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Midterm 01-solutions

# Midterm 01-solutions - Version 012 Midterm 01 florin(56930...

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Version 012 – Midterm 01 – florin – (56930) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points An 800 kg car skidding due north on a level frictionless icy road at 100 km / h collides with a 2110 kg car skidding due east at 44 . 8 km / h in such a way that the wreckage sticks to- gether and skid as shown in the picture. 2110 kg 44 . 8 km / h 100 km / h 800 kg v f θ N At what angle East of North do the two coupled cars skid? 1. 36.8846 2. 61.7675 3. 36.1832 4. 50.8886 5. 37.3848 6. 49.7584 7. 55.7113 8. 46.2873 9. 59.1821 10. 54.4406 Correct answer: 49 . 7584 . Explanation: Let : m 1 = 800 kg , v 1 = 100 km / h , m 2 = 2110 kg , and v 2 = 44 . 8 km / h . m 2 v 2 p 2 p 1 m 1 v 1 p f = m f v f θ N During the collision, the total momentum of the two car system will be conserved: vectorp f = vectorp 1 + vectorp 2 = m 1 vectorv 1 + m 2 vectorv 2 p fx ˆ ı + p fy ˆ = m 1 v 1 ˆ + m 2 v 2 ˆ ı . Looking at the x and y components of mo- mentum, p fx = m 2 v 2 = (2110 kg) (44 . 8 km / h) × parenleftbigg 10 3 m km parenrightbigg parenleftbigg 1 h 3600 s parenrightbigg = 26257 . 8 kg m / s , and p fy = m 1 v 1 = (800 kg) (100 km / h) × parenleftbigg 10 3 m km parenrightbigg parenleftbigg 1 h 3600 s parenrightbigg = 22222 . 2 kg m / s . Since we are asked to find the angle from the y -axis instead of the x , tan θ = p fx p fy = 26257 . 8 kg m / s 22222 . 2 kg m / s = 1 . 1816 θ = arctan parenleftbigg 26257 . 8 kg m / s 22222 . 2 kg m / s parenrightbigg = 49 . 7584 . 002 10.0points A hanging wire made of gold with diameter 0 . 09 cm is initially 2 m long and a cubic atomic

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Version 012 – Midterm 01 – florin – (56930) 2 structure. When a 62 kg mass is hung from it, the wire stretches an amount 1 . 13 cm. A mole ( N A = 6 . 02 × 10 23 atoms) of gold has a mass of 197 g / mol, and its density is 19 . 3 g / cm 3 . Use g = 9 . 8 m / s 2 . Based on these experimental measure- ments, find the approximate value of the effec- tive spring stiffness of one interatomic bond in gold. 1. 46.7481 2. 54.6406 3. 66.3317 4. 40.2155 5. 51.9423 6. 60.4593 7. 41.8459 8. 43.4275 9. 40.9284 10. 42.5678 Correct answer: 43 . 4275 N / m. Explanation: Let M be the weight of the mass hung from the wire, let d be the diameter of the wire, and let l and x be the original length of the wire and the amount by which it stretches, respectively. The tension in the wire is equal to the weight of the hanging mass, since the wire’s mass is negligible. Therefore, using the definition of Young’s modulus, and letting Y = F A L Δ L = 4 M g π d 2 l x = 4(62 kg)(9 . 8 m / s 2 ) π (0 . 09 cm) 2 parenleftbigg 2 m 1 . 13 cm parenrightbigg = 1 . 69042 × 10 11 N / m 2 . To find the stiffness of an interatomic bond, we use the expression k = Y a, where k is the stiffness and a is the length of a single interatomic spacing. We find the value of a by first finding the volume of a cubic lattice of atoms in the gold, and then taking the cubed root. Converting the given quantities into the necessary units, we find that the density and the molar mass of gold are ρ = 19300 kg / m 3 M = 0 . 197 kg / mol , respectively. Now the volume of a cubic lat- tice of atoms is given by V = ( 19300 kg / m 3 ) 1 (0 . 197 kg / mol) parenleftbigg 1 mol N A parenrightbigg = 1 . 69556 × 10 29 m 3 .
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