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