Unit 8: Linear Momentum
description
Transcript of Unit 8: Linear Momentum
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Impulsive Force Model 1
Unit 8: Linear Momentum
Work on Unit 8 HW asapDemos: Collision carts and track
U7b: All students get perfect Score - answers online
What is implied?
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Agenda Review Dynamics for exam Operational Definitions
momentum units
Momentum Motion Map Elastic and Inelastic Collisions Conservation of Momentum
Mid Term Review
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Scratch off Practice The University of New England was founded
in… A) 1953 B) 1978 C) 1996 D) 1939 E) 1831
Add 147 to the above answer to determine when UNE was first established.
A) 1978 B) 1953 C) 1996 D) 1831 E) 1939
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Operational DefinitionsLinear momentum:
p mvUnits (kg.m/s)
Elastic Collision ~ “No-Stick”Inelastic Collision ~ “Stick”
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Clicker ?
1 2 3 4
0% 0%0%0%
1. Ekf = Eki, pf ≠ pi
2. Ekf = Eki, pf = pi
3. Ekf ≠ Eki, pf ≠ pi
4. Ekf ≠ Eki, pf = pi
During elastic collisions…
60
+x
t=0
t=0
1 2 3 4
3
4
1
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Clicker ?
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0% 0%0%0%
1. Ekf = Eki, pf ≠ pi
2. Ekf = Eki, pf = pi
3. Ekf ≠ Eki, pf ≠ pi
4. Ekf ≠ Eki, pf = pi
During inelastic collisions…
60
+x
t=0
t=0
1 2 3 4
1
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Elastic Collision Kinetic energy is conserved (Eki = Ekf) Linear momentum is conserved (pf=pi=p)
Ekf (J) pf (kg.m/s)
Eki (J)
1
1pi
(kg.m/s)
1
1
Ekf(J) = 0.95 Eki(J) pf(kg.m/s) = 0.95 p
i(kg.m/s)
0 0
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Inelastic Collision Kinetic energy is NOT conserved (Eki ≠
Ekf)
Linear momentum is ALWAYS conserved (pf=pi=p)Ekf (J) pf (kg.m/s)
Eki (J)
1
1pi
(kg.m/s)
1
1
Ekf(J) = 0.5 Eki(J) pf(kg.m/s) = 0.95 pi(kg.m/s)
0 0
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Elastic Collision Motion Map
Insert masses at positions proportional to those in the collision. No motion => no change in position. Time increases from the top down.
+x
t=0
t=0
1 2 3 4
3
4
1
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Elastic Collision MM1 Insert velocity vectors at each block’s
position. Identify the interaction point with a dashed circle. Define initial and final momenta for each object’s mass and velocity.
p2i=0
t=0
t=0
1 2
1
3 4
3
4
p2fp1i
p1f=0
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Elastic Collision MM2 Insert total initial and final momentum
as vector sums of either the initial or final momenta.
p2i=0
+ = + =
t=0
t=0
1 2
1
3 4
3
4
p1i + p2i = pi = p
p1f + p2f = pf = p
p2fp1i
p1f=0
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Elastic Collision MM3 Predictive Power: Since pf = p2f = pi = p1i
=> m2v2f = m1v1i
since m1=m2 => v2f = v1i
+ = + =
p2i=0
t=0
t=0
1 2
1
3 4
3
4
p2fp1i
p1f=0
p1i + p2i = pi = p
p1f + p2f = pf = p
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Alternative Representations
Bar graph to monitor before, after, and total momentum.
p2p1
pBefor
ep2p1
pAfter
Area of velocity vs. mass plot (same area).
m2
m1
v
Before
m2
m1
v
After
p
p
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Inelastic Collision MM1 Insert velocity vectors at each block’s
position. Identify the interaction point with a dashed circle. Define initial and final momenta for each particle based on masses and velocities.
p1i p1&2f
t=0
t=0
1 2 3 4
p2i=0 1
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Inelastic Collision MM2 Insert total initial and final momentum
as vector sums of either the initial or final momenta.
