hw 6 - sousse (res2468) HW 6 Kleinman (58225) 1 This...

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Unformatted text preview: sousse (res2468) HW 6 Kleinman (58225) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. This HW covers Chs. 10 and 11. 001 10.0 points An ostrich with a mass of 158 kg is running to the right with a velocity of 15 m / s. Find the momentum of the ostrich. Correct answer: 2370 kg m / s. Explanation: Let : m = 158 kg and v = 15 m / s to the right . p = mv = (158 kg) (15 m / s) = 2370 kg m / s to the right. 002 10.0 points Two blocks of masses m and M [ M = 1 . 62 m ] are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord holding them together is burned, after which the block of mass M moves to the right with a speed of 1 . 62 m / s. m M m M Before After (a) (b) v What is the speed of the block of mass m ? Correct answer: 2 . 6244 m / s. Explanation: From conservation of momentum p = 0, in our case we obtain 0 = M v M mv m . Therefore v m = M m v M = (1 . 62) v M = (1 . 62) (1 . 62 m / s) = 2 . 6244 m / s . 003 10.0 points A 64.9 kg astronaut is on a space walk when the tether line to the shuttle breaks. The astronaut is able to throw a 12.0 kg oxygen tank in a direction away from the shuttle with a speed of 14.9 m/s, propelling the astronaut back to the shuttle. Assuming that the astronaut starts from rest, find the final speed of the astronaut after throwing the tank. Correct answer: 2 . 75501 m / s. Explanation: Let toward the spacecraft be positive. Let : m 1 = 64 . 9 kg , m 2 = 12 . 0 kg , and v 2 ,f = 14 . 9 m / s . The wrench has the same initial speed as the astronaut, and v i = 0 m/s, so ( m 1 + m 2 ) vectorv i = m 1 vectorv f, 1 + m 2 vectorv f, 2 0 = m 1 vectorv f, 1 + m 2 vectorv f, 2 v f, 1 = m 2 v f, 2 m 1 = (12 kg) ( 14 . 9 m / s) 64 . 9 kg = 2 . 75501 m / s . 004 (part 1 of 2) 10.0 points Three particles of masses 83 kg, 47 kg, and 98 kg are located at the vertices of an equilateral triangle of sides 7 m. sousse (res2468) HW 6 Kleinman (58225) 2 83 kg 47 kg 98 kg 7m 7 m 7 m Find the x-coordinate of the center of grav- ity of this system with respect to a coordinate system with origin at the 83 kg particle and with the 47 kg particle located on the positive y-axis. Correct answer: 2 . 60567 m. Explanation: Let : m 1 = 83 kg , m 2 = 47 kg , m 3 = 98 kg , x 1 = 0 m , x 2 = 0 m , x 3 = 6 . 06218 m , y 1 = 0 m , y 2 = 7 m , and y 3 = 3 . 5 m . Basic Concepts: Center of Gravity x-coordinate of the center of gravity is given by the following formula x = n summationdisplay i =0 m i x i n summationdisplay i =0 m i Solution: x cm = 3 summationdisplay i =0 m i x i 3 summationdisplay i =0 m i = m 1 x 1 + m 2 x 2 + m 3 x 3 m 1 + m 2 + m 3 = (83 kg) (0 m) + (47 kg) (0 m) (83 kg) + (47 kg) + (98 kg) + (98 kg) (6 . 06218 m) (83 kg) + (47 kg) + (98 kg) = 2 . 60567 m ....
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hw 6 - sousse (res2468) HW 6 Kleinman (58225) 1 This...

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