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## Practice Exam 3 for General Physics 1

by: Ericah Notetaker

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# Practice Exam 3 for General Physics 1 PHYS113

Marketplace > Kansas State University > Art > PHYS113 > Practice Exam 3 for General Physics 1
Ericah Notetaker
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## About this Document

Speed, angle, force, distance, kinetic energy, velocity
COURSE
General Physics 1
PROF.
Horton Smith, Glenn A.
TYPE
Study Guide
PAGES
18
WORDS
CONCEPTS
Physics, General Physics, general, direction, acceleration, velocity, Force, time, practice exam, exam, PHYS113, Average acceleration, average, vertical, horizontal, Energy, kinetic energy, distance, angle
KARMA
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This 18 page Study Guide was uploaded by Ericah Notetaker on Tuesday March 29, 2016. The Study Guide belongs to PHYS113 at Kansas State University taught by Horton Smith, Glenn A. in Fall 2015. Since its upload, it has received 79 views. For similar materials see General Physics 1 in Art at Kansas State University.

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Date Created: 03/29/16
Quiz instructions You should have both an exam and a machine-gradeable answer card (Scantron). On the other side of this page is an official “physics fact sheet”. Detach it if you wish. Not all of the information on the “fact sheet” will be useful. Before you start: ● Fill in the bubbles for your Wildcat ID number. Scantron cards without ID number bubbles filled out won't be counted. ● Fill in the the test version number (written in big letters before the first question), and the answer sheet number (1). ● Also print your name, the course, and the date in the box at the top of the card. ● Also write your name on the first and last pages of the exam. (If you detach this page, write your name on the next page, and also the last page.) ● Take a deep breath. Relax and pay attention. It's really not that bad. Please note: ● Your score on the multiple choice questions will be determined by your answers on the answer card. However, it is a good idea to mark your answers on the exam sheet too, for your own reference. ● Do not fold or roll the card. Unreadable cards won't be counted. ● Mark your responses on the answer card by filling in the bubbles completely with a #2 pencil. ● Do not fill in more than one response for each question. ● If you erase an answer, make sure you erase it completely. ● If you need a new card, please ask. When you are done: ● Recheck your filled-in circles against your exam sheet. ● Be sure that you have answered all questions, that you have all of the pages of the exam, and that both the question number and your response match agree with your exam sheet. You should see the words “The End” after the last question. ● Turn in both the Scantron card and your exam at the front of the room. ● Your exam will be returned to you after all exams are scanned and checked and the written part is graded. ● Please write your name on your exam. Constants SI prefixes Gravity at Earth's surface g = 9.80 m/s , down -12 12 -11 3 -1 -2 pico 10-9 p tera 10 9 T Gravitational constant G = 6.674 × 10 5 kg 2 nano 10 -6 n giga 10 6 G Atmospheric pressure 1 atm = 1.013×10 N/m micro 10 μ mega 10 M Boltzmann's constant k = 1.38×102J/K milli 10-3 m kilo 10 3 k Ideal gas constant R = 8.31 J/(mole·K) centi 10 -2 c hecto 10 2 h Avagadro's number NA= 6.02×1023 Speed of light c = 2.998 × 10 m/s Trigonometry sin θ = “opposite” / “hypotenuse” Kinematics, dynamics, energy, momentum cos θ = “adjacent” / “hypotenuse” Δ ⃗ ω= Δθ hypotenuse = opposite + adjacent ⃗= Δt Δt Δω t hy a= Δ⃗v α= Δt s po e n Δt o use ⃗ θ(t)=θ +ω t+ α1 t2 p θ ∑ F=ma ⃗ 0 0 2 const o adjacent 1 2 2 2 ⃗(t)=⃗0+v⃗0+ 2 ⃗const ω =ω +0αΔθ 2 2 ∑ τ=I α Impulses and collisions v =v0+2aΔ x Impulse: F Δ t=Δ ⃗ 1 2 KE rotationω 2 KE linear mv 2 Collision: ∑ ⃗pinitial ⃗final 2 W=τΔθ W=F dcosθ Fd Simple harmonic motion and waves p=mv L=I ω ⃗ ⃗ Period: T=2π inertia √ restoring Some kinds of potential energy Uniform gravity: PE Gmgy Spring: F=k x Pendulum: F=mg x/L Spring: PE = kx 2 Frequency: f =1/T S 2 Oscillator: v0/ A=2π f Waves: v= f λ Geometry of circles and spheres Standing: f =n(v/2L),any n 360 degrees = 2π radians = 1 full circle n Circumference of a circle: 2πr f nn(v/4L),odd n Rolling distance = (radius)*(angle in radians) Small angles: (linear size of object) ≈ (arc length) Area of a sphere: 4πr Other definitions and relations Density: ρ=m/V Pressure: P=F/ A Circular or rotational motion v2 2 In a fluid: Δ P=ρ gΔ y v=rω, acentrip =rω Buoyancy: F Bρ fluidg r 2 τ=r Fsinθ, I= ∑ mr Gravity (sphere): F =G m 1 2 G r2 Static equilibrium Friction: F Fμ F N ax=a y0 → ∑ F x ∑ F y0 Ideal gas: PV=nRT=NkT 3 α=0 → ∑ τ=0 Avg transl. KE: KE= kT 2 Name: Date: General Physics Exam 2 Practice, version 1 1. Three paintballs are shot at the same speed but different angles at a wall. They hit the wall at different heights. Which has the highest speed when it hits the wall? A. The paintball in the middle. B. The paintball that hits highest on the wall. C. The paintball that hits lowest on the wall. D. There is not enough information given to answer the question. 2. A submarine moving through water experiences a constant force of 5.6 N. How much work does the submarine do in moving 7.3 m? A. 41 N B. 0.77 N/m C. 41 J D. 5.6 N A catapult. 3. A catapult flings a 13 kg stone from an "arm" 2 m long by applying a force of 330 N to the arm a distance 1.1 m from the pivot. (See figure.) What force is applied to the stone by the catapult arm? Ignore the mass of the arm itself. A. 3.3×10+02 N, calculated from (330 N) +02 B. 1.8×10 N, calculated from (1.1 m)*(330 N)/(2 m) C. 6×10+02 N, calculated from (330 N)*(2 m)/(1.1 m) 4. A 530 kg rock slides 8.3 m down a ramp, experiencing a constant friction force of 66 N. The change in height of the rock as it slides this distance is 0.53 m. What is the work done by friction? +03 2 A. 2.8×10 J, calculated from (530 kg)*(9.80 m/s )*(0.53 m) B. 35 J, calculated from (66 N)*(0.53 m) C. 4.3×10 +04 J, calculated from (530 kg)*(9.80 m/s )*(8.3 m) +05 2 2 D. 3.4×10 N , calculated from (66 N)*(9.80 m/s )*(530 kg) E. 5.5×10 +02 J, calculated from (66 N)*(8.3 m) 5. What is the work done by gravity on the rock in the previous problem? A. 35 J B. 2.8×10 +03 J +02 C. 5.5×10 J D. 3.4×10 +05 N 2 +04 E. 4.3×10 J 6. How much work is done by the normal force of the ramp on the rock? A. 469 J B. 137 J C. 0 J D. There is not enough information given in the problem to say. E. 233 J 7. What is the total work done on the rock? +04 A. 4.3×10 J B. 2.8×10 +03 J C. 35 J D. 3.4×10 +05 N 2 +03 E. 2.2×10 J 8. What is the change in the kinetic energy of the rock? A. The same as the total work done on the rock. B. The same as the work done by gravity on the rock. C. The same as the work done by friction on the rock. D. There is not enough information given to say. 9. A 3.011×10 kg train moves west at 5.333 m/s. A 1800 kg mass truck moves east at 21.05 m/s. What is the total kinetic energy of the train and truck together? +07 6 A. 1.602×10 kg*m/s east, calculated from (3.011×10 kg)*(5.333 m/s)-(1800 kg)* (21.05 m/s) +07 6 B. 1.602×10 N*s west, calculated