TIME OF COMPLETION_______________ NAME_______SOLUTION______________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Version 1 Total Weight: 100 points Section 1 November 2, 2005 1. Check your examination for completeness prior to starting. There are a total of nine (9) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 50 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are five (5) multiple choice and four (4) calculation problems. Work all multiple choice problems and 4 calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 6:00 p.m. 7:15 p.m PROBLEM POINTS 1-5 20 6 20 7 20 8 20 9 20 TOTAL 100 PERCENTAGE CREDIT CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. The quantity ВЅ kx2 is a. The kinetic energy of the object. b. The elastic potential energy of the object. (4) c. The work done on the object by the force. d. The power supplied to the object by the force. 2. The total mechanical energy of the system a. Is equally divided between kinetic energy and potential energy. b. Is either all potential or all kinetic energy. (4) c. Can never be negative. d. Is constant if there are no nonconservative forces. 3. Two objects collide and stick together. Kinetic energy a. Is definitely conserved. (4) b. Is definitely not conserved. c. Could be conserved or not conserved depending on the masses. d. Is conserved only if there is no friction present. 4. Angular velocity is expressed in units of a. Meters per second. b. Radians per second. (4) c. Omegas per second. d. Arcs per second. 5. Negative work done on an object by an applied force implies that a. The kinetic energy of the object increases. (4) b. The applied force is not constant. c. The applied force is perpendicular to the displacement. d. The applied force is directed opposite to the displacement. 6. An automobile engine slows down from 4500 rpm to 1200 rpm in 2.50 s. Given: пЃ·пЂ°пЂ пЂЅпЂ 4500 rpm пЂ пЂ пЂ пЂ пЂ пЂ пЂ пЂ пЂ пЂ пЂ пЂ пЃ·пЂ пЂЅпЂ 1200 rpm t = 2.50 s a. Express these angular velocities in terms of rad/s. (4500 rev/min)(2пЃ° rad/rev)(1 min/60s) = 471 rad/s (1200 rev/min)(2пЃ° rad/rev)(1 min/60s) = 126 rad/s b. Calculate the angular acceleration of the engine, assumed constant. пЃ·пЂ пЂЅпЂ пЂ пЃ·пЂ°пЂ пЂ«пЂ пЃЎпЂ t пЃЎпЂ пЂЅпЂ (пЃ·пЂ пЂпЂ пЂ пЃ·пЂ°пЂ©пЂ пЂЇпЂ t = ((126 rad/s) вЂ“ (471 rad/s))/(2.50 s) = -138 rad/s2 c. The total number of revolutions the engine makes in this time. пЃ·пЂІпЂ пЂЅпЂ пЂ пЃ·пЂ°пЂІпЂ пЂ«пЂ пЂІпЂ пЃЎпЂ пЂЁпЃ±пЂ пЂпЂ пЃ±пЂ°) пЂЁпЃ±пЂ пЂпЂ пЃ±пЂ°) = (пЃ·пЂІпЂ пЂпЂ пЂ пЃ·пЂ°пЂІпЂ©пЂЇпЂЁпЂІпЂ пЃЎ) пЂЁпЃ±пЂ пЂпЂ пЃ±пЂ°) = (пЂЁ126 rad/sпЂ©пЂІпЂ пЂпЂ пЂ пЂЁ471 rad/sпЂ©пЂІпЂ©пЂЇпЂЁпЂІпЂ пЂЁпЂ138 rad/s2пЂ©) = 746 rad = 119 rev 7. A 1.00-kg box is initially at rest on a horizontal road. The coefficient of kinetic friction between the box and the road is 0.200. The box is pushed a distance of 3.00 m by a force of 30.0 N applied to the box at an angle of 25.0o below the horizontal. (Hint: the magnitude of the normal force that the road exerts on the box is n = A sinпЃ± + mg) a. Find the work done by the applied force. WA = (30.0 N)(3.00 m)cos(25.0o) = 81.6 J b. Find the energy dissipated by friction. fk = пЃk n = пЃk(A sinпЃ± + mg) = (0.200)((30.0 N) sin(25.0o) + (1.00 kg)(9.81 m/s2)) = 4.50 N Wfk = (4.50 N)(3.00 m)cos(180o) = -13.5 J c. Find the change in the kinetic energy of the box. Wnet пЂЅпЂ пЃ„K Wnet = 81.6 N + (-13.5 N) = 68.1 J пЃ„K = 68.1 J Bonus (5 points): Find the speed of the box after it has traveled 3.00 m. пЃ„K = Kf вЂ“ Ki = Kf вЂ“ 0 = Kf = ВЅ mV2 ВЅ mV2пЂ пЂ пЂЅпЂ пЃ„K V2пЂ пЂ пЂЅпЂ пЂЁпЂІпЂ пЃ„K)/m VпЂ пЂ пЂЅпЂ пЂЁпЂЁпЂІпЂ пЃ„K)/m)1/2 V = 11.7 m/s 8. A 2.00-kg block is pushed against a spring that has a spring constant of 500 N/m, compressing the spring by 20.0 cm. The block is then released, and the spring projects it along a frictionless horizontal surface. Right before a frictionless incline of angle 45o, there is another 3.00-kg block. Two blocks collide and stick together. (20) a. What is the speed of the 2.00-kg block the moment it looses the contact with the spring? Ei = m g yi + ВЅ mVi2 + ВЅ kxi2 = 0 + 0 + ВЅ kxi2 = ВЅ kxi2 Ef = m g yf + ВЅ mVf2 + ВЅ kxf2 = 0 + ВЅ mVf2 + 0 = ВЅ mVf2 Ei = Ei ВЅ kxi2 = ВЅ mVf2 kxi2 = mVf2 Vf2 = kxi2 /m Vf = (kxi2 /m)1/2 Vf = ((500 N/m)(0.200 m)2 /(2.00 kg))1/2 = 3.16 m/s b. What is the speed of combined mass right after the collision? m1V1i + m2V2i = (m1 + m2 )Vf Vf = (m1V1i) / (m1 + m2 ) = (2.00 kg)(3.16 m/s) / (2.00 kg + 3.00 kg) = 1.26 m/s c. How far up the incline do the blocks travel before momentarily coming to rest? Ei = m g yi + ВЅ mVi2 + ВЅ kxi2 = 0 + ВЅ mVi2 + 0 = ВЅ mVi2 Ef = m g yf + ВЅ mVf2 + ВЅ kxf2 = m g yf + 0 + 0 = m g yf Ei = Ei ВЅ mVi2 = m g yf Vi2 = 2 g yf yf = Vi2 /(2 g) = 0.0816 m пЂ L = yf / sin(45.0o) = 0.115 m пЂ пЂ 9. A small car of mass 1.20 Mg (1.20 x 103 kg) traveling east at 60.0 km/h collides at an intersection with a truck of mass 3.00 Mg traveling west at 40.0 km/h. a. The car and truck stick together. Find the velocity of the wreckage just after the collision. m1V1i + m2V2i = (m1 + m2 )Vf Vf = (m1V1i + m2V2i)/ (m1 + m2 ) = ((1.20 x 103 kg)(60.0 km/h) + (3.00 x 103 kg)(-40.0 km/h))/ (1.20 x 103 kg + 3.00 x 103 kg) = -11.4 km/h = -3.17 m/s b. Calculate the decrease in kinetic energy that results from the collision. Ki = ВЅ m1V1i2 + ВЅ m2V2i2 = ВЅ (1.20 