Linear Momentum, Systems of Particles, and Collisions Chapter 9 Linear Momentum, Systems of Particles, and Collisions

Linear momentum (Ch. 4) Linear momentum (or, simply momentum) of a point-like object (particle) is SI unit of linear momentum is kg*m/s Momentum is a vector, its direction coincides with the direction of velocity

Newton’s Second Law revisited (Ch. 4) Originally, Newton formulated his Second Law in a more general form The rate of change of the momentum of an object is equal to the net force acting on the object For a constant mass

Center of mass In a certain reference frame we consider a system of particles, each of which can be described by a mass and a position vector For this system we can define a center of mass:

Center of mass of two particles A system consists of two particles on the x axis Then the center of mass is

Center of mass of a rigid body For a system of individual particles we have For a rigid body (continuous assembly of matter) with volume V and density ρ(V) we generalize a definition of a center of mass:

Chapter 9 Problem 41 Find the center of mass of the uniform, solid cone of height h, base radius R, and constant density shown in the figure. (Hint: Integrate over disk-shaped mass elements of thickness dy, as shown in the figure.)

Newton’s Second Law for a system of particles For a system of particles, the center of mass is Then

Newton’s Second Law for a system of particles From the previous slide: Here is a resultant force on particle i According to the Newton’s Third Law, the forces that particles of the system exert on each other (internal forces) should cancel: Here is the net force of all external forces that act on the system (assuming the mass of the system does not change)

Newton’s Second Law for a system of particles

Linear momentum for a system of particles We define a total momentum of a system as: Using the definition of the center of mass The linear momentum of a system of particles is equal to the product of the total mass of the system and the velocity of the center of mass

Linear momentum for a system of particles Total momentum of a system: Taking a time derivative Alternative form of the Newton’s Second Law for a system of particles

Conservation of linear momentum From the Newton’s Second Law If the net force acting on a system is zero, then If no net external force acts on a system of particles, the total linear momentum of the system is conserved (constant) This rule applies independently to all components

Chapter 9 Problem 17 A popcorn kernel at rest in a hot pan bursts into two pieces, with masses 91 mg and 64 mg. The more massive piece moves horizontally at 47 cm/s. Describe the motion of the second piece.

Impulse During a collision, an object is acted upon by a force exerted on it by other objects participating in the collision We define impulse as: Then (momentum-impulse theorem)

Elastic and inelastic collisions During a collision, the total linear momentum is always conserved if the system is isolated (no external force) It may not necessarily apply to the total kinetic energy If the total kinetic energy is conserved during the collision, then such a collision is called elastic If the total kinetic energy is not conserved during the collision, then such a collision is called inelastic If the total kinetic energy loss during the collision is a maximum (the objects stick together), then such a collision is called perfectly inelastic

Elastic collision in 1D

Elastic collision in 1D: stationary target Stationary target: v2i = 0 Then

Perfectly inelastic collision in 1D

Collisions in 2D

Chapter 9 Problem 86 In a ballistic pendulum demonstration gone bad, a 0.52-g pellet, fired horizontally with kinetic energy 3.25 J, passes straight through a 400-g Styrofoam pendulum block. If the pendulum rises a maximum height of 0.50 mm, how much kinetic energy did the pellet have after emerging from the Styrofoam?

Questions?

Answers to the even-numbered problems Chapter 9 Problem 12 2.5 m

Answers to the even-numbered problems Chapter 9 Problem 16 4680 km

Answers to the even-numbered problems Chapter 9 Problem 18 – 10.6 iˆ – 2.8 jˆ m/s

Answers to the even-numbered problems Chapter 9 Problem 78 7.95 s