# Physics of the Car Accident. Building a Safe Campus by Solving Physics

20 Slides915.50 KB 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 What normal force does the surface exert? mgsin mgcos W mg y x Forces on an inclined road F F x mg sin y n mg cos Forces on an inclineed road F F x mg sin ma y n mg cos 0 Equilibrium Forces on an inclined road If the car is just stationary on the incline what is the (max) coefficient of static friction? Fx mg sin s n ma 0 F y n mg cos 0 mg sin s n s mg cos sin s tan cos Horizontal (Flat) Curve The force of static friction supplies the centripetal force v2 f S m r 2 v f S n S mg m r Solving for the maximum speed at which the car can negotiate the curve gives: v gr Fr 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: 2 v tan rg mv 2 n sin r (1) n cos mg ( 2) 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? n mg mv 2 Fy may n mg r 2 2 8.33 m s v 2 n m g 1 800 kg 9.8 m s r 20.4 m 1 h 1 000 m v 30 km h 8.33 m s 3 600 s 1 km 1.15 10 4 N up n 0 mv 2 mg r v gr 9.8 m s 20.4 m 14.1 m s 2 50.9 km h “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 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 μs 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? If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is μs 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. Fy may : n mg 0 Fx max : mv 2 f s n s mg r v s gr 0.8 9.8 m s 2 30 m 15.3 m s 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 p i and p f 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 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 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 1400 0.8 tan 800 54.50 (1400)(20) (2200)v f sin 54.50 v f (9.07iˆ)m / s (12.72 ˆj )m / s v f 15.63m / s 12.75 0 tan 89.6 9.07 1