1. Chapter 9 Systems of Particles

2. Section 9.2: Center of Mass in a Two Particle System Center of Mass is the point at which all forces are assumed to act. For a two body system, this force is calculated using the equation: Where m represents the mass of each body and x represents its position. Equation 1

3. Velocity and Acceleration of the Center of Mass The Velocity and Acceleration of the center of mass of a system is calculated using a similar equation: Equation 2 Equation 3

4. Section 9.2: Many Particle Systems Expanding equations 1,2,& 3 from a two particle system to a many particle system is simply a matter of expanding the terms to as many particles as the system holds, thus the general form of these equations are:

5. Analyzing Multi-particle Systems Since we can calculate the center of mass of a system and its velocity and acceleration, we can then analyze the system using the following rule: The overall translational motion of a system of particles can be analyzed using Newton’s Laws as if all the mass were concentrated at the center of mass and the total external force were applied at that point. A corollary which follows is: If the net external force on a system of particles is zero, then the center of mass of the system move with a constant velocity.

6. The Center of Mass of Solid Objects If an object has spherical symmetry – the center of mass is at the geometric center If an object is not symmetrical then you calculate the center of mass along each axis and this determines the center of mass of that object for translational motion in that direction.

7. Section 9.3: Newton’s Second Law for a System of Particles F net = Ma cm where:  F net – the sum of all external forces acting on the system.  M – total mass of the system  a cm – the acceleration of the center of mass of the system. This is equivalent to the three equations:  F net,x = Ma cm,x F net,y = Ma cm,y F net,z = Ma cm,z

8. Section 9.4 Linear Momentum of a Particle Linear Momentum is the product of the mass of an object and its linear velocity. ρ = mv where ρ (roe) is the momentum, m is the mass in kilograms, and v is velocity in m/s.

9. Section 9.5: Linear Momentum in System of Particles In a system of particles:  So

10. Section 9:6 Change in Momentum: Impulse The rate of change of momentum of a body (impulse) is equal to the resultant force acting on the body and is the direction of that force.  Since ρ = mv and F = ma and a = Δv/t  Δρ =mΔv = FΔt Hence change in momentum is a product of the force acting on an object and the time during which it acts.

11. Section 9.6 Conservation of Linear Momentum Like energy – momentum in a system is also conserved. For example, prior to firing a shotgun, neither the gun, nor the shell inside are moving, hence the momentum of the system is zero. m g v g + m s v s = 0 After the gun is fired, the shell moves forward, and by conservation of momentum, the gun moves backward m g v g = -(m s v s )

12. Sample Problem A 3 kg shotgun contains a.140 kg shell. When the shotgun is fired, the shell leaves the barrel with a velocity of 400 m/s. What is the recoil velocity of the shotgun? If the Shotgun impacts the hunter’s shoulder for 0.03 seconds, what force does it exert on her shoulder?

13. Solution By Conservation of Momentum: Hence (3 kg + 0.140 kg)(0 m/s) = (3 kg)(V g ) + (0.140 kg)(400 m/s) -(3 kg)(Vg) = 56 kgm/s

14. Alta High AP Physics Section 9.8: Momentum and Kinetic Energy in Collisions Two type of Collisions Elastic Two objects collide and there are two objects move away In a perfectly elastic collision kinetic energy is conserved. Inelastic collision  Two objects collide and stick together- hence they move off as one object after the collision  The converse of an inelastic collision is when one object breaks into two pieces and both move off after the breaking (often an explosion)  While total energy is conserved in the collision, often it is changed into heat energy or energy of deformation so kinetic energy is not conserved.

15. Alta High AP Physics Section 9.9 Inelastic Collisions Inelastic Collisions can be described mathematically by the equation: m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v’ since kinetic energy is not conserved this is the only equation for an inelastic collision

16. Alta High AP Physics Sample Problem A car with a mass of 1000 kg moving at a velocity of 25 m/s, strikes a second car with a mass of 1500 kg that is at rest. The bumpers lock and the cars travel together after the collision. What is the velocity of the two car system immediately after the collision?

17. Alta High AP Physics Sample Problem Solution m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v’ (1000 kg)(25 m/s) + (1500 kg)(0 m/s) = (2500 kg)(v’) 25000 kg m/s = (2500 kg)(v’) V’ = 10 m/s

18. Alta High AP Physics Section 9.10 Elastic Collisions  Elastic Collisions can be described mathematically by the following: Conservation of Momentum m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ Conservation of kinetic energy v 1 - v 2 = v 2 ’ - v 1 ’

19. Alta High AP Physics Sample Problem A ball with a mass of 0.5 kg rolling at a velocity of 10 m/s, strikes a second ball with a mass of 0.75 kg that is at rest in an elastic collision. What is the velocity of each ball after the collision?

