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Unformatted text preview: Physics 121 Honors Common Exam 1 Spring 2003 Name: Section: Closed book exam. Only one 8 . 5 00 11 00 formula sheet (front and back side) can be used. Calculators are allowed. Use the scantron forms (pencil only!) for the multiple choice problems. Circle the answers on the examination sheet as well, and return it together with the scantron form. Use the back of these pages, or attach your own pages with solutions for problems that require calculations. The multiplechoice problems are 5 points each. Partial credit (up to 3 points) will be given for multiple choice problems, if you provide a detailed solution on the examination sheets. The workout problems are 10 points each. The nominal number of points is 100. Clearly print your first and last name and indicate your section number on both the scantron form and the examination sheet. Good luck! MultipleChoice Problems Problem 1: Two tiny conducting balls of identical mass m and identical charge q = 3 . 20 10 8 C hang from nonconducting threads of length L = 140 cm. The angle in the equilibrium position as indicated in the figure is = 2 . 8 . What is the mass m of the conducting balls? A) m = 1 g B) m = 2 g C) m = 10 g D) m = 20 g E) m = 40 g F e = T sin mg = T cos tan = F e mg and sin = x 2 L m = q 2 4 x 2 g tan = q 2 16 gL 2 tan sin 2 = . 001 kg Problem 2: The radius of the nucleus of a Gold atom is about r Au 1 . 4 10 15 3 A m, where A = 197 u is the atomic mass of Gold. What is the magnitude of the repulsive electrostatic force between two protons separated by r Au ? A) F = . 60 N B) F = 3 . 5 N C) F = 8 . 4 N D) F = 120 N E) F = 1 . 8 10 5 N F = 1 4 e 2 r 2 Au = e 2 4 ( 1 . 4 10 15 m ) 2 A 2 / 3 = 3 . 47 N Problem 3: Three point charge lie along the xaxis. The positive charge q 1 = 12 . C is at x 1 = 1 . 50 m, the positive charge q 2 = 5 . 20 C is at the origin, and the resultant force on the negative charge q 3 is zero. What is the xcoordinate of q 3 ? A) . 90 m B) . 60 m C) + . 60 m D) + . 90 m E) + 2 . 40 m 1 4 q 2 q 3 x 2 = 1 4 q 1 q 3 ( x 1 x ) 2 q 2 ( x 1 x ) = q 1 x x = q 2 q 1 + q 2 x 1 = + . 595 m Problem 4: What is the direction and magnitude of the electric field due to the three point charges at point...
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This note was uploaded on 04/18/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.
 Spring '08
 Opyrchal
 Physics

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