Solution to Problem. Part A. x m. x o = 0, y o = 0, t = 0. Part B m m. range

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1 PRACTICE PROBLEMS: Final Exam, December 4 Monday, GYM, 6 to 9 PM Problem A Physics Professor did a daredevil stunt in his spare time. In the figure below he tries to cross a river from a 53 ramp at an unknown initial speed v o. The river is 40 m wide, and the far bank is 5 m lower than the top of the ramp. The river itself is 00 m below the ramp. A) What should his speed at the top of the ramp be for him to just make it to the edge of the far bank. B) If his speed were only half the value found in part A, where would he land in the river? Solution to Problem y x v 0? 53.0 Part A x o 0, y o 0, t m Part B 00.0 m 40.0 m range

2 (A) Here follow the solid trajectory in the diagram: Initial components of the velocity x-component: v 0 x v 0 cos53 0.6v 0, () y-component: v 0 y v 0 sin53 0.8v 0, () To just make it across, he must travel 40.0 m horizontally: 40.0m v 0 x t 0.6v 0 t t 40.0m 66.7m, (3), where we used equation (). 0.6v 0 v 0 In the same period, he must fall to y -5.0 m: 5.0m v 0 y t gt. Using equations () and (3): 66.7m 5.0m 0.8v 0 ( 4.9m / s ) v m v m.777m3 / s v 0.. v 0 777m3 / s 68.33m v 0 7.8m / s. (B) For this part follow the dashed trajectory in the diagram, and let v m/s Initial components of the velocity x-component: v 0 x v 0 cos m / s y-component: v 0 y v 0 sin m / s ( ) 5.34m / s ( ) 7.m / s He must falls 00.0 m to y m Use y v 0 y t gt 00.0m ( 7.m / s)t ( 4.9m / s )t, which gives 4.9 m s t 7. m s t 00.0m 0 with the solution ( )( 00) t 7. ± ( 4.9) The physical solution is t 5.3 s. s ( 0.77 ± 4.57)s 5.3s, 3.84s ( )( 5.3s) 8.3m The distance traveled (range in the diagram) range v 0 x t 5.34m / s into the water (see dashed line) Problem In diagram below a 65 kg bucket of concrete hangs from a cable connected to a 80 kg box with 50 kg of gravel on top. The friction between the box and the floor is as shown. A) If the system is at rest find the friction force on the box and the tension in the rope. B) If the gravel were removed the bucket will start to move down. Find the friction force and the acceleration and tension in the rope.

3 C B a A The rope is massless and there is no friction in the pulley (A) Free Body diagram on object A (bucket) Free-Body diagram on composite B+C object (box + bag of gravel) +y f s m A g T n +x m B g m C g a Since the bag of gravel rest on the box without moving the friction force, the only force that can act on it, on it is zero. (B) If the gravel were removed, there would not be enough friction to keep the box stationary. The box and the bucket will move downward and leftward respectively. The free body-diagram (FBD) of Box A would be the same leading to the equation F y net T m A g m A a T 637N m A a [] Make sure you understand why it is m A a FBD of Box B is as below Kinetic Friction F N m B g 784N Newton s nd Law x-component f k F N µ k 33.6N B T F net x T f k m B a a T 33.6N + m B a [] m B g 784N Equate [] and [] 637N m A a 33.6N + m B a Solving ( m A + m B )a 33.4N a 33.4N 45kg.3mi s Substituting into [] T 637N 65kg.3mi s 49N. Or [] T 33.6N +80kg.3mi s 49N. T System at equilibrium (at rest) so net force is zero F y NET T m A g m A g T m A g m A a (65.0 kg)(9.8 m/s ) T 637 N y-component Using Newton s first law n (m B +m C )g x-component f s < nµ s (m B +m C )gµ s (80kg+50kg)(9.8m/s )(0.7) 89 N System at equilibrium so net force must be zero F NET x T f s 0 Since the tension is less than the maximum value of static friction, T 637 N < 89 N nµ s, friction equals the tension f s T 637 N

Problem 3 In the diagram to the left, the coefficient of kinetic friction between the table and the 8kg crate is 0.50.

4 Problem 3 In the diagram to the left, the coefficient of kinetic friction between the table and the 8kg crate is Use the work energy theorem to calculate the speed of the 6kg block after it has fallen.50 m. Solution uses diagram below: If the blocks is released from rest find the speed of the blocks after the 6.00 kg block has descended.50 m. µ k This problem can be solved by combining Newton s law and kinematic equations, but it is easiest to use the work-energy method: Note that the work done by the tension T on the 8.00 kg block is positive and is equal in magnitude to work done by the tension T on the 6.00 kg block, which is negative. So the work done by the tension T in the rope is zero. Make sure you understand this. Hence we need only consider the work done by the friction f k and gravitational force, mg. Work done by friction on 8 kg block is f k s mgµ k s 8.00kg ( ) 0.50 ( ) 9.8m / s ( )(.50m) 9.4J Work done by gravitational force on 6.00 kg block is ( ) 9.8m / s mgs 6.00kg ( ).50m ( ) 88.J The total work done on the system is W mgs f s s 88.J 9.4J 58.8J Using the work-energy theorem and the fact that the initial speeds are zero W ΔK mv F v F W m Since the 8.00 kg and 6.00 kg blocks move together with the same speed, m is the combined mass of the 6.00 kg and 8.00 kg block. Hence m 4.00kg. Again make sure you understand this. So we obtain ( ) v F 58.8J 4.00kg.89m / s