+ = =p1&2f = pi =
p
p1i p1&2f
t=0
t=0
1 2 3 4
p2i=0 1
p1i + p2i = pi = p
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Inelastic Collision MM3 Predictive Power: Since pf = p1&2f = pi = p1i
=> (m1+m2)v1&2f = m1v1i
since m1=m2 => v1&2f = (1/2)v1i
+ = =
t=0
t=0
1 2 3 4
p1i p1&2f
p1&2f = pi = p
p1i + p2i = pi = p
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Alternative Representations
Bar graph to monitor before, after, and total momentum.
Area of velocity vs. mass plot (same area!).
m2
m1
v
Before
m1+m
2
vAfter
p
p
p2p1
pBefor
ep1&2
pAfter
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Work out problem Create an inelastic MM for m1=m striking
m2=3m.
Predict v1&2f if v1i = +4m/s, v2i = 0
t=02
t=0
p1i
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Inelastic Collision MM5 Predictive Power: Since pf = p1&2f = pi = p1i
=> (m1+m2)v1&2f = m1v1i
m1=m, m2=3m => v1&2f=(1/4)v1i =(1/4)(+4m/s) =+1m/s
p1i
p2i=0
+ = =
p1&2f
t=0
t=0
1 2
1
3 4
p1&2f = pi = p
p1i + p2i = pi = p
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Alternative Representations
Bar Graph to represent initial, final, and total momentum
p2p1
pBefor
ep1&2
pAfter
Area of velocity vs. mass plot (same area!!!).
m2
m1
v
Before
m1+m
2
vAfter
p
p
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Clicker ?
1 2 3 4
0% 0%0%0%
1. 2m rebounds, m moves slowly2. 2m stops, m moves at medium pace 3. 2m continues slow, m moves quickly4. None of the above
If mass “2m” collides ELASTICALLY with mass “m” then…
60
p1i p2i=0
t=0
1 2
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Elastic Collision MM4 Create an elastic MM for m1=2m striking
m2=m. Predict v1f and v2f if v1i = +3m/s, v2i=0. Very challenging because we have two
unknowns so that we need two equations. Solved generically. Details are provide
below but will not be discussed.
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Elastic Collision Prediction Because momentum is conserved, pf=pi:
m1v1f + m2v2f = m1v1i
Because the collision is elastic, Ekf = Eki
(1/2)m1v1f2 + (1/2)m2v2f
2 = (1/2)m1v1i2
Two equations with two unknowns: (v1f and v2f given m1 = 2m, m2 = m, v1i =
+3m/s). To simplify let v1i = v, v2f = v2, v1f = v1 then:
m1v1+m2v2=m1v Solve for v1:
v1 = (m1v-m2v2)/m1
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Crunch Time 1 Square v1 and multiply by m1…
v12= [(m1v-m2v2)/m1]2 =
v2-2vv2m2/m1+v22m2
2/m12
=> m1v12=m1v2-2vv2m2+v2
2m22/m1
Dropping (1/2) in Ek:
m1v12 + m2v2
2 = m1v2 Substitute m1v1
2 into Ek …
m1v2-2vv2m2+v22m2
2/m1 + m2v22 = m1v2
Cancel m1v2 on both sides and divide by v2…
-2vm2 + v2m22/m1 + m2v2 = 0
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Crunch Time 2 Divide by m2 and isolate v2…
v2(m2/m1 + 1) = 2v Solve for v2…
v2 = 2v/(1+m2/m1) = v2f
Solve for v1…
v1 = v-m2v2/m1 = v - 2m2v/m1(1+m2/m1)
= v(m1+m2-2m2)/(m1+m2) = v(m1-m2)/(m1+m2)
v1 = v(1-m2/m1)/(1+m2/m1) = v1f
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m1 > m2 => m1 Plows on Since m1=2m, m2=m and v1i=3m/s v1f = v(1-m2/m1)/(1+m2/m1) = +1m/s v2f = 2v/(1+m2/m1) = +4m/s
p1fp1i
p2i=0
+ = + =p1i + p2i = pi
t=0
t=0
1 2
1
3 43 4
p1f + p2f = pf
p2f
4
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Alternative Representations
Bar graphs to represent initial, final and total momentum
p2p1
p
Before
p2
p
After
p1
Area of velocity vs. mass plot (same area!!!).
m2
m1
v
Before
vAfter
p
p
m2
m1
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Clicker ?