from (3.011×10 kg)*(5.333 m/s)-(1800 kg)* (21.05 m/s) C. 1.61×10 +07 N*s west, calculated from (3.011×10 kg)*(5.333 m/s)+(1800 kg)* (21.05 m/s) D. 4.242×10 +07 J west, calculated from (1/2)*(3.011×10 kg)*(5.333 m/s) -(1/2)* 2 (1800 kg)*(21.05 m/s) E. 4.322×10 +07 J, calculated from (1/2)*(3.011×10 kg)*(5.333 m/s) +(1/2)* 2 (1800 kg)*(21.05 m/s) 10. What is the total momentum of the train and truck? A. 4.322×10 +07 J B. 1.602×10 +07 kg*m/s west +07 C. 1.61×10 kg*m/s west D. 4.322×10 +07 kg*m/s north +07 E. 1.602×10 J east 11. The truck and the train collide head-on and stick together. (Fortunately, the truck and train are both runaways, with no people or dangerous substances on board.) What happens to the total momentum and kinetic energy of the system? A. The total momentum is unchanged, but the kinetic energy is decreased. B. The total momentum and kinetic energy increase. C. The total momentum and kinetic energy are unchanged. D. The kinetic energy is unchanged, and the total momentum decreases. E. The total momentum is unchanged, but the kinetic energy is increased. 12. What is the velocity of the train-truck combination after the collision? A. 5.317 m/s west, calculated from ((3.011×10 kg)*(5.333 m/s)-(1800 kg)* 6 (21.05 m/s))/((3.011×10 kg)+(1800 kg)) B. 5.342 m/s east, calculated from ((3.011×10 kg)*(5.333 m/s)+(1800 kg)* (21.05 m/s))/((3.011×10 kg)+(1800 kg)) C. 5.342 m/s west, calculated from ((3.011×10 kg)*(5.333 m/s)+(1800 kg)* (21.05 m/s))/((3.011×10 kg)+(1800 kg)) 6 D. 5.317 m/s east, calculated from ((3.011×10 kg)*(5.333 m/s)-(1800 kg)* (21.05 m/s))/((3.011×10 kg)+(1800 kg)) E. 13.19 m/s east, calculated from ((5.333 m/s)+(21.05 m/s))/2 13. Suppose the truck were traveling north instead of east. Which way would the total momentum vector point? A. Slightly east of north. B. Nowhere, since momentum is not a vector. C. Slightly north of west. D. East. E. Slightly north of east. 14. A force of 58 N is applied to a 8.1 kg block on a frictionless surface for a time 3.3 s. What is the change in momentum of the block? A. 1.9×10 +02 N*s B. 27 kg*m/s C. 18 N/s D. 24 m/s E. 7.2 m/s2 15. A carnival ride lifts a car containing 10 passengers high in the air, then drops the car so it falls freely. Neglecting friction, what is the velocity of the car after it has fallen a distance 3.3 m? A. 65 m/s, calculated from 2*(9.80 m/s )*(3.3 m) 2 B. 8 m/s, calculated from sqrt(2*(9.80 m/s )*(3.3 m)) C. 0.58 s, calculated from sqrt((3.3 m)/(9.80 m/s )) D. 3 1/s, calculated from (9.80 m/s )/(3.3 m) 16. Another ride lets people slide down a nearly frictionless slide. If the slide is 3.3 m high, what is their speed at the bottom of the slide? A. It depends on the mass of the person. B. The same as found in the previous problem. C. It depends on the slope of the slide. D. One half the answer of the previous problem. 17. A 0.311 kg mass ball is dropped from a height 1.51 m. Air friction is negligable, but it loses an energy of 1.23 J when it bounces. What height does it return to on its first bounce? [Hint: use conservation of energy.] A. 2.6 m/s B. 5.4 m/s C. 4.1 m D. 1.1 m E. 1.5 m 18. A string is wrapped around a wheel of radius 0.38 m. If the rope is pulled with force 868 N tangent to the wheel, what is the torque on the wheel? +02 A. 3.3×10 m*N B. 3.3×10+02 N C. 0.00044 m/N D. 1.6×10+02 J E. 8.7×10+02 N 19. Which of the following is NOT a unit of energy? A. N*m B. J C. W 2 2 D. kg*m /s 20. Which of the following is NOT a unit of momentum? A. N*s B. J C. kg*m/s Cartoon of a person climbing a wall using a rope. 21. A 50 kg climber stands perpendicular to a vertical wall, hanging on to a rope