02/24/2003 physics 103, spring 2003, u. wisconsin1 physics 103: lecture 10 impulse and momentum l...
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02/24/2003 Physics 103, Spring 2003, U. Wisconsin 1
Physics 103: Physics 103: Lecture 10Lecture 10Impulse and MomentumImpulse and Momentum
Today’s lecture will cover the following new Concepts:ImpulseMomentumImpulse-Momentum TheoremMomentum Conservation
02/24/2003 Physics 103, Spring 2003, U. Wisconsin 2
Impulse and MomentumImpulse and Momentum Impulse = average force times time
I = Fave t
€
rF =m
r a =m
Δr v
Δt=
Δr p
Δtsince
Δr p =
r p f −
r p i
I =FaveΔt
(nothing new….. Newton’s second law)
Assumption: mass m of the object is constant
not necessary, but otherwise there are complications.
QuickTime™ and aAnimation decompressorare needed to see this picture.
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You drop an egg onto
a) the floor
b) a thick piece of foam rubber
In both cases, the egg does not bounce.
In which case is the impulse greater?
A) case 1
B) case 2
C) the same
In which case is the average force greater
A) case 1
B) case 2
C) the same
correct
correct
QuestionsQuestions
02/24/2003 Physics 103, Spring 2003, U. Wisconsin 4
Preflight Lecture 10 No.1Preflight Lecture 10 No.1
The impulse delivered to a body by a force is
1. defined only for interactions of short duration.
2. equal to the change in momentum of the body.
3. equal to the area under an F versus x graph.
4. defined only for elastic collisions.
•Often useful in this situation, but not ONLY
•equal to the area under an F versus t graph!
•it is defined for all collisions elastic and inelastic!
0
10
20
30
40
50
60
70
Preflight 10.1
ABCD
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Preflight Questions 2 & 3Preflight Questions 2 & 3Two identical balls are dropped from the same height onto the floor. In case 1 the ball bounces back up, and in case 2 the ball sticks to the floor without bouncing. In which case is the impulse given to the ball by the floor the biggest?
1. Case 1
2. Case 2
3. The same
the impulse is greater for case one because ... the change in momentum of the object ... is proportional to the change in velocity which is greater in case one because it has a greater final velocity (down then up) than case 2 (which is only from down to zero). Impulse must be greater for case 1.
Example: suppose m=1 kg, v(initial)=-1 m/s
mv(initial)= -1 kg-m/s
Case 1 mv(final)= +1 kg-m/s Impulse = 1- (-1)=2
Case 2 mv(final)= 0 Impulse = 1 - 0 = 1
correct
0
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45
Preflight 10.2
ABC
Pretty SureNot Quite SureJust Guessing
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Preflight Question 4 & 5Preflight Question 4 & 5
In both cases of the above question, the direction of the impulse given to the ball by the floor is the same. What is this direction?
1. Upward
2. Downward
time
correct
25%
75%
0% 20% 40% 60% 80%Pretty SureNot Quite SureJust Guessing
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Preflight Questions 6, 7 & 8Preflight Questions 6, 7 & 8
Is it possible for a system of two objects to have zero total momentum while having a non-zero total kinetic energy?
1. YES
2. NO in an isolated system, two ice skaters starting at rest and pushing on one another will move in opposite directions thus the momenta of the two are equal and opposite and total momentum is zero. but they are moving apart after the push and therefore the KE is non-zero.
two hockey pucks moving towards each other with the same speed on a collision course have zero total momentum, but a non zero total kinetic energy
correct
47%
53%
0% 20% 40% 60%
Pretty SureNot Quite SureJust Guessing
02/24/2003 Physics 103, Spring 2003, U. Wisconsin 8
Impulse and MomentumImpulse and Momentum Momentum-Impulse Theorem
Favet I = pf - pi = p
For single object….
F = 0 momentum conserved (p = 0)
• For a collection of objects …
o Fext = 0 total momentum conserved
€
r
p ∑ =r p
f
∑ −r p
i
∑ = 0
“External” is a VERY important qualifier
o Fext = mtotal a
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Some TerminologySome Terminology
• Elastic Collisions:
collisions that conserve kinetic energy
• Inelastic Collisions:
collisions that do not conserve kinetic energy
* Completely Inelastic Collisons:
objects stick together
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Elastic Collision in 1-DimensionElastic Collision in 1-Dimension
QuickTime™ and aCinepak decompressorare needed to see this picture.