20. Alta High AP Physics Sample Problem Solution m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ (0.5 kg)(10 m/s) + (0.75 kg)(0 m/s) = (0.5 kg)(v 1 ’) + (0.75 kg)(v 2 ’) v 1 - v 2 = v 2 ’ - v 1 ’ 10 m/s - 0 m/s = v 2 ’ - v 1 ’ V 2 ’ = 10m/s + v 1 ’ 5 kg m/s = (0.5 kg)(v 1 ’) + 7.5 kg m/s + 0.75v 1 ’ -2.5 kg m/s = 1.25 kg(v 1 ’) V 1 ’ = -2 m/s V 2 ’ = 10 m/s + -2 m/s = 8 m/s

21. Alta High AP Physics Section 9.11 Two Dimensional Collisions When analyzing a collision in two dimensions, the key is to remember that momentum is conserved in each direction. ρ ix = ρ fx and ρ iy = ρ fy Remembering that since ρ = mv and v is a vector.

22. Alta High AP Physics Sample Problem A gas molecule having a speed of 322 m/s collides elastically with another molecule having the same mass which is initially at rest. After the collision, the first molecule move at an angle of 30° with respect to its initial direction. Find the velocity of each molecule after the collision and the angle made with the incident direction by the recoiling target (second) molecule.

23. Alta High AP Physics Solution m 1 v 1x + m 2 v 2x = m 1 v’ 1x + m 2 v’ 2x and m 1 v 1y + m 2 v 2y = m 1 v’ 1y + m 2 v’ 2y Since m 1 =m 2 mass cancels so you have 322 m/s + 0 m/s = v’ 1x + v’ 2x 0 m/s + 0 m/s = v’ 1y + v’ 2y θ β

24. Alta High AP Physics Solution (Continued) From previous work with vectors we remember: V’ 1x = V’ 1 cos θ and V’ 1y = V’ 1 sin θ V’ 2x = V’ 2 cos β and V’ 2y = V’ 2 sin β So the equations become: V 1x = V’ 1 cos θ + V’ 2 cos β -V’ 1 sin θ = V’ 2 sin β (negative sign denotes opposite direction) Rearranging the first equation we get: V 1x - V’ 1 cos θ = V’ 2 cos β

25. Alta High AP Physics Solution (Continued) Squaring both equations and adding them together you get: V 1x 2 -2V 1x V’ 1 cos θ + V’ 1 2 sin 2 θ + V’ 1 2 cos 2 θ = V’ 2 2 cos 2 β + V’ 2 2 sin 2 β Using a well know trig identity this becomes: V 1x 2 -2V 1x V’ 1 cos θ + V’ 1 2 = V’ 2 2

26. Alta High AP Physics Solution (Continued) Remember that this is an elastic collision where energy is conserved so: V 1 2 = V’ 1 2 + V’ 2 2 Combined with V 1x 2 -2V 1x V’ 1 cos θ + V’ 1 2 = V’ 2 2 And the fact that V 1x = v 1 (since there was no velocity in the Y direction initially)

27. Alta High AP Physics Solution (Continued) V’ 2 2 +V’ 1 2 -2V 1 V’ 1 cos θ + V’ 1 2 = V’ 2 2 2V’ 1 2 = 2V 1 V’ 1 cos θ V’ 1 = V 1 cos θ = (322 m/s)(cos 30°)= 279 m/s Remember: V 1 2 = V’ 1 2 + V’ 2 2 so: (322 m/s) 2 = (279 m/s) 2 + V’ 2 2 V’ 2 = 161 m/s

28. Alta High AP Physics Solution (continued) Finally, since: V’ 1 sin θ = V’ 2 sin β Sin β = V’ 1 sin θ/V’ 2 Sin β = 279 m/s sin 30°/161 m/s Β = 60°

29. Alta High AP Physics Momentum and Energy Combined Momentum is often combined in problems with energy. One of the most common problems used to illustrate this is the ballistic pendulum problem. Ballistics is the study of projectiles and a ballistic pendulum is often used to measure the speed of a bullet. This is done by firing the bullet into a wooden block suspended on a string (hence the pendulum) and using the change in the pendulum’s height to back calculate the bullet’s speed.

30. Alta High AP Physics Sample Problem A bullet with a mass of 14 grams is fired into a suspended wooden block with a mass of 2 kg. If the block travels to a maximum height of 0.75 meters above its rest position, what was the velocity of the bullet when it was fired?

31. Alta High AP Physics Sample Problem Solution In order to solve this problem you work backward. At the top of the arc the pendulum-bullet combination has only potential energy. At the bottom of the arc, immediately after the bullet is embedded in the block the combination has only kinetic energy. So: ½ (m 1 + m 2 )v’ 2 = (m 1 + m 2 )gh v’ = SQRT (2gh) = SQRT(2 x 9.8 m/s 2 x 0.75 m) = 3.83 m/s Since the bullet embedding itself into the block is an inelastic collision the equation: m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v’ applies, hence (0.014 kg)(v 1 ) + (2 kg)(0 m/s) = (2.014 kg)(3.83 m/s) V 1 = 551.5 m/s

32. Alta High AP Physics Problem Types Linear momentum of a single object Impulse Collisions  Elastic Conservation of Momentum and Kinetic Energy  Inelastic Conservation of Momentum but not Kinetic Energy  Locked Bumpers, explosions Two dimensional Collisions  Energy Momentum Combinations