with makes an angle 75 deg from horizontal. (See figure.) The climber's center of gravity is 0.75 m from the wall, and his shoulders are 1.4 m from the wall. What is the tension in the rope? [Hint: use the feet of the climber as the center of rotation when calculating torque.] +02 2 A. 2.6×10 N, calculated from (50 kg)*(9.80 m/s )*(0.75 m)/(1.4 m) B. 2.7×10 +02 N, calculated from (0.75 m)*(50 kg)* 2 (9.80 m/s )/((1.4 m)*sin((75 deg)*pi/180)) C. 9.1×10 +02 N, calculated from (50 kg)*(9.80 m/s )*(1.4 m)/(0.75 m) +02 2 D. 4.9×10 N, calculated from (50 kg)*(9.80 m/s ) 22. What force does the climber feel in his* legs? (I.e., what is the normal force from the wall on the climber?) 2 A. 70 N, calculated from (0.75 m)*(50 kg)*(9.80 m/s )/((1.4 m)*tan((75 deg)*pi/180)) B. 2.6×10 +02 N, calculated from (50 kg)*(9.80 m/s )*(0.75 m)/(1.4 m) C. 4.9×10 +02 N, calculated from (50 kg)*(9.80 m/s ) +02 2 D. 9.1×10 N, calculated from (50 kg)*(9.80 m/s )*(1.4 m)/(0.75 m) 23. What force of friction on the climber's feet is required? A. 4.9×10 +02 N, calculated from (50 kg)*(9.80 m/s ) +02 2 B. 9.1×10 N, calculated from (50 kg)*(9.80 m/s )*(1.4 m)/(0.75 m) C. 2.3×10 +02 N, calculated from (50 kg)*(9.80 m/s ) - (0.75 m)*(50 kg)* 2 (9.80 m/s )/(1.4 m) D. 2.6×10 +02 N, calculated from (50 kg)*(9.80 m/s )*(0.75 m)/(1.4 m) 24. In which direction does the force of friction act on the climber's feet? A. up B. out from the wall C. down D. into the wall 25. If the climber slowly walks up the wall and continues to stand perpendicular to the wall, the rope becomes more and more horizontal. Does the tension in the rope increase, decrease, or stay the same? A. Increase B. Decrease C. Stay the same 26. If the climber hung straight down from the rope, would the tension in the rope be greater than, less than, or the same as it is when the climber stands perpendicular to the wall? A. Greater B. The same C. Less Short written answer questions A physics professor holds a 7 kg bowling ball in front of his nose, 1.7 m above the floor, and releases it. Unlike the demo done in class, the bowling ball is not attached to anything, so it falls straight down. (i) What is the kinetic energy of the bowling ball when it reaches the floor? (ii) If the professor's sneaker-clad foot stops the bowling ball over a distance of 1 cm, what is the average force applied by the bowling ball to the sneakers? The End Quiz instructions You should have both an exam and a machine-gradeable answer card (Scantron). On the other side of this page is an official “physics fact sheet”. Detach it if you wish. Not all of the information on the “fact sheet” will be useful. Before you start: ● Fill in the bubbles for your Wildcat ID number. Scantron cards without ID number bubbles filled out won't be counted. ● Fill in the the test version number (written in big letters before the first question), and the answer sheet number (1). ● Also print your name, the course, and the date in the box at the top of the card. ● Also write your name on the first and last pages of the exam. (If you detach this page, write your name on the next page, and also the last page.) ● Take a deep breath. Relax and pay attention. It's really not that bad. Please note: ● Your score on the multiple choice questions will be determined by your answers on the answer card. However, it is a good idea to mark your answers on the exam sheet too, for your own reference. ● Do not fold or roll the card. Unreadable cards won't be counted. ● Mark your responses on the answer card by filling in the bubbles completely with a #2 pencil. ● Do not fill in more than one response for each question. ● If