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

2. 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

3. 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

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. 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:

7. 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.)

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

9. 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)

10. Newton’s Second Law for a system of particles

11. 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

12. 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

13. 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

14. 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.

15. 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)

16. 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

17. Elastic collision in 1D

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

19. Perfectly inelastic collision in 1D

20. Collisions in 2D

21. 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?

22. Questions?

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

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

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

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