Physics of the Car Accident. Building a Safe Campus by Solving Physics Problem Service-Learning Component of General Physics Course Elena Flitsiyan Department of Physics

The Problem: Students have difficulty seeing connections between physics class and the “real world” Opportunities to help students these connections are not being fully realized in the current course design Implementing service learning elements in introductory physics course will result in improving the interactive learning component and also educate student community how prevent the car accidents

Forces on an inclined road  Often when solving problems involving Newton’s laws we will need to deal with resolving acceleration due to gravity on an inclined surface

Forces on an inclined road x y W = mg   mgsin  mgcos  What normal force does the surface exert?

Forces on an inclined road

Forces on an inclineed road Equilibrium

Forces on an inclined road  If the car is just stationary on the incline what is the (max) coefficient of static friction?

Horizontal (Flat) Curve  The force of static friction supplies the centripetal force  Solving for the maximum speed at which the car can negotiate the curve gives: FrFr Note, this does not depend on the mass of the car

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:

Suppose that a 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?

“ “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 centrifugal force  The outward force is often called a centrifugal force  It is a fictitious force due to the acceleration associated with the car’s change in direction

If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is μ s = 0.800, how fast can you drive on a horizontal roadway around a right turn of radius 30.0 m before the cup starts to slide? If you go too fast, in what direction will the cup slide 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. If you go too fast the cup will begin sliding

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 collision

Impulse-Momentum: Crash Test Example  Categorize  Assume force exerted by wall is large compared with other forces  Gravitational and normal forces are perpendicular and so do not effect the horizontal momentum  Can apply impulse approximation

Crash Test Example   Analyze  The momenta before and after the collision between the car and the wall can be determined  Find  Initial momentum  Final momentum  Impulse  Average force reasonable  Check signs on velocities to be sure they are reasonable

Two-Dimensional Collision Example  Conceptualize  See picture  Choose East to be the positive x-direction and North to be the positive y-direction  Categorize  Ignore friction  Model the cars as particles  The collision is perfectly inelastic  The cars stick together

Two dimensional collision m 1 = kg m 2 = kg Find v f.

Two dimensional collision m 1 = 800.0kg m 2 = kg Find v f. m 1 v 1 + m 2 v 2 = (m 1 + m 2 ) v f (800kg) (25m/s) + 0 = (2200kg) v f cosθ – x-component (1400kg) (20m/s) + 0 = (2200kg) v f sinθ - y-component