1. Physics 101: Lecture 12, Pg 1 Physics 101: Lecture 12 Collisions and Explosions l Today’s lecture will cover Textbook Sections 7.5 - 7.8 Exam II

2. Physics 101: Lecture 12, Pg 2 Overview of Semester l Newton’s Laws   F = m a l Work-Energy   F = m a multiply both sides by d   W =  KE Energy is “conserved” è Useful when know Work done by forces l Impulse-Momentum   F = m a multiply both sides by  t   I =  p Momentum is “conserved” è Useful when know about EXTERNAL forces è Works in each direction independently 08

3. Physics 101: Lecture 12, Pg 3 Collisions “before” “after” m1m1 m2m2 m1m1 m2m2 Explosions “before” “after” M m1m1 m2m2 Draw “before”, “after” Define system so that F ext = 0 Set up axes Compute P total “before” Compute P total “after” Set them equal to each other Procedure 11

4. Physics 101: Lecture 12, Pg 4 ACT A railroad car is coasting along a horizontal track with speed V when it runs into and connects with a second identical railroad car, initially at rest. Assuming there is no friction between the cars and the rails, what is the speed of the two coupled cars after the collision? A. V B. V/2 C. V/4 D. 0 Demo with gliders  P initial =  P final M V = M V f + M V f V = 2V f V f = V/2 15

5. Physics 101: Lecture 12, Pg 5 ACT What physical quantities are conserved in the above collision? A. Only momentum is conserved B. Only total mechanical energy is conserved C. Both are conserved D. Neither are conserved CORRECT Mechanical Energy = Kinetic Energy + Potential E = ½ m v 2 + 0 K initial = ½ m v 2 K final = ½ m (v/2) 2 + ½ m (v/2) 2 = ¼ m v 2 20 Elastic Collisions: collisions that conserve mechanical energy Inelastic Collisions: collisions that do not conserve mechanical energy * Completely Inelastic Collisons: objects stick together

6. Physics 101: Lecture 12, Pg 6 Preflight 1 & 2 Is it possible for a system of two objects to have zero total momentum and zero total kinetic energy after colliding, if both objects were moving before the collision? 1. YES 2. NO If two pieces of chewed gum (gross, yes, but effective for example) of equal mass and velocity collided after moving from opposite directions, their total momentum would equal the difference between the initial momentum of the positive direction gum and the initial momentum of the negative direction gum (thus =0). And the kinetic energy would also be 0 when the two pieces of gum stopped. CORRECT Demo with gliders 22 I think you could have a zero total momentum, but having both zero total momentum and zero total kinetic energy after colliding is impossible.

7. Physics 101: Lecture 12, Pg 7 H LL LL m M A projectile of mass m moving horizontally with speed v strikes a stationary mass M suspended by strings of length L. Subsequently, m + M rise to a height of H. Given H, M and m what is the initial speed v of the projectile? M + m v V V=0 See I.E. 1 in homework Ballistic Pendulum demo 29 Collision Conserves Momentum 0+m v = (M+m) V After, Conserve Energy ½ (M+m) V 2 +0 = 0+(M+m) g H V = sqrt(2 g H) Combine:

8. Physics 101: Lecture 12, Pg 8 Explosions “before” “after” M m1m1 m2m2 Example: m 1 = M/3 m 2 = 2M/3 Which block has larger |momentum|? * Each has same |momentum| Which block has larger speed? * mv same for each  smaller mass has larger velocity Which block has larger kinetic energy? * KE = mv 2 /2 = m 2 v 2 /2m = p 2 /2m *  smaller mass has larger KE Is mechanical (kinetic) energy conserved? * NO!! v1v1 v2v2 35 A=1, B=2, C=same 0 = p 1 +p 2 p 1 = -p 2

9. Physics 101: Lecture 12, Pg 9 Collisions or Explosions in Two Dimensions y x before after P total,x and P total,y independently conserved P total,x,before = P total,x,after P total,y,before = P total,y,after 37

10. Physics 101: Lecture 12, Pg 10 Explosions ACT “before” M A B Which of these is possible? (Ignore friction and gravity) A B C =both D = Neither 42 “after” P x = 0 and P y = 0  P x = 0, but  P y > 0  P x = 0, and  P y = 0

11. Physics 101: Lecture 12, Pg 11 l Assuming è Collision is elastic (KE is conserved) è Balls have the same mass è One ball starts out at rest l Then the angle between the balls after the collision is 90 o ppfppf ppippi F PPfPPf beforeafter v v cm Shooting Pool... 43

12. Physics 101: Lecture 12, Pg 12 Center of Mass Center of Mass = Balance point 46 l Shown is a yummy doughnut. Where would you expect the center of mass of this breakfast of champions to be located? Center of Mass! i would expect it to be exactly in the center, but i bet thats not correct b/c physics is tricky like that. The center of mass will be nonexistent after I eat that yummy doughnut. The center is in a bag at Dunkin' Donuts labeled Munchkin

13. Physics 101: Lecture 12, Pg 13 Center of Mass Center of Mass of a system behaves in a SIMPLE way - moves like a point particle! - velocity of CM is unaffected by collision if F ext = 0 P tot = M tot V cm F ext  t =  P tot = M tot  V cm So if F ext  = 0 then V cm is constant Also: F ext = M tot a cm 48

14. Physics 101: Lecture 12, Pg 14 Summary Collisions and Explosions Draw “before”, “after” Define system so that F ext = 0 Set up axes Compute P total “before” Compute P total “after” Set them equal to each other Center of Mass (Balance Point) 50