the airbag are the
same amount because they both cause you to
stop. However, the airbag applies a smaller force
to your body over a larger time, therefore
keeping you safer.
Journal # 32
•
Order these objects from the most
momentum to the least.
A.
B.
C.
D.
A bullet shot from a rifle
An elephant standing still
A bowling ball rolling
A fly buzzing by your ear
Impulse and Momentum
• Newton’s 2nd Law of
motion can be
rewritten by using the
definition of
acceleration as the
change in velocity over
the change in time.
F  ma
v 
F  m 
t 
Impulse and Momentum
• If the change in time is multiplied out of the
denominator, we are left with the following:
Ft  mv
Ft  mv

• The product of force and change in time is
called the Impulse (symbol is J).
• Impulse is a vector quantity and is
measured in Newton-seconds (Ns)
Impulse
• If a car hits a haystack or the same car hits a wall,
momentum is decreased by same impulse – the same
products of force and time.
• However, impact force is greater into the wall than it is
into the haystack as the haystack extends impact time,
lessening the impact force.
• Impact time is the time during which momentum is
brought to zero.
Momentum
• Can increase with increase in either mass or in
velocity or both.
– Ex: a rolling bowling ball has greater momentum than a
tennis ball rolling at the same speed because its mass
is greater
– Ex: a racecar going forward at 120 mi/hr has greater
momentum than the same size car going 90 mi/hr due
to its velocity.
• If an object is not moving (no matter how big it
is), the momentum = zero

  mv
• The product of the mass and the velocity is called the
momentum (symbol  -“rho”) of an object.
• Momentum is also a vector quantity and is measured
in kg*m/s.
• Note that the units for impulse and momentum
appear different, but they are actually the same unit
when simplified.

Ft  mv f  mvi
• The impulse-momentum theorem states that
the impulse on an object is equal to the
object’s final momentum minus the object’s
initial momentum.
• Can also be written as:
Ft   f  i
Example 1
• Tiger Woods hits a 0.050kg golf ball, giving it a
speed of 75m/s. What is the impulse given to
the ball?
Example 1 Answer
m  0.050kg
Ft  mv
v  75m /s
impulse  0.050  75
impulse  ?
impulse  3.75  3.8N  s

Example 2
• Shane hits a stationary 0.12kg hockey puck
with a force that lasts for 1.0x10-2s and makes
the puck shoot across the ice with a speed of
20.0m/s, scoring a goal for the team. With
what force did Shane hit the puck?
Example 2 Answer
m  0.12kg
Ft  mv
2
t  1.0 10 s
v  20.0m / s
F ?
mv
F
t
0.12  20.0
F
2
1.0 10
F  240N
Example 3
• Diana, whose mass is 50.0kg, leaves a ski jump
with a velocity of 21.0m/s. What is her
momentum as she leaves the ski jump?
Homework
Attempt as many as you can and set ALL of them
up in your notebook with a left-hand lineup!
• P. 233 #1-4 Due Thursday
• P. 251 #56-64 Even Due Friday
• PAY CAREFUL ATTENTION TO THE UNITS GIVEN
AND CONVERT THEM WHEN
NECESSARY!!!!!!!!
Conservation of Momentum
• A system is the environment and all of the objects
examined in a problem.
• A closed system is a system in which no mass is
gained or lost.
• An isolated system is a system in which the net
external force is zero… no forces acting outside of
the system have an effect inside of it.
Conservation of Momentum
• The law of conservation of momentum states
that the sum of momentum of any closed,
isolated system does not change… or that the
sum of the momentum of the objects in that
system is constant.
Conservation of Momentum
• Mathematically, we can view this as a BEFORE
and AFTER situation.
• For any two objects A and B:
Ai  Bi  Af  Bf
Types of Collisions
•
If two objects bounce apart when they collide, it is called an elastic
collision and can be written:
m1v1i  m2v 2i  m1v1 f  m2v 2 f
•
If two objects stick together when they collide, it is called an inelastic
collision and can be written:
m1v1i  m2v 2i  (m1  m2 )v f
Example 1
• Tubby and his twin brother Chubby have a combined
mass of 200.0kg and are zooming along in a 100.0kg
amusement park bumper car at 10.0m/s. They
bump Melinda’s car, which is sitting still. Melinda
has a mass of 25.0kg. After the elastic collision, the
twins continue ahead with a speed of 4.12m/s. How
fast is Melinda’s car bumped across the floor?
Example 1 Picture
Before Collision
After Collision
T&C
Mel
T&C
Mel
m1  300.0kg
m2 125.0kg
m1  300.0kg
v1i 10.0m /s
v 2i  0m /s
v1 f  4.12m /s
m2  125.0kg
v2 f  ?



Example 1 Answer
m1v1i  m2v 2i  m1v1 f  m2v 2 f
m1v1i  m2v 2i  m1v1 f  m2v 2 f
m1v1i  m2v 2i  m1v1 f
m2
 v2 f
Example 1 Answer
[m1v1i  m2v 2i  m1v1 f ]
m2
 v2 f
[(300.0 10.0)  (125.0  0)  (300.0  4.12)]
 v2 f
125.0
14.1m /s  v 2 f
Example 2
• If an 800.kg sports car slows to 13.0m/s to check out
an accident scene and the 1200.kg pick-up truck
behind him continues traveling at 25.0m/s, with
what velocity will the two move if they lock bumpers
after a rear-end collision?
Example 2 Picture
Before Collision
After Collision
m1 1200.kg
m2  800.kg
v1i  25.0m /s
v 2i 13.0m /s


(m1  m2 )  2000.kg
vf  ?
Example 2 Answer
m1v1i  m2v 2i  (m1  m2 )v f
(m1v1i  m2v 2i )
 vf
(m1  m2 )
Example 2 Answer
(m1v1i  m2v 2i )
 vf
(m1  m2 )
[(1200  25.0)  (800 13.0)]
 vf
(1200  800)
20.2 m/s forward  v f
Homework
• P. 238 #13-18
• P. 252 #73-77 (All Due Monday)
Homework
• P. 238 #13-18
13. 1.1 m/s
14. 0.034 m/s
15. 1.2 X 103 m/s
16. 2.8 m/s
17. 6.7 m/s
18. 2.0 m/s
• P. 252 #73-77
75. -0.30 m/s
76. -4.94 m/s or 4.94
backward
77.0.35 m/s
Homework
• P. 251 #56-64 Even
56. 0.013 s
58. 74 kg* m/s; 10. m/s
3

1.2
X
10
N
60.
62. 4.8 N
64. 42 m/s