you erase an answer, make sure you erase it completely. ● If you need a new card, please ask. When you are done: ● Recheck your filled-in circles against your exam sheet. ● Be sure that you have answered all questions, that you have all of the pages of the exam, and that both the question number and your response match agree with your exam sheet. You should see the words “The End” after the last question. ● Turn in both the Scantron card and your exam at the front of the room. ● Your exam will be returned to you after all exams are scanned and checked and the written part is graded. ● Please write your name on your exam. Constants SI prefixes Gravity at Earth's surface g = 9.80 m/s , down -12 12 -11 3 -1 -2 pico 10-9 p tera 10 9 T Gravitational constant G = 6.674 × 10 5 kg 2 nano 10 -6 n giga 10 6 G Atmospheric pressure 1 atm = 1.013×10 N/m micro 10 μ mega 10 M Boltzmann's constant k = 1.38×102J/K milli 10-3 m kilo 10 3 k Ideal gas constant R = 8.31 J/(mole·K) centi 10 -2 c hecto 10 2 h Avagadro's number NA= 6.02×1023 Speed of light c = 2.998 × 10 m/s Trigonometry sin θ = “opposite” / “hypotenuse” Kinematics, dynamics, energy, momentum cos θ = “adjacent” / “hypotenuse” Δ ⃗ ω= Δθ hypotenuse = opposite + adjacent ⃗= Δt Δt Δω t hy a= Δ⃗v α= Δt s po e n Δt o use ⃗ θ(t)=θ +ω t+ α1 t2 p θ ∑ F=ma ⃗ 0 0 2 const o adjacent 1 2 2 2 ⃗(t)=⃗0+v⃗0+ 2 ⃗const ω =ω +0αΔθ 2 2 ∑ τ=I α Impulses and collisions v =v0+2aΔ x Impulse: F Δ t=Δ ⃗ 1 2 KE rotationω 2 KE linear mv 2 Collision: ∑ ⃗pinitial ⃗final 2 W=τΔθ W=F dcosθ Fd Simple harmonic motion and waves p=mv L=I ω ⃗ ⃗ Period: T=2π inertia √ restoring Some kinds of potential energy Uniform gravity: PE Gmgy Spring: F=k x Pendulum: F=mg x/L Spring: PE = kx 2 Frequency: f =1/T S 2 Oscillator: v0/ A=2π f Waves: v= f λ Geometry of circles and spheres Standing: f =n(v/2L),any n 360 degrees = 2π radians = 1 full circle n Circumference of a circle: 2πr f nn(v/4L),odd n Rolling distance = (radius)*(angle in radians) Small angles: (linear size of object) ≈ (arc length) Area of a sphere: 4πr Other definitions and relations Density: ρ=m/V Pressure: P=F/ A Circular or rotational motion v2 2 In a fluid: Δ P=ρ gΔ y v=rω, acentrip =rω Buoyancy: F Bρ fluidg r 2 τ=r Fsinθ, I= ∑ mr Gravity (sphere): F =G m 1 2 G r2 Static equilibrium Friction: F Fμ F N ax=a y0 → ∑ F x ∑ F y0 Ideal gas: PV=nRT=NkT 3 α=0 → ∑ τ=0 Avg transl. KE: KE= kT 2 Name: Date: General Physics Exam 2 Practice, version 2 1. Three paintballs are shot at the same speed but different angles at a wall. They hit the wall at different heights. Which has the highest speed when it hits the wall? A. The paintball that hits lowest on the wall. B. There is not enough information given to answer the question. C. The paintball in the middle. D. The paintball that hits highest on the wall. 2. A submarine moving through water experiences a constant force of 6.6 N. How much work does the submarine do in moving 8.3 m? A. 55 J B. 0.8 N/m C. 55 N D. 6.6 N A catapult. 3. A catapult flings a 11 kg stone from an "arm" 1.8 m long by applying a force of 370 N to the arm a distance 1.1 m from the pivot. (See figure.) What force is applied to the stone by the catapult arm? Ignore the mass of the arm itself. A. 2.3×10+02 N, calculated from (1.1 m)*(370 N)/(1.8 m) +02 B. 3.7×10 N, calculated from (370 N) C. 6.1×10+02 N, calculated from (370 N)*(1.8 m)/(1.1 m) 4. A 790 kg rock slides 4.3 m down a ramp, experiencing a constant friction force of 86 N. The change in height of the rock as it slides this distance is 0.58 m. What is the work done by friction? +03 2 A. 4.5×10 J, calculated from (790 kg)*(9.80 m/s )*(0.58 m) B. 50 