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Unformatted text preview: PC1431 Physics IE 20042005 TERM TEST 23 September Time Allowed: ONE hour INSTRUCTIONS 1. This is a close book test. 2. This paper contains 15 multiple choice questions. 3. Answer all questions. 4. Answer questions by shading the appropriate bubbles on the answer sheet only. Answers written anywhere else will not be marked. 5. Only the answer sheet will be collected at the end of the test. 6. Use 2B pencil only. Using any other type of pencil or pen may result in answers unrecognizable by machine. SOME USEFUL INFORMATION I CM = MR 2 (for a ring) I CM = 1 2 MR 2 (for a uniform solid cylinder) I CM = 1 12 ML 2 (for a uniform rod) Z x n dx = x n +1 n + 1 g = 9 . 81 m/s 2 1. A rifle is aimed horizontally at the center of a large target 60 m away. The initial speed of the bullet is 240 m/s. What is the distance from the center of the target to the point where the bullet strikes the target? Ignore air resistance. (A) 0.17 m (B) 0.31 m (C) 0.48 m (D) 0.52 m (E) 0.69 m Solution: D= 60 m h The motion of the bullet consists of a constant velocity ( v x = 240 m/s) motion in the horizontal direction and a free fall in the vertical direction. The time it takes the bullet to reach the target: T = D v x = 60m 240m/s = 0 . 25s The fall in the vertical direction during this time is h = 1 2 gT 2 = 1 2 × 9 . 81m/s 2 × . 25s = 0.31 m 2 2. A 0.20km wide river has a uniform flow speed of 3.0 m/s toward the east. A boat with a speed of 8.0 m/s relative to the water leaves the south bank and heads in such a way that it crosses to a point directly north of its departure point. How long does it take the boat to cross the river? (A) 23 s (B) 25 s (C) 27 s (D) 29 s (E) 67 s Solution: 3.0 m/s 8.0 m/s υ Speed of boat relative to the bank is v = √ 8 . 2 3 . 2 = 7 . 42 m/s Time it takes the boat to cross the river is T = 200 m 7 . 42 m / s = 27.0 s 3 3. Assume that the magnitude of the resistive force acting on an object moving through a liquid with a speed v is proportional to v 3 / 2 , i.e. , R = bv 3 / 2 , where b is a constant. Consider a object of mass m released from rest in a liquid. If the only forces acting on the object are the above resistive force and the gravitational force, determine the terminal speed of the object. (A) mg b (B) s 2 mg DρA (C) mg b 2 / 3 (D) mg b 3 / 2 (E) mg b 1 e bt/m Solution: Apply Newton’s 2nd law to the vertical motion of the object mg bv 3 / 2 = ma As the object falls down, v and therefore the resistive force increase and the accel eration decreases (from g ). When a = 0, the terminal speed is given by mg bv 3 / 2 = 0 or v = mg b 2 / 3 4 4. The graph below shows the velocity versus time graph for a ball. Which explanation best fits the motion of the ball as shown by the graph?...
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This note was uploaded on 04/17/2008 for the course PC 1431 taught by Professor Andrewwee during the Summer '04 term at National University of Singapore.
 Summer '04
 AndrewWee

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