A Review of Some Newtonian Mechanics
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Transcript of A Review of Some Newtonian Mechanics
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A Review of Some Newtonian Mechanics
Gernot LaicherREFUGES Summer Bridge Course
2014
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rPosition :
t) small(for :
tr
dtrdvVelocity
t) small(for :
tv
dtvdaonAccelerati
Position, Velocity, Acceleration
x
y
z
r
x
y
z
r
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m : velocity)sit'in change toe(resistanc inertia sobject'an of measure a is Mass
Mass, Force
forces). atic(electrost repulsiveor attractive becan forces Someforces). onal(gravitati attractive are forces Some
F :objectsbetween forcesexist therenatureIn
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vmpMomentum :
Momentum
m v
p
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sN
ssmkg
smkg
smkg
milem
shour
hourmileskg
hourmileskgvmp
directionignoringmphgoingcaraofMomentum
5.670
5.6705.670
447.01500
3.16093600
11500
11500
)(1
2
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Nssmkg
smkg
sm
gkgg
smgvmp
rifleMsmgoingbulletaofMomentum
7.3
7.3
9480039.0
9481000
9.3
9489.3
)16(/948
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Sir Isaac NewtonGodfrey Kneller’s 1689 portrait of Isaac Newton (age 46)
(Source: http://en.wikipedia.org/wiki/Isaac_Newton)
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Newton's First Law : An object at rest tends to stay at rest and an object in uniform motion tends to stay in uniform motion unless acted upon by a net external force.
Newton's Third Law: For every action there is an equal and opposite reaction.
Newton's Second Law: An applied force on an object equals the rate of change of its momentum with time.
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Newton’s Second Law
i
inet FF
mmass
t)enough small(for
tp
dtpd
dtpdFF
iinet
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amvdtdm
dtvdmv
dtdm
vmdtddtpdFF
iinet
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For a non-changing mass:
netFm
a
amFnet
0dtdm
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pddtFdtpdF netnet
Rewriting Newton’s Second Law
)Impulse"("dtFnet
m)Momentum"in Change("pd
Impulse = Change in Momentum
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Example: A constant force of 5N applied for 0.5 seconds on a mass of 0.2kg, which is initially at rest.
s
s
p
pnetnet
final
in itial
pddtFpddtF5.0
0
sFdtFdtF net
s
snet
s
snet 5.0
5.0
0
5.0
0
initialfinal
p
p
pppdfina l
initial
smvvkgNsmvmvsN finalfinalinitialfinal 5.122.05.25.05
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In general, the force may depend on time
final
initia l
final
initia l
t
t
p
pnetnet pddttFpddttF
t
)(tFnet final
initial
t
tnet dttF
Impulse = “Area” under the F(t) curve (Integrate F(t) over time to get the impulse)
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A Two-body Collision
1v
1m
2112 FF
Before the collision
2v
2m
During the collision
12 onfromF
2v
1m 2m 1221 onfromonfrom FF
(Newton’s third law)
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During the Collision
t
)(tF
)(21 tF
)()( 2112 tFtF Impulse on m2
Impulse on m1
Impulse on m2 = - Impulse on m1
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Impulse on m2 = - Impulse on m1
Change of momentum of m2 = - Change of momentum of m1
initialfinalinitialfinal pppp 1122
12 pp
initialinitialfinalfinal pppp 2121
Total momentum after collision = Total momentum before collision
The total momentum of the system is “conserved” during the collision. (This works as long as there are no external forces acting on the system)
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1u
1m
After the collision
2u
2m
initialinitialfinalfinal pppp 2121
22112211 vmvmumum
For a 1-dimensional collision we can replace thevector with + or – signs to indicate the direction.
Note: u1, u2, v1, v2 may be positive or negative, depending on direction and depending on your choice of coordinate system.
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22112211 vmvmumum
Given: m1 , m2 , v1 , v2
What can we find out about the final velocities u1 and u2 ?
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22112211 vmvmumum
Given: m1 , m2 , v1 , v2
What can we find out about the final velocities u1 and u2 ?
Answer: In general there are an infinite number of possible solutions (combinations of the final velocities u1 and u2 thatfulfill the conservation of momentum requirement). Which ofthose solutions really happens depends on the exact natureof the collision.
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Excel Program – Practicing the Solver1) Solving a simple math problem (2x + 6y =14) infinitely many solutions
2) Adding a constraint (a second equation) (x+y=3) to get a single solution.
