HW #1-3-solutions

HW #1-3-solutions - pasha (sep635) HW #1-3 Antoniewicz...

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pasha (sep635) – HW #1-3 – Antoniewicz – (56445) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 2) 5.0 points PowerFul sports cars can go From zero to 25 m/s (about 60 mph) in 5 seconds. What is the magnitude oF the acceleration? Answer in m / s 2 . Correct answer: 5 m / s 2 . Explanation: To fnd the magnitude oF a constant accel- eration, we can simply create the Following ratio: v fnal - v inital t fnal - t initial = 25 m / s - 0 5s - 0 =5m / s 2 . 002 (part 2 of 2) 5.0 points How does this compare with the acceleration oF a Falling rock? 1. It is about twice the acceleration oF a Falling rock. 2. It is much less than the acceleration oF a Falling rock. 3. The acceleration oF a Falling rock increases as it Falls, so the question is not meaningFul. 4. The two accelerations are about the same. 5. It is about halF the acceleration oF a Falling rock. correct 6. It is much greater than the acceleration oF a Falling rock. Explanation: Objects near the surFace oF the earth all Fall at g 10 m / s 2 .T h e r e F o r e t h e c a r s acceleration is about halF that oF a Falling rock. 003 (part 1 of 2) 5.0 points AtennisballoFmass57gtravelswithvelocity ~ v i = h 51 , 0 , 0 i m / s toward a wall. AFter bouncing o f the wall, the tennis ball is observed to be traveling with velocity ~ v f = h- 46 , 0 , 0 i m / s . ~ v i ~ v f BeFore AFter Notice that this ball only has motion in the x direction, so the change in momentum will be oF the Form Δ ~ p = h Δ ~ p x , 0 , 0 i . ±ind Δ ~ p x .An swe rinkg · m / s. Correct answer: - 5 . 529 kg · m / s. Explanation: To fnd the change in momentum, we pro- ceed as Follows: Δ ~ p = ~ p f - ~ p i = m ~ v f - m ~ v i = m ( ~ v f - ~ v i ) = m ( h- 46 , 0 , 0 i m / s -h 51 , 0 , 0 i m / s) =(0 . 057 kg) ×h- 97 , 0 , 0 i m / s = h- 5 . 529 , 0 , 0 i kg · m / s . So Δ ~ p x = - 5 . 529 kg · m / s 004 (part 2 of 2) 5.0 points What is the change in the magnitude oF the
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pasha (sep635) – HW #1-3 – Antoniewicz – (56445) 2 we want the change in magnitude, we need to take the magnitudes of ~ p i and ~ p f and Fnd the change. We would usually use the Pythagorean theorem to Fnd the magnitude of a vector, but in this case, our vectors have only one component each, so Fnding the mag- nitudes is easy. ±or ~ p i , | ~ p i | = | (0 . 057 kg) ×h 51 , 0 , 0 i m / s | = | h 2 . 907 , 0 , 0 i kg · m / s | =2 . 907 kg · m / s . And for ~ p f , ± ± ~ p f ± ± = | (0 . 057 kg) ×h- 46 , 0 , 0 i m / s | = | h- 2 . 622 , 0 , 0 i kg · m / s | =2 . 622 kg · m / s . So the change in magnitude is
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This note was uploaded on 11/17/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

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HW #1-3-solutions - pasha (sep635) HW #1-3 Antoniewicz...

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