PHYS 1410, 11 Nov 2015, 12:30pm.

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1 PHYS 40, Nov 205, 2:30pm. A B = AB cos φ x = x 0 + v x0 t + a 2 xt 2 a ad = v2 2 m(v2 2 v) 2 θ = θ 0 + ω 0 t + 2 αt2 L = p fs µ s n 0 + αt K = 2 Iω2 cm = m +m m +m p = m v and L = I ω ω 2 = ω α(θ θ 0 ) P = F P + ρgy + A 2 ρv2 = constant PHYS 40: PHYSICAL SCIENCE Depatment of Physics and Astonomy Yok Univesity Fall test #2 six pieces of pape (double-sided) Wednesday Novembe, 205, 2:30pm-:30pm This test will be maked out of 6 points. The test contains 4 multiple choice questions ( point each) and 4 poblems (3 points each). No aids ae pemitted except a pen, a pencil, an ease and a simple calculato. Please don t panic. This test is woth 6% of you final mak in the couse. Please ente you student numbe hee: Row numbe: Seat numbe:

2 PHYS 40, Nov 205, 2:30pm. A B = AB cos φ x = x 0 + v x0 t + a 2 xt 2 a ad = v2 2 m(v2 2 v) 2 θ = θ 0 + ω 0 t + 2 αt2 L = p fs µ s n 0 + αt K = 2 Iω2 cm = m +m m +m p = m v and L = I ω ω 2 = ω α(θ θ 0 ) P = F P + ρgy + A 2 ρv2 = constant Opening a doo. [ point] As you ae leaving a building, the doo opens outwad. If the hinges on the doo ae on you ight, what is the diection of the angula velocity of the doo as you open it? (a) up (b) down (c) to you left (d) to you ight (e) fowad (f) backwad Answe: b A cosmic collision. [ point] A comet collides with an asteoid and the two objects emain stuck togethe. The asteoid s mass is 300 times lage than the comet s mass. Compae the comet s change of momentum p c with the asteoid s change in momentum p a. (a) p c = 300 p a (b) p c = p 300 a (c) p c = p a (d) p c = p a (e) Each object keeps the same momentum thoughout the collision. (f) Not enough infomation was given because momentum is not conseved in this situation. Answe: d

3 PHYS 40, Nov 205, 2:30pm. A B = AB cos φ x = x 0 + v x0 t + a 2 xt 2 a ad = v2 2 m(v2 2 v) 2 θ = θ 0 + ω 0 t + 2 αt2 L = p fs µ s n 0 + αt K = 2 Iω2 cm = m +m m +m p = m v and L = I ω ω 2 = ω α(θ θ 0 ) P = F P + ρgy + A 2 ρv2 = constant A ock in a tie. [ point] A bus is taveling with a speed v. A ock is stuck in the tead of one tie. When the ock is at the top of the tie, its speed with espect to the oad is (a) 2v (b) v (c) v/2 (d) zeo (e) The speed depends on the adius of the tie. (f) none of the above Answe: a A full cup. [ point] A cup of wate containing an ice cube at 0 C is filled to the bim. The ice cube is floating with its top in the ai above the suface. As the ice melts, you obseve that (a) the cup oveflows. (b) the cup might oveflow but it depends on the actual mass of the ice cube. (c) the wate level emains the same. (d) the wate level goes down. (e) Thee is not enough infomation to answe this question. Answe: c