5 Problem 4 The diagram below shows a crate, held at rest, compressing a spring.00 m from its equilibrium position. The mass of the crate is 5.0 kg and the elastic constant of the spring is 988 N/m. The coefficient of kinetic friction between the crate and the floor is µ k 0.0. Initially the crate is located a distance 3.0 m from a 40 incline. There is no friction between the crate and the surface of the incline. Use the work-energy theorem or conservation of energy to determine the distance the box travels up the incline. Spring is compressed.00 m by the crate. Elastic constant of spring is k 988 N/m Initial speed v 0 0 There is no friction between the crate and the incline surface. 40 The coefficient of kinetic friction between the crate and the floor is µ k m Initial Position (0) Initial speed v m There is no friction between the crate and the incline surface. 40 Spring is at.0 m v.0 m Position PART I: Begin by finding the speed of the crate at the bottom of the incline (position ). Use the work-energy theorem between position 0 and W el +W friction ΔK mv mv 0 mv, where we ve used v0 0 W el ΔU ( U U 0 ).

6 It is easy to see U 0 kx 988N /m W el 494 J. Using f k µ k n µ k mg 0. And hencew friction f k 3.0m This gives W el +W friction mv v W el +W friction m PART II Position to ( )( m) 494 J and U 0 ( )( 5.0kg )( 9.8m/ s ) 9.4N. ( ) ( 9.4N )( 3.0m) 88.N. ( ).Hence ( 494 J 88.N ) 7.36m/ s 5.0kg Position v 0 L h cos 40 h v Position 40 Use the work-energy theorem between position and W grav ΔK mv mv mv, where we ve used v 0. U U. Taking U 0, we have U mgh mglsin40. But W grav ΔU ( ) This givesw grav U mglsin40 mv v L ( gsin m/ s) ( 9.8m/ s )sin40 4.3m Alternative Solution See diagram on next page Use Conservation of total energy between position 0 and U el 0 +U grav 0 + K 0 +W friction U el +U grav + K. It is easy to see U el 0 kx ( 988N /m) m ( ) 494 J and el 0 U.

7 Taking U 0 grav 0, we have U grav mgh mglsin40. Using f k µ k n µ k mg 0. And hencew friction f k 3.0m ( )( 5.0kg )( 9.8m/ s ) 9.4N. ( ) ( 9.4N )( 3.0m) 88.N. Finally it is easy to see that K 0 0and K 0. Putting this altogether we obtain U 0 el +W friction U grav 494 J 88.J mglsin40. This gives L J mgsin J 5.0kg ( )sin40 4.3m. ( ) 9.8m/ s Position v 0 L h cos 40 Position 0 h v m m Problem 5 Two skaters Daniel (mass 65 kg) and Rebecca (mass 45 kg) are practicing. Rebecca moving a 3m/s collides with a stationary Daniel. After colliding Rebecca moves at 8 m/s at 53. w.r.t. her initial direction. What are the magnitude and direction Daniel s velocity after the collision? What is the change in kinetic energy as the result of the collision? Is this an elastic collision? 40

8 Before After Rebecca Daniel +y α 53. β? + x (a) The simplest way to solve the problem is to assume that the +x direction is the same as Rebecca s initial velocity, as shown in the diagram. There must be conservation of momentum for both the x and y components. x-component m R v R0 m R v R cos m D v D cos β ( 45kg) 3.0m / s v D cos β 5.68m / s () ( ) ( 45kg) ( 8.0m / s) ( 0.6) + 65kg y-component 0 mrv R sin 53. mdv D ( )v D cos β ( 65kg)v D cos β 369kg m / s sin β Take note of the negative sign ( 45kg)( 8.0m / s)( 0.8) ( 65kg) v sin β ( 65kg) v sin 88kg m s 0 D D β / v D sin β 4.43m / s () vd sin β 4.43m / s vd sin β Divide equation () by () tan β. 78 vd cos β 5.68m / s vd cos β tan 5.68m / s β ( 0.78) 38. Substituting into equation () v D 7.m / s. Daniel cos38 velocity is 7. m/s, 38 below the horizontal as shown in the above diagram. (b) The initial kinetic energy is K 0 mrvr0 ( 45kg)( 3m / s) 3800J (only Rebecca is moving). The final kinetic energy (Rebecca and Daniel are moving) is K m v R R + m v D D ( 45kg) ( 8m / s) + ( 65kg) ( 7.m / s) 35J. The change in kinetic energy is Δ K K K 0 675J. Note that kinetic energy is not conserved, and this is an inelastic collision. Problem 6 In the diagram below a 50 kg cart rolls to the left at 5 m/s and a 5 kg box slides down an incline and leaves the edge at 3 m/s, 4 m above the bottom of the cart. A) What is the speed of the package just before it lands in the cart? B) What is the speed of the cartpackage after the collision?

