physics of the car accident. building a safe campus by solving physics problem service-learning...
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Physics of the Car AccidentPhysics of the Car Accident. . Building a Safe Campus by Solving Building a Safe Campus by Solving Physics Problem Physics Problem
Service-Learning Component of General Service-Learning Component of General Physics CoursePhysics Course
Elena FlitsiyanElena FlitsiyanDepartment of PhysicsDepartment of Physics
The Problem:The Problem:
Students have difficulty seeing connections between Students have difficulty seeing connections between physics class and the “real world”physics class and the “real world”
Opportunities to help students these connections are Opportunities to help students these connections are not being fully realized in the current course designnot being fully realized in the current course design
Implementing service learning elements in introductory Implementing service learning elements in introductory physics course will result in improving the interactive physics course will result in improving the interactive learning component and also educate student learning component and also educate student community how prevent the car accidentscommunity how prevent the car accidents
Forces on an inclined road
Often when Often when solving problems solving problems involving involving Newton’s laws we Newton’s laws we will need to deal will need to deal with resolving with resolving acceleration due acceleration due to gravity on an to gravity on an inclined surfaceinclined surface
Forces on an inclined road
x
y
W = mg
mgsin
mgcos
What normal forcedoes the surface exert?
Forces on an inclined road
cos
sin
mgn
mg
y
x
F
F
Forces on an inclineed road
0cos
sin
mgn
mmg
y
x
F
aF
Equilibrium
Forces on an inclined road
If the car is just stationary on the incline what is the (max) coefficient of static friction?
0cos
0sin
mgn
mnmg
y
sx
F
aF
tancos
sin
cossin
s
ss mgnmg
Horizontal (Flat) Curve The force of static friction The force of static friction
supplies the centripetal supplies the centripetal forceforce
Solving for the maximum Solving for the maximum speed at which the car can speed at which the car can negotiate the curve gives:negotiate the curve gives:
v gr
Fr
r
vmfS
2
Note, this does not depend on the Note, this does not depend on the mass of the carmass of the car
r
vmmgnf SS
2
Banked Curve These are designed with
friction equaling zero There is a component of the
normal force that supplies the centripetal force (1), and component of the normal force that supplies the gravitational force (2).
Dividing (1) by (2) gives:
)2(cos
)1(sin2
mgn
r
mvn
2
tanv
rg
Suppose that a 1 800-kg car passes over a bump in a roadway that follows the arc of a circle of radius 20.4 m as in Figure. (a) What force does the road exert on the car as the car passes the highest point of the bump if the car travels at 30.0 km/h? (b) What is the maximum speed the car can have as it passes this highest point without losing contact with the road?
1000 m1 h30 km h 8.33 m s
3600 s 1 kmv
y yF ma2mv
n mgr
222
4
8.33 m s1800 kg 9.8 m s
20.4 m
1.15 10 N up
vn m g
r
n
mg
0n 2mvmg
r 29.8 m s 20.4 m 14.1 m s 50.9 km hv gr
““Centrifugal” Force From the frame of the passenger (b), a
force appears to push her toward the door From the frame of the Earth, the car
applies a leftward force on the passenger The outward force is often called a centrifugalcentrifugal force force It is a fictitious force due to the acceleration It is a fictitious force due to the acceleration
associated with the car’s change in directionassociated with the car’s change in direction
If the coefficient of static friction between your coffee cup If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is and the horizontal dashboard of your car is μμss = 0.800 = 0.800, ,
how fast can you drive on a horizontal roadway around a how fast can you drive on a horizontal roadway around a right turn of radius right turn of radius 30.0 m30.0 m before the cup starts to slide? before the cup starts to slide? If you go too fast, in what direction will the cup slide If you go too fast, in what direction will the cup slide relative to the dashboard?relative to the dashboard?
If the coefficient of static friction between your coffee cup If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is and the horizontal dashboard of your car is μμss = 0.800 = 0.800, ,
how fast can you drive on a horizontal roadway around a how fast can you drive on a horizontal roadway around a right turn of radius right turn of radius 30.0 m30.0 m before the cup starts to slide? before the cup starts to slide? If you go too fast, in what direction will the cup slide If you go too fast, in what direction will the cup slide relative to the dashboard?relative to the dashboard?
We adopt the view of an inertial observer. If it is on the verge of sliding, the cup is moving on a circle with its centripetal acceleration caused by friction.
2
: 0
:
y y
x x s s
F ma n mg
mvF ma f n mg
r
20.8 9.8 m s 30 m 15.3 m ssv gr
If you go too fast the cup will begin sliding
straight across the dashboard to the left.
Impulse Approximation In many cases, one force acting on a particle
acts for a short time, but is much greater than any other force present
When using the Impulse Approximation, we will assume this is true Especially useful in analyzing collisions
The force will be called the impulsive force The particle is assumed to move very little
during the collision represent the momenta
immediately before and after the collisioni fandp p
Impulse-Momentum: Impulse-Momentum: Crash Test ExampleCrash Test Example
CategorizeCategorize Assume force exerted by Assume force exerted by
wall is large compared wall is large compared with other forceswith other forces
Gravitational and normal Gravitational and normal forces are perpendicular forces are perpendicular and so do not effect the and so do not effect the horizontal momentumhorizontal momentum
Can apply impulse Can apply impulse approximationapproximation
Crash Test Example Analyze
The momenta before and after the collision between the car and the wall can be determined
Find Initial momentumFinal momentum ImpulseAverage force
Check signs on velocities to be sure they are reasonablereasonable
Two-Dimensional Collision ExampleTwo-Dimensional Collision Example
ConceptualizeConceptualize See pictureSee picture Choose East to be the Choose East to be the
positive positive xx-direction and North -direction and North to be the positive to be the positive yy-direction-direction
CategorizeCategorize Ignore frictionIgnore friction Model the cars as particlesModel the cars as particles The collision is perfectly The collision is perfectly
inelasticinelastic The cars stick togetherThe cars stick together
Two dimensional collision
m1 = 1500.0kg
m2 = 2500.0 kg
Find vf .
Two dimensional collision
m1 = 800.0kg
m2 = 1400.0 kg
Find vf .
m1 v1 + m2 v2 = (m1 + m2) vf
(800kg) (25m/s) + 0 = (2200kg) vf cosθ – x-component(1400kg) (20m/s) + 0 = (2200kg) vf sinθ - y-component
01
0
0
6.8907.9
75.12tan/)ˆ72.12(/)ˆ07.9(
/63.155.54sin)2200()20)(1400(
5.54tan800
14008.0
smjsmiv
smvv
f
ff