PAP111_Lecture10 - ,allwith thesameinitialspeed...

Info icon This preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Three identical balls are thrown from the top of a building, all with  the same initial speed.  A: horizontally, B: inclined upwards, C: inclined downwards Rank the speeds of the balls at the instant each hits the ground. 1 2 3 4 5 6 0% 0% 0% 0% 0% 0% 1. A, B, C 2. A, C, B 3. B, A, C 4. C, B, A 5. C, A, B 6. B, C, A
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Active figure 1
Image of page 2
PAP 111  Mechanics and Relativity  Lecture 10 Linear Momentum The Conservation of Linear Momentum Impulse and Momentum The Collision Problem Center of Mass and the Motion of a System of Particles Rocket Propulsion
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Linear Momentum v p m = The linear momentum of a particle (or an object  that can be modeled as a particle) of mass  m   moving with a velocity  v  is defined to be the  product of the mass and velocity:  (10.1) Remarks:  The SI units of momentum are kg ∙ m / s .  Momentum can be expressed in component form as follows: z z y y x x mv p mv p mv p = = = , ,
Image of page 4
Original and General Form of  Newton’s Second Law Newton’s Second Law states that: ( 29 dt d dt m d p v F = = The time rate of change of the linear momentum of  a particle is equal to the net force acting on the  particle: (10.2) Remarks: This general formulation allows for mass change, and is more powerful  when applies to a system of particles. When the mass is constant, Eq. (10.2) reduces to Newton called  m v , the quantity of motion. = = . / a v F m dt md
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Two Particles Interaction in an  Isolated System 0 12 21 = + F F ( 29 ( 29 0 2 2 1 1 = + dt m d dt m d v v ( 29 0 2 2 1 1 = + v v m m dt d By Newton’s Third Law: By Newton’s Second Law: ( 29 0 2 1 = + p p dt d Then, Note:  c  is a constant c tot = + = 2 1 p p p (10.3)
Image of page 6
Conservation of Linear  Momentum Equation (10.3) tells us that the total momentum of  an isolated system is constant    Law of Conservation of Linear Momentum f f i i 2 1 2 1 p p p p + = + Whenever two or more particles in an isolated system  interact, the total momentum of the system remains  constant. (10.4)
Image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Remarks on Conservation of  Linear Momentum The momentum of the system is conserved, not  necessary the individual particles. The total momentum of an isolated system at all  times equal its initial momentum. Conservation of momentum applies to systems with  any number of particles. In component form, the total momenta in each  direction are independently conserved: fz iz fy iy fx ix p p p p p p = = = , ,
Image of page 8
The Concept of Impulse The concept of Impulse relates to a net force  F  that may vary with time.
Image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern