HG_Solution

HG_Solution - Write down your recitation time: PHYS 172 -...

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Unformatted text preview: Write down your recitation time: PHYS 172 - Fall 2011 Day (either Wed, Th, Fri): Name (Print): , Time: Signature: PUID: Show as much work as possible to get full credit: list what you know, draw diagrams, define the system, list the relevant physical principle, and then solve the equation. You throw a 0.2 kg ball from shoulder height into the air. Right after it leaves your hand, the ball’s position is = (1,2,0) m and its initial velocity is 171. = (10,2,0) m / s . We want to determine the position of the ball after 0.3 seconds. Your feet are located at <0,0,0>. Take the ball to be the “system” for parts 1-3, and treat air resistance as negligible. 1. [3 points] Draw the initial state of the system, using a vector to indicate the ball’s initial position, a vector to indicate the ball’s initial velocity, and a vector to indicate the direction of the gravitational force due to the Earth on the ball. IIIIIHIIIIII IIIIIIII,fi IIII!!=i IIIW'“ VIII Illllll IIIIfiMflw » III-III..- III-III..- III-EIEIUI 2. [8 points] What is the momentum of the ball after 0.3 seconds? What principle will you use in order to answer this question? Momentum Principle: A}? = FMAI é” Z 9+5 mi. + F VIE! At , and M << 0 6 Z?“ {if=m<v[,x,vi‘y,0>+<0,—mg,0>At=\<2,—0.2,0>kg'm/s 3. [8 points] What is the position of the ball after 0.3 seconds? What principle will you use in order to answer this question? \7.+t7 ‘ fAt Position update principle A? = 17mm, F, = + 2 This expression for v is appropriate for a constant force or a time interval over which the force can be treated as constant. ave _. 1 pf pf, : ‘a '90 +_ hixa ' + J ,0 rf (x1 yr > 2<vt,r m v1.y m > 1 2 — . 7f=(1,2,0)+—<10+——,2+£3,0>-0.3m 2 0.2 .2 7f=(4,2.15,0)m 4. [11 points] Now assume that you are wearing ice-skates and are standing on frictionless ice. What is your velocity right after releasing the ball? Give the full vector form. Take your mass, M, to be 50 kg. Identify the object(s) that are in the system and in the surroundings and the forces that act on the object(s). What principle will you use to answer this question? ‘System 1: You, ball & Earth, ice. Forces: All forces are internal to the system. Reciprocity leads us to describe the forces in terms of pairs: Force you exert on ball — force ball exerts on you, components in x and y-direction. Force Earth exerts on ball - force ball exerts on Earth —y-component only. Force Earth exerts on you — force you exert on Earth — y-component only. Force ice exerts on you — force you exert on ice — y—component only. There is nothing in the surroundings acting on the system. Principle: Conservation of Momentum. The momentum of the system is constant (zero). You do not crash through the ice upon releasing the ball, so the net force on you in the y-direction is zero. (Because the mass of the Earth is so much greater than that of the ball, we need only concern ourselves with the forces acting in the x- direction.) pball,x = —pyau.x phallx = ' m/s : _50'v you ,x v =—.04m/s you .x = <—0.04,0,0) m/s Vyou System 2: You and Ball Surroundings: Earth, ice Forces acting between objects within the system: Force you exert on ball upon throwing it, force of ball on you (x and y-components). Forces due to interactions of surroundings with objects in the system: Gravitational force of Earth on You and on Ball - y-direction. Force of ice on you (y-direction) You don’t acquire momentum in the y-direction, so the net force acting on you in the y-direction is zero. No interaction of surroundings has a component of force acting on you in the x-direction. The Momentum Principle requires that momentum in the X-direction be conserved because the sum of the forces in the x-direction is zero (reciprocity). When the ball leaves your hand, it has an initial momentum. You must recoil with x-momentum of an equivalent magnitude. 1—5,," = <_pball,x,i ’0’0> aw = <§oo> m / s = (—0.04,0,0) m / s System 3: You Surroundings: Earth, Ice, Ball Force acting on you: x-direction: Force of ball on you due to change of ball’s x-component of momentum. y-direction: Normal force (feet on ice), gravity, force of ball on you (in y-direction) You don’t accelerate in the y-direction, so the net force on you in the y-direction must be zero. Only the x-component of the net force on you need be considered. This must be equal and opposite the force in the x-direction that you exerted on the ball. pyau = <_Fl7all,x ’0’0> At = <_pbally\' ’0’0> v, = <§§,0,0> m / s = (—0.04,0,0) m / s ...
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HG_Solution - Write down your recitation time: PHYS 172 -...

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