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m1v1i + m2v2i = m1v1 f + m2v2 f
1
2m1v1i
2 +1
2m2v2i
2 =1
2m1v1 f
2 +1
2m2v2 f
2
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Elastic CollisionElastic Collision
€
m1v1i + m2v2i = m1v1 f + m2v2 f
1
2m1v1i
2 +1
2m2v2i
2 =1
2m1v1 f
2 +1
2m2v2 f
2
m1(v1i2 − v1 f
2 ) = m2(v2 f2 − v2i
2 )
m1(v1i − v1 f )(v1i + v1 f ) = m2(v2 f − v2i)(v2 f + v2i)
m1(v1i − v1 f ) = m2(v2 f − v2i)
v1i + v1 f = v2i + v2 f
v1i − v2i = −(v1 f − v2 f )
Magnitude of relative velocity is conserved.
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Preflight Lecture 10 No.9Preflight Lecture 10 No.9
In an elastic collision
1. kinetic energy is conserved.
2. momentum is conserved.
3. the magnitude of the relative velocity is conserved.
4. all of the above are correct.
•True by definition of elastic
•True by definition of collision
• Total momentum is unchanged75%
3%
12%
10%
0% 20% 40% 60% 80%
02/24/2003 Physics 103, Spring 2003, U. Wisconsin 13
Preflight Lecture 10 No.10Preflight Lecture 10 No.10
In an inelastic collision
1. both kinetic energy and momentum are conserved.
2. only kinetic energy is conserved.
3. only momentum is conserved.
4. neither kinetic energy nor momentum are conserved.
•False by definition of inelastic collision
•False by definition of inelastic collision
•False by definition of collision21%
56%
14%
9%
0% 20% 40% 60%
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CollisionsCollisions
“before”m1 m2
“after”m1
m2
ExplosionsExplosions
“before”M
“after”m1
m2
• Draw “before”, “after”
• Define system so that Fext = 0
• Set up axes
• Compute Ptotal “before”
• Compute Ptotal “after”
• Set them equal to each other
Procedure
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Conceptual ExampleConceptual Example
Consider two blocks (mass m and M) sliding toward one another (with velocities vm and vM) on a frictionless plane.
m MvmvM
What happens after the collision?
What does your result depend on?
How would the answer change if there was friction between the blocks and the plane?
Pm + PM =Pm + PM
Is the collision elastic or?
Energy would be lost during the motion. We would need initial distances to calculate!
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Draw “before” and “after” “before”
“after”2
mvf
System = two blocks
Axis: postive to right
• Before: Ptotal,before = mv + (-mv) = 0 !
• After: Ptotal,after = (2m)vf
• Ptotal,before = Ptotal,after
• 0 = (2m)vf
• vf = 0 !
• Therefore KEafter = 0
mmv v
Totally Inelastic Head-on CollisionTotally Inelastic Head-on Collision
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Preflight Question 11 & 12Preflight Question 11 & 12
Movies often show someone firing a gun loaded with blanks. In a blank cartridge the lead bullet is removed and the end of the shell casing is crimped shut to prevent the gunpowder from spilling out. When a gun fires a blank, is the recoil greater than, the same as, or less than when the gun fires a standard bullet?
1. greater than
2. same as
3. less than
correctImpulse is the same
in the two cases
45%
36%
18%
0% 10% 20% 30% 40% 50%
Pretty SureNot Quite SureJust Guessing
02/24/2003 Physics 103, Spring 2003, U. Wisconsin 18
ExplosionsExplosions
“before”M
• Example: m1 = M/3 m2 = 2M/3
• Which block has larger momentum?
* Each has same momentum
• Which block has larger velocity?
* mv same for each smaller mass has larger velocity
• Which block has larger kinetic energy?
* KE = mv2/2 = m2v2/2m = p2/2m smaller mass has larger KE
• Is kinetic energy conserved?
*NO!!NO!!*NO!!NO!!
“after”m1
m2v1
v2
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ExampleExample
A rocket of mass 200 kg is fired at with the vertical component of the initial velocity 1000 m/s and the horizontal component of 300 m/s. When it reaches the highest point on the trajectory an explosion occurs. The rocket is split in half and part A develops a vertical component to its velocity of 100 m/s. How far from the launching point does part A hit the ground? (Assume the ground is flat and use g = 10 m/s2.)
a) 60.0 km
b) 65.8 km
c) 63.15 kmtdown
t1
t2
02/24/2003 Physics 103, Spring 2003, U. Wisconsin 20
SolutionSolution
€
ymax =v0
2
2g=
10002
20= 50,000m
t =v0
g=
1000
10=100s
after the explosion part A has 10% of the velocity
t =10s
Y =500m
so tup =110s and y =50,500m
y = y0 +v0yt +1
2ayt
2 0=50,500−1
210 t 2
tdown = 10,100 =100.5s
x =300 110+100.5( )= 63.15km
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RepriseReprise
3) Kinetic energy may be conserved (elastic) or may not (inelastic)
2) Conservation of Momentum is useful. Depends (as always) on the situation.
1a) Impulse is average force times time of action
1b) Momentum is mass times velocity.