1. Physics 201: Lecture 15, Pg 1 Lecture 15 l Goals  Employ conservation of momentum in 1 D & 2D  Introduce Momentum and Impulse  Compare Force vs time to Force vs distance  Introduce Center-of-Mass Note: 2 nd Exam, Monday, March 19 th, 7:15 to 8:45 PM

2. Physics 201: Lecture 15, Pg 2 Comments on Momentum Conservation l More general than conservation of mechanical energy l Momentum Conservation occurs in systems with no net external forces (as a vector quantity)

3. Physics 201: Lecture 15, Pg 3 Explosions: A collision in reverse l A two piece assembly is hanging vertically at rest at the end of a 20 m long massless string. The mass of the two pieces are 60 and 20 kg respectively. Suddenly you observe that the 20 kg is ejected horizontally at 30 m/s. The time of the “explosion” is short compared to the swing of the string. l Does the tension in the string increase or decrease after the explosion? l If the time of the explosion is short then momentum is conserved in the x-direction because there is no net x force. This is not true of the y-direction but this is what we are interested in. Before After

4. Physics 201: Lecture 15, Pg 4 Explosions: A collision in reverse l A two piece assembly is hanging vertically at rest at the end of a 20 m long massless string. The mass of the two pieces are 60 and 20 kg respectively. Suddenly you observe that the 20 kg mass is ejected horizontally at 30 m/s. l Decipher the physics: 1. The green ball recoils in the –x direction (3 rd Law) and, because there is no net external force in the x-direction the x-momentum is conserved. 2. The motion of the green ball is constrained to a circular path…there must be centripetal (i.e., radial acceleration) Before After

5. Physics 201: Lecture 15, Pg 5 Explosions: A collision in reverse l A two piece assembly is hanging vertically at rest at the end of a 20 m long massless string. The mass of the two pieces are 60 & 20 kg respectively. Suddenly you observe that the 20 kg mass is suddenly ejected horizontally at 30 m/s. l Cons. of x-momentum p x before = p x after = 0 = - M V + m v V = m v / M = 20*30/ 60 = 10 m/s T before = Weight = (60+20) x 10 N = 800 N  F y = m a cy = M V 2 /r = T – Mg T = Mg + MV 2 /r = 600 N + 60x(10) 2 /20 N = 900 N After Before

6. Physics 201: Lecture 15, Pg 6 Exercise 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?

7. Physics 201: Lecture 15, Pg 7 Exercise 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

8. Physics 201: Lecture 15, Pg 8 Exercise Momentum is a Vector (!) quantity Initial Final P x : MV x + mv x = (M+m) V ’ x M 0 + mv x = (M+m) V ’ x V ’ x = m v x / (M + m) = 2 (8.7)/ 12 m/s V ’ x = 1.4 m/s l x-direction: No net force so P x is conserved l y-direction: v y of the cart + block will be zero and we can ignore v y of the block when it lands in the cart. 5.0 m 30 ° 7.5 m N mg 1) a i = g sin 30°  = 5 m/s 2 2) d = 5 m / sin 30° = ½ a i  t 2 10 m = 2.5 m/s 2  t 2 2s =  t v = a i  t = 10 m/s v x = v cos 30° = 8.7 m/s i j x y 30 °

9. Physics 201: Lecture 15, Pg 9 Impulse (A variable external force applied for a given time) l Collisions often involve a varying force F(t): 0  maximum  0 I l We can plot force vs time for a typical collision. The impulse, I, of the force is a vector defined as the integral of the force during the time of the collision. l The impulse measures momentum transfer

10. Physics 201: Lecture 15, Pg 10 Force and Impulse (A variable force applied for a given time) F l J a vector that reflects momentum transfer t titi tftf tt Impulse I = area under this curve ! (Transfer of momentum !) Impulse has units of Newton-seconds

11. Physics 201: Lecture 15, Pg 11 Force and Impulse l Two different collisions can have the same impulse since I depends only on the momentum transfer, NOT the nature of the collision. tt F t F t tt same area F  t big, F small F  t small, F big

12. Physics 201: Lecture 15, Pg 12 Average Force and Impulse tt F t F t tt  t big, F av small  t small, F av big F av

13. Physics 201: Lecture 15, Pg 13 Exercise Force & Impulse A. heavier B. lighter C. same D. can’t tell l Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second. Which box has the most momentum after the force acts ? FF light heavy

