Momentum Balances & Quadratic Equations ABET Course Outcomes: C.1 formulate and solve engineering problems using linear and quadratic equations By the end of this class you should be able to: Solve problems involving quadratic equations Explain the momentum balance for a projectile problem Reading: Chapter 2 (finish next week) Class Plan Handouts: Quadratic equations worksheet (pass out at beginning) Momentum Balance summary (have students take as they leave) Outline: 1. Initial note on the different representations we have been using. 2. Example Problem (metric version of 1 st eg in Ch.2) worksheet (go through after students try, one part at a time) 1. Problem Setup and basic equation for time to 24 m 2. Approach 1: Factoring 3. Aproach 2: Quadratic eqn. 4. Approach 3: Completing the square 5. Additional questions (max ht, time of flight, graph of curve) 2. Momentum Balance to derive equation for above. Guided Example (handout): A ball launched vertically whose height is described by the equation below. h(t) = 29.4 t 4.9 t 2 h(t) = height at time t (m) t = time (s) Problem: At what time is the ball 24.5 m above the ground? Setup: notice on worksheet Part 1: Set up the quadratic equation for this case with an x 2 coefficient of 1 Q1 finding time to reach 24 m (worksheet) have students try one part at a time, consult with each other, and then go through on board Equation Setup: set h(t) = 24 h(t) = 29.4 t 4.9 t 2 = 24.5 rearrange so = 0 4.9 t 2 29.4 t + 24.5 = 0 divide by first coefficient Approach 1: Factoring factor into a product of binomials have students try (t 5)(t 1) = 0 t = 1 and t =5 Why 2? Mathematically a quadratic has two roots Physically one point is the way up, one the way down.
Q1 finding time to reach 24 m Solution Approach 1 (part 2): Factoring Solution Approach 2 (part 3): Quadratic Formula for ax 2 + bx + c = 0 Solution Approach 3 (part 4): Completing the Square Q1 finding time to reach 24 m (work through on board with student s working on parts) a quadratic eqn. Approach 1: Factoring factor into a product of binomials Approach 2: Quadratic eqn. give have students use a = 1, b = 6 (don t miss the negative), c =5 = t = 1 or 5 sec. (as before) = = 3 ±2 Approach 3: Completing the square (next) Q1 finding time to reach 24 m (work through on board with student s working on parts) a quadratic eqn. Approach 3: Completing the square explain, work answer a. Move constant to RHS b. Add to both sides Part 5: Additional questions a. What is the maximum height? (hint: first determine the time of the max height) b. How long will it take until the projectile returns to the ground? c. Sketch a graph of the height vs. time. c. LHS should be a perfect square, write the binomial that is squared d. solve t = 1 or 5 sec.
What is the maximum height? The equation is symmetric, so the maximum with be halfway between the two roots we have found: t max = (1 + 5)/2 = 3 s plug this into the original equation h max = 29.4 (3) 4.9 (3) 2 = 88.2 44.1 h max = 44.1 m How long will it take until the projectile returns to the ground? For this case h(t) = 0 or 0 = 29.4 t 4.9 t 2 We could use any of the previous techniques. However, factoring is quite easy in this case 0 = t (29.4 4.9t) or 0 = t (6 t) Roots: 0, & 6 sec. The first root is the starting time, The second is the time to return to the ground (flight time) Time (s) Height (m) 0 0 1 24.5 3 44.1 5 24.5 6 0 These are the points we have calculated, we can graph the curve Can you now sketch the graph? Review from last time Newton s 1 st Law: tendency for a body to stay in motion Momentum = mass x velocity (P = m v) Newton's 2 nd law: Force = mass x acceleration (F = m a) Momentum Balance: Start (t =0) Change End (t = t) Momentum Balance for Projectile How do we represent momentum at times 0 and t? How do we represent the change in momentum due to the force of gravity? t= t p y (t) = m v y (t) p y = m (g y ) t m x0 v x0 + m a x t = mv x v x0 + a x t = v x for: m = constant a = constant Can you put these into an equation and simplify? t= 0 p y0 = m v y0
Ask students each question, give a short time for them to confer in small groups Questions on previous slide will appear one at a time (answers for first two appear with next question) Representing Motion with Graphs How do we represent momentum at times 0 and t? How do we represent the change in momentum due to the force of gravity? Can you put these into an equation and simplify for constant g & m. t=t t=0 P y (t) = m v y (t) P y = m (g y ) t P y0 = m v y0 a = a = Slope of Velocity Distance Traveled = Area under Velocity Curve In this case: d = v t Do last answer on board: mv y (t) = mv y0 + mg y t v y (t) = v y0 + g y t 1. What is the equation of this line? For v 0 = 29.4 m/s a = g y = 9.8 m/s 2. What is the equation for h as a function of v & t? Remember: h(t) = Area 3. What is the equation for h as a function of t only? Previous slide will show questions one at a time. Have students work on and then solve on the board. 1. What is the equation of this line? For v y0 = 29.4 m/s a = g = 9.8 m/s v y (t) = 29.4 9.8 t m y 2. h(t) = Area =? = ½ (v y0 + v y (t)) t 3. General Case h(t) = ½(v 0 + v 0 + at) t h(t) = v 0 t+ ½ at 2 For this case h(t) = 29.4t + ½ gt 2 h(t) = 29.4t 4.9 t 2
Vertical Projectile Motion (Review) h(t) = v y0 t + g y t 2 Based on area under v vs. t graph v y (t) = v y0 + g y t Based on a momentum balance for constant g and constant m g y = 9.8 m/s or 32 ft/s Note: negative sign shows the vector is pointed down Solving Quadratic Equations: 1. Put in standard form (ax 2 + bx +c = 0) 2. Solve by: factoring, quadratic formula, completing the square.