x 103 kg) (16.7 m/s)2 + ВЅ (3.00 x 103 kg) (-11.1 m/s)2 = 352000 J = 352 kJ Kf = ВЅ (m1 + m2)Vf2 = ВЅ (4.20 x 103 kg) (-3.17 m/s)2 = 21000 J = 21.0 kJ пЃ„K = 21.0 kJ вЂ“ 352 kJ = -330 kJ c. What happens to the вЂњlostвЂќ kinetic energy? The kinetic energy was transformed into the energy of sound waves, energy of deformation, internal energy, etc. TIME OF COMPLETION_______________ NAME___SOLUTION__________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Version 1 Total Weight: 100 points Section 1 November 6, 2006 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 6:00 p.m. 7:15 p.m PROBLEM POINTS 1-6 20 7 20 8 20 9 20 10 20 TOTAL 100 CREDIT PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. What is the SI unit of momentum? a. Nв€™m. b. N/s. (4) c. Nв€™s. d. N/m. 2. Which of the following is an accurate statement? a. The momentum of a projectile is constant. b. The momentum of a moving object is constant. (4) c. If an object is acted on by a non-zero net external force, its momentum will not remain constant. d. If the kinetic energy of an object is doubled, its momentum will also double. 3. A boy and a girl are riding on a merry-go-round which is turning at a constant rate. The boy is near the outer edge, and the girl is closer to the center. Who has the greater angular displacement? a. The boy. b. The girl. (4) c. Both have the same non-zero angular displacement. d. Both have zero angular displacement. 4. Rolling without slipping depends on a. Kinetic friction between the rolling object and the ground. b. Static friction between the rolling object and the ground. (4) c. Tension between the rolling object and the ground. d. The force of gravity between the rolling object and the Earth. 5. If the net work done on an object is zero, then the object's kinetic energy a. Decreases. b. Remains the same. (4) c. Increases. d. Is zero. 6. A ball falls from the top of a building, through the air (air friction is present), to the ground below. How does the kinetic energy (K) just before striking the ground compare to the potential energy (U) at the top of the building? a. K is equal to U. b. K is greater than U. (4) c. K is less than U. d. It is impossible to tell. 7. An automobile engine slows down from 4500 rpm to 1200 rpm in 2.50 s. a. Express the initial and final angular velocities in rad/s. пЃ·0 = (4500 rev/min)(2пЃ° rad/rev)(1 min/60 s) = 471 rad/s пЃ· = (1200 rev/min)(2пЃ° rad/rev)(1 min/60 s) = 126 rad/s b. Calculate engineвЂ™s angular acceleration, assumed constant, and пЃ· = пЃ·0 + пЃЎ t пЃЎ = (пЃ· - пЃ·0) /пЂ t пЃЎ = (126 rad/s вЂ“ 471 rad/s) / (2.50 s) = - 138 rad/s2 c. the total number of revolutions the engine makes in this time. пЂ пЃ·пЂІ = пЃ·02+ 2пЃЎ (пЃ±пЂ пЂпЂ пЃ±пЂ°) (пЃ±пЂ пЂпЂ пЃ±пЂ°) = (пЃ·пЂІ - пЃ·02)/ (2пЃЎпЂ© (пЃ±пЂ пЂпЂ пЃ±пЂ°) = ((126 rad/s)2 - (471 rad/s)2)/(2пЂ пЂЁпЂ138 rad/s2пЂ©пЂ© = 746 rad = 119 rev 8. A 6.0-kg block is pushed 8.0 m up a frictionless 37Вє inclined plane by a horizontal force of 75 N. If the initial speed of the block is 2.2 m s up the plane calculate a. The initial kinetic energy of the block; Ki = ВЅ m Vi 2 = ВЅ (6.00 kg) (2.20 m/s) 2 = 14.5 J b. The work done by the 75-N force; WA = A d cos (пЃ±) = (75.0 N)(8.00 m)cos (37.0o) = 479 J c. The work done by gravity; Ww = w d cos (127.0o) = (m g) d cos (127.0o) = (6.00 kg) (9.81 m/s2) (8.00 m) cos (127.0o) = -283 J d. The work done by the normal force; Wn = n d cos (90o) = 0 e. The final kinetic energy of the block. Wnet = 479 J + (-283 J) + 0 = 196 J Wnet = Kf - Ki Kf = Ki + Wnet = 14.5 J + 196 J = 211 J 9. A bullet is fired vertically into a 1.40-kg block of wood at rest directly above it. If the bullet has a mass of 29.0 g and a speed of 510 m s , how high will the block rise after the bullet becomes embedded in it? Inelastic collision: m1 V1i + m2 V2i = (m1 + m2)Vf Vf = (m1 V1i + m2 V2i)/(m1 + m2) = [(0.029 kg)(510 m/s) + (1.40 kg)(0)]/(0.029 g + 1.40 kg) = 10.3 m/s Upward motion (total energy is conserved): m g yi + ВЅ m Vi 2 = m g yf + ВЅ m Vf 2 ВЅ Vi 2 = g yf yf = ВЅ Vi 2 / g = ВЅ (10.3 m/s)2 / (9.81 m/s2) = 5.41 m 10. Tarzan swings on a 30.0-m long vine initially inclined at an angle of 37.0o with the vertical. What is his speed at the bottom of the swing a. If he starts from rest? m g yi + ВЅ m Vi 2 = m g yf + ВЅ m Vf 2 m g yi = m g yf + ВЅ m Vf 2 g yi = g yf + ВЅ Vf 2 Vf 2 = 2 (g yi - g yf) Vf = [2 (g yi - g yf)]1/2 yi = L ( 1- cos (пЂіпЂ·пЂ®пЂ°пЃЇ)) = 6.04 m yf = 0 Vf = 10.9 m/s b. If he pushes off with a speed of 4.00 m/s? m g yi + ВЅ m Vi 2 = m g yf + ВЅ m Vf 2 g yi + ВЅ Vi 2 = g yf + ВЅ Vf 2 Vf 2 = 2 (g yi - g yf) + ВЅ Vi 2 Vf = [2 (g yi - g yf) + ВЅ Vi 2 ]1/2 yi = L ( 1- cos (пЂіпЂ·пЂ®пЂ°пЃЇ)) = 6.04 m yf = 0 Vf = 11.2 m/s TIME OF COMPLETION_______________ NAME____SOLUTION_________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Version 1 Total Weight: 100 points Section 1 October 29, 2007 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 6:00 p.m. 7:15 p.m PROBLEM POINTS 1-6 20 7 20 8 20 9 20 10 20 CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. A mass is attached to one end of a string. The other end is attached to a rigid support. The mass is released at A and swings in a vertical arc to points B, C, and D, as shown below. Neglect air resistance. 