Tuesday, June 20, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu1PHYS 1443 – Section 001Lecture #11Tuesday, June 20, 2006Dr. Jaehoon Yu• Linear Momentum• Linear Momentum and Forces• Conservation of Momentum• Impulse and Momentum Change• Collisions• Two Dimensional Collision s• Center of MassToday’s homework is HW #6, due 7pm, Friday, June 23!!Tuesday, June 20, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu2Announcements• Quiz this Thursday– Eerly in the class– Covers Ch. 8.5 – Ch. 9• Mid-term grade discussions tomorrow– Bottom half of the classTuesday, June 20, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu3Linear MomentumThe principle of energy conservation can be used to solve problems that are harder to solve just using Newton’s laws. It is used to describe motion of an object or a system of objects.A new concept of linear momentum can also be used to solve physical problems, especially the problems involving collisions of objects.p≡GLinear momentum of an object whose mass is m and is moving at a velocity of v is defined as What can you tell from this definition about momentum?What else can use see from the definition? Do you see force?The change of momentum in a given time intervalpt∆=∆G0mv mvt−=∆GG()0mv vt−=∆GGvmt∆=∆GF∑Gma=G1. Momentum is a vector quantity.2. The heavier the object the higher the momentum3. The higher the velocity the higher the momentum4. Its unit is kg.m/smvGTuesday, June 20, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu4Linear Momentum and ForcesWhat can we learn from this Force-momentum relationship?Something else we can do with this relationship. What do you think it is?F =∑GThe relationship can be used to study the case where the mass changes as a function of time.Can you think of a few cases like this?Motion of a meteoriteMotion of a rocket • The rate of the change of particle’s momentum is the same as the net force exerted on it.• When net force is 0, the particle’s linear momentum is constant as a function of time.• If a particle is isolated, the particle experiences no net force. Therefore its momentum does not change and is conserved.dpFdt=∑GG()dmvdt=Gdmvdt=Gdvmdt+GdpdtGTuesday, June 20, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu5Conservation of Linear Momentum in a Two Particle SystemConsider an isolated system with two particles that does not have any external forces exerting on it. What is the impact of Newton’s 3rdLaw?Now how would the momentaof these particles look like?If particle#1 exerts force on particle #2, there must be another force that the particle #2 exerts on #1 as the reaction force. Both the forces are internal forces and the net force in the entire SYSTEM is still 0. Let say that the particle #1 has momentum p1and #2 has p2at some point of time.Using momentum-force relationship121dpFdt=GGAnd since net force of this system is 021ppconst+=GGThereforeF∑GThe total linear momentum of the system is conserved!!!and12 21FF=+GG21dp dpdt dt=+GG()21dppdt=+GG0=212dpFdt=GGTuesday, June 20, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu6Linear Momentum Conservation12iipp+=GG12ffpp+=GG11 2 2mv mv+GG11 2 2mv mv′′+GGTuesday, June 20, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu7More on Conservation of Linear Momentum in a Two Particle SystemWhat does this mean?As in the case of energy conservation, this means that the total vector sum of all momenta in the system is the same before and after any interactionMathematically this statement can be written as Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant.p=∑GFrom the previous slide we’ve learned that the total momentum of the system is conserved if no external forces are exerted on the system.21iipp+=GGThis can be generalized into conservation of linear momentum in many particle systems.∑∑=systemxfsystemxiPP∑∑=systemyfsystemyiPP∑∑=systemzfsystemziPP21ffpp+GG21pp+=GGconstTuesday, June 20, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu8Example for Linear Momentum ConservationEstimate an astronaut’s resulting velocity after he throws his book to a direction in the space to move to a direction.ipGFrom momentum conservation, we can writevAvBAssuming the astronaut’s mass is 70kg, and the book’s mass is 1kg and using linear momentum conservationAv =GNow if the book gained a velocity of 20 m/s in +x-direction, the Astronaut’s velocity isAv=GAABBmv mv=+GGBBAmvm−=G170Bv−G()12070i−=G()0.3 /ims−G0=fp=GTuesday, June 20, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu9Impulse and Linear Momentum By integrating the above equation in a time interval tito tf, one can obtain impulse I.Effect of the force F acting on a particle over the time interval ∆t=tf-tiis equal to the change of the momentum of the particle caused by that force. Impulse is the degree of which an external force changes momentum.The above statement is called the impulse-momentum theorem and is equivalent to Newton’s second law. dpFdt=GGNet force causes change of momentum Newton’s second lawSo what do you think an impulse is?What are the dimension and unit of Impulse? What is the direction of an impulse vector? Defining a time-averaged force 1iiFFtt≡∆∆∑GGImpulse can be rewritten IFt≡∆GGIf force is constant IFt≡∆GGIt is generally assumed that the impulse force acts on a short time but much greater than any other forces present.dp Fdt=GGfittdp=∫Gfipp−=GGp∆=GfittFdt=∫GIGTuesday, June 20, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu10Example 9-6(a) Calculate the impulse experienced when a 70 kg person lands on firm ground after jumping from a height of 3.0 m. Then estimate the average force exerted on the person’s feet by the ground, if the landing is (b) stiff-legged and (c) with bent legs. In the former case, assume the body moves 1.0cm during the impact, and in the second case, when the legs are bent, about 50 cm.We don’t know the force. How do we do this?Obtain velocity of the person before striking the ground.KE=212mv=()img y y−−=imgyv=Solving the above for velocity v, we obtain2igy=2 9.8 3 7.7 /ms⋅⋅=Then as the person strikes the ground, the momentum becomes 0 quickly giving the impulseIFt=∆=70 7.7 / 540kgms Ns=−⋅ =− ⋅p∆=fipp−=0 mv−=PE−∆Tuesday, June 20, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu11Example 9 – 6 cont’dIn coming to rest, the body decelerates from 7.7m/s to 0m/s in a distance
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