PHY131 Midterm1 Solutions

University Physics with Modern Physics with Mastering Physics (11th Edition)

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PHY 131 MIDTERM 1 Version Do all problems, and show all work for partial credit ! Note: make sketches where appropriate and show your axis conventions. Do not forget units, number of significant digits, and check your results for consistency. This exam will last 1.5 hr. Success! 1. Consider the vectors A = 2.00 i – 1.00 j and B = –2.00 i + 2.00 j a. Make an x-y sketch of vectors A , B , and C = B – A b. Calculate the magnitude of vector C and the angle it makes with the positive x -axis. C = –4.00 i + 3.00 j (B: –4.00 i + 3.00 j ; C: –3.00 i + 4.00 j ; D: 5.00 i + 4.00 j) thus: C = (4 2 + 3 2 ) = 5.00 (B: 5.00; C: 5.00; D: 6.40) and θ C = arctan(3/–4) = 180° –36.9° = 143.1° (B: 143.1°; C: 126.7°; D: 38.7°) y C B A x c. Calculate the angle between vectors A and B cos θ AB = A·B /( AB ) = (–4–2)/ (5×8) = –0.949 thus θ AB = 162° (B: ; C: ; D: ) 2. A late commuter student bursts onto the Long Island train platform running at her top speed of 6.0 m/s. Upon arrival on the platform the nearest open train door is still 32 m away, and the train just starts to move (from rest) away from her, with constant acceleration 0.50 m/s 2 ; she keeps running… a. Calculate if she will make it to the door. A simple yes/no is not accepted; prove your answer! (12 points) Take the x-direction parallel to the train and platform, from the student towards the door. In order to get onto the train, there should be a time t for which the positions of the student and the door coincide: x student = v 0 t = x door = x 0 + ½ at 2 . Thus: t 1,2 = v 0 / a ± ( v 0 2 / a 2 – 2 x 0 / a ) = 12 ± (12 2 – 128) = 12 ± 4 = 8 s or again at 16 s. (B: ; C: ; D: ) Therefore a time exists when student and door coincide (at x = 48 m), and the student can jump aboard. b. Graph x ( t ) vs. t for both the student and the train in the same plot; and label the curves and axes. Indicate what types of curves you have drawn (e.g. sine, straight line, hyperbola, parabola, etc.). (8 points) Student: straight line through origin, slope 6 m/s; train: parabola, starting at x 0 = 32 m. 8 s 32 m x ( t ) t Student (line) Train (parabola) 3. A projectile is launched from a cliff 18.0 m above sea level, under an angle of +30° with respect to the horizontal. After 6.0 s it lands out in the water. Ignore drag and friction. In any convenient order , calculate the following (Show all work!): a. The initial speed of the projectile. (7 points) y L = 0 = y 0 + v 0 y t L – ½ gt L 2 , thus v 0 y = ½ gt L y 0 / t L = 26.4 m/s; thus: v 0 = v 0 y /sin30° = 52.8 m/s b. The maximum height (with respect to sea level) of the projectile. (8 points) at y = h the component v y = 0, thus: v y 2 v 0 y 2 = –2 g ( h – y 0 ) , thus: h = y 0 + ½ v 0 y 2 / g = 53.6 m c. The horizontal range of the projectile. (5 points) Range R = v 0 x t L = v 0 cos(30°)× t L = 274 m

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4. A 4.0 kg book is placed on a long straight plank. One end of the plank is slowly raised, to the point where the book just begins to slip. The angle at that point is 17°. a. Make a sketch showing all forces acting on the book. (5 points) Three forces: Weight w , (static) friction F f , and normal force N b. Calculate the coefficient of static friction. (5 points) When the block just starts to slip, the plank-parallel component of the weight w , mg sin θ , just equals the force of static friction F fs . Moreover, the plank-perpendicular acceleration is zero, and
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This homework help was uploaded on 02/04/2008 for the course PHY 131 taught by Professor Rijssenbeek during the Fall '03 term at SUNY Stony Brook.

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PHY131 Midterm1 Solutions - PHY 131 MIDTERM 1 Version A Do...

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