This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: sousse (res2468) HW 6 Kleinman (58225) 1 This printout should have 29 questions. Multiplechoice 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 xcoordinate 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 yaxis. 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 xcoordinate 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 ....
View Full
Document
 Spring '08
 Turner
 Physics

Click to edit the document details