4 PHYS 40, Nov 205, 2:30pm. A B = AB cos φ x = x 0 + v x0 t + a 2 xt 2 a ad = v2 2 m(v2 2 v) 2 θ = θ 0 + ω 0 t + 2 αt2 L = p fs µ s n 0 + αt K = 2 Iω2 cm = m +m m +m p = m v and L = I ω ω 2 = ω α(θ θ 0 ) P = F P + ρgy + A 2 ρv2 = constant Woking with a cate. [3 points] A 20.0 kg cate sits at est at the bottom of a 5. m long amp that is inclined at 36.0 degees above the hoizontal. A constant hoizontal foce of 290 N is applied to the cate to push it up the amp. While the cate is moving, the amp exets a constant fictional foce on it that has magnitude 65.0 N. (a) Daw the fee body diagam fo the cate. (b) What is the total wok done on the cate as it moves fom the bottom to the top of the amp? (c) How much time does it take the cate to tavel to the top of the amp? (a) simple sketch: F=290N fee body diagam: n fiction o o F mg (b) Wok is only done by the component of each foce that is along the diection of motion. The total foce in that diection is F x = (290N) cos(36.0 ) 65.0N (20.0kg)(9.80m/s 2 ) cos( ) = 54.4N so the wok done is W = (54.4N)(5.m) = 822Nm = 822J. (c) The cate begins fom est and acceleates. x = x 0 + v x0 t + 2 a xt 2 5.m = F x 2 m t2 2(5.m)(20.0kg) t = 54.4N = 3.33s

5 PHYS 40, Nov 205, 2:30pm. A B = AB cos φ x = x 0 + v x0 t + a 2 xt 2 a ad = v2 2 m(v2 2 v) 2 θ = θ 0 + ω 0 t + 2 αt2 L = p fs µ s n 0 + αt K = 2 Iω2 cm = m +m m +m p = m v and L = I ω ω 2 = ω α(θ θ 0 ) P = F P + ρgy + A 2 ρv2 = constant

6 PHYS 40, Nov 205, 2:30pm. A B = AB cos φ x = x 0 + v x0 t + a 2 xt 2 a ad = v2 2 m(v2 2 v) 2 θ = θ 0 + ω 0 t + 2 αt2 L = p fs µ s n 0 + αt K = 2 Iω2 cm = m +m m +m p = m v and L = I ω ω 2 = ω α(θ θ 0 ) P = F P + ρgy + A 2 ρv2 = constant Rolling basketballs. [3 points] The mass of a basketball is kg and the moment of inetia of a basketball is I = 2 3 MR2. (a) The fist basketball is eleased with an initial speed of 8.0 m/s. It olls patway up a wooden amp and then back down. Its speed is once again 8.0 m/s at the bottom. Find the maximum vetical height incease of the fist ball as it olls up the amp. (b) The second basketball is eleased with an initial speed of 8.0 m/s. It olls patway up a capeted amp and then back down. Its speed is then only 4.0 m/s at the bottom. Assume that the wok done by fiction is the same when the ball moves up o down the amp. Find the maximum vetical height incease of the second ball on the amp. (a) Enegy is conseved, so 2 mv Iω2 0 = mgh h = = = 2 mv Iω2 0 mg 2 mv mv2 0 mg ( 2 3 mr2) v 2 0 R 2 mg = 5v2 0 6g = 5(8.0m/s)2 6(9.80m/s 2 ) = 5.58m (b) Again enegy is conseved but now fiction does wok (name it W ), so 2 mv Iω2 0 W = mgh and mgh = 2 mv2 + 2 Iω2 + W 2mgh = 2 m(v2 0 + v 2 ) + 2 I(ω2 0 + ω 2 ) = 2 m(v2 0 + v 2 ) + ( ) ( ) 2 v mr2 R + v2 2 R 2 = 2 m(v2 0 + v 2 ) + 3 m(v2 0 + v 2 ) = 5 6 m(v2 0 + v 2 ) h = 5(v2 0 + v 2 ) 2g = 5[(8.0m/s)2 + (4.0m/s) 2 ] 2(9.80m/s 2 ) = 3.50m

7 PHYS 40, Nov 205, 2:30pm. A B = AB cos φ x = x 0 + v x0 t + a 2 xt 2 a ad = v2 2 m(v2 2 v) 2 θ = θ 0 + ω 0 t + 2 αt2 L = p fs µ s n 0 + αt K = 2 Iω2 cm = m +m m +m p = m v and L = I ω ω 2 = ω α(θ θ 0 ) P = F P + ρgy + A 2 ρv2 = constant