1. At what point does the mass have the most gravitational potential energy? (A) A. (B) B. (C) C. (D) D. (E) None of the given points. 2. At what point does the mass have the most kinetic energy? (A) A. (B) B. (C) C. (D) D. (E) None of the given points. 3. At what point does the mass have its highest speed? (A) A. (B) B. (C) C. (D) D. (E) None of the given points. 4. A container of water is lifted vertically 3.00 m then returned to its original position. If the total weight is 30 N, how much work was done? (A) 45.0 J. (B) 90.0 J. (C) 180 J. (D) No work was done. (E) Negative work is done. 5. What is the SI unit of momentum? (A) Nв€™m. (B) N/s. (C) Nв€™s. (D) N/m. (E) W. 6. The area under the curve on a Force versus time (F vs. t) graph represents (A) Impulse. (B) Momentum. (C) Work. (D) Kinetic energy. (E) Does not represent a physical quantity. 7. The force on a 10.0 kg-mass object, acting along the x axis, varies as shown below. Determine the work done by this force to move the object (a) from x пЂЅ 0.0 to x пЂЅ 4.00 m, and W = ВЅ (4.00 m)(500 N) = 1000 J (b) from x пЂЅ 0.0 to x пЂЅ 6.00 m. W = ВЅ (4.00 m)(500 N) - ВЅ (1.00 m)(200 N) вЂ“ (1.00 m)(200 N) = 700 J (c) If the object is at rest initially, what is the final speed of the object? W = пЃ„K Kf = ВЅ mVf 2 = W Vf 2 = 2 W /m Vf = (2 W /m)1/2 = 11.8 m/s 8. A sled of 20.0 kg mass is initially given a shove up a 28.0Вє incline. Its initial speed is 5.15 m/s. (a) How far would it move along the slope before coming to rest if the slope were frictionless? Ei = Ef Ei = ВЅ m Vi 2 Ef = m g yf ВЅ m Vi 2 = m g yf ВЅ Vi 2 = g yf yf = (ВЅ Vi 2)/g = 1.35 m d = yf /sin (28.0Вє) = 2.88 m (b) You find that the sled stops after traveling through a distance of 2.30 m (measured along the slope). How much work is done on the sled by friction? Wf = Ef вЂ“ Ei = m g yf - ВЅ m Vi 2 yf = d sin (28.0Вє) = 1.08 m Wf = -53.3 J (c) Find the average friction force that acts on the sled. Wf = f d cos (180o) = -f d f = - Wf / d = - (-53.3 J)/(2.30 m) = 23.2 N 9. A 15.0-kg object moving in the пЂ« x direction at 5.5 m s collides head-on with a 10.0-kg object moving in the пЂ x direction at 4.0 m s . Find the final velocity of each mass if: (a) the objects stick together; m1V1i + m2V2i = (m1 + m2 )Vf Vf = (m1V1i + m2V2i) /(m1 + m2 ) = 1.7 0m/s (b) the 15.0-kg object is at rest after the collision; m1V1i + m2V2i = m2 Vf Vf = (m1V1i + m2V2i) /(m2 ) = 4.25 m/s (c) the 10.0-kg object is at rest after the collision. m1V1i + m2V2i = m1 Vf Vf = (m1V1i + m2V2i) /(m1 ) = 2.83 m/s 10. A 6.00-kg block is pushed 8.00 m up a rough 37Вє inclined plane by a horizontal force of 75 N. If the initial speed of the block is 2.2 m s up the plane and a constant kinetic friction force of 25 N opposes the motion, calculate (a) the initial kinetic energy of the block; Ki = ВЅ mVi2 = ВЅ (6.00 kg)(2.20 m/s)2 =14.5 J (b) the work done by the 75-N force; WA = F d cos (37.0o) = (75.0 N) (8.00 m) cos (37.0o) = 479 J (c) the work done by the friction force; Wf = f d cos (180o) = (25.0 N) (8.00 m) cos (180o) = -200 J (d) the work done by gravity; Wg = w d cos (180o) = (58.9 N) (8.00 m) cos (127o) = -283 J (e) the work done by the normal force; Wn = 0 (f) the final kinetic energy of the block. Wnet= пЃ„K = Kf вЂ“ Ki Kf = Ki + Wnet Kf = 14.5 J + (-4.00 J) = 10.5 J TIME OF COMPLETION_______________ NAME_________SOLUTION____________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Version 1 Total Weight: 100 points Section 1 March 30, 2005 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on nine (9) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Do all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the room. 7. Start: Stop: 10:30 a.m. 11:45 a.m. PROBLEM POINTS 1-6 24 7 19 8 19 CREDIT 9 19 10 19 TOTAL 100 POINTS EARNED CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. Wt. No. 04 1. 04 04 2. 5. A block of mass, m, is released from rest at point A and slides down the frictionless surface in the figure below. The kinetic energy of the block at point B is equal to a. (0.300)mgh. b. (0.700)mgh. c. mgh. d. None of the above. If the work done by all the external forces acting on an object is 5.80 J, and the object is initially at rest, its final kinetic energy is a. zero. b. 2.90 J c. 5.80 J d. cannot be determined. A wheel is initially rotating with an angular velocity of 20.0 radians/s. If the wheel rotates through an angular displacement of 100 radians while slowing down to rest, the constant angular acceleration was a. -2.00 radians/s2. пЃ·пЂІ = пЃ·o2 + 2 пЃЎ (пЃ±пЂ пЂпЂ пЃ±пЂ°) b. 0.200 radians/s2. пЂЁпЃ·пЂІ - пЃ·o2)/ (2(пЃ±пЂ пЂпЂ пЃ±пЂ°)) = пЃЎ rad/s) )/(2 x 100 rad) пЃЎпЂ пЂЅпЂ пЂЁпЂ°пЂ пЂ (20.0 2 ` Wt. No. 04 4. 