3) Solving a quadratic equation (4x2 + 2x =20) how many solutions do you get?
4) Pick different initial conditions for x to find all the solutions.
5) Find a solution to the following set of equations: 3x2 + 4y + 12z = 294x - y2 – z = 12x + 2y - 5z = 5
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Excel Program – Using the Solver to determine 1-dimensional collision outcome
1) Write an Excel program that has the following input fields: m1, m2 , v1 , v2 , u1 , u2 . 2) Create (and label) fields that calculate p1initial , p2initial , p1final , p2final .
3) Create (and label) fields that calculate ptotal initial , ptotal final.
4) Create (and label) fields that calculate ptotal final - ptotal initial.
5) Fill in these values: m1=1 (kg), m2=1 (kg), v1=1 (m/s), v2=-1 (m/s), u1=some value (m/s), u2=some value (m/s).
6) Use a solver that changes u1 and u1 and makes ptotal final - ptotal initial = 0. How many possible solutions can you find?
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Mechanical Energy of Two Mass System
222
211 2
121 vmvmEinitial Before the collision
222
211 2
121 umumE final After the collision
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Totally Elastic Collision: initialfinal EE
(no mechanical energy is lost)
021
21
21
21 2
22211
222
211
vmvmumum
0 initialfinal EE
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Excel Program – Using a SolverAdd the following to your Excel “Collision-Solver”:
Create fields that calculate Etotal initial , Etotal final , Etotal final - Etotal initial ,% Energy change.
100ChangeEnergy %
initial
initialfinal
EEE
Now solve your collision problem again. This time, use the constraint Etotal final - Etotal initial = 0Find all possible solutions and interpret their meaning.
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Totally Inelastic Collision: finalfinalfinal uuu 21
(The two masses stick together after the collision and move with the same velocity)
Task:
Determine a mathematical formula for the final velocity after a totally inelastic collision as a function of the masses and the initial velocities of two colliding objects.
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Totally Inelastic Collision: finalfinalfinal uuu 21
(The two masses stick together after the collision and move with the same velocity)
221121:momentum ofon Conservati vmvmumm
21
2211
mmvmvmu
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Is there Mechanical Energy Conservation in Totally Inelastic Collisions?
222
211 2
121 vmvmEinitial
2
21
221121
221 )(
21)(
21
mmvmvmmmummE final
21
2211
mmvmvmu
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Excel Program – Using a SolverName the Excel worksheet tab for the previously created elasticcollision “Elastic Collision”.
Make a copy of this tab and rename it “Inelastic Collision”
Modify the “Inelastic Collision” tab so that it calculates “u1” from theinitial conditions (masses, initial velocities) according to the formulafor u on the previous page. Then you can make u2=u1. So now both u1 and u2 will always have the same value.
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Use a lab notebook to record your findings and investigate several scenarios of collisions
For both elastic and inelastic collisions you should find solutions forthese cases. Imagine each scenario and describe what is happening.
1) m1=5kg, m2=5kg, v1=15m/s, v2=0 (one mass initially at rest)
2) m1=5kg, m2=5kg, v1=15m/s, v2=-15m/s (one mass initially at rest)(head on collision).
3) m1=5000kg, m2=0.01kg, v1=15m/s, v2=0 (very unequal masses, lighter mass at rest).
4) m1=5000kg, m2=0.01kg, v1=0, v2=-15m/s (very unequal masses, heavier mass at rest).
5) m1=5000kg, m2=0.01kg, v1= 15m/s, v2=-15m/s(head on collision)
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Quick Review: Vectors and Scalars
• Vectors: Magnitude and Direction(e.g., Force, Displacement, Velocity, Acceleration)
• Scalars: Magnitude only(e.g., Temperature, Altitude, Age)
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Adding Vectors
1F
2F
Two forces are acting on a mass m:How do we find the total force ?
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1F
2F
totalF
1F
2F
21 FFFFi
itotal
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Subtracting Vectors
initialv
An object changes it’s velocity.What is it’s change in velocity?
finalv
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initialv
finalv
initialfinalinitialfinal vvvvv
initialv
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initialv
initialfinalinitialfinal vvvvv
finalv
v
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1F
xxxtotal FFF 21
x
y
xF1
yF1
x
y
xF2
yF2
2F
yyytotal FFF 21
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xF1
yF1
x
y
xF2
yF2
xxxtotal FFF 21
yyytotal FFF 21
xtotalF
ytotalF
(Note: F2y had a negative value)
totalF