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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 13, 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 » IIIIII..
IIIIII..
IIIEIEIUI 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 iceskates 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 ydirection.
Force Earth exerts on ball  force ball exerts on Earth —ycomponent only. Force Earth exerts on you — force you exert on Earth — ycomponent 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 ydirection 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 ycomponents). Forces due to interactions of surroundings with objects in the system:
Gravitational force of Earth on You and on Ball  ydirection.
Force of ice on you (ydirection) You don’t acquire momentum in the ydirection, so the net force acting on you in
the ydirection is zero. No interaction of surroundings has a component of force
acting on you in the xdirection. The Momentum Principle requires that momentum in the Xdirection be conserved
because the sum of the forces in the xdirection is zero (reciprocity). When the ball
leaves your hand, it has an initial momentum. You must recoil with xmomentum 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: xdirection: Force of ball on you due to change of ball’s xcomponent of momentum.
ydirection: Normal force (feet on ice), gravity, force of ball on you (in ydirection) You don’t accelerate in the ydirection, so the net force on you in the ydirection
must be zero. Only the xcomponent of the net force on you need be considered. This
must be equal and opposite the force in the xdirection 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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