04 5. c. 5.00 radians/s2. d. -200 radians/s2. A person riding a bicycle is traveling due North, and decreases her linear velocity from 10.0 m/s to 3.00 m/s. The direction of the angular acceleration vector of the front wheel points a. East. b. West. c. North. d. downward. The curve shown below represents the magnitude of the force exerted on a ball during a collision with a wall plotted against time. The area under the curve represents: a. the force on the ball. b. the work done on the ball. c. the change in linear momentum of the ball. d. the power delivered by the wall to the ball. PHYS 1111 Exam 2, Version 1 Spring 2005 2 Wt. No. 19 7. An applied force of magnitude A = 50.0 N, oriented at an angle of 35.0п‚° with respect to the vertical as shown in the diagram below, pushes a block of mass, m = 3.00 kg, up a wall a distance of 1.80 m. The coefficient of kinetic friction between the block and the wall is 0.230. (04) a. Draw a free body diagram for the block. (04) b. Calculate the work done by the applied force. PHYS 1111 Exam 2, Version 1 Spring 2005 3 WA = A d cos(пЃ±) = (50.0 N)(1.80 m)cos (35.0o) =73.7J (04) c. Calculate the work done by the weight of the block. Ww = w d cos(пЃ±) = (3.00 kg)(9.80 m/s2)(1.80 m)cos (180o) = -52.9 J (07) d. Calculate the work done by the kinetic friction force. Wf = fk d cos(пЃ±) = пЃk n d cos(пЃ±) n = A sin (35.0o) = 28.7 N Wf = (0.230)(28.7 N)(1.80 m)cos (180o) = -11.9 J Wt. No. 19 9. (05) A bullet of mass, m1 = 0.0200 kg, is fired horizontally into a block of wood of mass, m2 = 8.00 kg. The initial speed of the bullet is 300 m/s. The block rests on a horizontal surface. a. If the collision between the bullet and the block is perfectly inelastic, calculate the final speed of the masses immediately after the collision. m1V1i = (m1 + m2)Vf Vf = m1V1i /(m1 + m2) = 0.748 m/s PHYS 1111 Exam 2, Version 1 Spring 2005 4 (14) b. A wooden block, m2 = 8.00 kg, as in part a above is now attached to an unextended horizontal spring of spring constant of 300 N/m. A bullet, m1 = 0.0200 kg, is again fired horizontally into the block and again undergoes a perfectly inelastic collision. Assume the speed of the masses immediately after the collision is 0.748 m/s. If the maximum compression of the spring is 0.0668 m, calculate the coefficient of kinetic friction between the block with the bullet and the horizontal surface. WNC = TEf - TEi = ВЅ k x2 вЂ“ ВЅ (m1 + m2) Vf2 WNC = fk x cos (180o) = - (m1 + m2) g пЃk x - (m1 + m2) g пЃk x = ВЅ k x2 вЂ“ ВЅ (m1 + m2) Vf2 пЃk = - (ВЅ k x2 вЂ“ ВЅ (m1 + m2) Vf2)/[(m1 + m2) g x] пЃk = 0.300 Wt. No. 19 10. A coin is located 0.250 m from the center of a rotating turntable. The coefficient of static friction between the coin and the turntable is 0.500. PHYS 1111 Exam 2, Version 1 Spring 2005 5 (06) a. Draw a complete free body diagram for the coin. (13) b. Calculate the magnitude of the maximum tangential velocity of the coin so that it does not slip off of the turntable. Show your work. FC = m aC = mV2/R FC = fS = пЃS n = пЃS mg mV2/R = пЃS mg PHYS 1111 Exam 2, Version 1 Spring 2005 6 V2 = пЃS g R V = ( пЃS g R)ВЅ V = 1.11 m/s TIME OF COMPLETION_______________ NAME____SOLUTION_________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Version 1 Total Weight: 100 points Section 2 March 29, 2006 1. Check your examination for completeness prior to starting. There are a total of nine (9) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are five (5) multiple choice and four (4) calculation problems. Work all multiple choice problems and 4 calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 6:00 p.m. 7:15 p.m PROBLEM POINTS 1-5 20 6 20 PHYS 1111 Exam 2, Version 1 Spring 2005 7 CREDIT 7 15 8 20 9 15 TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. When is kinetic energy conserved? a. In elastic collisions b. In inelastic collisions (4) c. In any collision in which the objects do not stick together d. In all collisions 2. Angular acceleration is expressed in units of a. Meters per second squared. b. Radians per second squared. (4) c. Alphas per second squared. d. Arcs per second squared. 3. A container of water is lifted vertically 3.0 m then returned to its original position. If the total weight is 30 N, how much work was done? PHYS 1111 Exam 2, Version 1 Spring 2005 8 a. 45 J b. 90 J (4) c. 180 J d. No work was done. 4. The quantity 1/2 mv2 is a. The kinetic energy of the object. b. The potential energy of the object. (4) c. The work done on the object by the force. d. The power supplied to the object by the force. 5. Describe the energy of a car driving up a hill. a. Entirely kinetic. b. Entirely potential. (4) c. Both kinetic and potential. d. Gravitational. 6. A pitcher throws a baseball with mass 0.145 kg straight upward with an initial speed of 25.0 m/s. a. How much work has gravity done on the baseball when it reaches a height of 20.0 m above the pitcherвЂ™s hand? Wg = m g yi - m g yf = 0 вЂ“ (0.145 kg)(9.81 m/s2)(20.0 m) = -28.4 J PHYS 1111 Exam 2, Version 1 Spring 2005 9 b. Use the energy approach to calculate the speed of the baseball at a height of 20.0 m above the pitcherвЂ™s hand. You can ignore air resistance. m g yi + ВЅ m Vi 2 = m g yf + ВЅ m Vf 2 ВЅ Vi 2 = g yf + ВЅ Vf 2 Vi 2 = 2 g yf + Vf 2 Vf = (Vi 2 - 2 g yf) 1/2 Vf = 15.3 m/s c. Does the answer depend on whether the baseball is moving upward or downward at a height of 20.0 m? Explain. No, the speed of the baseball will not depend on whether the ball is moving up or down, the velocity though will. d. Bonus (5 points): Assuming that the ball reaches the height of 20.0 m with the speed of 10.0 m/s, how much work was done on the baseball by air resistance? WNC = Ef вЂ“ Ei = (m g yf + ВЅ m Vf 2) вЂ“ (m g yi+ ВЅ m Vi 2) = -9.61 J 7. A bicycle wheel has an initial angular velocity of 1.50 rad/s. a. If its angular acceleration is constant and equal to 0.300 rad/s2, what is its angular velocity at t = 2.50 s? пЃ·o = 1.50 rad/s пЃЎ = 0.300 rad/s2 t = 2.50 s пЂ пЃ· =пЂ пЃ·o +пЂ пЃЎпЂ t = (1.50 rad/s) + (0.300 rad/s2)(2.50 s) = 2.25 rad/s PHYS 1111 Exam 2, Version 1 Spring 2005 10 b. Through what angle has the wheel turned between t = 0 and t = 2.50 s? пЃ± =пЂ пЃ±0 +пЂ пЃ·0 t +пЂ пЂ±пЂЇпЂІпЂ пЃЎпЂ t2 = (1.50 rad/s)(2.50 s) + ВЅ (0.300 rad/s2)(2.50 s)2 = 4.69 rad c. Draw the wheel and indicate the direction of its angular velocity and its angular acceleration. Assuming that the wheel rotates in counterclockwise direction, both angular velocity and angular acceleration point out of the page. 8. A 2.00 kg piece of wood slides on the surface shown below. The curved sides are perfectly smooth, but the rough horizontal bottom is 10.0 m long and has a coefficient of kinetic friction with the wood of 0.200. The piece of wood starts from rest 4.00 m above the rough bottom. a. Find the magnitude of the kinetic friction that acts on the wood as it moves over the rough bottom. fk = пЃk n = пЃk m g = (0.200)(2.00 kg)(9.81 m/s2) = 3.92 N b. How much work is done by friction on the wood as the wood crosses the rough bottom? PHYS 1111 Exam 2, Version 1 Spring 2005 11 Wf = F d cos (пЃ±) = (3.92 N)(10.0 m)cos (180o) = -39.2 J c. Where will this wood eventually come to rest? Wf = Ef вЂ“ Ei = (m g yf) вЂ“ (m g yi) Wf /(m g) = yf вЂ“ yi yf = Wf /(m g) + yi = 2.00 m пЂ 9. Your 1050-kg sports car, parked on a hill without the parking brake set, has rolled to the bottom of the hill and is moving westward on a level road at a speed of 15.0 m/s. A truck driver driving east on the same road sees the sports car coming, and decides to stop the car by running his 6320-kg truck into it. The two vehicles remain locked together after the collision. a. At what speed should the truck be moving so that the truck and car are both stopped in the collision? m1 V1i + m2 V2i = (m1 + m2)Vf Vf = 0 m1 V1i + m2 V2i = 0 m1 V1i = - m2 V2i V1i = - (m2 V2i )/m1 V1i = 2.49 m/s b. How much energy will be lost in the collision? Ei = ВЅ m1 V1i 2 + ВЅ m2 V2i 2 = 137, 717 J Ef = 0 пЃ„E = -137, 717 J PHYS 1111 Exam 2, Version 1 Spring 2005 12 TIME OF COMPLETION_______________ NAME___SOLUTION__________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Version 1 Total Weight: 100 points Section 1 March 26, 2007 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 6:00 p.m. 7:15 p.m PROBLEM POINTS 1-6 20 7 20 8 20 9 20 10 20 PHYS 1111 Exam 2, Version 1 Spring 2005 13 CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. In a two-body collision, if the momentum of the system is conserved, then which of the following best describes the kinetic energy after collision? a. Must be less. b. Must also be conserved. (4) c. May also be conserved. d. Is doubled in value. 2. The impulse experienced by a body is equivalent to its change in: a. Velocity. b. Kinetic energy. (4) c. Momentum. d. None of the above choices are valid. 3. The unit of power, Watt, is dimensionally the same as: a. Joule-second. b. Joule/second. (4) c. Joule-meter. PHYS 1111 Exam 2, Version 1 Spring 2005 14 d. Joule/meter. 4. As an object is lowered into a deep hole in the surface of the earth, which of the following is true in regard to its gravitational potential energy? a. Potential energy increases. b. Potential energy decreases. (4) c. Potential energy stays constant. d. Not enough information given. 5. A very light cart holding a 300 N box is moved at a constant velocity across a 15 m level surface. What is the net work done in the process? a. Zero. b. 1/20 J. (4) c. 20 J. d. 2000 J. 6. You drop a 60.0 g golf ball from 2.00 m high. It rebounds to 1.50 m. How much energy is lost? a. 0.294 J. b. 0.500 J. PHYS 1111 Exam 2, Version 1 Spring 2005 15 (4) c. 0.883 J. d. 1.00 J. 7. A 55-kg hiker starts at an elevation of 1600 m and climbs to the top of a 3300-m peak. a. What is the hikerвЂ™s change in potential energy? Ui = m g yi Uf = m g yf пЃ„U = m g yf - m g yi = m g (yf - yi) = (55.0 kg)(9.81 m/s2)(3300 m вЂ“ 1600 m) = 917,235 J b. What is the minimum work required of the hiker? Wmin = 917,235 J, in the perfect world without friction, air resistance, etc. c. Can the actual work done be more than this? Explain why. Yes, because he will work against friction, air resistance, as well as possibly increase his kinetic energy. 8. A 65-kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.0 m s . a. How fast is he going as he lands on the trampoline, 3.0 m below? yi = 3.00 m yf = 0 PHYS 1111 Exam 2, Version 1 Spring 2005 16 Vi = 5.00 m/s Vf = ? Ei = Ef m g yi + ВЅ m Vi 2 = m g yf + ВЅ m Vf 2 g yi + ВЅ Vi 2 = ВЅ Vf 2 2 g yi + Vi 2 = Vf 2 Vf = (Vi 2 - 2 g yf) 1/2 Vf = 9.16 m/s 4 b. If the trampoline behaves like a spring with spring stiffness constant 6.2 п‚ґ10 N m , how far does he depress it? m g yi + ВЅ m Vi 2 = m g yf + ВЅ m Vf 2 + ВЅ k yf 2 now Vf = 0, m g yi + ВЅ m Vi 2 = m g yf + ВЅ k yf 2 This is a quadratic equation which you need to solve for yf yf = - 0.286 m 8. A skier traveling 12.0 m s reaches the foot of a steady upward 18.0Вє incline and glides 12.2 m up along this slope before coming to rest. a. How much work was done by friction? Wf = Ef вЂ“ Ei = m g yf вЂ“ ВЅ m Vi 2 PHYS 1111 Exam 2, Version 1 Spring 2005 17 yf = (12.2 m)sin (18.0Вє) = 3.77 m, Vi = 12.0 m/s Wf = -2,100,978 J b. What was the magnitude of the average friction force acting on the skier? Wf = f d cos (180o) = - f d f = - Wf /d = - (- 2,100,978 J)/(12.2 m) = 172 N 10. A 9300-kg boxcar traveling at 15.0 m s strikes a second boxcar at rest. The two stick together and move off with a speed of 6.0 m s . a. What is the mass of the second car? m1 V1i + m2 V2i = (m1 + m2)Vf Vf = 6.00 m/s, V1i = 15.0 m/s, V2i = 0 m1 V1i = (m1 + m2)Vf m1 V1i = m1 Vf + m2 Vf m2 = (m1 V1i - m1 Vf )/Vf m2 = (9300 kg)(15.0 m/s) вЂ“ 6.00 m/s)/(6.00 m/s) = 13, 950 kg b. How much kinetic energy was lost during collision? Ei = ВЅ m1 V1i 2 = 1,046, 250 J PHYS 1111 Exam 2, Version 1 Spring 2005 18 Ef = ВЅ m1 Vf 2 + ВЅ m2 Vf 2 = 418,500 J пЃ„E = -627, 750 J TIME OF COMPLETION_______________ NAME__SOLUTION___________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Version 1 Total Weight: 100 points Section 1 April 2, 2008 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 6:00 p.m. 7:15 p.m PROBLEM POINTS 1-6 20 7 20 8 20 PHYS 1111 Exam 2, Version 1 Spring 2005 19 CREDIT 9 20 10 20 TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. A mass is attached to one end of a string. The other end is attached to a rigid support. The mass is released from rest at A and swings in a vertical arc to points B, C, and D, as shown below. Neglect air resistance. 1. At what point does the mass have the least gravitational potential energy? (A) A. (B) B. (C) C. (D) D. (E) None of the given points. 2. At what point does the mass have the least kinetic energy? (A) A. (B) B. (C) C. (D) D. (E) None of the given points. PHYS 1111 Exam 2, Version 1 Spring 2005 20 3. At what point does the mass have negative gravitational potential energy? (A) A. (B) B. (C) C. (D) D. (E) Answer depends on the choice of zero level for gravitational potential energy. 4. In an inelastic collision, if the momentum is conserved, then which of the following statements is true about kinetic energy? (A) Kinetic energy is also conserved. (B) Kinetic energy is gained. (C) Kinetic energy is lost. (D) Kinetic energy can either be gained or lost. (E) None of the above is correct. 5. A person applies a constant force on an object of mass 20.0 kg that causes the object to move horizontally at a constant speed of 0.200 m/s through a distance of 0.800 m along a rough surface. What is the net work done on the object? (A) 160 J. (B) 10.0 J. (C) 16.0 J. (D) 0 J. (E) Cannot be determined without knowing the magnitude of the applied force. 6. A ball drops some distance and loses 30.0 J of gravitational potential energy. Do NOT ignore PHYS 1111 Exam 2, Version 1 Spring 2005 21 air resistance. How much kinetic energy did the ball gain? (A) More than 30.0 J. (B) Exactly 30.0 J. (C) Less than 30.0 J. (D) Exactly 0.0 J. (E) Exactly 60.0 J. 7. A 100 kg warthog (Phacochoerus aethiopics) squeals with delight as it slides from rest down the greased (frictionless) slope and plane as pictured below. a. How fast does the hog move as it just reaches the bottom of the slope? Ei = m g yi + ВЅ m Vi 2 = m g yi Ef = m g yf + ВЅ m Vf 2 = ВЅ m Vf 2 Ei = Ef m g yi = ВЅ m Vf 2 g yi = ВЅ Vf 2 Vf 2 = 2 g yi Vf = 9.90 m/s b. As the hog continues along the plane, it encounters another hog of 150 kg mass, initially at PHYS 1111 Exam 2, Version 1 Spring 2005 22 rest. The hogs collide and as a result of the collision the 150-kg hog acquires 5.20 m/s velocity to the right. How fast does the 100-kg hog move after the collision and in what direction? m1 V1i + m2 V2i = m1 V1f + m2 V2f m1 V1f = m1 V1i + m2 V2i - m2 V2f V1f = (m1 V1i + m2 V2i - m2 V2f )/m1 V1f = ((100 kg)(9.90 m/s) - (150 kg)(5.20 m/s))/(100 kg) = 2.10 m/s, to the right 10. You are pushing a 7.00 kg box along a rough ceiling with a force of 180 N directed at 35.0o to the horizontal. The magnitude of the friction force acting on the box is 15.0 N. a. Calculate the work done by each force on the box after you pushed it through 0.750 m. PHYS 1111 Exam 2, Version 1 Spring 2005 23 W = F пЃ„x cos (пЃ±) WA = A пЃ„x cos(пЃ±) = (180 N) (0.750 m) cos(35.0o) = 147 J Wn = n пЃ„x cos(пЃ±) = n (0.750) cos(90.0o) = 0 J Ww = wпЂ пЃ„x cos(пЃ±) = (7.00 kg)(9.81 m/s2) (0.750 m) cos(90.0o)= 0 J Wf = fпЂ пЃ„x cos(пЃ±) = (15.0 N)(0.750 m) cos(180o)= - 11.3 J b. Calculate the total work done on the box. WNET = WA + Wn + Ww + Wf = 147 J + 0 J + 0 J - 11.3 J = 136 J c. If initially the box was at rest, what is the final speed of the box after you pushed it through 0.750 m? WNET = ВЅ m Vf 2 вЂ“ ВЅ mVi 2 PHYS 1111 Exam 2, Version 1 Spring 2005 24 Vi = 0 m/s ВЅ mVf 2 = WNET Vf = в€љ( 2WNET/m) = 6.23 m/s 11. The force shown in the force-time diagram below acts on a 1.50 kg object. The force is directed along the x-axis. Find a. The impulse of the force. I = (2.00 N)(3.00 s) + ВЅ (2.00 N)(2.00 s) = 8.00 N-s b. The final velocity of the object if it is initially at rest. I = m Vf вЂ“ mVi Vi = 0 m/s Vf = I/m = 5.33 m/s c. The final velocity of the object if it is initially moving along the x axis with a velocity of -2.00 m/s. I = m Vf вЂ“ mVi PHYS 1111 Exam 2, Version 1 Spring 2005 25 Vi = -2.00 m/s Vf = (I + mVi )/m = 3.33 m/s 12. The launching mechanism of a toy gun consists of a spring of unknown spring constant, as shown below. If the spring is compressed a distance of 0.120 m and the gun is fired vertically as shown, the gun launches a 20.0-g projectile from rest to a maximum height of 20.0 m above the starting point. Neglecting all