midterm2-sol - Purdue Physics 172 Midterm II Solutions Fall...

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Purdue Physics 172 Midterm II Solutions Fall 2009 Matt Matolcsi Revision 20091027 Directions: Show as much work as possible to get full credit: list what you know, draw diagrams, de ne the system, list the relevant physical principle, and then solve the equation. Consider a basketball player making a jump shot. A jump shot is when a player jumps o the ground and then she releases the ball at the top of her jump. The ball then travels through the air, falls into the basket, and drops to the ground. Let's break this into a few di erent parts and apply the energy principle. The ball has mass m b and the player has mass m p . Assume that there is no heat loss or change in thermal energy. Choose the x - direction as towards the basket and the y - direction as straight up. 1 Energy Catalog 1.1 Problem First, let's apply the energy principle to the basketball player jumping. Let's also choose the ball and basketball player together as the system. Also choose the initial time as when she is standing on the ground, and the nal time as when she is at the greatest height of her jump (that is, height h 1 ). List all the kinds of internal energy of the ball + player system that might change as the player jumps. Then give reasons why some of these terms are zero. 1.2 Solution First, clarify for yourself what the chosen initial and nal points are as well as the choice of system and surroundings. The work-energy principle only concerns what happens at the initial and nal states of a system, so it's critical to understand what those are. 1. Initially, the player is standing on the ground (therefore, not moving.) 2. Finally, the player is at her maximum height during the jump. You know from intuition that at the maximum height of a jump, her v y = 0 . 3. System = Player, Ball 4. Surroundings = Earth What kinds of energy does a mechanical system like the ball and player have? 1. Kinetic, because they are both moving for most of the jump. 2. Rest energy, because they have mass. 3. Chemical energy, because the player must have gotten from the ground into the air somehow. We know that it couldn't have been kinetic energy because in both the initial and nal states these are zero. Also, there was nothing in the surroundings that pushed her up. Why are some of these terms zero? 1. Δ K = 0 by de nition of the problem. 2. Δ RE = 0 , because the player did not undergo a change in identity or simply, she lost no mass. 1
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1.3 Common Mistakes 1.3.1 Player has gravitational potential energy U g (that is, U g = mgh ) You have to ask yourself, Gravitational potential energy with respect to what? The system contains only the player and basketball, and while these two do have a very small attractive force between them (and hence, a potential energy,) it's much, much smaller than other energies in the system. What about the Earth? The problem chose to put the Earth into the surroundings, hence there will be no
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This note was uploaded on 04/21/2010 for the course PHYS 172 taught by Professor ? during the Spring '08 term at Purdue University-West Lafayette.

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midterm2-sol - Purdue Physics 172 Midterm II Solutions Fall...

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