1 2 3 4
0% 0%0%0%
1. m rebounds, 2m moves slowly2. m stops, 2m moves at medium pace 3. m continues slow, 2m moves quickly4. None of the above
If mass “m” collides ELASTICALLY with mass “2m” then…
60
p1i p2i=0
t=0
1 2
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m1 < m2 => m1 bounces back!
If m1=m, m2=2m and v1i=3m/s
v1f = v(1-m2/m1)/(1+m2/m1) = -1m/s
v2f = 2v/(1+m2/m1) = +2m/s
p1f
p1i
p2i=0
+ = + =p1i + p2i = pi
t=0
t=0
2
1
3 4
p1f + p2f = pf
p2f4 3
2
Before
After
p2p1 p2p1
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Unit 8a: Momentum ctd.
HW U8a due after ThanksgivingDemos: Collision carts and two air
pucks
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Inelastic Collision Motion Map
Insert masses and positions proportional to those in the collision. No motion => no change in position. Time increases from the top down.
+x
t=0
t=0
1 2 3 4
1
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Agenda Refined Operational Definitions:
Elastic Collision Inelastic Collision
Explosions Two objects in motion and colliding Two-Dimensional Collisions
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Operational DefinitionsLinear momentum:
p mvUnits (kg.m/s)
Elastic Collision: Eki = Ekf
Inelastic Collision: Eki ≠ Ekf
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What is your opinion?
1 2
56%
44%
1. True2. False
Explosions are examples of elastic collisions because kinetic energy before and after the explosion is conserved.
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Explosions 1 MMM an explosion of an object into two
pieces, one with mass m, the other with mass 2m. What is the pi? Therefore what must be pf? Is this by definition an elastic or inelastic event?t=
0
p2p1
Before
After
p2p1
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Explosions 2 Since Eki = 0 and Ekf≠0 => Inelastic “event”
Predictive Power: Since pf = p2f + pif = pi = 0
=> +m2v2f - m1v1f = 0
since m2=2m1 => v2f = (1/2)v1f
+ =
+ =
p1i + p2i = pi
p1f + p2f = pf
p2i=0
3 2 3
p2fp1f
p1i=0
t=01
44
p2p1
pBefore
pAfter
p2p1
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2 Moving Objects 1 MMM the situation when two carts of
equal mass m collide elastically, but in opposite directions and different velocities. Look at the individual and total momenta we observe that momentum is transferred.
t=0
12t=01
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2 Moving Objects 1 Prediction: Since momenta are transferred:
p1i = p2f and p2i = p1f
And since m1=m2=m v2f = v1i and v1f = v2i
p1i + p2i = pi
t=0
12 t=0
1p2i
p1i
p1f + p2f = pf
4 2 4p1f
p2f
33
Rigorous solution requires conservation of
kinetic energy, lots of algebra.
p2p1
pBefor
e pAfter
p2p1
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Two Dimensional Collisions Motivation: A 1000 kg eastbound car
traveling 30m/s runs a red light and center-punches a 10,000 kg truck traveling southbound at 4 m/s. The two collide inelastically.
Draw a top view of the incident and qualitatively estimate where the two vehicles end up, assuming no interactions with any other obstacles.
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Clicker ?
1 2 3 4
3% 7%76%14%
1. A2. B3. C4. D
Choose the correct path, either A, B, C, or D AND explain why the paths you did not choose are incorrect.
A
B
C
D
N
W E
S
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Because p = constant… … only “C” can be correct since it is the
only one with momenta both in eastern and southern directions.
But what is pf and ? A
B
C
D
N
W E
S
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Use Vector Tool: p=0 pf is the total momentum after the
collision. What is it comprised of? The momenta of the car (pc) and truck
(pt). What are these in terms of pf?
pf
N
W E
S
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Seek Resultant Vector The components of total momentum! pcar = mV(east) = 30,000kgm/s (east) ptruck = Mv(south) = 40,000kgm/s(south) pf = ?, = ? pf = 50,000 kgm/s = tan-1(4/3) = 53° What is vf =? vf = pf/(m+M) = 4.5m/s
pf
pcar
ptruck
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Arbitrary Impact Gets messy, though understandable
through tip to tail vector addition. For example, two billiard balls. Easy to solve if + = 90°
p1i
pi pf
p1f p2f
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Arbitrary Impact p1f = pfcos and p2f = pfcos COM: pf = p1f + p2f = pi
p1fy = p1fsin = -p2fy = -p2fsin The vertical components of the final
momenta must cancel.
p1i
pi pf
p1f p2f
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Clicker ?
