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Chapter 9 Linear Momentum and Collisions. Linear momentum Linear momentum (or, simply momentum) of a point-like object (particle) is SI unit of linear.

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Presentation on theme: "Chapter 9 Linear Momentum and Collisions. Linear momentum Linear momentum (or, simply momentum) of a point-like object (particle) is SI unit of linear."— Presentation transcript:

1 Chapter 9 Linear Momentum and Collisions

2 Linear momentum 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

3 Newton’s Second Law revisited 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

4 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:

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

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

7 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)

8 Newton’s Second Law for a system of particles

9 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

10 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

11 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

12 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:

13 Chapter 9 Problem 37 A uniform piece of steel sheet is shaped as shown. Compute the x and y coordinates of the center of mass of the piece.

14 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)

15 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

16 Elastic collision in 1D

17 Elastic collision in 1D: stationary target Stationary target: v 2i = 0 Then

18 Chapter 9 Problem 18 A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length l and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

19 Perfectly inelastic collision in 1D

20 Collisions in 2D

21 Chapter 9 Problem 57 A bullet of mass m is fired into a block of mass M initially at rest at the edge of a frictionless table of height h. The bullet remains in the block, and after impact the block lands a distance d from the bottom of the table. Determine the initial speed of the bullet.

22 Answers to the even-numbered problems Chapter 9 Problem kg ⋅ m/s upward

23 Answers to the even-numbered problems Chapter 9 Problem 16 (a) 2.50 m/s; (b) 3.75 × 10 4 J

24 Answers to the even-numbered problems Chapter 9 Problem 20 (a) 4.85 m/s; (b) 8.41 m

25 Answers to the even-numbered problems Chapter 9 Problem cm

26 Answers to the even-numbered problems Chapter 9 Problem 58 (a) – m/s; (b) m


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