2. 10/22/2015A.PH 105 PH 105-003/4 ----Monday, Oct. 8, 2007 Homework: PS8 due Wednesday: conservation of energy Chapter 9: momentum Review Wednesday & Friday: define p = mv Why? Then dp/dt=0 for isolated system  p =  F dt = impulse Elastic collision: KE conserved Completely inelastic collision: stick together (v 1f = v 2f = v f ) – solve C of Mom. eqn. for v f.

3. 10/22/2015A.PH 105 Elastic Collisions in 1D Equations: m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f (Cons. of p x ) ½m 1 v 1i 2 + ½ m 2 v 2i 2 = ½ m 1 v 1f 2 + ½ m 2 v 2f 2 (Conservation of E) Known: v 1i & v 2i Unknown: v 1f & v 2f – can solve 2 equations. But quadratic! (Wednesday, Oct. 10) Trick (only works for 1D elastic collisions): can show relative velocity reverses, (v 1f - v 2f ) = - (v 1i - v 2i ) Proof: factor  E 1 into  p 1...

4. Ungraded question (related to PS 7, Chapter 8) A mass m strikes a spring with spring constant k. There is a friction force of magnitude f =  K mg. In a process in which the spring moves a distance d, conservation of mechanical energy E = K + U can be written as A.E f = E i B.E f = E i + f d C.E f = E i - f d Answer is (c) -- the mass must lose energy to friction.

5. 10/22/2015A.PH 105 Example: Consider two equal masses m 1 = m 2 = m, m 1 moving to the right at velocity v and m 2 stationary. Find the velocities after an elastic collision. Solution: We know conservation of momentum, m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f and (ONLY because this is 1D elastic) relative velocity reverses, (v 1f - v 2f ) = - (v 1i - v 2i )..

6. Clicker question: A small mass moving at +8 m/s collides elastically with a stationary larger mass, which moves away at +2 m/s. What is the velocity of the smaller mass after the collision? -6 0.1

7. Clicker question: In the previous question, how much heavier was m 2 ? That is, find the ratio m 2 /m 1. 7 0.1

8. 10/22/2015A.PH 105 PH 105-003/4 ----Monday, Oct. 15, 2007 Homework: PS9 (short) due Wednesday: conservation of momentum Rest of Chapter 9: Center of Mass Why? If a system isn’t just a point mass, but we want to treat it as one (use F=ma, for example) what is its position x? Answer: use x cm where (for a simple system of 2 masses m 1 and m 2 ) Check: x 1 =x 2 case, m 1 =0 case, … m1m1 m2m2 x x1x1 x2x2 x cm

9. 10/22/2015A.PH 105 Example: Consider a system of two masses m1m1 m2m2 If they are equal (m 1 = m 2 ) then x 0 6 m x cm

10. Clicker question (not recorded): In the system of two masses, suppose m 1 =1 kg, m 2 =2 kg. Find the center of mass coordinate x cm. 4 0.1 m1m1 m2m2 x 0 6 m

11. 10/22/2015 A.9.7 Deformable bodies (skip) 9.8 Rockets: Serway derives vfvf v i =0 mimi mfmf exhaust v 0

12. Clicker question (didn’t have time): Use the rocket velocity equation to calculate the final velocity v f (in m/s), given that m f = 1000 kg, m i = 7389 kg, and the rocket exhaust velocity v 0 = 200 m/s, 400 20 10/22/2015

13. A.PH 105-003/4 ----Wednesday, Oct. 17, 2007 Homework: PS9 (short) due tonight: conservation of momentum Test next Friday Recoil problems: rocket, gun, hose, …. Please turn in filled-out evaluations in box at front. Talk Thursday: “Energy for All in the 21st Century ”Energy for All in the 21st Century Linear vs. Rotational motion: x  etc. Torque: F=ma   = I  where I =  m r 2 Kinetic energy: ½ m v 2  ½ I  2

14. Clicker Question: Two cylindrical drums (on axles through their centers) have string wound around them. One string has an object of mass m attached to it, the other is pulled on by a constant force T = mg. Which drum will turn faster? A.The one with constant tension mg. B.The one with the mass m attached C.They will turn at the same rate. T = mg A m B

15. Clicker Q [not done]: Find the acceleration of the falling mass in the case m disk = m = 2 kg, r = 0.2 m. 6.7 0.4 m r

16. A.PH 105-003/4 ---- Friday, Oct. 19, 2007 Homework (PS 9, Ch 9&10) is on WebAssign: Test next Friday Did: Linear vs. Rotational motion: x  etc. F=ma   = I  where I =  m r 2 Torque:  = F r More generally,  = F r sin  Line of action Calculating I for various shapes (make table) Parallel Axis Theorem Kinetic energy: ½ m v 2  ½ I  2

17. A.PH 105-003/4 ---- Monday, Oct. 22, 2007 Homework (PS 9, Ch 9&10) is due Wednesday 11PM Test this Friday Review: F=ma   = I  where I =  m r 2 Torque:  = F d = F r sin  d = r sin  = moment arm) Line of action Calculating I for various shapes (made table) Parallel Axis Theorem: I = I cm + Mr 2

18. A.Example: An airplane has wingspan 20 m, length 20 m, and mass 20,000 kg. It snags a weather balloon at the end of its wing, which exerts a drag force 10 5 N on the wing. Will the plane crash? Assume it crashes if it rotates > 1 radian in 1 second (the pilot’s reaction time). To answer this, answer (a)what is the torque about the center of mass? (aa) where is the center of mass? model: 2 rods, each 10,000 kg (ab) what is the torque? (b)What is the moment of inertia? (c)What is the angular acceleration? (d)How far does it turn in 1 second?

19. Clicker question: What is the moment of inertia of a rod of length L about a point on the rod, L/8 from the center? For simplicity, use m = 1 kg and L = 20 m, and give the answer in kg m 2. Hint: for a rod of length L rotating about its center of mass, I/mL 2 = 1/12. 39.6 1

20. A.Monday, continued Kinetic energy: K total = K trans + K rot = ½ m v cm 2 + ½ I  2 Rolling motion -- calculate v at bottom of incline Conservation of energy: mgh = ½ mv 2 + ½ I  2 & v =  r give v 2 = 2gh/(1+I/MR 2 )

21. A.PH 105-003/4 ---- Wednesday, Oct. 24, 2007 Homework (PS 9, Ch 9&10) is due tonight 11PM Test this Friday Review: Parallel Axis Theorem: I = I cm + Mr 2 Kinetic energy: K total = K trans + K rot = ½ m v cm 2 + ½ I  2 Rolling motion -- calculate v at bottom of incline Conservation of energy: mgh = ½ mv 2 + ½ I  2 & v =  r give v 2 = 2gh/(1+I/MR 2 ) v sphere >v disk > v hoop Angular momentum