Transcript Slide 1 - Cloudfront.net

MOMENTUM
Linear Momentum
 Impulse
 Conservation of Momentum
 Types of Collision
 2-D Momentum
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MOMENTUM
How do you chop through cement
blocks with a bare hand?
Why does falling on a wooden floor
hurt less than onto a cement floor?
Why do people in larger vehicles
end up with fewer injuries in
accidents?
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Linear Momentum
Measure of how hard it is to stop a
moving object or change the
motion’s direction
 How a mass moves in a straight path
 Momentum = mass times velocity
 p = m v
 Units: kg m/s (SI)
 Vector; same
direction as velocity
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IMPULSE
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Directly proportional to force
and time
Force exerted over time
Derived from Newton’s 2nd Law of
motion
F = ma = mv/t = p/t
I=Ft
Units: Ns (SI)
IMPULSE
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Area under the curve of the F vs t
IMPULSE-MOMENTUM
THEOREM
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Impulse is the change in momentum
I = Δp = pf - p°
Ft = mvf – mv° = m (vf - v°)
Units : Ns = kg m/s
Momentum is in the same direction
as Force
IMPULSE MOMENTUM THEOREM
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Bouncing causes a greater change in
momentum and impulse.
EFFECT OF COLLISION TIME
UPON FORCE
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Air bags
Seat Belts
Boxing
Padding
Baseball
Throwing an egg on the bed sheet
Car collisions…crumple zones
IMPULSE-MOMENTUM THEOREM
IMPULSE-MOMENTUM THEOREM
CONSERVATION OF
MOMENTUM
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For a collision in an isolated system, the
total momentum before the collision =
the total momentum after collision.
If one object gains momentum then the
second object has lost the same amount.
Momentum is ALWAYS conserved so
constant
pbefore = pafter
Number of momentum equations =
number of drawings
CONSERVATION OF
MOMENTUM
• p° = pf
• ptruck + pcar = ptruck’ + pcar’
• mtruckvtruck + mcarvcar = mtruckvtruck’ + mcarvcar’
CONSERVATION OF
MOMENTUM
• p° = pf
• pbigfish + plittlefish = ptotal
• mbigfishvbigfish + mlittlefishvlittlefish= mbigfish+littlefishvt
TYPES OF COLLISION
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Types of collision
Elastic collision
• Momentum is conserved
• KE is conserved
• Bounce off of each other
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Inelastic collision
• Momentum is conserved
• KE is NOT conserved
• Damaged or stick together
TYPES OF COLLISION
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Two types of inelastic collision
Perfectly inelastic collision
• Objects collide and stick together
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Inelastic collision
• Objects collide and damage is present
LINEAR MOMENTUM
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Jocko, who has a mass of 60 kg and stands at
rest on ice, catches a 20 kg ball that is thrown to
him at 10 km/h. How fast does Jocko and the ball
move across the ice?
The momentum before the catch is all in the ball,
20 kg x 10 km/h = 200 kg km/h
This is also the momentum after the catch,
where the moving mass is 80 kg
60 kg for Jocko and 20 kg for the caught ball.
LINEAR MOMENTUM
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The roads in Dr. J's neighborhood are slightly more crowded
in the mornings these days since he has taken up jogging.
The crowding comes from the crew that helps Dr. J get
through this physical fit that is overrunning the country. J.
Jr. marks off a new course each day while Timex mans the
stopwatch. Tripod is there at the end supporting a tray of
cereal, fruit, and bacon on his nose (his idea of a balanced
breakfast). Dr. J does have one quirk - he doesn't use
shoelaces. As a result, his shoes always look like they are
ready to come apart when he finishes. (Don't most joggers
finish with their tongues hanging out?) But alas, this is the
day that Dr. J finally gets tired of it all. His course takes him
through the local park, but a thick fog has decreased
visibility. As a result, he runs into a swing made from an old
tire suspended by a long rope. Dr. J experiences a new high
as he and the tire rise 0.3 m above their initial level. If Dr.
J weighs 750 N and the mass of the tire is 10 kg, how fast
was Dr. J running?
Chapter 9: R pg 178
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1) A compact car, mass 725 kg,
is moving at +100 km/h.
a) Find its momentum.
b) At what velocity is the
momentum of a larger car, mass
2175 kg , equal to that of the
smaller car?
2.02 x 10 4 kgm/s; 33.4 km/h
Chapter 9: R pg 178
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2) A snowmobile has a mass of 2.50
x 102 kg. A constant force is exerted
on it for 60.0 s. The snowmobile’s
initial velocity is 6.00 m/s and its
final velocity 28.0 m/s.
a) What is its change in momentum?
b) What is the magnitude of the
force exerted on it?
