* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Version 072 – Midterm 2
Survey
Document related concepts
Newton's theorem of revolving orbits wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Faster-than-light wikipedia , lookup
Equations of motion wikipedia , lookup
Coriolis force wikipedia , lookup
Classical mechanics wikipedia , lookup
Specific impulse wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Hunting oscillation wikipedia , lookup
Fictitious force wikipedia , lookup
Jerk (physics) wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Center of mass wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Seismometer wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Transcript
Version 072 – Midterm 2 – OConnor (05141) This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. V1:1, V2:1, V3:3, V4:5, V5:3. BE SURE TO CIRCLE THE ANSWERS ON THE FRONT PAGES AND ATTACH YOUR WORK TO THE BACK. YOU MUST SHOW YOUR WORK! DO THE EASY PROBLEMS FIRST AND SAVE THE HARD ONES FOR LAST. 1 10. 5.08593 s Explanation: The initial x and y components of the velocity are vx0 = v0 cos θ0 = 9.18446 m/s vy0 = v0 sin θ0 = 3.95546 m/s . To find t, we can use the relation 1 2 gt 2 y = vy0 t − 001 (part 1 of 2) 10 points A stone is thrown from the top of a building upward at an angle of 23.3◦ to the horizontal and with an initial speed of 10 m/s, as in the figure. The height of the building is 52.7 m. The acceleration of gravity is 9.8 m/s2 . y v0 θ0 x with y = −52.7 m and vy0 = 3.95546 m/s (we have chosen the top of the building as the origin): 1 (3.95546 m/s) t − (9.8 m/s2 ) t2 2 = −52.7 m . 2 2 2 (4.9 m /s )t − (3.95546 m/s) t −52.7 m = 0 Applying the quadratic formula, since (−3.95546 m/s)2 − 4(4.9 m/s2 )(−52.7 m) = 1048.57 m2 /s2 , h then How long is the stone “in flight”? 1. 3.17932 s 2. 3.56725 s 3. 3.63387 s p 3.95546 m/s ± 1048.57 m2 /s2 t= 9.8 m/s2 = 3.70786 s 002 (part 2 of 2) 10 points What is the speed of the stone just before it strikes the ground? 1. 26.0056 m/s 4. 3.70786 s correct 2. 27.5092 m/s 5. 3.98715 s 3. 30.052 m/s 6. 4.15309 s 4. 33.1771 m/s 7. 4.3606 s 5. 33.6589 m/s correct 8. 4.46242 s 6. 34.7925 m/s 9. 4.78976 s 7. 35.8627 m/s Version 072 – Midterm 2 – OConnor (05141) 8. 36.8744 m/s 4. 42.8758 N 9. 38.0433 m/s 5. 47.8125 N 10. 40.1408 m/s Explanation: The y component of the velocity just before the stone strikes the ground can be obtained using the equation 8. 52.8437 N vy = vy0 − g t 9. 54.3434 N with t = 3.70786 s : 2 6. 48.9223 N correct 7. 49.7272 N 10. 56.8695 N Explanation: vy = (3.95546 m/s) − (9.8 m/s2 ) (3.70786 s) = −32.3816 m/s . Since vx = vx0 = 9.18446 m/s, the required speed is q v = vx2 + vy2 q = (9.18446 m/s)2 + (−32.3816 m/s)2 = 33.6589 m/s . 003 (part 1 of 3) 10 points A block of mass 2.29967 kg lies on a frictionless table, pulled by another mass 4.99207 kg under the influence of Earth’s gravity. The acceleration of gravity is 9.8 m/s2 . Given : m1 = 2.29967 kg , m2 = 4.99207 kg , µ = 0. a T T m1 N and m1 g m2 a m2 g The net force on the system is simply the weight of m2 . Fnet = m2 g = (4.99207 kg) (9.8 m/s2 ) = 48.9223 N . 2.29967 kg 004 (part 2 of 3) 10 points What is the magnitude of the acceleration a of the two masses? 