2. Physics 207: Lecture 13, Pg 1 Lecture 13 Goals: Assignments: l HW5, due tomorrow l For Wednesday, Read all of Chapter 10 Chapter 9 Chapter 9  Employ conservation of momentum in 1 D & 2D  Examine forces over time (aka Impulse) Chapter 10 Chapter 10  Understand the relationship between motion and energy

3. Physics 207: Lecture 13, Pg 2 Momentum is conserved if no external force

4. Physics 207: Lecture 13, Pg 3 Inelastic collision in 1-D: Example 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 final speed V. l In terms of m, M, and V : What is the momentum of the bullet with speed v ? v V beforeafter x

5. Physics 207: Lecture 13, Pg 4 Inelastic collision in 1-D: Example What is the momentum of the bullet with speed v ?  Key question: Is there a net external x-dir force ?  If not, then momentum in the x-direction is conserved! v V beforeafter x P Before P After

6. Physics 207: Lecture 13, Pg 5 Momentum is a vector

7. Physics 207: Lecture 13, Pg 6 A perfectly inelastic collision in 2-D l Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). v1v1 v2v2 before m1m1 m2m2 l If no external force momentum is conserved. l Momentum is a vector so p x, p y and p z V after m 1 + m 2 

8. Physics 207: Lecture 13, Pg 7 A perfectly inelastic collision in 2-D v1v1 v2v2 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   p1p1 p2p2 p

9. Physics 207: Lecture 13, Pg 8 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?

10. Physics 207: Lecture 13, Pg 9 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

11. Physics 207: Lecture 13, Pg 10 Exercise Momentum is a Vector (!) quantity 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 °

12. Physics 207: Lecture 13, Pg 11 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 i j x y 30 °

13. Physics 207: Lecture 13, Pg 12 Elastic Collisions l Elastic means that the objects do not stick. l There are many more possible outcomes but, if no external force, then momentum will always be conserved l Start with a 1-D problem. BeforeAfter

14. Physics 207: Lecture 13, Pg 13 Billiards l Consider the case where one ball is initially at rest. p a  pbpb F Pa Pa  before after The final direction of the red ball will depend on where the balls hit.

15. Physics 207: Lecture 13, Pg 14 Billiards: Without external forces, conservation of momentum (and energy Ch. 10 & 11) 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 after  ppbppb F P after  before after

16. Physics 207: Lecture 13, Pg 15 Home Exercise 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?

17. Physics 207: Lecture 13, Pg 16 Home exercise Inelastic Collision in 1-D v f v f =? finally m v = 0 ice M = 2m V0V0V0V0 (no friction) initially 2V o /3 = 20 m/s

18. Physics 207: Lecture 13, Pg 17 Explosions: A collision in reverse l A two piece assembly is hanging vertically at rest at the end of a 2.0 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? Before After

19. Physics 207: Lecture 13, Pg 18 Explosions: A collision in reverse l A two piece assembly is hanging vertically at rest at the end of a 2.0 m long massless string. The mass of the two pieces are 60 and 20 kg respectively. Suddenly you observe that the 10 kg 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 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

20. Physics 207: Lecture 13, Pg 19 Explosions: A collision in reverse l A two piece assembly is hanging vertically at rest at the end of a 20.0 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 y (radial) = M V 2 /r = T – Mg T = Mg + MV 2 /r = 600 N + 60x(10) 2 /20 N = 900 N Before After

21. Physics 207: Lecture 13, Pg 20 Impulse (A variable force applied for a given time) l Gravity: At small displacements a “constant” force Springs often provide a linear force (- k  x) towards its equilibrium position (Chapter 10) l Collisions often involve a varying force F(t): 0  maximum  0 l We can plot force vs time for a typical collision. The impulse, J, of the force is a vector defined as the integral of the force during the time of the collision.

22. Physics 207: Lecture 13, Pg 21 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 J = area under this curve ! (Transfer of momentum !) Impulse has units of Newton-seconds

23. Physics 207: Lecture 13, Pg 22 Force and Impulse J l Two different collisions can have the same impulse since J 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

24. Physics 207: Lecture 13, Pg 23 Average Force and Impulse tt F t F t tt  t big, F av small  t small, F av big F av

25. Physics 207: Lecture 13, Pg 24 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

26. Physics 207: Lecture 13, Pg 25 Boxing: Use Momentum and Impulse to estimate g “force”

27. Physics 207: Lecture 13, Pg 26 Back of the envelope calculation (1) m arm ~ 7 kg (2) v arm ~7 m/s (3) Impact time  t ~ 0.01 s Question: Are these reasonable?  Impulse J =  p ~ m arm v arm ~ 49 kg m/s  F avg ~ J/  t ~ 4900 N (1) m head ~ 7 kg  a head = F / m head ~ 700 m/s 2 ~ 70 g ! l Enough to cause unconsciousness ~ 40% of a fatal blow l Only a rough estimate!

28. Physics 207: Lecture 13, Pg 27 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 = J 0,1 + J 1,2 + J 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

29. Physics 207: Lecture 13, Pg 28 Ch. 10 : Kinetic & Potential energies l Kinetic energy, K = ½ mv 2, is defined to be the large scale collective motion of one or a set of masses l Potential energy, U, is defined to be the “hidden” energy in an object which, in principle, can be converted back to kinetic energy l Mechanical energy, E Mech, is defined to be the sum of U and K.

30. Physics 207: Lecture 13, Pg 29 Lecture 13 Assignment: l HW6 up soon l For Wednesday: Read all of chapter 10