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: Bodet (ngb299) – homework 19 – turner – (57340) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What velocity must a car with a mass of 1070 kg have in order to have the same mo mentum as a 2390 kg pickup truck traveling at 24 m / s to the east? Correct answer: 53 . 6075 m / s. Explanation: Let : m 1 = 1070 kg , m 2 = 2390 kg , and v 2 = 24 m / s to the east . !p = m 1 !v 1 = m 2 !v 2 v 1 = m 2 v 2 m 1 = (2390 kg) (24 m / s) 1070 kg = 53 . 6075 m / s to the east. 002 10.0 points Note: Take East as the positive direction. A(n) 82 kg fisherman jumps from a dock into a 139 kg rowboat at rest on the West side of the dock. If the velocity of the fisherman is 3 . 5 m / s to the West as he leaves the dock, what is the final velocity of the fisherman and the boat? Correct answer: 1 . 29864 m / s. Explanation: Let West be negative: Let : m 1 = 82 kg kg , m 2 = 139 kg kg , and v i, 1 = 3 . 5 m / s m / s . The boat and fisherman have the same final speed, and v i, 2 = 0 m/s, so m 1 !v i, 1 + m 2 !v i, 2 = ( m 1 + m 2 ) !v f m 1 !v i, 1 = ( m 1 + m 2 ) !v f v f = m 1 v i m 1 + m 2 = (82 kg) ( 3 . 5 m / s) 82 kg + 139 kg = 1 . 29864 m / s , which is 1 . 29864 m / s to the West. 003 10.0 points A(n) 598 N man stands in the middle of a frozen pond of radius 14 . 8 m. He is unable to get to the other side because of a lack of fric tion between his shoes and the ice. To over come this difficulty, he throws his 1 kg physics textbook horizontally toward the north shore, at a speed of 5 . 3 m / s. The acceleration of gravity is 9 . 81 m / s 2 . How long does it take him to reach the south shore? Correct answer: 170 . 223 s. Explanation: Let : W m = 598 N , r = 14 . 8 m , m b = 1 kg , and v b = 5 . 3 m / s . The mass of the man is m m = W m g . From conservation of momentum, m m v m + m b v b = m m v m + m b v b 0 = m m v m + m b v b v m = m b m m v b = g m b W m v b = ( 9 . 81 m / s 2 ) (1 kg) 598 N (5 . 3 m / s) = . 0869448 m / s . Bodet (ngb299) – homework 19 – turner – (57340) 2 The time to travel the 14 . 8 m to shore is t = Δ x  v m  = 14 . 8 m . 0869448 m / s = 170 . 223 s . 004 10.0 points An 73 . 7 kg object moving to the right at 36 . 4 cm / s overtakes and collides elastically with a second 55 . 4 kg object moving in the same direction at 20 . 7 cm / s. Find the velocity of the second object after the collision. Correct answer: 38 . 6255 cm / s. Explanation: Let : m 1 = 73 . 7 kg , m 2 = 55 . 4 kg , v 1 = 36 . 4 cm / s , and v 2 = 20 . 7 cm / s . Momentum conservation gives m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f (1) For headon elastic collisions the centerof mass motion remains constant ( v 1 f v 2 f ) = v 1 i v 2 i , that is v 2 f = v 1 i v 2 i + v 1 f , or v 1 f = v 2 i v 1 i + v 2 f . (2) Substituting Eq. 2 into Eq. 1, we have m 1 v 1 i + m 2 v 2 i = m 1 v 2 i m 1 v 1 i +( m 1 + m 2 ) v 2 f 2 m 1 v...
View
Full
Document
This note was uploaded on 10/06/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Mass, Work

Click to edit the document details