1. PHY 151: Lecture 9B 9.5 Collisions in Two Dimensions 9.6 The Center of Mass 9.7 Systems of Many Particles 9.8 Deformable Systems 9.9 Rocket Propulsion

2. PHY 151: Lecture 9B Linear Momentum and Collisions 9.5 Collisions in Two Dimensions

3. Two-Dimensional Collisions The momentum is conserved in all directions Use subscripts for –Identifying the object –Indicating initial or final values –The velocity components If the collision is elastic, use conservation of kinetic energy as a second equation –Remember, the simpler equation can only be used for one-dimensional situations

4. Two-Dimensional Collision, example - 1 Particle 1 is moving at velocity and particle 2 is at rest. In the x-direction, the initial momentum is m 1 v 1i In the y-direction, the initial momentum is 0

5. Two-Dimensional Collision, example - 2 After the collision, the momentum in the x-direction is m 1 v 1f cos  + m 2 v 2f cos  After the collision, the momentum in the y-direction is m 1 v 1f sin  - m 2 v 2f sin  –The negative sign is due to the component of the velocity being downward If the collision is elastic, apply the kinetic energy equation This is an example of a glancing collision

6. Two-Dimensional Collision Example A 1500 kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2500 kg truck traveling north Find the direction and magnitude of the velocity of the wreckage after the collision, assuming the vehicles stick together after the collision  p xi =  p xf m 1 v 1i = (m 1 + m 2 )v f cos   p yi =  p yf m 2 v 2i = (m 1 + m 2 )v f sin  m 2 v 2i /m 1 v 1i = sin  /cos  = tan   = tan -1 (m 2 v i2 /m 1 v 1i )  = tan -1 [(2500)(20)/(1500)(25)] = 53.1 0 v f = m 2 v 2i /(m 1 +m 2 )sin  v f = (2500)(20)/(1500+2500)sin53.1 = 15.6 m/s

7. PHY 151: Lecture 9B Linear Momentum and Collisions 9.6 Center of Mass

8. The Center of Mass There is a special point in a system or object, called the center of mass, that moves as if all of the mass of the system is concentrated at that point The system will move as if an external force were applied to a single particle of mass M located at the center of mass –M is the total mass of the system This behavior is independent of other motion, such as rotation or vibration, or deformation of the system –This is the particle model

9. Center of Mass, Coordinates The coordinates of the center of mass are –M is the total mass of the system.

10. Center of Mass, Extended Object Similar analysis can be done for an extended object Consider the extended object as a system containing a large number of small mass elements. Since separation between the elements is very small, it can be considered to have a constant mass distribution

11. Center of Mass, position The center of mass in three dimensions can be located by its position vector, –For a system of particles, is the position of the i th particle, defined by –For an extended object,

12. Center of Mass, Symmetric Object The center of mass of any symmetric object of uniform density lies on an axis of symmetry and on any plane of symmetry

13. Center of Gravity Each small mass element of an extended object is acted upon by the gravitational force The net effect of all these forces is equivalent to the effect of a single force acting through a point called the center of gravity –If is constant over the mass distribution, the center of gravity coincides with the center of mass

14. Finding Center of Gravity, Irregularly Shaped Object Suspend the object from one point Then, suspend from another point The intersection of the resulting lines is the center of gravity and half way through the thickness of the wrench

15. Center of Mass, Rod Find the center of mass of a rod of mass M and length L Rod has a uniform mass per unit length Location is on the x-axis y CM = z CM = 0  = M/L linear mass density Mass of length dx is dm = dx

16. PHY 151: Lecture 9B Linear Momentum and Collisions 9.7 Systems of Particles

17. Motion of a System of Particles Assume the total mass, M, of the system remains constant We can describe the motion of the system in terms of the velocity and acceleration of the center of mass of the system We can also describe the momentum of the system and Newton’s Second Law for the system

18. Velocity and Momentum of a System of Particles The velocity of the center of mass of a system of particles is The momentum can be expressed as The total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass

19. Acceleration and Force in a System of Particles The acceleration of the center of mass can be found by differentiating the velocity with respect to time The acceleration can be related to a force If we sum over all the internal force vectors, they cancel in pairs and the net force on the system is caused only by the external forces

20. Newton’s Second Law for a System of Particles Since the only forces are external, the net external force equals the total mass of the system multiplied by the acceleration of the center of mass: The center of mass of a system of particles of combined mass M moves like an equivalent particle of mass M would move under the influence of the net external force on the system