1. 5.19695 m/s2 µ=0 4.99207 kg What is the magnitude of the net external force F acting on the two masses? 1. 29.6699 N 2. 5.80805 m/s2 3. 6.27197 m/s2 4. 6.36673 m/s2 2. 33.3338 N 5. 6.49314 m/s2 3. 40.1959 N 6. 6.60275 m/s2 Version 072 – Midterm 2 – OConnor (05141) 7. 6.70927 m/s2 correct 3 006 (part 1 of 3) 10 points A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are µs = 0.75 and µk = 0.64, respectively. The acceleration of gravity is 9.8 m/s2 . 8. 6.86492 m/s2 9. 7.87789 m/s2 10. 8.06403 m/s2 Explanation: From Newton’s second law, 42 Fnet = m2 g = (m1 + m2 ) a . Solving for a, m2 g a= m1 + m 2 4.99207 kg = (9.8 m/s2 ) 2.29967 kg + 4.99207 kg = 6.70927 m/s2 . 005 (part 3 of 3) 10 points What is the magnitude of the tension T of the rope between the two masses? 1. 9.62636 N 2. 9.75315 N kg µ 33◦ What is the frictional force acting on the 42 kg mass? 1. 60.5674 N 2. 63.8114 N 3. 89.3359 N 4. 91.7786 N 5. 95.9001 N 3. 10.0116 N 6. 113.806 N 4. 12.0003 N 7. 164.617 N 5. 14.3197 N 8. 181.3 N 6. 14.5665 N 9. 224.174 N correct 7. 15.4291 N correct 10. 261.536 N Explanation: 8. 16.3735 N 9. 16.5868 N N Ff 10. 17.2402 N Explanation: Analyzing the horizontal forces on block m1 , we have X Fx : T = m 1 a 2 = (2.29967 kg) (6.70927 m/s ) = 15.4291 N . 33◦ mg Version 072 – Midterm 2 – OConnor (05141) The forces acting on the block are shown in the figure. Since the block is at rest, the magnitude of the friction force should be equal to the component of the weight on the plane of the incline Ff = M g sin θ = (42 kg) (9.8 m/s2 ) sin 33◦ = 224.174 N . 007 (part 2 of 3) 10 points What is the largest angle which the incline can have so that the mass does not slide down the incline? 1. 19.7989 ◦ 2. 20.8068 ◦ 3. 24.2277 ◦ 4. 26.1048 ◦ 5. 27.0216 ◦ 6. 30.1137 ◦ 7. 33.0239 ◦ 8. 34.992 9. 36.1294 4 the incline if the angle of the incline is 44◦ ? 1. 0.854362 m/s2 2. 1.12114 m/s2 3. 1.1512 m/s2 4. 1.24553 m/s2 5. 1.47742 m/s2 6. 1.49405 m/s2 7. 1.59314 m/s2 8. 1.67388 m/s2 9. 2.06274 m/s2 10. 2.29596 m/s2 correct Explanation: When θ exceeds the value found in part 2, the block starts moving and the friction force is the kinetic friction Fk = µk N = µk M g cos θ. Newton’s equation for the block then becomes M a = M g sin θ − Ff = M g sin θ − µk M g cos θ ◦ and ◦ 10. 36.8699 ◦ correct Explanation: The largest possible value the static friction force can have is Ff,max = µs N , where the normal force is N = M g cos θ. Thus, since Ff = M g sin θ, M g sin θm = µs M g cos θm tan θm = µs θm = tan−1 (µs ) = tan−1 (0.75) = 36.8699◦ . 008 (part 3 of 3) 10 points What is the acceleration of the block down a = g [sin θ − µk cos θ] = (9.8 m/s2 ) [sin 44◦ − (0.64) cos 44◦ ] = 2.29596 m/s2 . 009 (part 1 of 1) 10 points As viewed by a bystander, a rider in a “barrel of fun” at a carnival finds herself stuck with her back to the wall. ω Version 072 – Midterm 2 – OConnor (05141) Which diagram correctly shows the forces acting on her? 1. correct 5 The acceleration of gravity is 980 cm/s2 . What is the coefficient of static friction between the coin and the turntable? 1. 0.30012 2. 0.340136 3. 0.364431 2. None of the other choices 4. 0.398288 5. 0.445269 3. 6. 0.473996 correct 7. 0.489796 8. 