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physics hw 3

# physics hw 3 - homework 03 HILL WILLIAM Due 1:00 am...

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homework 03 – HILL, WILLIAM – Due: Sep 21 2007, 1:00 am 1 Question 1, chap 4, sect 3. part 1 of 2 10 points A particle travels horizontally between two parallel walls separated by 18 . 4 m. It moves toward the opposing wall at a constant rate of 9 m / s. Also, it has an acceleration in the direction parallel to the walls of 4 . 6 m / s 2 . It hits the opposite wall at the same height. 18 . 4 m 4 . 6 m / s 2 9 m / s a) What will be its speed when it hits the opposing wall? Correct answer: 13 . 017 m / s (tolerance ± 1 %). Explanation: Let : d = 18 . 4 m , v x = 9 m / s , a = 4 . 6 m / s 2 , Basic Concepts Kinematics equations v = v o + g t s = s o + v o t + 1 2 g t 2 d a 9 . 40444 m / s 9 m / s 13 . 017 m / s 43 . 7411 9 . 61343 m The horizontal motion will carry the parti- cle to the opposite wall, so d = v x t f and t f = d v x = (18 . 4 m) (9 m / s) = 2 . 04444 s . is the time for the particle to reach the oppo- site wall. Horizontally, the particle reaches the maxi- mum parallel distance when it hits the oppo- site wall at the time of t = d v x , so the final parallel velocity v y is v y = a t = a d v x = (4 . 6 m / s 2 ) (18 . 4 m) (9 m / s) = 9 . 40444 m / s . The velocities act at right angles to each other, so the resultant velocity is v f = radicalBig v 2 x + v 2 y = radicalBig (9 m / s) 2 + (9 . 40444 m / s) 2 = 13 . 017 m / s . Question 2, chap 4, sect 3. part 2 of 2 10 points

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homework 03 – HILL, WILLIAM – Due: Sep 21 2007, 1:00 am 2 b) At what angle with the wall will the particle strike? Correct answer: 43 . 7411 (tolerance ± 1 %). Explanation: When the particle strikes the wall, the ver- tical component is the side adjacent and the horizontal component is the side opposite the angle, so tan θ = v x v y , so θ = arctan parenleftbigg v x v y parenrightbigg = arctan parenleftbigg 9 m / s 9 . 40444 m / s parenrightbigg = 43 . 7411 . Note: The distance traveled parallel to the walls is y = 1 2 a t 2 = 1 2 (4 . 6 m / s 2 ) (2 . 04444 s) 2 = 9 . 61343 m . Question 3, chap 4, sect 3. part 1 of 1 10 points A quarterback takes the ball from the line of scrimmage, runs backward for 11 . 0 yards, then runs sideways parallel to the line of scrimmage for 20 . 0 yards. At this point, he throws a 15 . 0 yard forward pass straight down the field. What is the magnitude of the football’s resultant displacement? Correct answer: 20 . 3961 yards (tolerance ± 1 %). Explanation: Basic Concepts: Δ x tot = Δ x Δ y tot = Δ y 1 + Δ y 2 The displacements are perpendicular, so d tot = radicalBig x tot ) 2 + (Δ y tot ) 2 Given: Δ y 1 = 11 . 0 yards Δ x = ± 20 . 0 yards Δ y 2 = 15 . 0 yards Solution: Δ y tot = 11 yards + 15 yards = 4 yards Δ x tot = ± 20 yards Thus d = radicalBig ( ± 20 yards) 2 + (4 yards) 2 = 20 . 3961 yards Question 4, chap 4, sect 4. part 1 of 1 10 points A ball is thrown and follows the parabolic path shown. Air friction is negligible. Point Q is the highest point on the path. Points P and R are the same height above the ground. Q R P How do the speeds of the ball at the three points compare? 1. bardbl vectorv P bardbl = bardbl vectorv R bardbl < bardbl vectorv Q bardbl 2. bardbl vectorv Q bardbl < bardbl vectorv P bardbl = bardbl vectorv R bardbl correct 3. bardbl vectorv P bardbl < bardbl vectorv Q bardbl < bardbl vectorv R bardbl 4. bardbl vectorv P bardbl = bardbl vectorv R bardbl = bardbl vectorv Q bardbl
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