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hw2solutions - Niu(qn269 HW02 Tsoi(58020 This print-out...

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Niu (qn269) – HW02 – Tsoi – (58020) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points Given: The acceleration of gravity on Earth is 9 . 8 m / s 2 . Consider a ball thrown up from the ground (the point O). It passes a window (the seg- ment AB) in the time interval 0 . 262 s (see the figure). The points in the figure represent the sequential order, and are not drawn to scale. The distance AB = 1 . 17 m. O C B A 1 . 17 m x y Find the average speed as the ball passes the window. Correct answer: 4 . 46565 m / s. Explanation: Basic Concepts: vectora = Δ vectorv Δ t For constant acceleration ¯ v = Δ s Δ t = v i + v f 2 v = v 0 + a t . For the acceleration of gravity ( a = g ) v = v 0 g t . Solution: The average velocity is given by ¯ v = Δ s Δ t = | AB | Δ t = 1 . 17 m 0 . 262 s = 4 . 46565 m / s . Alternative Part 1: d = y B y A = v A t 1 2 g t 2 v A = d + 1 2 g t 2 t = 1 . 17 m + 1 2 (9 . 8 m / s 2 ) (0 . 262 s) 2 (0 . 262 s) = 5 . 74945 m / s v B = v A g t = 5 . 74945 m / s (9 . 8 m / s 2 ) (0 . 262 s) = 3 . 18185 m / s v = v A + v B 2 = 5 . 74945 m / s + 3 . 18185 m / s 2 = 4 . 46565 m / s . 002 (part 2 of 3) 10.0 points What is the magnitude of the decrease of the velocity from A to B? Correct answer: 2 . 5676 m / s. Explanation: For a constant acceleration vectora = Δ vectorv Δ t , = ⇒| Δ vectorv | = | vectora Δ t | = g Δ t . Thus the decrease in velocity is Δ v = g Δ t = (9 . 8 m / s 2 )(0 . 262 s) = 2 . 5676 m / s . Alternative Part 2: Δ v = v A v B = 5 . 74945 m / s 3 . 18185 m / s = 2 . 5676 m / s .
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Niu (qn269) – HW02 – Tsoi – (58020) 2 003 (part 3 of 3) 10.0 points If the ball continues its path upward without obstruction, find the travel time between B and C, where point C is at the ball’s maximum height. Correct answer: 0 . 324678 s. Explanation: The velocity change is v A v B = Δ v = v A = v B + Δ v and the average velocity is ¯ v = v A + v B 2 = 2 v B + Δ v 2 . Thus v B = ¯ v Δ v 2 = 4 . 46565 m / s 2 . 5676 m / s 2 = 3 . 18185 m / s . The ball reaches its maximum height when its velocity is zero. Based on v = v 0 + a t v C = 0 = v B g t t = v B g = 3 . 18185 m / s 9 . 8 m / s 2 = 0 . 324678 s . 004 (part 1 of 2) 10.0 points A ball is thrown upward. Its initial verti- cal speed v 0 , acceleration of gravity g , and maximum height h max are shown in the figure below. Neglect: Air resistance. The acceleration of gravity is 9 . 8 m / s 2 . v 0 9 . 8 m / s 2 h max What is its maximum height, h max (in terms of the initial speed v 0 )? 1. h max = v 2 0 2 g correct 2. h max = 5 v 2 0 2 2 g 3. h max = v 2 0 2 g 4. h max = 3 v 2 0 2 g 5. h max = 3 v 2 0 2 2 g 6. h max = v 2 0 g 7. h max = 5 v 2 0 8 g 8. h max = v 2 0 4 g 9. h max = 3 v 2 0 4 g Explanation: Basic Concept: For constant accelera- tion, we have v 2 = v 2 0 + 2 a ( x x 0 ) . (1) Solution: The velocity at the top is zero. Since we know velocities and acceleration, Eq. 1 containing v , a , y 0 , and y (no time) is the easiest one to use. Choose the positive direction to be up; then a = g and 0 = v 2 0 + 2 ( g ) ( h max 0)
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Niu (qn269) – HW02 – Tsoi – (58020) 3 or h max = v 2 0 2 g .
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