1 2 3
89% 7%4%
1. A2. B3. C
Two identical billiard balls collide; what will be their post-collision paths?
A CB
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Unit 8b: Impulse ForceHW U8b due after Thanksgiving
Demos: Force sensor and motion detector with collision cart, eggs and blanket, small and big balls.
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Agenda Newton’s Third Law Impulsive Force and change of velocity Examples
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Clicker ?
1 2 3 4
0% 0%0%0%
1. VW Bug2. Mac Truck3. Both feel same
force4. No Clue
Which feels the greater force, a VW Bug colliding at 60 mph with the Mac Truck or the Mac Truck colliding at 30 mph with the VW Bug?
60Pow
!
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Clicker ?
1 2 3 4
0% 0%0%0%
1. VW Bug2. Mac Truck3. Both interact
over same t4. Texting buddies
Which vehicle interacts over a longer time during the collision?
60Pow
!
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Recall Newton’s Third Law Recall, the forces between two colliding
carts, regardless of mass, are equal and opposite.
Examine the units of F vs. t, break them down into their simplest form.
t (s)
Pow!
F (N)
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Clicker ?
1 2 3 4 5
4%6%
39%39%
13%
1. 32 Ns2. 16 kgm/s3. 3.2 kgm/s4. 1.6 Ns5. 0.8 kgm/s
Estimate the numerical value of the area of the F vs. t curve below.
F (N)
t (s)
4
0 0.8
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Area under F vs. t Plot N.s = kg.m.s/s2 = kg.m/s Same units as p = mv Count the blocks Area ≈ 8 blocks x 0.1kg.m/s = 0.8kg.m/s
F (N)
t (s)
4
0 0.8
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Impulse vs. Momentum Examine the force over time and the
change is velocity of a cart.
t (s)
v(m/s)
t (s)vf
vi
F (N)
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Compare F vs. t w/ v Constant: mass of cart = 0.33 kg vi = - 1.2m/s, vf = + 1.2m/s v = vf-vi =(+1.2m/s-(-1.2m/s)) =
+2.4kg.m/s
t (s)
4
0 0.8
v(m/s)
t (s)
+1.2
F (N)
-1.2
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Impulse - Momentum Ft = (0.33kg)vFt = mvMass constantFt = (mv)Ft = porF = p/t
F∆t(N.s)
v(m/s)
0.8
0 2.4
Really just a restatement of Newton’s Second Law. In fact, what Newton really discovered was: F = p/t
Slope = 0.33Ns/m/s = 0.33kg = mcart
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Impulsive Force Particle Page 57
Impulse - Momentum F = p/t (mass constant) => F = mv/t => F = ma The form of Newton’s Second Law we
have been using for the past month. Also note that if there is no external
force acting on m, Fext = 0 = p Which is the conservation of linear
momentum!
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Clicker ?
1 2 3 4
0% 0%0%0%
1. Smashes2. Cracks3. Doesn’t break4. Has fear
A raw egg is thrown 25 m/s (50 mph) at a blanket. The egg…
60
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Applications I Reduce F and increase t. Give six examples of engineering
designed to make “soft landings.”
t (s)
4
0 8
F (N)
Hard Landing
Soft Landing
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PRS ?
1. Golfball high, Volleyball low2. Golfball low, Volleyball high3. Golfball and volleyball stay together4. Cataclysm causing the earth to vanish.
What will happen if a 0.03kg golfball atop a 0.3kg volleyball fall 1 meter in concert? Each ball separately rebounds to a medium height. What will happen to the balls if they are stacked on top of each other and allowed to fall?
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Impulsive Force Particle Page 61
Make a predictionHurley, at height “zi” above the ground, swings on a rope and collides inelastically with Burley at rest. What height will the pair reach? Start by diagramming the conservation principles needed to solve this problem.
