1 / 7

Physics 101: Lecture 15 Impulse and Momentum

Physics 101: Lecture 15 Impulse and Momentum. Today’s lecture will be a review of Chapters 7.1 - 7.2 and New material: Collisions and Center of Mass, Chapters 7.3-7.5 Rotational Motion and Angular Displacement, Chapter 8.1 . Conservation of Linear Momentum.

dolan
Download Presentation

Physics 101: Lecture 15 Impulse and Momentum

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Physics 101: Lecture 15Impulse and Momentum • Today’s lecture will be a review of Chapters 7.1 - 7.2 and • New material: Collisions and Center of Mass, Chapters 7.3-7.5 Rotational Motion and Angular Displacement, Chapter 8.1

  2. Conservation of Linear Momentum • Consider a system of two colliding objects with masses m1 and m2 and initial velocities v01 and v02 and final velocitiesvf1 and vf2 : If the sum of the average external forces acting on the two objects is zero ( = isolated system), the total momentum of the system is conserved: SFave,ext Dt = Pf - P0 => Pf = P0 if SFave,ext = 0 Pf and Po arethetotal momenta of the system: Pf = pf1 + pf2 and P0 = p01 + p02 This is true for any number of colliding objects.

  3. Applying the Principle of Momentum Conservation • Decide which objects are included in the system. • Identify external and internal forces acting on the system. • Verify that the system is isolated. • Initial and final momenta of the isolated system can be considered to be equal. Example for an application: Determination of velocities of colliding objects after collision.

  4. Impulse and Momentum Summary Fave tJ = pf – p0 = p • For a single object…. •  Fave = 0 momentum conserved(p = 0) • For collection of objects … • Fave,ext = 0  total momentum conserved(P = 0)

  5. Collisions • If colliding objects constitute an isolated system (= no average external force), the total linear momentum is conserved. Sometimes also the kinetic energy is conserved. Elastic collision: Total kinetic energy before and after the collision is the same. Inelastic collision: Total kinetic energy is not conserved, i.e. part (or all) of the kinetic energy of the objects is converted into another form of energy. Collisions in two dimensions: SFave,ext,x Dt = Pf,x - P0,x => Pf,x = P0,x if SFave,ext,x = 0 SFave,ext,y Dt = Pf,y - P0,y => Pf,y = P0,y if S Fave,ext,y = 0 x and y components of the total linear momentum are separately conserved.

  6. Center of Mass • The center of mass of a system of objects is defined as the average location of the total mass. Consider two interacting objects (in 1-dim.) with masses m1 and m2 at the positions x1 and x2: xcm = (m1 x1 + m2 x2)/(m1+m2) Displacement of center of mass: Dxcm = (m1 Dx1 + m2Dx2)/(m1+m2) Velocity of center of mass: vcm = (m1 v1 + m2 v2)/(m1+m2) In an isolated system vcm does not change.

  7. Rotational Kinematics • The motion of a rigid body about a fixed axis is described by using the same concept as for linear motion (see C&J Chapter 2): Displacement, Velocity, Acceleration Angular Displacement: Identify the axis of rotation and choose a line perpendicular to this axis. Observe the motion of a point on this line. How can one define the change of position of this point during rotation about an axis ? Answer: Change of angle the line makes with a reference line: Dq

More Related