8 PHYS 40, Nov 205, 2:30pm. A B = AB cos φ x = x 0 + v x0 t + a 2 xt 2 a ad = v2 2 m(v2 2 v) 2 θ = θ 0 + ω 0 t + 2 αt2 L = p fs µ s n 0 + αt K = 2 Iω2 cm = m +m m +m p = m v and L = I ω ω 2 = ω α(θ θ 0 ) P = F P + ρgy + A 2 ρv2 = constant A safe amp fo ski jumpes. [3 points] You ae designing a ski jump amp fo the next Winte Olympics. The skies push off had with thei ski poles at the stat, just above the stating gate, so they typically have a speed of 2.0 m/s as they each the gate. Fo safety, the skies should have a speed of no moe than 30.0 m/s when they each the bottom of the amp. You detemine that fiction and ai esistance will do total wok of magnitude 4300 J on each skie duing the tip down the slope. (a) Fo an 83 kg skie, calculate the maximum vetical height h fom the stating gate to the bottom of the amp. (b) Is the height fom pat (a) safe fo skies with othe masses? If yes, then calculate which masses ae safe. If no, then explain why. (a) Enegy is conseved, so 2 mv2 0 + mgh = 2 mv2 + W h = = 2 m(v2 v0) 2 + W mg 2 (83kg)[(30.0m/s)2 (2.0m/s) 2 ] J (83kg)(9.80m/s 2 ) = 5m (b) Using a lage mass in pat (a) gives a smalle maximum height h. Fo example, m = 66 kg gives h = 48 kg. Theefoe the height in pat (a) is not safe fo skies with a lage mass. Using a smalle mass in pat (a) gives a lage maximum height h. Fo example, m = 4.4 kg gives h = 56 kg. Theefoe the height in pat (a) is safe fo skies with a smalle mass.

9 PHYS 40, Nov 205, 2:30pm. A B = AB cos φ x = x 0 + v x0 t + a 2 xt 2 a ad = v2 2 m(v2 2 v) 2 θ = θ 0 + ω 0 t + 2 αt2 L = p fs µ s n 0 + αt K = 2 Iω2 cm = m +m m +m p = m v and L = I ω ω 2 = ω α(θ θ 0 ) P = F P + ρgy + A 2 ρv2 = constant

10 PHYS 40, Nov 205, 2:30pm. A B = AB cos φ x = x 0 + v x0 t + a 2 xt 2 a ad = v2 2 m(v2 2 v) 2 θ = θ 0 + ω 0 t + 2 αt2 L = p fs µ s n 0 + αt K = 2 Iω2 cm = m +m m +m p = m v and L = I ω ω 2 = ω α(θ θ 0 ) P = F P + ρgy + A 2 ρv2 = constant Fake gold. [3 points] The Bank of Canada has hied you to find lead bas among thei collection of gold bas. All of thei bas have identical volumes. You plan is to measue the tension in a sting that holds each ba suspended in wate. Calculate the atio of the tensions, T lead T gold. IF YOU NEED THEM: density of gold = 9300 kg/m 3, density of lead = 300 kg/m 3, density of wate = kg/m 3. Tgold gold T lead lead Newton s second law gives T lead + B m lead g = 0 and T gold + B m gold g = 0 whee B = m wate g. Note that m wate is the mass of the wate displaced by a ba. Theefoe T lead = m leadg B T gold m gold g B = m leadg m wate g m gold g m wate g = m lead m wate m gold m wate = ρ lead ρ wate ρ gold ρ wate = = 0.563

11 PHYS 40, Nov 205, 2:30pm. A B = AB cos φ x = x 0 + v x0 t + a 2 xt 2 a ad = v2 2 m(v2 2 v) 2 θ = θ 0 + ω 0 t + 2 αt2 L = p fs µ s n 0 + αt K = 2 Iω2 cm = m +m m +m p = m v and L = I ω ω 2 = ω α(θ θ 0 ) P = F P + ρgy + A 2 ρv2 = constant

12 PHYS 40, Nov 205, 2:30pm. In case you need even moe space:

PHYSICS 1210 Exam 2 University of Wyoming 14 March ( Day!) points

PHYSICS 1210 Exam 2 University of Wyoming 14 March ( Day!) points PHYSICS 1210 Exam 2 Univesity of Wyoming 14 Mach ( Day!) 2013 150 points This test is open-note and closed-book. Calculatos ae pemitted but computes ae not. No collaboation, consultation, o communication