Physics 101: Lecture 15, Pg 1 Physics 101: Lecture 15 Impulse and Momentum l Today’s lecture will be a review of Chapters 7.1 - 7.2 and l New material:

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Presentation transcript:

Physics 101: Lecture 15, Pg 1 Physics 101: Lecture 15 Impulse and Momentum l Today’s lecture will be a review of Chapters and l New material: Collisions and Center of Mass, Chapters Rotational Motion and Angular Displacement, Chapter 8.1

Physics 101: Lecture 15, Pg 2 Conservation of Linear Momentum l Consider a system of two colliding objects with masses m 1 and m 2 and initial velocities v 01 and v 02 and final velocities v f1 and v f2 : 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:  F ave,ext  t = P f - P 0 => P f = P 0 if  F ave,ext = 0 P f and P o are the total momenta of the system: P f = p f1 + p f2 and P 0 = p 01 + p 02 This is true for any number of colliding objects.

Physics 101: Lecture 15, Pg 3 Applying the Principle of Momentum Conservation l Decide which objects are included in the system. l Identify external and internal forces acting on the system. l Verify that the system is isolated. l 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.

Physics 101: Lecture 15, Pg 4 Impulse and Momentum Summary  F ave  t  J = p f – p 0 =  p l For a single object…. è  F ave = 0  momentum conserved (  p = 0) l For collection of objects … è  F ave,ext = 0  total momentum conserved (  P = 0)

Physics 101: Lecture 15, Pg 5 Collisions l 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:  F ave,ext,x  t = P f,x - P 0,x => P f,x = P 0,x if  F ave,ext,x = 0  F ave,ext,y  t = P f,y - P 0,y => P f,y = P 0,y if  F ave,ext,y = 0 x and y components of the total linear momentum are separately conserved.

Physics 101: Lecture 15, Pg 6 Center of Mass l 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 m 1 and m 2 at the positions x 1 and x 2 : x cm = (m 1 x 1 + m 2 x 2 )/(m 1 +m 2 ) Displacement of center of mass:  x cm = (m 1  x 1 + m 2  x 2 )/(m 1 +m 2 ) Velocity of center of mass: v cm = (m 1 v 1 + m 2 v 2 )/(m 1 +m 2 ) In an isolated system v cm does not change.

Physics 101: Lecture 15, Pg 7 Rotational Kinematics l 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: 