QuickTime™ and aTIFF (Uncompressed) decompressor
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QuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.
zi
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zf=?
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Make a prediction1. Conservation of Energy
2. Conservation of linear momentum
3. Conservation of Energy
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zi
Eg Ek
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pfpi
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zf=?
EkEintEgEint
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Oscillations & Waves
Oscillating Particle Model
Source of Waves:Simple Harmonic Oscillator
Linear Restoring ForceSHO Kinematics
SHO Energy Conservation
Mechanical Waves:Sound Waves
Energy PropagationSuperposition Principle
Doppler Shift
Light Waves & InterferenceDiffraction/Refraction
Polarization/Colors
Classical Mechanics
Particle
Model
KinematicsConstant Velocity
Constant Acceleration
Conservation of Energy
E = W + Q + R
DynamicsFree Particle F=0
Constant Force F=ma
Impulse/MomentumFt = p
ptotal = mv = constantConservation of Linear
Momentum
Central ForceF = (mv2/r) inwardRotational Mechanics
1st Semester
Electricity & Magnetism
Field Model
Particle of Mass mGravitational Field g
Gravitational Force Fg
Potential Energy Eg
Potential Vg &Tools
Particle of Charge qElectric Field E
Electrostatic Force FE
Potential Energy EE
Potential V & Tools
Magnetic dipole µMagnetic Field B
Magnetic Force FB
RHR & Induction
E -> current IOhms Law, Circuits
I -> Magnetic Field B
Multiple Particles
2nd SemesterFluid Statics/Dynamics
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Momentum Transfer Use energy conservation to determine initial and
final v before and after the collision with the floor. Ef = mgz = Ei = (1/2)mv2 => v = (2gz)1/2
v=0
m
M
z
z(m) M z
(m) m
vi(m/s)
1.0 -4.5 1.0 -4.5
vfs(m/s)
0.7 +3.7 0.3 +2.4
ps(Ns) +2.4 +.21
vfc(m/s)
? ? 2.0 +6.3
pc(Ns) +2.3 +.32
pc-s(Ns) -0.1 +0.1
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Bouncing Ball Data
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Impulsive Force Model 67
Unit 8: Consensus
MomentumImpulse
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Elastic Collision MM4 Alternative representation: Bar graph to
monitor before, after, and total momentum.
+ = + =
p1i + p2i = pi p1f + p2f = pf
t=0
t=0
1 2
1
3 4
3
4
p2p1
pBefor
ep2p1
pAfter
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Inelastic Collision MM4 Alternative representation: Bar graph to monitor
before, after, and total momentum.
+ = =p1i + p2i = pi p1&2f = pf
t=0
t=0
1 2 3 4
p2i=0
p2p1
pBefor
ep1&2
pAfter
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Elastic Collision Kinetic energy is conserved (Eki = Ekf) Linear momentum is conserved (pf = pi)
Ekf (J) pf (kg.m/s)
Eki (J)
1
1pi
(kg.m/s)
1
1
Ekf(J) = 0.95 Eki(J) pf(kg.m/s) = 0.95 p
i(kg.m/s)
0 0
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Inelastic Collision Kinetic energy is NOT conserved (Eki ≠ Ekf)
Linear momentum IS conserved (pf = pi)
Ekf (J) pf (kg.m/s)
Eki (J)
1
1pi
(kg.m/s)
1
1
Ekf(J) = 0.5 Eki(J) pf(kg.m/s) = 0.95 pf
i(kg.m/s)
0 0
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Compare F vs. t w/ v Constant: mass of cart = 0.33 kg vi = - 1.2m/s, vf = + 1.2m/s v = vf-vi =(+1.2m/s-(-1.2m/s)) =
+2.4kg.m/s
t (s)
4
0 0.8
v(m/s)
t (s)
+1.2
F (N)
-1.2
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Impulse - Momentum Ft = (0.33kg)vFt = mvMass constantFt = (mv)Ft = porF = p/t
F∆t(N.s)
v(m/s)
0.8
0 2.4
Really just a restatement of Newton’s Second Law. In fact, what Newton really discovered was: F = p/t
Slope = 0.33Ns/m/s = 0.33kg = mcart