0.522573 4. 5. 9. 0.565149 10. 0.663265 Explanation: The normal force on the coin is N = m g . The force provided by friction immediately before slippage is given by f = µN = µmg. 6. Explanation: The normal force of the wall on the rider provides the centripetal acceleration necessary to keep her going around in a circle. The downward force of gravity is equal and opposite to the upward frictional force on her. Note: Since this problem states that it is viewed by a bystander, we assume that the free-body diagrams are in an inertial frame. 010 (part 1 of 1) 10 points A coin is placed 31 cm from the center of a horizontal turntable, initially at rest. The turntable then begins to rotate. When the speed of the coin is 120 cm/s (rotating at a constant rate), the coin just begins to slip. At this moment the centripetal acceleration (there is no tangential acceleration) provides a force m v2 F = . r Summation of forces yields m v2 , therefore r v2 µ= rg (120 cm/s)2 = (31 cm) (980 cm/s2 ) = 0.473996 . F = f = µmg = 011 (part 1 of 2) 10 points Consider the “loop-a-loop” setup, where a mass m is sliding along a frictionless track and the radius of the loop is R. Version 072 – Midterm 2 – OConnor (05141) C m R A D Given h = 5 R, determine the speed of the mass at A. p 1. |vA | = 4 2 g R p 2. |vA | = 3 2 g R p 3. |vA | = 2 2 g R correct p 4. |vA | = 3 g R p 5. |vA | = 4 g R p 6. |vA | = 2 g R Explanation: Applying the work energy theorem from C to A we get mgh − mgR = 1 2 m vA 2 mg(5 R − R) = 4 R m g = m v2 = mg. R Conservation of energy between C and B implies m g (h − 2 R) = 012 (part 2 of 2) 10 points Find the critical initial height of the mass, such that it would just barely pass the point B. 1 1 m v2 = m g R 2 2 m v2 R 5 R h = 2R + = R 2 2 mg = 013 (part 1 of 1) 10 points A(n) 63.2 g ball is dropped from a height of 65.8 cm above a spring of negligible mass. The ball compresses the spring to a maximum displacement of 4.25828 cm. The acceleration of gravity is 9.8 m/s2 . h 1 2 m vA 2 2 vA = 8gR p vA = 2 2 g R 1 1. h = R 3 5 2. h = R correct 2 3 3. h = R 2 2 R 3 1 5. h = R 2 Explanation: The critical condition at B implies that 4. h = B h 6 x Calculate the spring force constant k. 1. 364.509 N/m 2. 478.591 N/m correct 3. 511.447 N/m 4. 537.872 N/m 5. 560.028 N/m 6. 588.055 N/m 7. 621.622 N/m Version 072 – Midterm 2 – OConnor (05141) 8. 702.946 N/m 9. 851.065 N/m 7 9. 1.48144 m/s 10. 1.5 m/s Explanation: 10. 1028.17 N/m Explanation: Let : Let : m = 63.2 g , h = 65.8 cm , and x = 4.25828 cm . Using conservation of energy, we have 1 2 k x = m g (h + x) 2 m = 16.3 kg , M = 2000 kg , and v = 140 m/s . The cannon’s velocity immediately after it was fired is found by using conservation of momentum along the horizontal direction: M V + mv = 0 m v M from which where M is the mass of the cannon, V is the 2 m g (h + x) velocity of the cannon, m is the mass of the k= x2 cannon ball and v is the velocity of the cannon 2 (0.0632 kg)(9.8 m/s2 )(0.658 m + 0.0425828 m) ball. Thus, the cannon’s speed is = (0.0425828 m)2 m |v| |V | = = 478.591 N/m . M 16.3 kg = (140 m/s) 2000 kg 014 (part 1 of 2) 10 points A revolutionary war cannon, with a mass of 2000 kg, fires a 16.3 kg ball horizontally. The cannonball has a speed of 140 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired? 1. 