PHY302K-Sai-Midterm 3-solutions

# PHY302K-Sai-Midterm 3-solutions - Version 191 Midterm 3 sai...

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Version 191 – Midterm 3 – sai – (58015) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A uniForm beam oF weight W b and length \$ has weights W 1 and W 2 at two positions. W 1 W 2 cm x \$ d OP The beam is resting at two points. ±or what value oF x will the beam be balanced at P such that the normal Force at O is zero? 1. x = ( W 1 + W b ) d + W 2 ± \$ 2 ² W 1 2. x = ( W 1 + W b ) d + W 1 ± \$ 2 ² W 2 correct 3. x = ( W 1 + W b ) d + W 2 ± \$ 2 ² ( W 1 + W 2 ) 4. x = ( W 1 + W 2 ) d + W 1 ± \$ 2 ² ( W 1 + W b ) 5. x = ( W 2 + W b ) d + W 1 ± \$ 2 ² W 2 6. x = ( W 1 + W b ) d + W 2 ± \$ 2 ² ( W 1 + W b ) 7. x = ( W 1 + W 2 ) d + W 1 ± \$ 2 ² W 2 8. x = ( W 1 + W b ) d + W 2 ± \$ 2 ² W 2 Explanation: In equilibrium, ³ % F = 0 and ³ =0 . Take the torques about the point P . ³ τ P = - N O ± \$ 2 + d ² + W 1 ± \$ 2 + d ² + W b d - W 2 x . We want to fnd x For which N O x = ( W 1 + W b ) d + W 1 ± \$ 2 ² W 2 . keywords: 002 (part 1 oF 2) 10.0 points In a particular crash test, an automobile oF mass 1745 kg collides with a wall and bounces back o² the wall. The x components oF the initial and fnal speeds oF the automobile are 15 m / s and 2 . 4m / s, respectively. IF the collision lasts For 0 . 13 s, fnd the magnitude oF the impulse due to the collision. 1. 24531.4 2. 43916.6 3. 30363.0 4. 26061.7 5. 35622.2 6. 28556.0 7. 22337.5 8. 20959.5 9. 21853.8 10. 36080.0 Correct answer: 30363 kg · m / s. Explanation: Let : m = 1745 kg , v i = 15 m / s , v f =2 . / s , and Δ t . 13 s . The initial and fnal momenta oF the auto- mobile are %p i = m%v i

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Version 191 – Midterm 3 – sai – (58015) 2 = (1745 kg) ( - 15 m / s) =( - 26175 kg · m / s)ˆ ı %p f = m%v f = (1745 kg) (2 . 4m / s) = (4188 kg · m / ı. Thus the impulse is % I = Δ = f - i = [(4188 kg · m / s) - ( - 26175 kg · m / s)] ˆ ı = (30363 kg · m / ı, with magnitude 30363 kg · m / s . 003 (part 2 of 2) 10.0 points Calculate the magnitude of the average force exerted on the automobile during the colli- sion. 1. 233562.0 2. 405000.0 3. 188259.0 4. 260651.0 5. 131397.0 6. 135284.0 7. 185422.0 8. 411480.0 9. 255635.0 10. 139855.0 Correct answer: 2 . 33562 × 10 5 N. Explanation: The average force exerted on the automo- bile during the collision is ¯ F = Δ Δ t = (30363 kg · m / s) (0 . 13 s) ˆ ı = (2 . 33562 × 10 5 N) ˆ ı with magnitude 2 . 33562 × 10 5 N . 004 (part 1 of 2) 10.0 points A Fgure skater rotating on one spot with both arms and one leg extended has moment of inertia I i . She then pulls in her arms and the extended leg, reducing her moment of inertia to 0 . 75 I i . What is the ratio of her Fnal to initial ki- netic energy? 1. 1 2. 3/8 3. 16/9 4. 9/16 5. 4/3 correct 6. 1/2 7. 8/3 8. 3/4 9. 2 Explanation: The angular momentum was conserved, so L i = L f Since L = I ω and KE r = 1 2 2 , r = L 2 2 I So, f i = L 2 f 2 I f L 2 i 2 I i = I i I f = I i 0 . 75 I i = 4 3 005 (part 2 of 2) 10.0 points
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## This note was uploaded on 02/22/2012 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas at Austin.

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PHY302K-Sai-Midterm 3-solutions - Version 191 Midterm 3 sai...

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