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HW #9-2-solutions

# HW #9-2-solutions - pasha(sep635 – HW 9-2 – Antoniewicz...

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Unformatted text preview: pasha (sep635) – HW 9-2 – Antoniewicz – (56445) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Two disks are initially at rest, each of mass M , connected by a string between their centers, as shown in the figure below. The disks slide on low-friction ice as the center of the string is pulled by a string with a constant force F through a distance d . The disks collide and stick together, having moved a distance b horizontally. F F b d String String String M M What is the final speed of the stuck- together disks? 1. F b M 2. r F b 2 M 3. r F d M 4. F b 2 M 5. r F b M correct 6. r F d 2 M Explanation: Apply the Energy Principle to the point- particle system: Δ K trans = ~ F net · Δ ~ r CM = F Δ x CM 1 2 (2 M ) v 2 CM,f = F b ⇒ v 2 CM,f = r F b M . 002 (part 2 of 2) 10.0 points When the disks collide and stick together, their temperature rises. Calculate the in- crease in internal energy of the disks, assum- ing that the process is so fast that there is insu ffi cient time for there to be much trans- fer of energy to the ice due to a temperature di ff erence. (Also ignore the small amount of energy radiated away as sound produced in the collisions between the disks.) 1. F ( d + b ) 2. F ( b- d ) 2 3. F ( d- b ) correct 4. F ( d- b ) 2 5. F ( b- d ) 6. F ( d + b ) 2 Explanation: Apply the Energy Principle to the real sys- tem: Δ E = W Δ K trans + Δ E int = F Δ x F F b + Δ E int = F d ⇒ Δ E int = F ( d- b ) . 003 20.0 points Consider two masses of 3 kg and 6 . 9 kg con- nected by a string passing over a pulley having a moment of inertia 7 . 9 g · m 2 about its axis of rotation, as in the figure below. The string does not slip on the pulley, and the system is released from rest. The radius of the pulley is . 37 m. pasha (sep635) – HW 9-2 – Antoniewicz – (56445) 2 47 cm . 37 m 7 . 9 g · m 2 ω 6 . 9 kg 3 kg Find the linear speed of the masses after the 6 . 9 kg mass descends through a distance 47 cm. Assume mechanical energy is con- served during the motion. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 89946 m / s. Explanation: Let : I = 7 . 9 g · m 2 = 0 . 0079 kg · m 2 , R = 0 . 37 m , m 1 = 3 kg , m 2 = 6 . 9 kg , and h = 47 cm = 0 . 47 m ....
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HW #9-2-solutions - pasha(sep635 – HW 9-2 – Antoniewicz...

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