0.797431 m/s 2. 0.872308 m/s 3. 1.12371 m/s ⇒ −V = = 1.141 m/s . 015 (part 2 of 2) 10 points The same explosive charge is used, so the total energy of the cannon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all other parameters remained the same? 1. 0.397969 m/s 4. 1.141 m/s correct 2. 0.435308 m/s 5. 1.16174 m/s 3. 0.560479 m/s 6. 1.22479 m/s 4. 0.569342 m/s correct 7. 1.37624 m/s 5. 0.579618 m/s 8. 1.42581 m/s 6. 0.611034 m/s Version 072 – Midterm 2 – OConnor (05141) r 8 m → v . M man 7. 0.686424 m/s 3. VEarth = + 8. 0.71115 m/s 4. VEarth = − v man . µ ¶ M → v man . 5. VEarth = − m → 9. 0.738726 m/s 10. 0.747671 m/s Explanation: By knowing the speeds of the cannon and the cannon ball, we can find out the total kinetic energy available to the system Knet = 1 1 m v2 + M V 2 . 2 2 This is the same amount of energy available as when the cannon is fixed. Let v 0 be the speed of the cannon ball when the cannon is held fixed. Then, 1 1 m v 02 = (m v 2 + M V 2 ) . 2 2 r M 2 v2 + V m r m =v 1+ M s ⇒ v0 = = (140 m/s) 1+ 16.3 kg 2000 kg = 140.569 m/s . Thus, the velocity difference is v 0 − v = 140.569 m/s − 140 m/s = 0.569342 m/s . 016 (part 1 of 1) 10 points Bill (mass m) plants both feet solidly on the ground and then jumps straight up with ve→ locity v . The earth (mass M ) then has velocity µ ¶ M → 1. VEarth = + v man . m ³m´ → v man . correct 2. VEarth = − M → 6. VEarth = + v man . r m → 7. VEarth = − v . M man ³m´ → v man . 8. VEarth = + M Explanation: The momentum is conserved. We have → → m v man + M V Earth = 0 So → V Earth = − ³m´ M → v man . 017 (part 1 of 1) 10 points A child bounces a 49 g superball on the sidewalk. The velocity change of the superball is from 21 m/s downward to 17 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the force exerted on the superball by the sidewalk? 1. 1414.4 N 2. 1452 N 3. 1489.6 N correct 4. 1545.6 N 5. 1579.2 N 6. 1640 N 7. 1680 N 8. 1760 N 9. 1804 N 10. 1856 N Explanation: Version 072 – Midterm 2 – OConnor (05141) Let : m = 49 g = 0.049 kg , vu = 17 m/s , vd = 21 m/s , and ∆t = 0.00125 s Choose the upward direction as positive. Then the impulse is I = F ∆t = ∆P = m vu − m (−vd ) = m (vu + vd ) m (vu + vd ) F = ∆t (49 g) (17 m/s + 21 m/s) = 0.00125 s = 1489.6 N . 018 (part 1 of 1) 10 points As shown in the top view above, a disc of mass m is moving horizontally to the right with speed v on a table with negligible friction when it collides with a second disc of mass 6 m. The second disc is moving horizontally v at the moment of to the right with speed 6 impact. The two discs stick together upon impact. v 6 v before m 6m vf after 7m The speed of the composite body immediately after the collision is 1. vf = 3 v. 10 9 1 v. 3 13 v. 3. vf = 45 1 4. vf = v. 2 4 5. vf = v. 9 2. vf = 6. None of these are correct. 2 v. correct 7 11 v. 8. vf = 27 3 9. vf = v. 5 3 10. vf = v. 7 Explanation: The total momentum of the system is conserved because there is no exterior force. So what we have is v (m + 6 m) vf = m v + 6 m 6¶ µ 6 7 m vf = m v 1 + 6 µ ¶ 6 6 7 vf = + v 6 6 12 v 7 vf = 6 (12) vf = v (6) (7) 12 v , so = 42 2 = v . 7 7. vf =