# SC PHYS 201 - Exam 3 Study Guide (6 pages)

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## Exam 3 Study Guide

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- Pages:
- 6
- Type:
- Study Guide
- School:
- University Of South Carolina-Columbia
- Course:
- Phys 201 - General Physics I
- Edition:
- 1

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Phys 201 1st Edition Exam 3 Study Guide Lectures 15 20 Lecture 15 October 27 A 70kg soccer player kicks a 5kg ball giving it a speed of 14m s The contact time of the foot on the ball is 0 1s a What is the force on the ball b What is the force on the foot Answer a 70N b 70N Explaination a Kicking the ball changes the balls momentum Therefore we can use the momentum equation to solve for the force F t f i mVf mVi 14 5 0 7 F t F t 7kgm s F 7 t 7 1 70N b There is no need to calculate the force on the foot Consider Newton s third law For every action there is an equal and opposite reaction This means that the force on the foot will be the same as the force on the foot just in the opposite direction Lecture 15 October 27 A 1668 0 g mass is on a horizontal surface with k 0 37 and is in contact with a massless spring with a force constant of 619 0 N m which is compressed When the spring is released it does 2 51 J of work on the mass while returning to its equilibrium position a Calculate the distance the spring was compressed b What is the velocity of the mass as it loses contact with the spring Answer a 09 m b 1 534 m s Explanation a PEspring 5kx2 2 51 5 619 x2 X2 0081 X 09 b The easiest was to calculate the velocity of the mass is from its kinetic energy which is equal to the work done by the compressed minus the work done by friction The work done by friction is equal to the force of friction times the distance that the spring was compressed KE Ws Wf 5mV2 Ws fx F 37 1 668 9 81 F 6 04N 5 1 668 V2 2 51 6 04 09 V2 2 3758 V 1 535 Lecture 16 October 29 A croquet mallet delivers an impulse of 8 77 N s to a 0 64 kg croquet ball initially at rest What is the speed of the ball immediately after being struck Answer 13 7m s Explanation Impulse p Ft p mV 8 77 64 V V 13 7m s Lecture 16 October 29 A 0 145 kg baseball traveling 33 50 m s is stopped in 0 193 s by a catcher s mitt A What is the average acceleration of the ball b What is the average force on the catcher s mitt Answer a 173 58m s b 25 17N Explanation a a V t a 33 5 193 a 173 575m s b The force that the ball is thrown is the same as the force the ball exerts on the mitt Plug your value for acceleration into the equation F ma F 145 173 575 F 25 168N Lecture 16 October 29 A 7 85 g bullet from a 9 mm pistol has a velocity of 374 0 m s It strikes the 0 745 kg block of a ballistic pendulum and passes completely through the block If the block rises through a distance h 13 96 cm what was the velocity of the bullet as it emerged from the block Answer 216 93m s Explanation Use the conservation of momentum equation for this problem m1V1i m2V2 m1V1f where m1 and V1 I are the mass and initial speed of the bullet m2 and V2 are the mass and velocity of the block and V1f is the final speed of the bullet which is what you re trying to find Before you solve for V1f you need to solve for V2 Use conservation of energy KE PE 5mV2 mgh 5 745 V2 745 9 81 1396m V 2 739 1 655m s Plug this value into the conservation of momentum equation to solve for Vf m1V1i m2V2 m1V1f 00785 374 745 1 655 00785 Vf Vf 216 93m s Lecture 17 November 3 An Escort and a Camaro traveling at right angles collide and stick together The Escort has a mass of 1200 kg and a speed of 20 km h in the positive x direction before the collision The Camaro has a mass of 1500 kg and was traveling in the positive y direction After the collision the two move off at an angle of 54 o to the x axis What was the speed of the Camaro Answer 2 20 101 km h Explanation Use the mass and velocity of the cars before the collision to find the velocity of the cars after the collision The angle describes the direction that they move off at after the collision Use the x and y components of this final velocity to figure out the initial speed of the Camaro M1 Vf V1 Vy Vx M1 1200kg V1 5 56m s M2 1500 P1 Px m1V1 m1 m2 Vx m1 m1 m2 V1 Vx V2 M2 Vx 1200 1500 1200 5 56 2 471m s Tan Opp adj Tan 54 Vy 2 471 Vy 3 4m s Piy Pfy M2V2 m1 m2 Vfy 1500V2 1200 1500 3 4 V2 6 12m s 22 032 Km h Lecture 17 November 3 A 30 caliber rifle with a mass of 3 84 kg fires a 7 13 g bullet with a speed of 606 m s with respect to the ground What kinetic energy is released by the explosion of the gunpowder that fires the bullet Answer 1312J Explanation m1V1 m2V2 3 84V1 00713 606 V1 1 125m s KE 5m1V12 5m2V22 KE 5 3 84 1 1252 5 00713 6062 KE 1311 626 Lecture 17 November 3 An 84 kg ice hockey player standing on a frictionless sheet of ice throws a 5 4 kg bowling ball horizontally with a speed of 3 3 m s With what speed does the hockey player recoil Answer 212m s Explanation Use conservation of momentum m1V1 m2V2 The hockey player s recoil is V1 m1V1 m2V2 84 V1 5 4 3 3 V1 17 82 84 V1 212m s Lecture 18 November 05 A distant planet has a mass of 0 6800 ME and a radius of 0 8000 RE What is the ratio of the escape speed from this planet to the escape speed from the earth Answer 922 Explanation The equation for the escape velocity on Earth is V esc 2MG Re To find the ratio of this new planet just substitute these values in for M and R All other variables in this equation are the same and therefore irrelevant M R 68 8 922 Lecture 19 November 10 Two children are playing on a balanced seesaw One child with a mass of 39 kg sits 1 3 m from the center Where on the other side must the second child sit if her mass is 32 kg Answer 1 58m Explanation Torque is equal to force times radius or distance from the center For the seesaw to balance the torques of the two children must be equal T1 T2 F1r1 …

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