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PHY2053 Exam 1 PHY 2053 Fall13 Exam 1 80 Number 60 Number of Students = 638 Average = 11.8 Median = 12 High = 20 Low = 3 40 20 0 Grade Average = 11.8 High = 20 (6 students) Low = 3 R. Field 10/08/2013 University of Florida PHY 2053 Very Good! Page 1 Estimated Course Grades 26 students have 100 points! PHY 2053 Fall13 Estimated Course Grades ? 40 D? 45 D ? 50 D+ ? 55 C? 60 C ? 65 C+ ? 70 B? 75 B ? 82 B+ ? 87 A? 92 A 30 Number 25 20 15 10 5 20% After One Exam C Average = 74.2 High = 100 B A Percent of Students 35 PHY 2053 Fall13 Estimated Course Grades 15% After One Exam Number = 650 A or A- = 25.1% >=B = 51.8% >=C = 82.2% Very Good! 11.5% 3.4% 3.1% 4.3% 15.4% 10.5% 9.5% 9.2% 10% 5% 16.3% 9.7% 5.4% 1.7% 0% 0 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Points (100 max) E D- D D+ C- C C+ B- B B+ A- A Grade • Estimated Grades: 1. Assumes that you get the same grade on Exam 2 and the Final Exam that you received on Exam 1. 2. Include the first 4 quizzes and assumes that you get the same average on all your remaining quizzes that you have for the first 4 quizzes. 3. Includes the first 5 WebAssign HW assignments and assumes that you get the same average on all your remaining homework assignments that you have for the first 5 assignments. 4. Includes your HITT scores through 9/26/13 and assumes you maintain the same average. 5. Includes your first 4 Sakai HW assignments and assumes you maintain the same average. R. Field 10/08/2013 University of Florida PHY 2053 Page 2 Linear Momentum Units = kg∙m/s • Single Particle: The linear momentum of a particle is its mass times its velocity vector. dv d (mv ) dp Fnet ma m dt dt dt p mv dp Fnet dt The time rate of change of the momentum is equal to the net force acting on the particle. If Fnet = 0 then p is constant in time (i.e. does not change). • System of Particles: Center-of-Mass The overall linear momentum of a system of particles is the vector sum of the momentums of all the particles. N Ptot pi m1v1 m2 v2 m3v3 mN v N M totvcm i 1 • Momentum Conservation: dPtot Fnet M tot acm dt (system of particles) If the net external force acting on a system of particles is zero (i.e. isolated system) then Ptot is constant in time (i.e. does not change) and hence the total momentum at some initial time ti is equal to the total momentum at some final time tf. tot tot Pi Pf R. Field 10/08/2013 University of Florida (conservation of linear momentum, isolated system) PHY 2053 Page 3 Momentum Conservation: Example • Example: Near the surface of the Earth, a bullet with mass m moving directly upward at speed v = 1,000 m/s strikes and passes through the center of mass of a block of mass M initially at rest as shown in the figure. The bullet then emerges from the block moving directly upward at speed v‘ = 500 m/s. If the mass of the block is 250 times the mass of the bullet, to what maximum height does the block then rise above its initial position? Answer: 20.4 cm Initial Mometum y-axis Final Mometum y-axis Initial Energy y-axis Final Energy v' V =0 M m V x-axis h x-axis x-axis v pi p f mv MV mv' m V (v v ' ) M R. Field 10/08/2013 University of Florida Ei E f 2 1 2 MV Mgh V 2 m (v v ' ) 2 h 2g M 2g 2 2 1 (1000m / s 500m / s ) 20.4cm 2 250 2 ( 9 . 8 m / s ) 2 PHY 2053 Page 4 Exam 2 Spring 2012: Problem 10 q v x-axis • A cannon on a railroad car is facing in a direction parallel to the tracks as shown in the figure. The cannon can fires a 100-kg cannon ball at a muzzle speed of 150 m/s at an angle of q above the horizontal as shown in the figure. The cannon plus railway car have a mass of 5,000 kg. If the cannon and one cannon ball are travelling to the right on the railway car a speed of v = 2 m/s, at what angle q must the cannon be fired in order to bring the railway car to rest? Assume that the track is horizontal and there is no friction. Answer: 48.2o ( px )i ( M car M Ball )Vi ( p x ) f M BallVball M Ball (Vi Vmuzzle cos q ) % Right: 57% cos q ( px )i ( px ) f (M car M Ball )Vi M Ball (Vi Vmuzzle cos q ) M carVi (5000kg)( 2m / s) 0.6667 M BallVmuzzle (100kg)(150m / s ) R. Field 10/08/2013 University of Florida PHY 2053 q 48.2 Page 5 Momentum Change: Impulse The time rate of change of the momentum is equal to the net force acting on an object. dp F (t ) dt dp F (t )dt p p f pi Ft • Impulse: The change in the linear momentum is equal to the impulse. The impulse J is defined as follows: J Ft (constant force) p p f pi J (definition of impulse for constant force) • Average Constant Force: Often one is interested in the average constant force necessary to produce an impulse J within the time interval t = t2 – t1. J Favet Fave J t Example: A ball of mass M drops vertically onto the floor, hitting with speed v1. If the ball is in contact with the floor for a time t and if it rebounds with speed v2, what is the magnitude of the average force on the floor from the ball? R. Field 10/08/2013 University of Florida PHY 2053 Fave M (v1 v2 ) / t Page 6 Impulse: Example Problem A 0.5-kg rubber ball is dropped from rest a height of 19.6 m above the surface of the Earth. It strikes the sidewalk below and rebounds up to a maximum height of 4.9 m. What is the magnitude of the impulse due to the collision with the sidewalk? Answer: 14.7 N∙s 1 2 y-axis v1 = 0 MgH 12 Mv22 H 1 2M p22 y-axis v'1 J p p f pi p'1 p2 M J M v'2 = 0 h p2 M 2 gH yˆ v2 (impulse) Mv12 MgH 12 Mv22 2 gh 2 gH 1 2 Mv '12 12 Mv '22 Mgh 1 2 Mv '12 p'12 Mgh p'1 M 2 gh yˆ 2 gh 2 gH yˆ (0.5kg) 2(9.8m / s 2 )( 4.9m) 2(9.8m / s 2 )(19.6m) 1 2M (0.5kg) (9.8m / s) 2 2(9.8m / s) 2 3(0.5kg)(9.8m / s) 14.7 N s R. Field 10/08/2013 University of Florida PHY 2053 Page 7