Momentum Balances & Quadratic Equations

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1 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 = t t = 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.

2 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.

3 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 = 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 ( t) 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) 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

represent momentum at times 0 and t? How do we represent the change in momentum due to the force of gravity?

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

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 t only? Previous slide will show questions one at a time.

4 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) = 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

5 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.

Physics 11 Course Plan

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