14. Physics 201: Lecture 15, Pg 14 Discussion Exercise l The only force acting on a 2.0 kg object moving along the x- axis. Notice that the plot is force vs time. l If the velocity v x is +2.0 m/s at 0 sec, what is v x at 4.0 s ?  p = m  v = Impulse m  v = I 0,1 + I 1,2 + I 2,4 m  v = (-8)1 N s + ½ (-8)1 N s + ½ 16(2) N s m  v = 4 N s  v = 2 m/s v x = 2 + 2 m/s = 4 m/s

15. Physics 201: Lecture 15, Pg 15 A perfectly inelastic collision in 2-D l Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). vv1vv1 vv2vv2 V beforeafter m1m1 m2m2 m 1 + m 2 l If no external force momentum is conserved. l Momentum is a vector so p x, p y and p z 

16. Physics 201: Lecture 15, Pg 16 A perfectly inelastic collision in 2-D vv1vv1 vv2vv2 V beforeafter m1m1 m2m2 m 1 + m 2 x-dir p x : m 1 v 1 = (m 1 + m 2 ) V cos  y-dir p y : m 2 v 2 = (m 1 + m 2 ) V sin  l If no external force momentum is conserved. l Momentum is a vector so p x, p y and p z are conseved 

17. Physics 201: Lecture 15, Pg 17 2D Elastic Collisions l Perfectly elastic means that the objects do not stick and, by stipulation, mechanical energy is conservsed. l There are many more possible outcomes but, if no external force, then momentum will always be conserved BeforeAfter

18. Physics 201: Lecture 15, Pg 18 Billiards l Consider the case where one ball is initially at rest. p p a  ppbppb F PPa PPa  before after The final direction of the red ball will depend on where the balls hit. v v cm

19. Physics 201: Lecture 15, Pg 19 Billiards: Without external forces, conservation of both momentum & mech. energy l Conservation of Momentum x-dir P x : m v before = m v after cos  + m V after cos  y-dir P y : 0 = m v after sin  + m V after sin  p p after  ppbppb F P P after  before after If the masses of the two balls are equal then there will always be a 90 ° angle between the paths of the outgoing balls

20. Physics 201: Lecture 15, Pg 20 Center of Mass l Most objects are not point-like but have a mass density and are often deformable. l So how does one account for this complexity in a straightforward way? Example l In football coaches often tell players attempting to tackle the ball carrier to look at their navel. l So why is this so?

21. Physics 201: Lecture 15, Pg 21 System of Particles: Center of Mass (CM) l If an object is not held then it will rotate about the center of mass. l Center of mass: Where the system is balanced !  Building a mobile is an exercise in finding centers of mass. m1m1 m2m2 + m1m1 m2m2 + mobile

22. Physics 201: Lecture 15, Pg 22 System of Particles: Center of Mass l How do we describe the “position” of a system made up of many parts ? l Define the Center of Mass (average position):  For a collection of N individual point-like particles whose masses and positions we know: (In this case, N = 2) y x r2r2 r1r1 m1m1 m2m2 R CM

23. Physics 201: Lecture 15, Pg 23 Momentum of the center-of-mass is just the total momentum l Notice l Impulse and momentum conservation applies to the center-of-mass

24. Physics 201: Lecture 15, Pg 24 Sample calculation: l Consider the following mass distribution: (24,0) (0,0) (12,12) m 2m m R CM = (12,6) X CM = (m x 0 + 2m x 12 + m x 24 )/4m meters Y CM = (m x 0 + 2m x 12 + m x 0 )/4m meters X CM = 12 meters Y CM = 6 meters

25. Physics 201: Lecture 15, Pg 25 A classic example l There is a disc of uniform mass and radius r. However there is a hole of radius a a distance b (along the x-axis) away from the center. l Where is the center of mass for this object?

26. Physics 201: Lecture 15, Pg 26 System of Particles: Center of Mass l For a continuous solid, convert sums to an integral. y x dm r where dm is an infinitesimal mass element (see text for an example).

27. Physics 201: Lecture 15, Pg 27 Recap l Thursday, Review for exam l For Tuesday, Read Chapter 10.1-10.5