Newton’s Laws of Motion

Newton’s Laws of Motion

Newton’s First Law of Motion http://www.wisc-online.com/objects/tp1202/tp1202.swf

Newton’s First Law of Motion Demo using the Barbie car with Barbie in the front seat and a baby in the back seat without child restraint.

Newton’s Second Law of Motion The acceleration of an object is directly proportional to the force applied to it and inversely proportional to the mass of the object. Note: both acceleration and force have magnitude and direction, so both are vectors! The newton is the unit of force N = kg*m/s2 a = F m F = ma

Newton’s Second Law of Motion • Two horses are pulling a barge with a mass of 2.00 x 103 kg along a canal. The cable connecting to the first horse makes an angle of 30.0 with respect to the direction of the canal, while the cable connected to the second horse makes an angle of 45.0. Find the initial acceleration of the barge, starting at rest, if each horse exerts a force of magnitude 6.00 x 102 N on the barge. Ignore forces of resistance on the barge. Horse 1  F1 Barge = 2.00 x 103 kg q1 q2  F2 Horse 2 Problem modified from Serway and Vuille’s College Physics 8th Edition

Newton’s Third Law of Motion http://www.cpo.com/home/portals/2/Media/post_sale_content/newtons%20third%20law.swf

Newton’s Laws of Motion Here is a neat quiz that you can use with your students! http://www.softschools.com/quizzes/science/newtons_laws/quiz384.html

Newton’s Laws of Motion Let’s do some problems.

Momentum The linear momentum of an object is dependent upon the mass and the velocity of the object. Notice that both momentum and velocity are vectors and, thus, have both magnitude and direction! Changing the momentum of an object requires that a force be applied. p = m*v Dp = Dt*F

Momentum The impulse is defined as the force applied to an object over a given period of time. Substituting in the equation from the previous slide, we can get I = Dt*F I = DP = mvf- mvi

Momentum Let’s try a problem! In a crash test, a car of mass 1.50 x 103 kg collides with a wall and rebounds. The initial and final velocities of the car are vi= -15.0 m/s and vf = 2.60 m/s, respectively. If the collision lasts for 0.15 s, what is the impulse delivered to the car due to the collision? What is the size and direction of the average force exerted on the car?

Conservation of Momentum In a closed system, momentum must be conserved. If 2 objects in motion collide, there can be a shift in momentum, but the total sum must remain the same. http://www.youtube.com/watch?v=mFNe_pFZrsA

Conservation of Momentum If 2 objects in motion collide, there can be a shift in momentum, but the total sum must remain the same. V1i V2i V1f V2f F21Dt = - F12Dt V1fm1 – V1im1 = V2fm2– V2im2 V1im1 + V2im2 = V2fm2+ V1fm1

Inelastic Collisions Inelastic collisions – when momentum is conserved but kinetic energy is not. Some energy lost when the one object deforms to “stick” to the other. KE = p2 2m V1im1 + V2im2 = Vf(m2+ m1) V1im1 + V2im2= Vf (m2 + m1)

Elastic Collisions Elastic collisions – when momentum is conserved and kinetic energy is also conserved. V1im1 + V2im2 = V2fm2+ V1fm1 ½ m1V1i2 + ½ m2V2i2 = ½ m1V1f2 +½ m2V2f2 Which can simplify to V1i- V2i = - (V1f-V2f)

Collisions Let’s do some problems! • An SUV with a mass of 1.80 x 103 kg is traveling eastbound at 15.0 m/s, while a compact car with mass 9.00 x 103 kg is traveling westbound at – 15.0 m/s. The cars collide head-on, becoming entangled. • Find the speed of the entangled cars after the collision. • Find the change in the velocity of each car. • Find the change in the kinetic energy of the system consisting of both cars.

Collisions Let’s do some problems! Two billiard balls of identical mass move toward each other. Assume that the collision between them is perfectly elastic. If the initial velocities of the balls are + 30 cm/s and – 20 cm/s, respectively, what is the velocity of each ball after the collision? Assume that friction and rotation are unimportant.

Newton’s Laws of Motion