21. Impulse and Momentum of System of Particles The impulse imparted to the system by external forces is The total linear momentum of a system of particles is conserved if no net external force is acting on the system For an isolated system of particles, both the total momentum and the velocity of the center of mass are constant in time

22. Motion of the Center of Mass, Example A projectile is fired into the air and suddenly explodes With no explosion, the projectile would follow the dotted line After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion

23. PHY 151: Lecture 9B Linear Momentum and Collisions 9.8 Deformable Systems

24. Deformable Systems To analyze the motion of a deformable system, use Conservation of Energy and the Impulse-Momentum Theorem –If the force is constant, the integral can be easily evaluated

25. Deformable System (Spring) Example - 1 Two blocks are at rest on a frictionless, level table Both blocks have the same mass m They are connected by a spring of negligible mass The separation distance of the blocks when the spring is relaxed is L During a time interval  t, a constant force of magnitude F is applied horizontally to the left block, moving it through a distance x 1 During this time interval, the right block moves through a distance x 2 At the end of this time interval, the force F is removed (a) Find the resulting speed v CM of the center of mass of the system

26. Deformable System (Spring) Example - 2 Apply impulse-momentum theorem to the system of two blocks F  t = p f – p i = (2m)(v CM – 0) = 2mv CM During the time interval  t, the center of mass of the system moves a distance ½ (x 1 + x 2 ) ½ (x 1 + x 2 ) = v CM,avg  t  t = ½ (x 1 + x 2 ) / v CM,avg Average velocity of the center of mass is the average of the initial velocity, which is zero, and the final velocity v CM  t = ½ (x 1 +x 2 )/ ½ (0+v CM ) = (x 1 +x 2 )/v CM F(x 1 +x 2 )/v CM = 2mv CM vCM = sqrt[F(x 1 + x 2 )/2m]

27. Deformable System (Spring) Example - 3 (b) Find the total energy of the system associated with vibration relative to its center of mass after the force F is removed The vibrational energy is all the energy of the system other than the kinetic energy associated with the translational motion of the center of mass Apply conservation of energy K = K CM + K vib K vib is the kinetic energy of the blocks relative to the center of mass due to their vibration The potential energy of the system is U vib, which is the potential energy stored in the spring when the separation of the blocks is some value other than L

28. Deformable System (Spring) Example - 4  K CM +  K vib +  U vib = W K vib + U vib = E vib  K CM +  E vib = W Initial values of the kinetic energy of the center of mass and the vibrational energy of the system are zero Work on the system by F is Fx 1  K CM +  E vib = W = Fx 1 E vib = Fx 1 – K CM = Fx 1 – ½(2m)v CM 2 E vib = F(x 1 – x 2 )/2

29. PHY 151: Lecture 9B Linear Momentum and Collisions 9.9 Rocket Propulsion

30. Rocket Propulsion - 1 When ordinary vehicles are propelled, the driving force for the motion is friction –The car is modeled as an non-isolated system in terms of momentum –An impulse is applied to the car from the roadway, and the result is a change in the momentum of the car The operation of a rocket depends upon the law of conservation of linear momentum as applied to an isolated system, where the system is the rocket plus its ejected fuel As the rocket moves in free space, its linear momentum changes when some of its mass is ejected in the form of exhaust gases –Because the gases are given momentum when they are ejected out of the engine, the rocket receives a compensating momentum in the opposite direction –In free space, the center of mass of the system moves uniformly

31. Rocket Propulsion - 2 The initial mass of the rocket plus all its fuel is M +  m at time t i and speed v The initial momentum of the system is At some time t +  t, the rocket’s mass has been reduced to M and an amount of fuel,  m has been ejected The rocket’s speed has increased by  v p i = p f (M+  m)v=M(v+  v)+  m(v–v e )

32. Rocket Propulsion - 3 M  v = v e  m Mdv = v e dm = -v e dM The increase in rocket speed is proportional to the speed of the escape gases (ve) –So, the exhaust speed should be very high The increase in rocket speed is also proportional to the natural log of the ratio M i /M f –So, the ratio should be as high as possible, meaning the mass of the rocket should be as small as possible and it should carry as much fuel as possible

33. Thrust The thrust on the rocket is the force exerted on it by the ejected exhaust gases The thrust increases as the exhaust speed increases The thrust increases as the rate of change of mass increases –The rate of change of the mass is called the burn rate