2. Physics 207: Lecture 11, Pg 1 Lecture 11 Goals: Assignment: l Read through Chapter 10 l MP HW5, due Wednesday 3/3 Chapter 9: Momentum & Impulse Chapter 9: Momentum & Impulse  Understand what momentum is and how it relates to forces  Employ momentum conservation principles  In problems with 1D and 2D Collisions  In problems having an impulse (Force vs. time)  Chapter 8: Use models with free fall

3. Physics 207: Lecture 11, Pg 2 Problem 7.34 Hint Suggested Steps l Two independent free body diagrams are necessary l Draw in the forces on the top and bottom blocks l Top Block  Forces: 1. normal to bottom block 2. weight 3. rope tension and 4. friction with bottom block (model with sliding) l Bottom Block  Forces: 1. normal to bottom surface 2. normal to top block interface 3. rope tension (to the left) 4. weight (2 kg) 5. friction with top block 6. friction with surface 7. 20 N Use Newton's 3rd Law to deal with the force pairs (horizontal & vertical) between the top and bottom block.

4. Physics 207: Lecture 11, Pg 3 Locomotion: how fast can a biped walk?

5. Physics 207: Lecture 11, Pg 4 How fast can a biped walk? What about weight? (a)A heavier person of equal height and proportions can walk faster than a lighter person (b)A lighter person of equal height and proportions can walk faster than a heavier person (c)To first order, size doesn’t matter

6. Physics 207: Lecture 11, Pg 5 How fast can a biped walk? What about height? (a)A taller person of equal weight and proportions can walk faster than a shorter person (b)A shorter person of equal weight and proportions can walk faster than a taller person (c)To first order, height doesn’t matter

7. Physics 207: Lecture 11, Pg 6 How fast can a biped walk? What can we say about the walker’s acceleration if there is UCM (a smooth walker) ? Acceleration is radial ! So where does it, a r, come from? (i.e., what external forces are on the walker?) 1. Weight of walker, downwards 2. Friction with the ground, sideways

8. Physics 207: Lecture 11, Pg 7 Orbiting satellites v T = (gr) ½

9. Physics 207: Lecture 11, Pg 8 Geostationary orbit

10. Physics 207: Lecture 11, Pg 9 Geostationary orbit l The radius of the Earth is ~6000 km but at 36000 km you are ~42000 km from the center of the earth. l F gravity is proportional to r -2 and so little g is now ~10 m/s 2 / 50 n v T = (0.20 * 42000000) ½ m/s = 3000 m/s At 3000 m/s, period T = 2  r / v T = 2  42000000 / 3000 sec = = 90000 sec = 90000 s/ 3600 s/hr = 24 hrs l Orbit affected by the moon and also the Earth’s mass is inhomogeneous (not perfectly geostationary) l Great for communication satellites (1 st pointed out by Arthur C. Clarke)

11. Physics 207: Lecture 11, Pg 10 Impulse & Linear Momentum l Transition from forces to conservation laws Newton’s Laws  Conservation Laws Conservation Laws  Newton’s Laws They are different faces of the same physics NOTE: We have studied “impulse” and “momentum” but we have not explicitly named them as such Conservation of momentum is far more general than conservation of mechanical energy

12. Physics 207: Lecture 11, Pg 11 Collisions are a fact of life

13. Physics 207: Lecture 11, Pg 12 Forces vs time (and space, Ch. 10) l Underlying any “new” concept in Chapter 9 is (1) A net force changes velocity (either magnitude or direction) (2) For any action there is an equal and opposite reaction l If we emphasize Newton’s 3 rd Law and look at the changes with time then this leads to the Conservation of Momentum Principle

14. Physics 207: Lecture 11, Pg 13 Example 1 A 2 kg block, initially at rest on frictionless horizontal surface, is acted on by a 10 N horizontal force for 2 seconds (in 1D). What is the final velocity? l F is to the positive & F = ma thus a = F/m = 5 m/s 2 v = v 0 + a  t = 0 m/s + 2 x 5 m/s = 10 m/s (+ direction) Notice: v - v 0 = a  t  m (v - v 0 ) = ma  t  m  v = F  t If the mass had been 4 kg … now what final velocity? - + F (N) 10 0 2 time (s)

15. Physics 207: Lecture 11, Pg 14 Twice the mass l Same force l Same time Half the acceleration (a = F / m ’ ) l Half the velocity ! ( 5 m/s ) F (N) 10 0 2 Time (sec) Before

16. Physics 207: Lecture 11, Pg 15 Example 1 l Notice that the final velocity in this case is inversely proportional to the mass (i.e., if thrice the mass….one-third the velocity). l Here, mass times the velocity always gives the same value. (Always 20 kg m/s.) F (N) 10 0 2 Time (sec) Area under curve is still the same ! Force x change in time = mass x change in velocity

