CHAPTER 9 NOTES - School District of La Crosse

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Transcript CHAPTER 9 NOTES - School District of La Crosse

CHAPTER 9 NOTES
KONICHEK
• I. Impulse and change in momentum
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A. Newton wrote his 3 laws in terms of momentum- quantity of
motion
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B. Momentum is the product of the mass and the velocity of an
object.
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1. symbol is p, so p=mv vector quantity
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a. units: Kgm/s
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C. If a single body has no net forces acting on it, its momentum is
conserved.
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D. Impulse theorem- Ft=p Newton’s 2nd law can be derived
from this
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1. (F)Δt= M(V2 – V1)
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2. Often the force isn’t constant during exertion. Therefore
average force is used in the equation
• 3. A large change in momentum occurs
when there is a large impulse force
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a. a large impulse can result can
result from a large force acting over a
short time
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1. a car crash-The air bags
increase the contact time against the
dash, so the impulse force isn’t as great.
• SAMPLE PROBLEM- A BASEBALL OF MASS
.14Kg moving at 35m/s
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a. find the momentum of the baseball
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p=mv- p= (.14Kg)35m/s=4.9Kgm/s
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b. find the velocity at which a bowling ball
mass of 7.26Kg would have to equal the
momentum of the baseball
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4.9Kgm/s=7.26KgVbb =.67m/s
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ANOTHER PROBLEM
• An engine of the orbital maneuvering system (OMS) on a
space shuttle exerts a force of 30000 N j for 4.00 s,
exhausting a negligible mass of fuel relative to the
95,000 kg mass of the shuttle.
(a) What is the impulse of the force for this 4.00 s?
• (b) What is the shuttle's change in momentum from this
impulse?
(c) What is the shuttle's change in velocity from this
impulse?
(d) Why can't we find the resulting change in kinetic
energy of the shuttle?
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• Answer to Example Impulse Problems
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• a. Impulse = Ft = 30000(4) = 120000 N-s
b. change in momentum = impulse(FΔt) = 120000 N-s
c. FΔt = m∆v
120000 = (95000)∆v
∆v = 1.26 m/s
d. We do not know the initial velocity.
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SAMPLE PROBLEM- IMPULSE
AND MOMENTUM
• A .02Kg golf ball strikes a guy in the head
with a velocity of 90m/s. The ball strikes
and bounce off in .06s. Find the impulse
forces off their head.
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FΔt= MΔV F= M( V2-V1)/Δt
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F= .2Kg(0-90m/s)/.06s= -300N
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1. like getting hit with a 13 pound
club in the head
• II. ANGULAR MOMENTUM
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A. the quantity of angular motion that is similar
to linear momentum
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1. No torque. The angular momentum is
constant
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2. angular momentum is the product of the
mass , velocity, and the distance from the center
of rotaion, and the distance perpendicular to that
distance.
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a. IiI= Iff I= angular momentum,  is
angular velocity