1. Chapter 9 Linear Momentum and Collisions

2. Momentum 2 Linear momentum of object: mass × velocity Magnitude: p=mv, Direction: same as velocity Quantity of motion, ability to acting force Newton’s second law of motion (original edition) The changing rate of momentum of an object is equal to the net force applied on it.

3. Impulse 3 Change in momentum is the integral of the net force over the time, the integral is called impulse Integral form of Newton’s second law momentum ~ impulse energy ~ work Eq. ① is called impulse-momentum theorem ① 1) any force3)2) vector

4. Projectile problem 4 Example1: Object m is fired with v at angle 45°to the horizontal, no air friction. Determine: a) Initial momentum b) Final momentum c) Impulse during the motion Solution: a) x y v 45° b) same magnitude different direction c)

5. Conical pendulum 5 Example2: Known: m, L, , v, T. Determine the impulse of tension and gravity respectively during one-half period. O m L Solution: FTFT G So: F T is not a constant force! Direction of these vectors

6. Collisions and impulsive force 6 Concept of momentum and impulse is useful when dealing with collision problems Key points: 1) Time interval is usually very short 2) Interaction of collision, or impulsive force is changing fast and may become very large The effect is shown by its impulse 3) Impulses of other forces can be ignored

7. Bend your knees 7 Example3: A 70-kg person jumps from h=3.0m and lands on firm ground. a) Calculate the impulse experienced; b) Estimate the average force by the ground if he is stiff- legged (s=1.0cm); c) average force with bent legs (s=50cm). Solution: a) Landing speed Impulse b) c) impulsive force x

8. Washing a car 8 Example4: Water is jetting at rate of R=1.5kg/s with v=20m/s, and is stopped by a car (no splashing back). What is the force exerted on the car? Solution: In a short time interval dt, How much water hits on the car: dm=Rdt Change in momentum: dp=vdm=Rvdt Collision: ignore gravity, only force by car What if the water splashes back?

9. Falling rope 9 Example5: One end of a hanged rope just touch the table, then it is released. Prove: Normal force acting on table is always 3 times of weight of that part already on the table. Solution: Linear density λ, falling height h Falling speed m Mass of part already on table m = λ h In time interval dt,dm = λ · vdt Impulse F·dt = dm·v = λ ·v 2 dt F = λ ·v 2 = λ ·2gh =2mg F mg dm v

10. Momentum of a system 10 Consider a system of n interacting particles f 1n f n1 f i1 f 1i FiFi F1F1 mimi m1m1 FnFn f ni f in mnmn Total momentum of the system Change in total momentum Internal forces are always in pairs Only external forces can change the total momentum!

11. Conservation of momentum 11 If the net external force is 0, then This is the law of conservation of momentum: When the net external forces on a system is zero, the total momentum remains constant. Or: The total momentum of an isolated system of bodies remains constant. 1) isolated2) system3) ignoring forces

12. 12 Rifle recoiling Example6: Calculate the recoil speed of a 5kg rifle that shoots a 10g bullet at v=700m/s. Solution: Total momentum is conserved The rifle moves back and applies a recoil force

13. 13 Billiard ball collision Example7: A billiard ball with v in the +x direction strikes an identical resting ball. Directions after collision are shown in figure, what are the speeds? Solution: conservation of momentum 45  x o y component form: kinetic energy is also conserved

14. 14 30  m v Conservation in component form Example8: Bullet (20g) hits into hanging ball (980g) with v=400m/s. Determine the speed after collision. M Solution: No net force in horizontal Horizontal component P x is conserved T (M+m)g If, but → x component of momentum is conserved

15. 15 Homework A gun carrier M moves on a frictionless incline, its speed reduces from v to 0 after shooting a canon-ball m in the horizontal direction. Is the total momentum of system (M and m) conserved in this process, and why? Find out the speed of canon-ball. M  v mv’

16. 16 Distance traveled Example9: Two objects start from rest, what is the distance traveled by m 2 when m 1 reaches the ground? No friction. a b m1m1 m2m2  x o Solution: conservation of momentum in horizontal where So distance traveled by m 2

