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HW3 - boland(kbb544 HW3 Mackie(10611 This print-out should...

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boland (kbb544) – HW3 – Mackie – (10611) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 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 7 . 8 m / s. Also, it has an acceleration in the direction parallel to the walls of 4 . 1 m / s 2 . 18 . 4 m 4 . 1 m / s 2 7 . 8 m / s What will be its speed when it hits the opposing wall? Correct answer: 12 . 4251 m / s. Explanation: Let : d = 18 . 4 m , v x = 7 . 8 m / s , a = 4 . 1 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 . 67179 m / s 7 . 8 m / s 12 . 4251 m / s 38 . 8851 11 . 4078 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) (7 . 8 m / s) = 2 . 35897 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 . 1 m / s 2 ) (18 . 4 m) (7 . 8 m / s) = 9 . 67179 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 (7 . 8 m / s) 2 + (9 . 67179 m / s) 2 = 12 . 4251 m / s . 002 (part 2 of 2) 10.0 points

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boland (kbb544) – HW3 – Mackie – (10611) 2 At what angle with the wall will the particle strike? Correct answer: 38 . 8851 . 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 7 . 8 m / s 9 . 67179 m / s parenrightbigg = 38 . 8851 . Note: The distance traveled parallel to the walls is y = 1 2 a t 2 = 1 2 (4 . 1 m / s 2 ) (2 . 35897 s) 2 = 11 . 4078 m . 003 (part 1 of 3) 10.0 points A ball of mass 0 . 6 kg, initially at rest, is kicked directly toward a fence from a point 20 m away, as shown below. The velocity of the ball as it leaves the kicker’s foot is 16 m / s at angle of 47 above the horizontal. The top of the fence is 4 m high. The ball hits nothing while in flight and air resistance is negligible. The acceleration due to gravity is 9 . 8 m / s 2 . 20 m 4 m 16 m / s 47 Determine the time it takes for the ball to reach the plane of the fence. Correct answer: 1 . 83285 N. Explanation: Let : θ = 47 and d = 20 m . The horizontal component of the velocity is constant, so v horiz = v 0 cos θ = (16 m / s) cos 47 = 10 . 912 m / s .
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