resistive forces, determine a. The spring constant, b. The speed of the projectile as it moves through the equilibrium position of the PHYS 1111 Exam 2, Version 1 Spring 2005 26 spring. TIME OF COMPLETION_______________ NAME____SOLUTION_________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Version 1 Total Weight: 100 points Section 1 July 18, 2005 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on eight (8) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 80 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 11:30 a.m. 12:50 p.m PROBLEM POINTS 1-6 30 7 20 8 20 9 15 PHYS 1111 Exam 2, Version 1 Spring 2005 27 CREDIT 10 15 TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. A work done by static friction is e. Always positive. (5) f. Always negative. g. Always zero. h. Can be either depending on circumstances. 2. For an object moving along a circular trajectory at a constant speed: a. Its velocity is constant. b. Its acceleration is zero. (5) c. Its acceleration is perpendicular to its instantaneous velocity. d. None of these. 3. When you sit on a chair, the resultant force on you is e. Zero. f. Up. (5) g. Down. PHYS 1111 Exam 2, Version 1 Spring 2005 28 h. Depends on your weight. 4. An object in motion must have: a. Acceleration. b. Momentum. (5) c. Potential energy. d. All of the above. 5. A garbage truck crashes head-on into a Volkswagen and the two come to rest in a cloud of flies. Which experiences the greater impact force? i. The truck. j. The Volkswagen. (5) k. Both experience the same force. l. Not enough information. 6. Angular displacement is usually express in units of e. Meters. (5) f. Radians. g. Revolutions. PHYS 1111 Exam 2, Version 1 Spring 2005 29 h. Arcs. . 7. A 1.00-kg object moving at 10.0 m/s to the right collides with a stationary 2.00-kg object. The collision is perfectly inelastic. Surface is frictionless. (20) a. What are the magnitude and direction of the combined objectвЂ™s velocity immediately after collision? m1V1i + m2V2i = (m1 + m2 )Vf Vf = (m1V1i) / (m1 + m2 ) = (1.00 kg)(10.0 m/s) / (1.00 kg + 2.00 kg) = 3.33 m/s b. What is the kinetic energy of the combined object immediately after collision? KE = ВЅ m Vf2 = ВЅ (3.00 kg) (3.33 m/s)2 = 16.6 J c. If we neglect friction, how far along the inclined plane will the combined object travel? KEi = PEf = (m1 + m2 ) g yf yf = KEi /((m1 + m2 ) g) = 0.566 m or 0.940 m along the incline. PHYS 1111 Exam 2, Version 1 Spring 2005 30 8. A skier with a mass of 80.0 kg starts from rest at the top of a slope and skis down from an elevation of 110 m. The speed of the skier at the bottom of the slope is 20.0 m/s. a. Find the change in the skierвЂ™s potential energy. пЃ„PE = m g yf - m g yf = - (80.0 kg)(9.80 m/s2)(110 m) = - 86,240 J b. Find the change in the skierвЂ™s kinetic energy. пЃ„KE = ВЅ m Vf2 - ВЅ m Vi2 = ВЅ (80.0 kg) (20.0 m/s)2 = 16,000 J c. How much work was done by friction force? Wnc = пЃ„KE +пЂ пЃ„PE = - 84,240 J + 16,000 J = -70,240 J d. What average friction force acted on the skier during his descent, if the total path length along the slope was 250 m? Wnc = -fk d fk = - Wnc /d= 281 N 9. An 2 18.0-kg box is released on a 37.0Вє incline and accelerates down the incline at 0.270 m s . a. Draw a free-body diagram. PHYS 1111 Exam 2, Version 1 Spring 2005 31 b. Find the friction force impeding its motion. nx = n cos (90o) = 0 ny = n sin (90o) = N fkx = fk cos (180o) = - fk fky = fk sin (180o) = 0 wx = m g sin (37o) wy = - m g cos (37o) пЃ“Fx = ax - fk + m g sin (37o) = m ax fk = m g sin (37o) - m ax = (18.0 kg) (5.90 m/s2 вЂ“ 0.270 m/s2) = 101 N c. Bonus: How large is the coefficient of kinetic friction? пЃ“Fy = 0 => n - m g cos (37o) = 0 n = m g cos (37o) = 140 N пЃk = fk / n = 0.721 пЂ PHYS 1111 Exam 2, Version 1 Spring 2005 32 10. A merry-go-round makes 24.0 revolutions in 3.00-minute interval. a. What is its angular velocity in rad/s? пЃ·пЂ° = (24.0 rev)(2пЃ° rad /1 rev)/(3.00 min)(1 min/60 sec) = 0.838 rad/s b. The merry-goвЂ“round starts to slow down uniformly and comes to a rest after completing 6.00 revolutions. What was its angular acceleration during that time? пЃ·пЂІпЂ пЂЅпЂ пЂ пЃ·пЂ°пЂІпЂ пЂ«пЂ пЂІпЂ пЃЎпЂ пЂЁпЃ±пЂ пЂпЂ пЃ±пЂ°) пЃЎпЂ пЂЅпЂ пЂЁпЃ·пЂІпЂ пЂпЂ пЃ·пЂ°пЂІпЂ пЂ©пЂЇпЂ пЂЁпЂІпЂ пЂЁпЃ±пЂ пЂпЂ пЃ±пЂ°)) пЃЎпЂ пЂЅпЂ пЂЁпЂ°пЂІпЂ пЂпЂ пЂЁпЂ°.838 rad/sпЂ©пЂІпЂ©пЂЇпЂ пЂЁ2 (6 xпЂ пЂІпЂ пЃ° rad)) = -0.00931 rad/s2 c. How long did it take to come to rest? пЃ·пЂ пЂЅпЂ пЂ пЃ·пЂ°пЂ пЂ«пЂ пЃЎпЂ t tпЂ пЂЅпЂ пЂЁпЃ·пЂ пЂпЂ пЂ пЃ·пЂ°пЂ©пЂЇпЃЎпЂ = 90.0 s TIME OF COMPLETION_______________ NAME___SOLUTION__________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Version 1 Total Weight: 150 points Section 1 July 16, 2007 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). PHYS 1111 Exam 2, Version 1 Spring 2005 33 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 150 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 1:30 p.m. 2:45 p.m PROBLEM POINTS 1-6 30 7 30 8 30 9 30 10 30 TOTAL 150 CREDIT PERCENTAGE NOTE: ALL MULTIPLE CHOICE PROBLEMS BELOW ARE WORTH 6 POINTS EACH. CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. The quantity mgy is (A) The kinetic energy of the object. (B) The elastic potential energy of the object. (C) The gravitational potential energy of the object. PHYS 1111 Exam 2, Version 1 Spring 2005 34 (D) The work done on the object by a force. (E) The power supplied to the object by the force. 