17. 17 Challenging question Question: Someone sits on the top of an ice half sphere on the ground, and then slips down. At what angle  will he leave the sphere? No friction. top Leave the half sphere: N = 0 Thinking: ? Relative speed & inertial frame Conservation of momentum and mechanical energy Velocity transformation m M

18. Elastic collisions 18 Conservation of momentum (1-dimension) It is called an elastic collision If total kinetic energy is conserved in a collision Ideal model in macro world Relative speed: equal but opposite head-on collision

19. 19 Final speeds Special cases: 1) Equal masses 2) Target m 2 at rest

20. 20 Baseball batting Example10: A baseball with speed 40m/s is hit by a bat with speed 30m/s in the opposite direction. Determine the speed of baseball if the collision is elastic and m bat >> m ball. Solution: final speed of ball Or in the frame of bat: The ball moves in 70m/s, and rebounds in the same speed Transform to the frame of ground, it’s 100m/s.

21. 21 Slingshot effect Example11: Spacecraft Voyager II approaches the Jupiter, it rounds the planet and departs in the opposite direction. What is the speed after this slingshot encounter? ( v S =10.4km/s, v J = - 9.6km/s ) Solution: Like an elastic collision speed after slingshot:

22. Inelastic collisions 22 It is called an inelastic collision If kinetic energy is not conserved in a collision Restitution coefficient e=1: elastic collision e<1: inelastic collision e=0: completely inelastic collision How energy transform? heat and sound …

23. Completely inelastic collision 23 Two objects stick together in collision: v’ 1 = v’ 2 If target m 2 is initially at rest Kinetic energy loss Depends on the mass ratio of two objects Almost no energy loss Hammer hit on nail … Maximum of energy loss Throw egg against rock …

24. Hit by bullet 24 Example12: m 1 =0.49kg and m 2 = 0.50kg are related by a massless spring (k=100N/m). The system is initially at rest on frictionless horizontal plane, then a bullet m 3 =0.01kg hits into m 1 with v =100m/s. Determine the maximum stretch. m1 m1 m2 m2 v Solution: Completely inelastic collision between m 1 and m 3 Conservation of momentum and mechanical energy

25. Collisions in 2-dimensions 25 x o y 22 11 Target initially at rest Conservation of momentum Discuss an elastic collision in 2 dimensions Conservation of kinetic energy It can be solved if one of variables is known

26. Equal-mass case 26 If it occurs between two objects of equal mass This is shown to be true in many experiments x o y 22 11 Angle between final velocities is always 90 ° You can try to prove it (Problem 57, Page 229)

27. Center of mass 27 Real body (instead of particle): general motion As a particle with same mass and same net force..... c c c c c It is called center of mass (CM) One point represents motion of object (system) General motion = Translational motion of CM + Other motion about CM (rotation, vibration…) Different from center of gravity (CG)

28. CM of several particles 28 Definition of CM: by weighted average 1-dimension or more-dimensions Example13: Determine the CM of stones, where the mass of black stone is twice as much as white stone. Solution: O x y

29. CM of continuous object 29 Continuous object: summations become integrals or Example14: CM of a thin rod with varying density. The total mass Solution: Choose a segment dx ox L x dx CM:

30. CM of more objects 30 Example15: CM of uniform thin right triangle. Solution: Choose a segment dm of the triangle xyO ab dx x Similarly we have Comparing with Example14 What about other objects?

31. CM and translational motion 31 Total momentum of a system is equal to the product of the total mass M and velocity of CM. The sum of all the forces acting on the system is equal to the total mass M times the acceleration of CM. Or: The center of mass of a system with total mass M moves like a single particle of mass M acted by the same net external force.

32. Man on boat 32 Example16: A man (m) moves from one side of a boat (M, L) to another side, determine the distance traveled by the boat. Initially at rest, no resistance. Solution: No net external force, CM stays at rest s y x o

33. Two stage rocket 33 Example17: A rocket is separated into two parts of equal mass at its highest point. Part A falls vertically to Earth from rest, where does part B lands? x y d A B ? Solution: CM still moves along the curve It lands at x=2d So part B will land at x=3d If part A had been given a kick up or down?

34. *Rocket propulsion 34 System of variable mass & ignore external forces Gas dm is jetting with relative speed v r dmvrvr m v Speed of rocket changes from v to v+dv Tsiolkovsky equation