Monday, June 27, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 1 PHYS 1443 – Section 001 Lecture #12 Monday, June 27, 2011 Dr. Jaehoon Yu • Linear Momentum and Forces • Linear Momentum Conservation • Impulse and Linear Momentum • Collisions Today’s homework is homework #7, due 10pm, Thursday, June 30!!Monday, June 27, 2011 2 Announcements • Quiz #3 tomorrow, Wednesday, June 29 – Beginning of the class – Covers CH8.1 through what we learn tomorrow, Tuesday, June 28 • Mid-term grade discussions – Second half of the class – I strongly urge you all to come and discuss your grades • Bring your special projects to the grade discussion if you haven’t already submitted PHYS 1443-001, Spring 2011 Dr. Jaehoon YuMonday, June 27, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 3 Linear Momentum The 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≡Linear momentum of an object of mass m moving at the velocity 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 interval 1. Momentum is a vector quantity. 2. The heavier the object the higher the momentum 3. The higher the velocity the higher the momentum 4. Its unit is kg.m/sMonday, June 27, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 4 Linear Momentum and Forces What can we learn from this force-momentum relationship? Something else we can do with this relationship. What do you think it is? F∑=Δt→0limΔpΔt=dpdtThe 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 meteorite Motion of a rocket • The rate of the change of particle’s momentum is the same as the net force exerted on it. • When the net force is 0, the particle’s linear momentum is a 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. F∑=dpdt =d mv( )dt =dmdtv +mdvdtMonday, June 27, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 5 Conservation of Linear Momentum in a Two Particle System Consider an isolated system with two particles that do not have any external forces exerting on it. What is the impact of Newton’s 3rd Law? Now how would the momenta of these particles look like? If particle#1 exerts force on particle #2, there must be a reaction force that the particle #2 exerts on #1. Both the forces are internal forces, and thus the net force in the entire SYSTEM is still 0. Let say that the particle #1 has momentum p1 and #2 has p2 at some point of time. Using momentum-force relationship And since net force of this system is 0 p2+ p1= constTherefore F∑The total linear momentum of the system is conserved!!! and = F12+ F21 =dp2dt+dp1dtMonday, June 27, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 6 More on Conservation of Linear Momentum in a Two Body System What 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 interactions Mathematically 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=∑From the previous slide we’ve learned that the total momentum of the system is conserved if no external forces are exerted on the system. p2i+p1i=This can be generalized into conservation of linear momentum in many particle systems. p2 f+p1 f p2+ p1= constMonday, June 27, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 7 Linear Momentum Conservation Initial FinalMonday, June 27, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 8 Example 9.4: Rifle Recoil Calculate the recoil velocity of a 5.0kg rifle that shoots a 0.020kg bullet at a speed of 620m/s. piFrom momentum conservation, we can write Solving the above for vR and using the rifle’s mass and the bullet’s mass, we obtain vR= =PR+PB = pf = mRvR+ mBvB PR+ PB vR= −2.5i m s( )The x-comp = mRvR+ mBvB = 0 mBmRvB= 0.0205.0⋅ 620 = −2.5m sMonday, June 27, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 9 Example for Linear Momentum Conservation Estimate an astronaut’s (M=70kg) resulting velocity after he throws his book (m=1kg) to a direction in the space to move to another direction. piFrom momentum conservation, we can write vA vB Assuming the astronaut’s mass is 70kg, and the book’s mass is 1kg and using linear momentum conservation vA=Now if the book gained a velocity of 20 m/s in +x-direction, the Astronaut’s velocity is vA= −mBvBmA= −170vB −17020i( )= −0.3i m / s( ) = pfMonday, June 27, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 10 There are many situations where the force on an object is not constant and in fact quite complicated during the motion!! ImpulseMonday, June 27, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 11 F∑( )Δt = = JBall Hit by a Bat Multiply either side by ΔtMonday, June 27, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu 12 Impulse and Linear Momentum By integrating the above equation in a time interval ti to tf, one can obtain impulse I. Effect of the force F acting on an object over the time interval Δt=tf-ti is equal to the change of the momentum of the object caused by that force. Impulse is the degree of which an external force changes an object’s momentum. The above statement is called the impulse-momentum theorem and is equivalent to Newton’s second law. F =dpdtNet force causes change of momentum Newton’s second law So 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 F ≡1ΔtFiΔti∑Impulse can be rewritten J≡ FΔtIf force is constant J≡ FΔtIt is generally assumed that the impulse force acts on
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