# quiz 3 - Version 105 Quiz 3 coker(58245 This print-out...

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Version 105 – Quiz 3 – coker – (58245) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This quiz covers topics from class lectures over Chs. 10, 11, 12, 13 and 14 in the text- book. 001 10.0 points A 80 . 1 kg ice skater, moving at 8 . 4 m / s, crashes into a stationary skater of equal mass. After the collision, the two skaters move as a unit at 4 . 2 m / s. Suppose the average force a skater can experience without breaking a bone is 4424 N. If the impact time is 0 . 11 s, what is the magnitude of the average force each skater experiences? 1. 4862.41 2. 3496.96 3. 7131.25 4. 5807.08 5. 2884.72 6. 1375.86 7. 3659.15 8. 5118.75 9. 3891.77 10. 3058.36 Correct answer: 3058 . 36 N. Explanation: Let : m = 80 . 1 kg , v i = 8 . 4 m / s , v f = 4 . 2 m / s , F = 4424 N , and Δ t = 0 . 11 s . From the impulse-momentum equation, F av Δ t = Δ p = m ( v f - v i ) . F av = m ( v f - v i ) Δ t = (80 . 1 kg) (4 . 2 m / s - 8 . 4 m / s) 0 . 11 s = - 3058 . 36 N , which has a magnitude of 3058 . 36 N . The average force on skater 2 has the same magni- tude but opposite direction by Newton’s third law. This force is not great enough to break a bone. 002 (part 1 of 2) 10.0 points Consider the rectangular block of mass m = 43 kg height h = 0 . 58 m, length l = 0 . 7 m. A force F is applied horizontally at the upper edge. The acceleration of gravity is 9 . 8 m / s 2 . l m h F What is the minimum force required to start to tip the block? 1. 86.45 2. 81.0133 3. 238.378 4. 123.284 5. 129.392 6. 112.531 7. 109.564 8. 75.5491 9. 155.018 10. 254.293 Correct answer: 254 . 293 N. Explanation: Basic Concepts: In equilibrium summationdisplay vector F = 0 summationdisplay vector τ = 0 Part 1: Let the right lower corner of the block be denoted as A . Then summationdisplay τ A = h F - l 2 m g = 0 Therefore F = m g l 2 h = (43 kg) (9 . 8 m / s 2 ) 0 . 7 m 2 (0 . 58 m) = 254 . 293 N .

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Version 105 – Quiz 3 – coker – (58245) 2 003 (part 2 of 2) 10.0 points What is the minimum coefficient of static fric- tion required for the block to tip with the application of a force of this magnitude? 1. 0.603448 2. 0.243333 3. 0.254167 4. 0.427083 5. 0.290909 6. 0.211538 7. 0.396907 8. 0.555556 9. 0.414894 10. 0.5 Correct answer: 0 . 603448. Explanation: summationdisplay F y = N A - m g = 0 Therefore N A = m g and summationdisplay F x = F - f = F - μN A = 0 . Solving for μ , μ = F N A = F m g = 254 . 293 N (43 kg) (9 . 8 m / s 2 ) = 0 . 603448 004 10.0 points A sticky blob strikes and sticks to a free rod, which is initially at rest, as shown. Let E be the mechanical energy of the sys- tem, vector P the linear momentum of the system, and vector L the angular momentum of the system. What is conserved? 1. vector P only 2. vector L and vector P correct 3. vector L and E 4. vector L , vector P , and E 5. vector L only 6. vector P and E Explanation: The mechanical energy of the system is not conserved because this is not an elastic colli- sion, but the linear momentum and angular momentum are always conserved in such free collisions. 005 10.0 points A 2 . 28 kg particle has a velocity of v x = 1 . 18 m / s and v y = 4 . 32 m / s. Find the magnitude of its total momentum. 1. 25.9406 2. 11.181 3. 19.6792 4. 8.89824 5. 11.8751 6. 13.0993 7. 15.1486 8. 23.9244 9. 10.2104 10. 15.8511 Correct answer: 10 . 2104 kg · m / s.
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