5.5 x 10 3 kgm/s; 91.7 N
Chapter 9: R pg 178
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3) The brakes exert a 6.40 x 102 N force
on a car weighing 15680 N and moving at
20.0 m/s. The car finally stops.
a) What is the car’s mass?
b) What is the initial momentum?
c) What is the change in the car’s
momentum?
d) How long does the braking force act on
the car to bring it to a halt?
1.60 x 10 3 kg; 3.20 x 10 4 kgm/s; - 3.20
x 10 4 kgm/s; 50.0 s
Chapter 9: R pg 178
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4) Figure 9-1 shows, as a function of time,
the force exerted by a ball that collided
with a box at rest. The impulse, Ft, is
the area under the curve.
a) Find the impulse given to the box by
the ball.
b) If the box has a mass of 2.4 kg, what
velocity did it have after the collision.
5.25 Ns; 2.2 m/s
Answers: R pg 193
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
351 kgm/s
4.8 kgm/s
42 m/s
60 Ns; 20.0 m/s
2.04 x 104 Ns; 300 N
2.35 x 104 kgm/s; 2.6 x 104 N
260 N
-250 N
1100 kg
1300 s
Momentum in 1-D
Objects bounce apart
A 0.15 kg blue billiard ball moving at 8.0
m/s to the right hits a similar red billiard
ball at rest. If the blue ball continues to
move to the right at 2.5 m/s, what is the
velocity of the red ball.
ptotal = ptotal’
pb + pr = pb’ + pr’
mbvb + mrvr = mbvb’ + mrvr’
0.15 kg(8.0 m/s) + 0.15 kg(0m/s) = 0.15 kg(2.5 m/s) +
0.15kg(vr’)
vr’ = 5.5 m/s right
Momentum in 1-D
Objects stick together
Two balls of clay, a blue one being 2.3 kg
and the second red one being 5.6 kg, hit
each other and stick together. If the blue
one was moving to the right at 12 m/s,
and the red was moving at 8.1 m/s to
the left, what is their final velocity?
ptotal = ptotal’
pb + pr = ptotal
mbvb + mrvr = v’ (mb + mr)
2.3 kg(12.0 m/s) + 5.6 kg(-8.1m/s) = v’(2.3 kg + 5.6
kg)
v’ = -2.2 m/s left
Chapter 9: R pg 185
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5) A 0.105 kg hockey puck moving at 48
m/s is caught by a 75 kg goalie at rest.
With what speed does the goalie slide on
the ice?
0.067 m/s
6) A 35.0 g bullet strikes a 5.0 kg
stationary wooden block and embeds
itself in the block. The block and bullet
fly off together at 8.6 m/s. What was the
original velocity of the bullet?
1200 m/s
Chapter 9: R pg 185
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7) A 35.0 g bullet moving at 475 m/s strikes a
2.5 kg wooden block. The bullet passes through
the block, leaving at 275 m/s. The block was at
rest when it was hit. How fast is it moving
when the bullet leaves?
2.8 m/s
8) A 0.50 kg ball traveling at 6.0 m/s collides
head-on with a 1.00 kg ball moving in the
opposite direction at a velocity of –12.0 m/s.
The 0.50 kg ball moves away at –14 m/s after
the collision. Find the velocity of the second
ball.
- 2.0 m/s
Chapter 9: R pg 188
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9) A 4.00 kg model rocket is launched,
shooting 50.0 g of burned fuel from its
exhaust at an average velocity of 625
m/s. What is the velocity of the rocket
after the fuel has burned?
7.91 m/s
10) A thread holds two carts together on
a frictionless surface as in the figure. A
compressed spring acts upon the carts.
After the thread is burned, the 1.5 kg
cart moves with a velocity of 27 cm/s to
the left. What is the velocity of the 4.5
kg cart?
9.0 cm/s to the right
Chapter 9: R pg 188
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11) Two campers dock a canoe.
One camper steps onto the dock.
This camper has a mass of 80.0 kg
and moves forward at 4.0 m/s.
With what speed and direction do
the canoe and the other camper
move if their combined mass is 110
kg?
2.9 m/s in the opposite direction
Chapter 9: R pg 188
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12) A colonial gunner sets up his 225 kg
cannon at the edge of the flat top of a
high tower. It shoots a 4.5 kg cannon
ball horizontally. The ball hits the ground
215 m from the base of the tower. The
cannon also moves, on frictionless
wheels, and falls off the back of the
tower, landing on the ground.
a) What is the horizontal distance of the
cannon’s landing, measured from the
base of the back of the tower?
b) Why do you not need to know the
width of the tower?
