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PHY2053 Exam 1

PHY2053 Exam 1. Average = 11.8 High = 20 (6 students) Low = 3. Very Good!. Estimated Course Grades. 26 students have 100 points!. Very Good!. Estimated Grades:. Assumes that you get the same grade on Exam 2 and the Final Exam that you received on Exam 1.

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PHY2053 Exam 1

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  1. PHY2053 Exam 1 • Average = 11.8 • High = 20 (6 students) • Low = 3 Very Good! PHY 2053

  2. Estimated Course Grades 26 students have 100 points! Very Good! • Estimated Grades: • Assumes that you get the same grade on Exam 2 and the Final Exam that you received on Exam 1. • 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. • 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. • Includes your HITT scores through 9/26/13 and assumes you maintain the same average. • Includes your first 4 Sakai HW assignments and assumes you maintain the same average. PHY 2053

  3. Linear Momentum Units = kg∙m/s • Single Particle: The linear momentum of a particle is its mass times its velocity vector. 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. (system of particles) • Momentum Conservation: 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. (conservation of linear momentum, isolated system) PHY 2053

  4. 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 PHY 2053

  5. Exam 2 Spring 2012: Problem 10 • 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 % Right: 57% PHY 2053

  6. Momentum Change: Impulse The time rate of change of the momentum is equal to the net force acting on an object. (constant force) • Impulse: The change in the linear momentum is equal to the impulse. The impulse J is defined as follows: (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 Dt = t2 – t1. 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 Dt and if it rebounds with speed v2, what is the magnitude of the average force on the floor from the ball? PHY 2053

  7. 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 (impulse) PHY 2053

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