J, calculated from (86 N)*(0.58 m) C. 3.3×10 +04 J, calculated from (790 kg)*(9.80 m/s )*(4.3 m) +02 D. 3.7×10 J, calculated from (86 N)*(4.3 m) E. 6.7×10 +05 N , calculated from (86 N)*(9.80 m/s )*(790 kg) 5. What is the work done by gravity on the rock in the previous problem? A. 50 J B. 3.3×10 +04 J +02 C. 3.7×10 J D. 6.7×10 +05 N 2 +03 E. 4.5×10 J 6. How much work is done by the normal force of the ramp on the rock? A. 233 J B. There is not enough information given in the problem to say. C. 0 J D. 137 J E. 469 J 7. What is the total work done on the rock? +04 A. 3.3×10 J B. 4.5×10 +03 J C. 50 J D. 6.7×10 +05 N 2 +03 E. 4.1×10 J 8. What is the change in the kinetic energy of the rock? A. There is not enough information given to say. B. The same as the work done by friction on the rock. C. The same as the total work done on the rock. D. The same as the work done by gravity on the rock. 9. A 5.033×10 kg train moves west at 4.333 m/s. A 1700 kg mass truck moves east at 23.09 m/s. What is the total kinetic energy of the train and truck together? +07 6 A. 2.177×10 kg*m/s east, calculated from (5.033×10 kg)*(4.333 m/s)-(1700 kg)* (23.09 m/s) +07 6 B. 2.177×10 N*s west, calculated from (5.033×10 kg)*(4.333 m/s)-(1700 kg)* (23.09 m/s) C. 4.77×10 +07 J, calculated from (1/2)*(5.033×10 kg)*(4.333 m/s) +(1/2)* 2 (1700 kg)*(23.09 m/s) D. 4.679×10 +07 J west, calculated from (1/2)*(5.033×10 kg)*(4.333 m/s) -(1/2)* 2 (1700 kg)*(23.09 m/s) E. 2.185×10 +07 N*s west, calculated from (5.033×10 kg)*(4.333 m/s)+(1700 kg)* (23.09 m/s) 10. What is the total momentum of the train and truck? A. 2.177×10 +07 kg*m/s west B. 2.177×10 +07 J east +07 C. 2.185×10 kg*m/s west D. 4.77×10 +07 J +07 E. 4.77×10 kg*m/s north 11. The truck and the train collide head-on and stick together. (Fortunately, the truck and train are both runaways, with no people or dangerous substances on board.) What happens to the total momentum and kinetic energy of the system? A. The total momentum is unchanged, but the kinetic energy is increased. B. The total momentum and kinetic energy increase. C. The total momentum and kinetic energy are unchanged. D. The kinetic energy is unchanged, and the total momentum decreases. E. The total momentum is unchanged, but the kinetic energy is decreased. 12. What is the velocity of the train-truck combination after the collision? A. 4.324 m/s east, calculated from ((5.033×10 kg)*(4.333 m/s)-(1700 kg)* 6 (23.09 m/s))/((5.033×10 kg)+(1700 kg)) B. 13.71 m/s east, calculated from ((4.333 m/s)+(23.09 m/s))/2 6 C. 4.339 m/s west, calculated from ((5.033×10 kg)*(4.333 m/s)+(1700 kg)* (23.09 m/s))/((5.033×10 kg)+(1700 kg)) 6 D. 4.339 m/s east, calculated from ((5.033×10 kg)*(4.333 m/s)+(1700 kg)* (23.09 m/s))/((5.033×10 kg)+(1700&bbsp;kg)) 6 E. 4.324 m/s west, calculated from ((5.033×10 kg)*(4.333 m/s)-(1700 kg)* (23.09 m/s))/((5.033×10 kg)+(1700 kg)) 13. Suppose the truck were traveling north instead of east. Which way would the total momentum vector point? A. Slightly north of east. B. East. C. Nowhere, since momentum is not a vector. D. Slightly north of west. E. Slightly east of north. 14. A force of 73 N is applied to a 5.7 kg block on a frictionless surface for a time 3.7 s. What is the change in momentum of the block? A. 21 kg*m/s B. 47 m/s +02 C. 2.7×10 N*s D. 20 N/s E. 13 m/s2 15. A carnival ride lifts a car containing 10 passengers high in the air, then drops the car so it falls freely. Neglecting friction, what is the velocity of the car after it has fallen a distance 3.1 m? A. 3.2 1/s, calculated from (9.80 m/s )/(3.1 m) 2 B. 0.56 s, calculated from sqrt((3.1 m)/(9.80 m/s )) C. 61 m/s, calculated from 2*(9.80 m/s )*(3.1 m) D. 7.8 m/s, calculated from sqrt(2*(9.80 m/s )*(3.1 m)) 16. Another ride lets people slide down a nearly frictionless slide. If the slide is 3.1 m high, what is their speed at the bottom of the slide? A. It depends on the slope of the slide. B. The same as found in the previous problem. C. It depends on the mass of the person. D. One half the answer of the previous problem. 