This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 106/ABCCC – midterm 02 – turner – (57340) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A car of mass 1500 kg is moving over a rounded hill of radius 100 m with a constant speed of 13 m / s. At the topmost point on the hill, what is the magnitude of the normal force exerted by the road on the car? Take g to be 9 . 8 m / s 2 . 1. 7440.0 2. 2940.0 3. 12885.0 4. 13740.0 5. 14160.0 6. 5325.0 7. 6060.0 8. 2085.0 9. 12165.0 10. 6765.0 Correct answer: 12165 N. Explanation: Given : M = 1500 kg , R = 100 m , v = 13 m / s , and g = 9 . 8 m / s 2 . Let F be the magnitude of the normal force exerted by the road on the car. Applying momentum principle at the top of the hill in the vertical direction (ycomponent) Mv 2 R = F Mg F = Mg Mv 2 R F = 1500 kg * parenleftbigg 9 . 8 m / s 2 (13 m / s) 2 100 m parenrightbigg F = 12165 N 002 10.0 points A circular pendulum of length l = 1 . 2 m goes around at an angle of θ = 36 ◦ to a line drawn perpendicular to the ground. Deter mine the speed of the mass at the end of the string. 1. 1.74148 2. 1.94193 3. 2.62135 4. 1.45754 5. 1.84402 6. 1.76623 7. 1.65986 8. 1.40724 9. 2.24101 10. 2.05973 Correct answer: 2 . 24101 m / s. Explanation: The equations of motion in the x and y directions are given by: summationdisplay vector F x = T sin θ = mv 2 l sin θ summationdisplay vector F y = T cos θ mg = 0 , where T is the tension along the string. Di viding the first equation by the second (to eliminate the unknown T ) yields tan θ = v 2 gl sin θ . Solving this for v yields v = radicalBigg gl sin 2 θ cos θ = 2 . 24101 m / s . Version 106/ABCCC – midterm 02 – turner – (57340) 2 003 10.0 points Two wires with equal length are made out of pure copper. The diameter of wire 1 is twice the diameter of wire 2. When the wire 1 is hung with a weight which is half of that hung on wire 2, find the ratio of the stretches Δ L 1 Δ L 2 . 1. 8 2. 2 3. 1 4. 4 5. 16 6. 1 16 7. 1 2 8. 1 8 correct 9. 1 4 Explanation: Young’s modulus is defined as Y = F A Δ L L .Since both wires are made out of the same materials, their Young’s moduli should be the same. Thus for equal length wires F 1 A 1 Δ L 1 L 1 = F 2 A 2 Δ L 2 L 2 Δ L 1 Δ L 2 = F 1 A 1 F 2 A 2 = F 1 A 2 F 2 A 1 = F 1 ( π d 2 2 ) F 2 ( π d 2 1 ) = F 1 F 2 parenleftbigg d 2 d 1 parenrightbigg 2 = F 1 2 F 1 parenleftbigg d 2 2 d 2 parenrightbigg 2 = 1 8 . 004 10.0 points A scientific probe of mass 420 kg is orbiting the asteroid Eros in a circular orbit of radius 44 . 2 km and period 1 . 06 days. Calculate the mass of Eros. ( G = 6 . 7 × 10 − 11 N · m 2 /kg 2 ) 1. 5.18654e+15 2. 5.81505e+15 3. 5.50635e+15 4. 6.36369e+15 5. 7.53009e+15 6. 6.53041e+15 7. 7.06e+15 8. 7.43518e+15 9. 7.67541e+15 10. 6.06614e+15 Correct answer: 6 . 06614 × 10 15 kg....
View
Full
Document
This note was uploaded on 04/13/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Mass

Click to edit the document details