New version page

UT Arlington PHYS 1441 - Lecture Notes

Documents in this Course
Velocity

Velocity

18 pages

Load more
Upgrade to remove ads
Upgrade to remove ads
Unformatted text preview:

Monday, Mar. 22, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu1PHYS 1441 – Section 004Lecture #14Monday, Mar. 22, 2004Dr. Jaehoon Yu• Linear Momentum• Linear Momentum Conservation• Impulse• Collisions: Elastic and Inelastic collisions• Center of MassMonday, Mar. 22, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu2Announcements• Quiz results– Average: 52/ 90 58/100– Top score: 80/90  89/100– How did we do compared to the other quizzes?•1stquiz: 38.5/100•2ndquiz: 41.8/100– Marked improvement • Second term exam next Monday, Mar. 29– In the class, 1:00 – 2:30pm in SH101– Sections 5.6 – 8.2– Mixture of multiple choices and numeric problems– Will give you exercise test problems WednesdayMonday, Mar. 22, 2004 PHYS 1441-004, Spring 2004Dr. 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.vmp =Linear momentum of an object whose mass is m and is moving at a velocity of v is defined as 1. 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/s 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∆=∆JG0mv mvt−=∆GG()0mv vt−=∆GGvmt∆=∆GF∑JGma=GMonday, Mar. 22, 2004 PHYS 1441-004, Spring 2004Dr. 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?pFt∆=∆∑JGJGThe 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.Monday, Mar. 22, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu5Linear Momentum Conservation221121vmvmppii+=+'22'1121vmvmppff+=+Monday, Mar. 22, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu6Conservation of Linear Momentum in a Two Particle SystemConsider a 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 SYSTEM is still 0. Let say that the particle #1 has momentum p1and #2 has p2at some point of time.Using momentum-force relationship121pFt∆=∆JGJGAnd since net force of this system is 0constpp =+12Therefore∑FThe total linear momentum of the system is conserved!!!and2112FF +=21pptt∆∆=+∆∆JGJG()21ppt∆=+∆JGJG0=212pFt∆=∆JGJGMonday, Mar. 22, 2004 PHYS 1441-004, Spring 2004Dr. 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.constppp =+=∑12From 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+=JGJGThis can be generalized into conservation of linear momentum in many particle systems.∑∑=systemxfsystemxiPP∑∑=systemyfsystemyiPP∑∑=systemzfsystemziPP21ffpp+JGJGMonday, Mar. 22, 2004 PHYS 1441-004, Spring 2004Dr. 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.ipFrom momentum conservation, we can writevAvBAssuming the astronaut’s mass if 70kg, and the book’s mass is 1kg and using linear momentum conservation=AvNow if the book gained a velocity of 20 m/s in +x-direction, the Astronaut’s velocity is=AvBBAAvmvm +==−ABBmvmBv701−()=− i20701()smi / 3.0−0=fp=Monday, Mar. 22, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu9Impulse and Linear Momentum By summing the above equation in a time interval tito tf, one can obtain impulse I.Impulse 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. pFt∆=∆JGJGNet force causes change of momentum Newton’s second lawSo what do you think an impulse is?IFt p≡∆=∆GJG JGWhat are the dimension and unit of Impulse? What is the direction of an impulse vector? Defining a time-averaged force 1iiFFtt≡∆∆∑JGJGImpulse can be rewritten tFI ∆≡If force is constant tFI ∆≡It is generally assumed that the impulse force acts on a short time but much greater than any other forces present.pFt∆=∆JG J Gfippp∆=−JGJGJGFt=∆JGMonday, Mar. 22, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu10Example 7-5(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.KEPE=−∆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−=Monday, Mar. 22, 2004 PHYS 1441-004, Spring


View Full Document
Download Lecture Notes
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture Notes and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture Notes 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?