17. Physics 207: Lecture 11, Pg 16 Example 1 l There many situations in which the sum of the products “mass times velocity” is constant over time l To each product we assign the name, “momentum” and associate it with a conservation law. (Units: kg m/s or N s) l A force applied for a certain period of time can be graphed and the area under the curve is the “impulse” F (N) 10 0 2 Time (sec) Area under curve : “impulse” With: m  v = F avg  t

18. Physics 207: Lecture 11, Pg 17 Force curves are usually a bit different in the real world

19. Physics 207: Lecture 11, Pg 18 Example 1 with Action-Reaction l Now the 10 N force from before is applied by person A on person B while standing on a frictionless surface l For the force of A on B there is an equal and opposite force of B on A F (N) 10 0 2 Time (sec) 0 -10 A on B B on A M A x  V A = Area of top curve M B x  V B = Area of bottom curve Area (top) + Area (bottom) = 0

20. Physics 207: Lecture 11, Pg 19 Example 1 with Action-Reaction M A  V A + M B  V B = 0 M A [V A (final) - V A (initial) ] + M B [V B (final) - V B (initial) ] = 0 Rearranging terms M A V A (final) +M B V B (final) = M A V A (initial) +M B V B (initial) which is constant regardless of M or  V (Remember: frictionless surface)

21. Physics 207: Lecture 11, Pg 20 Example 1 with Action-Reaction M A V A (final) +M B V B (final) = M A V A (initial) +M B V B (initial) which is constant regardless of M or  V Define MV to be the “momentum” and this is conserved in a system if and only if the system is not acted on by a net external force (choosing the system is key) Conservation of momentum is a special case of applying Newton’s Laws

22. Physics 207: Lecture 11, Pg 21 Applications of Momentum Conservation 234 Th 238 U Alpha Decay 4 He v1v1 v2v2 Explosions Radioactive decay: Collisions

23. Physics 207: Lecture 11, Pg 22 Impulse & Linear Momentum l Newton’s 2 nd Law: Fa F = ma l This is the most general statement of Newton’s 2 nd Law pv p ≡ mv pv (p is a vector since v is a vector) So p x = mv x and so on (y and z directions) l Definition: For a single particle, the momentum p is defined as:

24. Physics 207: Lecture 11, Pg 23 Momentum Conservation l Momentum conservation (recasts Newton’s 2 nd Law when F = 0) is an important principle l It is a vector expression (P x, P y and P z ).  And applies to any situation in which there is NO net external force applied (in terms of the x, y & z axes).

25. Physics 207: Lecture 11, Pg 24 Momentum Conservation l Many problems can be addressed through momentum conservation even if other physical quantities (e.g. mechanical energy) are not conserved l Momentum is a vector quantity and we can independently assess its conservation in the x, y and z directions (e.g., net forces in the z direction do not affect the momentum of the x & y directions)

26. Physics 207: Lecture 11, Pg 25 Exercise 1 Momentum is a Vector (!) quantity A. Yes B. No C. Yes & No D. Too little information given l A block slides down a frictionless ramp and then falls and lands in a cart which then rolls horizontally without friction l In regards to the block landing in the cart is momentum conserved?

27. Physics 207: Lecture 11, Pg 26 Exercise 1 Momentum is a Vector (!) quantity Let a 2 kg block start at rest on a 30 ° incline and slide vertically a distance 5.0 m and fall a distance 7.5 m into the 10 kg cart What is the final velocity of the cart? l x-direction: No net force so P x is conserved. l y-direction: Net force, interaction with the ground so depending on the system (i.e., do you include the Earth?) P y is not conserved (system is block and cart only) 5.0 m 30 ° 7.5 m 10 kg 2 kg

28. Physics 207: Lecture 11, Pg 27 Inelastic collision in 1-D: Example 2 l A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, and V : What is the momentum of the bullet with speed v ? v V beforeafter x

29. Physics 207: Lecture 11, Pg 28 Inelastic collision in 1-D: Example 2 What is the momentum of the bullet with speed v ?  Key question: Is x-momentum conserved ? v V beforeafter x BeforeAfter

30. Physics 207: Lecture 11, Pg 29 Example 2 Inelastic Collision in 1-D with numbers ice (no friction) Do not try this at home! Before: A 4000 kg bus, twice the mass of the car, moving at 30 m/s impacts the car at rest. What is the final speed after impact if they move together?

31. Physics 207: Lecture 11, Pg 30 Exercise 2 Momentum Conservation A. Box 1 B. Box 2 C. same l Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. l The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks.  Which box ends up moving fastest ? 1 2