Lecture 14 Goals More Energy Transfer and Energy Conservation Define and introduce power (energy per time) Introduce Momentum and Impulse Compare Force vs time to Force vs distance Employ conservation of momentum in 1 D & 2D Note: 2nd Exam, Monday, March 19th, 7:15 to 8:45 PM 1

Energy conservation for a Hooke’s Law spring m Associate ½ kx2 with the “potential energy” of the spring Ideal Hooke’s Law springs are conservative so the mechanical energy is constant if the spring and mass are the “system”

Hooke’s Law spring in the vertical m Gravity and perfect Hooke’s Law spring are conservative forces New equilibrium length at position where the gravitational force equals the spring force.

Energy (with spring & gravity) 1 3 2 h -x mass: m Given m, g, h & k, how much does the spring compress? Emech = constant (only conservative forces) At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0 Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0 Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2 Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0

Energy (with spring & gravity) 1 3 2 h -x mass: m Given m, g, h & k, how much does the spring compress? Emech = constant (only conservative forces) At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0 Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0 Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2 Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0 Given m, g, h & k, how much does the spring compress? Em1 = Em3 = mgh = -mgx + ½ kx2  Solve ½ kx2 – mgx - mgh = 0

Energy (with spring & gravity) 1 mass: m 2 h 3 -x When is the child’s speed greatest? (Hint: Consider forces & energy) (A) At y1 (top of jump) (B) Between y1 & y2 (C) At y2 (child first contacts spring) (D) Between y2 & y3 (E) At y3 (maximum spring compression)

Energy (with spring & gravity) 1 2 h mg 3 kx -x When is the child’s speed greatest? (D) Between y2 & y3 A: Calc. soln. Find v vs. spring displacement then maximize (i.e., take derivative and then set to zero) B: Physics: As long as Fgravity > Fspring then speed is increasing Find where Fgravity- Fspring= 0  -mg = kxVmax or xVmax = -mg / k So mgh = Ug23 + Us23 + K23 = mg (-mg/k) + ½ k(-mg/k)2 + ½ mv2  2gh = 2(-mg2/k) + mg2/k + v2  2gh + mg2/k = vmax2

Work & Power: Two cars go up a hill, a Corvette and a ordinary Chevy Malibu. Both cars have the same mass. Assuming identical friction, both engines do the same amount of work to get up the hill. Are the cars essentially the same ? NO. The Corvette can get up the hill quicker It has a more powerful engine.

1 W = 1 J / 1s Work & Power: Power is the rate at which work is done. Average Power is, Instantaneous Power is, If force constant in 1D, W = F Dx = F (v0 Dt + ½ aDt2) and P = F v = F (v0 + aDt) 1 W = 1 J / 1s

Exercise Work & Power P = dW / dt and W = F d = (Ff - mg sin q) d and d = ½ a t2 (constant acceleration) So W = F ½ a t2  P = F a t = F v (A) (B) (C) Power time Power Z3 time Power time

Work & Power: Power is the rate at which work is done. Example: A person of mass 80.0 kg walks up to 3rd floor (12.0m). If he/she climbs in 20.0 sec what is the average power used. Pavg = F h / t = mgh / t = 80.0 x 9.80 x 12.0 / 20.0 W P = 470. W

Ch. 9: Momentum & Impulse An alternative perspective (force vs time) Energy, Energy Conservation and Work Good approach if evolution with time is not needed. Energy is Conserved if only conservative (C) forces. Work relates applied forces (C and NC) along the path to energy transfer (in or out). Usually employed in situations with long times, large distances Are there any other relationships between mass and velocity that remain fixed in value (i.e. a new conservation law)?

Newton’s 3rd Law If object 1 and object 2 are the “system” then any change in the “momentum” of one is reflected by and equal and opposite change in the other.

Momentum Conservation Momentum conservation (recasts Newton’s 2nd Law when net external F = 0) is an important principle (most often useful when forces act over a short time) It is a vector expression so must consider px, py and pz if Fx (external) = 0 then px is constant if Fy (external) = 0 then py is constant if Fz (external) = 0 then pz is constant

A collision in 1-D 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. Because there is no external force momentum is conserved Because there is an internal non-conservative force energy conservation cannot be used unless we know the WNC So pxi = pxf In terms of m, M, and V, what is the momentum of the bullet with speed v ? V after x v before

A collision in 1-D What is the momentum of the bullet with speed v ? Key question: Is x-momentum conserved ? p Before p After v V before after x

A collision in 1-D: Energy What is the initial energy of the system ? What is the final energy of the system ? Is energy conserved? Examine Ebefore-Eafter V before after x v No! This is an example of an inelastic collison

Explosions: A collision in reverse 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. Does the tension in the string increase or decrease after the explosion? Before After

Explosions: A collision in reverse 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. Decipher the physics: 1. The green ball recoils in the –x direction (3rd 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

Explosions: A collision in reverse 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. Cons. of x-momentum px before= px after = 0 = - M V + m v V = m v / M = 20*30/ 60 = 10 m/s Tbefore = Weight = (60+20) x 10 N = 800 N SFy = m acy = M V2/r = T – Mg T = Mg + MV2 /r = 600 N + 60x(10)2/20 N = 900 N Before After

Impulse (A variable force applied for a given time) Collisions often involve a varying force F(t): 0  maximum  0 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. The impulse measures momentum transfer

Recap Read through all of Chapter 9