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Unformatted text preview: This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points An airplane of mass 998 kg flies level to the ground at a constant speed of 175 m / s relative to the Earth. An observer on the ground along the path of the plane sees the plane a distance 13003 . 7 m away at an angle above the horizontal of 57 . 77 . What is the magnitude of the airplanes angular momentum relative to a ground ob- server directly below the airplane? Correct answer: 1 . 92115 10 9 kg m 2 / s. Explanation: Basic Concepts: vector L = vector r vector p L = r mv sin( ) = (13003 . 7 m)(998 kg)(175 m / s) sin(57 . 77 ) = 1 . 92115 10 9 kg m 2 / s 002 (part 1 of 2) 10.0 points Given: Use counter-clockwise as the positive angular direction. The acceleration of gravity is 9 . 8 m / s 2 . A particle of mass m is shot with an initial velocity 2 m / s, making an angle 46 , with the horizontal as shown in figure. The particle moves in the gravitational field of the Earth. x y v R v o 2 m / s v o v h 46 R h O C A Using the origin as the pivot, find the an- gular momentum (along the z-axis, using a right-hand coordinate system) when the par- ticle is at the highest point of the trajectory. Correct answer:- . 440144 N m. Explanation: Basic Concepts vector L = vectorr vectorp vector L = I vector We also need the vector cross product | vectora vector b | = a b sin , where is the angle between the vectors vectora and vector b . Let : v o = 2 m / s , = 46 , and g = 9 . 8 m / s 2 . Solution: At the highest point (denote the position vector to that point with vector r h ) we know that vector v h = vector v x = v o cos and vector L = vector r h vector p h = vector r h m vector v h And, | vector L | = m | vector v h || vector r h | sin h . Let us analyze this expression. We see that vector r h sin h is the maximum height of y max = h h = | vector r h | sin h = v o 2 sin 2 2 g ....
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