4.3 m/s; speed remains constant
Answer: R pg 193
11) 30.0 s
12) 0.05 s; -4000 N; 410 kg: no; holding a child is
dangerous to the child
13) 888 kgm/s; 43.6° SE
14) 63 kgm/s; 63 Ns; 20000 N; 4000 N
15) 150 kgm/s; 150 Ns; 3000 N; 5W
16) 780 kgm/s; -780 kgm/s; 780 kgm/s; 6.1 m/s
17) 1.0 x 10-3 kgm/s; -6.0 x 10-4 kgm/s; 6.0 x 10-4
kgm/s; 1.6 x 10-3 kgm/s; 16 cm/s
18) –100 kgm/s; - 500 kgm/s
19) 11m/s
20) 340 m/s
21) 10.6 m/s
Momentum in 2-D
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A 1.20 kg red ball moving to the right at
17.1 m/s strikes a stationary 2.31 kg
blue ball. If the final velocity of the red
ball is 13.5 m/s at 23.0° above the
horizontal, determine the final velocity of
the blue ball.
Momentum in 2-D
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Momentum is conserved
Write a x and y equation.
For x:
prx + pbx = prx’ + pbx’
For y:
pry + pby = pry’ + pby’
Resolve v in vx and vy or determine the
resultant
Find angle
Blue ball is traveling at 3.66 m/s at an
angle of 48.4° below the horizontal
Momentum in 2-D
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A 1.20 kg red ball moving at 10.0
m/s strikes a 2.31 kg blue ball
moving at 15.0 m/s. If the final
velocity of the red ball is 13.5 m/s,
determine the final velocity of the
blue ball. Make use of the angles
drawn in the following diagram.
Momentum in 2-D
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Blue ball is moving at 10.5 m/s at
an angle of 23 degree above the
horizon
Chapter 9: R pg 191
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13) A 1325 kg car moving north at
27.0 m/s collides with a 2165 kg car
moving east at 17.0 m/s. They
stick together. Draw vector diagram
of the collision. In what direction
and with what speed do they move
after the collision?
44.2 ° NE, 14.7 m/s
Chapter 9: R pg 191
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14) A 6.0 kg object, A, moving at velocity 3.0
m/s, collides with 6.0 kg object, B, at rest.
After the collision, A moves off in a direction
40.0° to the left of its original direction. B
moves off in a direction 50.0° to the right of A’s
original direction?
a) Draw a vector diagram and determine the
momenta of object A and object B after the
collision.
b) What is the velocity of each object after the
collision?
pa’ = 14 kg m/s; pb’ = 12 kgm/s; va’ = 2.3 m/s
40° to left; vb’ = 2.0 m/s 60° to left
Chapter 9: R pg 191
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15) A stationary billiard ball, mass 0.17
kg, is struck by an identical ball moving
at 4.0 m/s. After the collision, the
second ball moves off at 60° to the left of
its original direction. The stationary ball
moves off at 30° to the right of the
second ball’s original direction. What is
the velocity of each ball after the
collision?
va’ = 3.5 m/s 30° to right; vb’ = 2.0 m/s
60° to left
Answers: R pg 194
22)
23)
24)
25)
26)
27)
28)
29)
30)
31)
34)
5.0 m/s west
1: -1.5
10 m/s
0.041 m/s, yes
–0.500 kgm/s; -0.995 kgm/s
0.22 m/s
3.1 m/s; 1.24 m/s; 1.6 s; 0.99 m
3.6 kgm/s 34° NW; 1.8 m/s 34° NW
5.4 Ns 22° from original direction
170 kg
1800 N; 3600 N
LAB WRITE UP
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Introduction (background, objective, theories
etc.)
Materials
Procedure (You must include step by step
directions including diagrams. Directions
should be written so that anyone can follow
the steps to rebuild your bridge.)
Data/Calculations
Mass of the bridge, Mass of the load, Mass
ratio of bridge/load
Conclusion (errors, theories summarize what
happened, what can you do better, etc)
GRADING
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2.
Construction: (50 pts)
A) 30 pts: building the bridge without any
violation of the rules and materials above.
B) 3 pts: lightest bridge
C) 3 pts: holds the heaviest load
D) 5 pts: lowest bridge to mass ratio in
each respective class, the remaining 9 points
will be distributed among the other teams in
the class based on their ratios.
Lab Report: (50 pts)
DUE DATES
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DUE DATE for the BRIDGE: Friday,
January 12, 2007
DUE DATE for the LAB REPORTS: The
day after you test your bridge!
ALL LONG TERM PROJECTS ARE DUE
ON THEIR DUE DATES!! A 10 %
DEDUCTION FOR EACH DAY WILL BE
ASSESSED ON LATE PROJECTS! THIS
INCLUDES THE WEEKENDS! IF YOU
PLAN TO BE ABSENT ON THAT DATE,
YOU NEED TO TURN IT IN EARLIER OR
HAVE SOMEONE BRING IT IN