121ss06m2_A

121ss06m2_A - Phy 121 Midterm 2 Exam Spring 2006A This is a...

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Phy 121 Midterm 2 Exam Spring 2006 A This is a closed book exam. Use your own formula sheet – 2 pages, 8x11 inches, handwritten . Problems 1 to 4 are “walkthroughs” (15 points each) with steps (1),(2) ,…These step numbers (in bold with points marked) are the question numbers you see on the screen below your pad number. Answer each step as a multiple choice question on the grid on paper and with your remote. Answers (A) ….(E) are possible. The last 12 steps are questions on a variety of topics in the format of the workshop. Answers (A),(B) and (C) are possible and each step is worth 3 points. Steps (1),….,(23): A,B,C,D,E 1. A rocket with a mass m = 12000 kg is shot straight up from the surface of the earth (it moves along a radial line through the center of the earth and the launch point). It loses 9 2.9 10 J × of energy via friction in the atmosphere and arrives with a velocity of ' 580 / v m s = at a height of h = 500 km above the surface of the earth. What is the launch velocity v ? (You cannot neglect the height h relative to the earth radius.) 11 2 2 24 (Use 6.67 10 for the gravitational constant, 6.0 10 for 6 the earth mass, 6.4 10 for the earth radius.) G Nm kg M kg R m - - = × = × = × (1) (2 point) The initial potential energy of the rocket is: (A) 0 (B) ) ( h R mg - (C) mgR - (D) R GmM / - (E) ) /( h R GmM - (2) (2 points) The final potential energy of the rocket is: (A) mgh (B) ) ( h R mg + (C) 2 / R GmM - (D) ) /( h R GmM - (E) ) /( h R GmM + - (3) (3 points) The initial total mechanical energy of the rocket is: 1

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(A) 9 /( ) 2.9 10 GmM R h J - + + × (B) 2 1 2 mv (C) 2 1 / 2 GmM R mv - + (D) 9 2.9 10 J × (E) ) /( h R GmM + - (4) (3 points) The final total mechanical energy of the rocket is: (A) 2 1/ 2 ' /( ) mv GmM R h - + (B) 2 1/ 2 ' / mv GmM R - (C) 9 2.9 10 J × (D) 2 9 1/ 2 ' 2.9 10 mv J - × (E) 2 1/ 2 ' / mv GmM R + (5) (3 points) The correct expression for the kinetic energy of the rocket at launch is: (A) ) /( / h R GmM R GmM + - (B) 2 9 1 ' /( ) / 2.9 10 2 mv GmM R h GmM R J + + - - × (C) 2 9 1 ' 2.9 10 2 mv J - × (D) 2 9 1 ' / 2.9 10 2 mv GmM R J + + × (E) 2 9 1 ' / /( ) 2.9 10 2 mv GmM R GmM R h J + - + + × 2
(6) (2 points) The correct value for the launch velocity is: (A) s m / 10 67 . 1 5 × (B) 2223 m/s (C) 3144 m/s (D) 0 (E) 240 m/s 2. A bullet with a mass m = 85 grams is incident on a ballistic pendulum of mass M = 2.2 kg . After the collision the bullet stays embedded in the pendulum mass. How high is the ballistic pendulum (the mass M ) lifted if the incident velocity v = 61 m/s ? Steps:

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This note was uploaded on 02/27/2011 for the course PHY 121 taught by Professor Stephens during the Spring '08 term at SUNY Stony Brook.

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121ss06m2_A - Phy 121 Midterm 2 Exam Spring 2006A This is a...

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