9 m v Bi-y v Bi-x v Bi 3.0 m/s initial just before collision final v TF v Ci 5.0 m/s v Ci 5.0 m/s v Bf The leftmost sketches show the box just before it leaves the chute with speed v Bi 3.0 m/s. First find its x and y components: vbi x ( 3.0m / s) cos37.4m / s v ( 3.0m / s) cos37.8m s Bi y / (A) Just before the collision between the box and the cart To find the speed of the box, v Bf, just before it hits the box use conservation of mechanical energy: U v Bf + K U gh + v + K Bi m B gh + m B v Bi 0 + m ( m / s )( 4m) + ( 3m / s) 9.35m / s 9.8 (B) Just after the collision between the box and the cart To find the speed of the cart plus box after the collision, v TF, use conservation of momentum in the horizontal component (can you see why the vertical component is not relevant?). Just before the collision the x-component of the cart s velocity is -v Ci -5.0 m/s and the x-component of the cart s velocity is v Bi-x.4 m/s mcvci + mbvbi x mcvci + mbvbi x ( mc + mb ) vtf vtf m + m ( 50.0kg)( 5.0m / s) + ( 5kg)(.4m / s) vtf 3.9m / s 65kg The cart plus box move with a velocity of 3.9 m/s to the left. B v Bf c B and

10 Problem 7 A computer disk starts rotating from rest at constant angular acceleration. If it takes s to complete its second revolution: a) How long does it take to complete the first complete revolution; b) What is the angular acceleration? (A) Let t be the time it takes to rotate through one revolution θ π. Then we obtain θ θ0 + ω0zt + α zt with 0 0 θ and ω 0 z 0, which gives π α z t (). Now the second revolution takes 0.75 s to complete, so it takes t s to rotate from θ 0 to θ 4π, which using Dividing () by () gives 4π π θ θ0 + ω0zt + α zt and 0 z 0 ( ) α z t s α t z 0 4π α z t s (). ω gives ( ) ( t s) t ( ) t.5t t.5 ±.5 + 4( ) ( 0.565) t t s t.8s, 0. 3s. Of course t.8 s is the physical solution. (b) Using () π α z t gives, which simplifies to or simply Problem 8 In the diagram below, two weights are connected by a very light string, which is passed over a 50 N solid-disk cylinder of radius 0.3 m. The 5 N accelerates downward, without the rope slipping. What force does the ceiling exerts on the pulley? F 8.0N F 6.0N F 3 4.0N

11 Clockwise is positive α +y F W r.3m a T 5 a 75N T 75 5 N F g 75N F g 5N T 75 F p 50N T 5 From FBD diagrams above: 75 kg box (Translation) 5N Box (Translation) Pulley (rotation) net F y T 75 75N m a net F y T 5 5N m a τ net T 5 r T 75 r Iα [] [] [3] Subtract [] by [], T 5 T 75 50N ( m + m )a [4] Using [3] I Mr (Solid disk), T 5 T 75 Mrα. Using the no-slip condition a rα, gives T 5 T 75 50N Ma, M 9.8 m 5.kg Substituting into [4] gives s Ma 00N ( m + m )a a 50N m + m + M 50N 7.65kg +.75kg +.55kg a.8 m s. Substituting into [] gives T 75 75N kg.8 m s 9.7N. Substituting into gives T 5 5N +.75kg.8 m s 53N. Using FBD of the pulley, the y-component of the net forces must be zero, otherwise, the pulley would translate. Problem 9 A diver comes off the board with straight arms and legs has a moment of inertia of 8kg m. She then tucks into a small ball decreasing her moment of inertia to 3.6kg m. While tucked in she makes two complete revolution in.0s. If she hadn t tucked in how many revolutions would she have made in.5 seconds?

12 This is a conservation of angular momentum problem. Initially the diver is straight arm with large moment of inertia, I 8kg m, and low angular velocity ω?. In the final scenario she tucks reducing the moment of inertia to I 3.6kg m, and rotates revolution in one seconds. Hence her final angular velocity is ω rev s π rad rev 4π rad / s.56 rad / s. initial angular momentum final angular momentum I ω I ω ω I 3.6kg m ω I 8kg m (.56rad / s).5 rad / s If she hadn t tucked her angular velocity would remainω.5rad / s. In.5 seconds she would rotate θ ωt.5rad / s ( ) θ 3.77rad ( )(.5s) 3.77rad.In revolution she would rotate rev π rad 0.600rev.

FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym

FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 13 pages. Make sure none are missing 2.