2. An object is held at a height h above the ground. A second object with four times the mass of the first is held at the same height. The potential energy of the second object compared to the first is (A) One-forth as much. (B) One-half as much. (C) Twice as much. (D) Four times as much. (E) Eight times as much. 3. An acorn falls from a tree. (A) Its kinetic energy increases and its gravitational potential energy decreases. (B) Its kinetic energy increases and its gravitational potential energy increases. (C) Its kinetic energy decreases and its gravitational potential energy increases. (D) Its kinetic energy decreases and its gravitational potential energy decreases. (E) Both kinetic and gravitational potential energies remain constant. 4. Two children are riding on a merry-go-round. Child A is at a greater distance from the axis of rotation than child B. Which child has the larger angular displacement? (A) Child A. (B) Child B. (C) They have the same zero angular displacement. (D) They have the same non-zero angular displacement. (E) There is not enough information given to answer the question. PHYS 1111 Exam 2, Version 1 Fall 2006 35 5. Identical forces act for the same time interval on two different masses. The change in momentum of the smaller mass is (A) Smaller than the change in momentum of the larger mass, but not zero. (B) Larger than the change in momentum of the larger mass. (C) Equal to the change in momentum of the larger mass. (D) Zero. (E) There is not enough information to answer the question. 6. It is more difficult to start moving a heavy box from rest than it is to keep pushing it with constant velocity, because (A) The normal force is greater when the box is at rest. пЂЁпЃ‚пЂ©пЂ пЃs < пЃk (C) Initially, the normal force is not perpendicular to the normal force. пЂЁDпЂ©пЂ пЃs > пЃk пЂЁпЃ…пЂ©пЂ пЃs = пЃk 7. A 6.00-kg block slides down a frictionless incline making an angle of 60.0o with the horizontal. a. List all the forces acting on the block, and find the work done by each force when the block slides 2.00 m (measured along the incline). Gravitational force and normal force Ww = (6.00 kg)(9.81 m/s2)(2.00 m)cos(30.0o) = 102 J Wn = 0 J PHYS 1111 Exam 2, Version 1 Fall 2006 36 b. What is the total work done on a block? Wnet = Ww + Wn = 102 J ` c. What is the speed of the block after it slid 2.00 m if it starts with an initial speed of 1.50 m/s? Wnet = Kf вЂ“ Ki = ВЅ mVf2 - ВЅ mVi2 ВЅ mVf2 = Wnet + ВЅ mVi2 Vf2 = 2 Wnet /m + Vi2 Vf = (2 Wnet /m + Vi2) 1/2 V f = 6.02 m/s 8. A 5.00 вЂ“kg object moving to the right with a speed of 4.00 m/s collides head-on with a 10.0kg object moving toward it with a speed of 3.00 m/s. The 10.0-kg object stops dead after the collision. a. What is the final velocity of the 5.00-kg object? Please specify both magniude and direction. m1V1i + m2V2i = m1V1f + m2 V2f V2f = 0 m1V1i + m2V2i = m1 V1f V1f = (m1V1i + m2V2i)/ m1 V1f = ((5.00 kg)(4.00 m/s) + (10.0 kg)(-3.00 m/s))/(5.00 kg) = -2.00 m/s (after collision the 5.00-kg object is moving with a speed of 2.00 m/s to the left) b. Is this collision elastic or inelastic? Ki = ВЅ m1V1i2 + ВЅ m2V2i2 = ВЅ (5.00 kg) (4.00 m/s)2 + ВЅ (10.0 kg) (3.00 m/s)2 = 85.0 J PHYS 1111 Exam 2, Version 1 Fall 2006 37 Kf = ВЅ m1V1f2 = ВЅ (5.00 kg) (2.00 m/s)2 = 10.0 J Kinetic energy is not conserved, collision is inelastic. 9. A compact disc rotates from rest to 500 rev/min in 5.50 s. a. What is the angular acceleration of the disc assuming it is constant? (500 rev/min)(2пЃ° rad/rev)(1 min/60s) = 52.4 rad/s пЃ·пЂ пЂЅпЂ пЂ пЃ·пЂ°пЂ пЂ«пЂ пЃЎпЂ t пЃЎпЂ пЂЅпЂ (пЃ·пЂ пЂпЂ пЂ пЃ·пЂ°пЂ©пЂ пЂЇпЂ t = ((52.4 rad/s) вЂ“ (0 rad/s))/(5.50 s) = 9.52 rad/s2 b. How many revolutions does it make in 5.50 s? пЃ·пЂІпЂ пЂЅпЂ пЂ пЃ·пЂ°пЂІпЂ пЂ«пЂ пЂІпЂ пЃЎпЂ пЂЁпЃ±пЂ пЂпЂ пЃ±пЂ°) пЂЁпЃ±пЂ пЂпЂ пЃ±пЂ°) = (пЃ·пЂІпЂ пЂпЂ пЂ пЃ·пЂ°пЂІпЂ©пЂЇпЂЁпЂІпЂ пЃЎ) пЂЁпЃ±пЂ пЂпЂ пЃ±пЂ°) = (пЂЁ52.4 rad/sпЂ©пЂІпЂ пЂпЂ пЂ пЂЁ0 rad/sпЂ©пЂІпЂ©пЂЇпЂЁпЂІпЂ пЂЁпЂ№пЂ®пЂµпЂІ rad/s2пЂ©) = 144 rad = 23.0 rev c. What is the centripetal acceleration of the point on the rim of the disc (6.00 cm from the center) at the moment the disc reaches angular velocity of 500 rev/min? acp = пЃ·пЂІпЂ R = (52.4 rad/s)2(0.0600 m) = 165 m/s2 10. You drop your physics textbook (of 4.00 kg mass) from the second floor window 7.00 m above the ground. a. Neglecting the air resistance, how fast do you expect it to move as it reaches the ground? Ei = m g yi + ВЅ mVi2 = m g yi PHYS 1111 Exam 2, Version 1 Fall 2006 38 Ef = m g yf + ВЅ mVf2 = ВЅ mVf2 Ei = Ef m g yi = ВЅ mVf2 g yi = ВЅ Vf2 Vf = (2 g yi)1/2 Vf = (2 (9.81 m/s2) (7.00 m))1/2 Vf = 11.7 m/s b. You measure the speed of the book just before it hits the ground and find it to be 9.00 m/s. You correctly assume that the air resistance slows the book down. How much work is done on the book by air resistance? Wf = Ef вЂ“ Ei = ВЅ m Vf 2 - m g yi Wf = Ef вЂ“ Ei = ВЅ (4.00 kg) (9.00 m/s) 2 - (4.00 kg) (9.81 m/s2)(7.00 m) = -113 J c. What is the average air resistance force that acts on the book during the fall? Wf = f d cos (180o) = - f d f = - Wf /d = - (- 113 J)/(7.00 m) = 16.1 N PHYS 1111 Exam 2, Version 1 Fall 2006 39 PHYS 1111 Exam 2, Version 1 Fall 2006 40 PHYS 1111 Exam 2, Version 1 Fall 2006 41