17. A 0.281 kg mass ball is dropped from a height 1.81 m. Air friction is negligable, but it loses an energy of 1.23 J when it bounces. What height does it return to on its first bounce? [Hint: use conservation of energy.] A. 1.8 m B. 1.4 m C. 6 m/s D. 4.2 m E. 2.4 m/s 18. A string is wrapped around a wheel of radius 0.55 m. If the rope is pulled with force 849 N tangent to the wheel, what is the torque on the wheel? +02 A. 8.5×10 N B. 2.3×10+02 J +02 C. 4.7×10 m*N D. 0.00065 m/N E. 4.7×10+02 N 19. Which of the following is NOT a unit of energy? A. kg*m /s B. W C. N*m D. J 20. Which of the following is NOT a unit of momentum? A. N*s B. kg*m/s C. J Cartoon of a person climbing a wall using a rope. 21. A 65 kg climber stands perpendicular to a vertical wall, hanging on to a rope with makes an angle 55 deg from horizontal. (See figure.) The climber's center of gravity is 0.85 m from the wall, and his/her shoulders are 1.4 m from the wall. What is the tension in the rope? [Hint: use the feet of the climber as the center of rotation when calculating torque.] +02 A. 4.7×10 N, calculated from (0.85 m)*(65 kg)* (9.80 m/s )/((1.4 m)*sin((55 deg)*pi/180)) +02 2 B. 3.9×10 N, calculated from (65 kg)*(9.80 m/s )*(0.85 m)/(1.4 m) C. 6.4×10 +02 N, calculated from (65 kg)*(9.80 m/s ) +03 2 D. 1×10 N, calculated from (65 kg)*(9.80 m/s )*(1.4 m)/(0.85 m) 22. What force does the climber feel in his/her* legs? (I.e., what is the normal force from the wall on the climber?) +03 2 A. 1×10 N, calculated from (65 kg)*(9.80 m/s )*(1.4 m)/(0.85 m) B. 6.4×10 +02 N, calculated from (65 kg)*(9.80 m/s ) C. 3.9×10 +02 N, calculated from (65 kg)*(9.80 m/s )*(0.85 m)/(1.4 m) +02 D. 2.7×10 N, calculated from (0.85 m)*(65 kg)* (9.80 m/s )/((1.4 m)*tan((55 deg)*pi/180)) 23. What force of friction on the climber's feet is required? +02 2 A. 2.5×10 N, calculated from (65 kg)*(9.80 m/s ) - (0.85 m)*(65 kg)* (9.80 m/s )/(1.4 m) +03 2 B. 1×10 N, calculated from (65 kg)*(9.80 m/s )*(1.4 m)/(0.85 m) C. 6.4×10 +02 N, calculated from (65 kg)*(9.80 m/s ) D. 3.9×10 +02 N, calculated from (65 kg)*(9.80 m/s )*(0.85 m)/(1.4 m) 24. In which direction does the force of friction act on the climber's feet? A. into the wall B. down C. up D. out from the wall 25. If the climber slowly walks up the wall and continues to stand perpendicular to the wall, the rope becomes more and more horizontal. Does the tension in the rope increase, decrease, or stay the same? A. Decrease B. Stay the same C. Increase 26. If the climber hung straight down from the rope, would the tension in the rope be greater than, less than, or the same as it is when the climber stands perpendicular to the wall? A. The same B. Greater C. Less Short written answer questions A physics professor holds a 7 kg bowling ball in front of his nose, 1.7 m above the floor, and releases it. Unlike the demo done in class, the bowling ball is not attached to anything, so it falls straight down. (i) What is the kinetic energy of the bowling ball when it reaches the floor? (ii) If the professor's sneaker-clad foot stops the bowling ball over a distance of 1 cm, what is the average force applied by the bowling ball to the sneakers? The End

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