PHY 102 1st EditionI. First law of motiona. Newton’s first lawb. Lawc. AristotleII. QuestionsIII. Types of ForcesIV. Where Aristotle went wrongV. Momentum and InertiaOutline of Current Lecture Current LectureI. Conservation of momentum – the momentum before a collision is the same as the momentum after a collisiona. Experiment – demonstrating the velocity of an object after a collisioni. Remember that momentum has direction and is mass*velocityii. Two carts are placed on a track, and will collide in between two monitors of each cart’s velocityiii. Momentum before: 1 kg + 3 m/s and p = mv so M = 3 kg m/siv. Momentum after: m1 + m2 = 2 kg and Vf = 1.5 m/s and p = mv so M = 3 kg m/sv. Pi = Pfb. ExampleThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.i. m1 = ¼ kg and m1 + m2 = 1 1/2 kgVi = 3.4 m/s¼ * 3.4 m/s = .85 kg m/sSo we know the end result needs to be .85 kg m/s because momentum is equal after the collision. SO:.85 kg m/s = (1.5 kg)*Vf Vf = (.85 kg m/s) /1.5 kg = .567 m/sII. Numerical problemsa. A truck with a mass of 3,000 kg and a velocity of 20 m/s collides head on with a car of 1500 kg and a velocity of -20 m/s. If the vehicles stick together, what is the resulting motion of the pair?We know: pi = pf(m1 + m2 )VV =m1∗v 1+m2∗v 2m1+m2¿(3000 kg)(20 ms)+(1500 kg)(−20 ms)3000 kg+1500 kg30,000 m/s/4500b. Two dynamic carts are separated by a compressed spring. The mass of cart 1 is twice the mass of cart 2. That is, m1 = 2m2. The spring is released and the carts flyapart. If cart 1 has a velocity of 3 m/s, what is the velocity of cart 2?pi = pf0 = m1 * v1 + m2 * v2- m1 * v1 = m2 * v2-2v1 = v2 v2 = 6 m/sTEST QUESTION: If two carts that start at 0 momentum and they roll awayfrom each other and one cart has a momentum of 100, what is the momentum of the other cart? ANSWER: -100c. A train car (mass 11,200 kg) is moving along at a speed of 2 m/s. A 5,000 kg mass of coal with no horizontal motion is dropped into the hopper of the train car. What is the resulting motion of the train car now loaded with coal?i. p1 = pfmtrain * vtrain = (mcoal + mtrain) VV = (mtrain * vtrain) / (mcoal + mtrain)(11,200)(2 m/s)/(5000+ 11,200 kg)1.38 m/sTEST QUESTION: What is the momentum of a pen in someone’s hand? ANSWER: 0. But before it hits a table when dropped, it has momentum. How is this possible? The mass ofthe whole world is too large to account for in the equation.III. Newton’s Second Lawa. The net instantaneous force acting on an object is precisely the instantaneous change of its momentum per unit time. IN symbols, the second law can be written as F = delta P/delta T with delta T being very small.b. Note that F and delta P are vectors.c. Experiment – cart is released on a track and acceleration is charted as mass of the system is constant but weights are moved from one side of the track to a force on the cart.Acceleration Force3.4 507.36 10013.3 15015.5 200Acceleration is proportional to the force.Then the experiment is redone by adding the weights to the cart to increase the mass.Acceleration Mass7.18 5505.40 6504.34 7503.64 850Acceleration is inversely proportional to the mass. (F = ma)IV. Second Law Reprisea. F = delta P/delta Tb. F = delta mv/delta Tc. F = m*deltaV/delta Td. F = mae. Both F and a are vectors; m is a scalar.f. The sum of forces acting on a body produce an acceleration inversely proportional to mass.g.∑F=ma (where F is expressed in Newtons)Weight = mgWhat is the weight of a person with a mas of 81 kg? Note that g = 9.8 m/s281 kg * -9.8 m/s^2-793.8