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Unformatted text preview: Physics 171.101 Exam 2 Solutions March 8, 2007
Prof. Barnett Do all problems. Everything should be written in ink.
You are allOwed to use a calculator and material written on one 3”x5" index card. Use Sin(370):0.6, Cos(37°)=0.8, = 10 urn/52, G = 6.7 X 10"11 N ' rag/1:92. 1. Two stationary 5 kg masses, called ‘Mlef; and Alright, are connected
together in a system with two perfect (massless, frictionless) pullies
and a perfect [massiess, ﬁxed length) rope. One end of the rope is
connected to the ceiling. One pulley is connected to the ceiling. One
pulley moves up and/ or down with Mgeft. I A) (5 pts) Draw a CLEAR picture of Mieﬁ and all of the forces acting on it. '
There are two ropes pulling up and gravity pulling down.
Call the + direction down. MLg — 2T = .MLaL. Jim‘s; B) (5 pts) Draw a CLEAR picture of Mright and all of the forces acting on it.
There is one rope pulling up and gravity pulling down.
Call the + direction down. MRg — T : ﬂfRoR. . C) (5 pts) What is the relation of the acceleration of Misﬁt to the acceleration of
Mﬁght? Hint: If M} goes up 1 cm how far and in which direction does Mg go?
If MR goes up 1 cm then M; goes down by 0.5 cm. So aL = —aR/2. D) (8 pts) What is the acceleration of .Mgeﬁ and the acceleration of Mrggm? Draw a
picture of the entire system and show which way each mass moves. ’ " ' ’
From Part 3)} T : MR(g — :13). Put that and {IL : wag/2 into the
equation for Part A) to get aR = 0.49 = 4 m/3802 downward and ’i‘ 61;, 2 0.29 = 2 m/sec2 upward. ' I, E) (7’ pts) What is the tension in the rope? T = MR(g — {1.3) —> T = 0.69M = 30 N. ‘2. Two masses, M1 = 10 kg and M2 = 20 kg, are located at R1 2 (2% + 23' + 3ft) meters
and R2 2 (5i — 2} w 9k) meters respectively relative to an origin. A) (10 pts) What is the gravitational ﬁeld, 3}}, produced by Ml at the location of M2?
R; — g; = (33 ~— 43 — 12%) meters. It’s length is 13 meters. The unit vector going
from M1 to M2 is r" = (R3 — 1351) AR} — R] = (0.23% — 0.3053 _ 0.923%) m. 57: —G—%Lr = —(3.96 x 1012)(o.23E — 0.305} — 0.923%) N/kg = (—0.91§+1.22}+ 3.66%) X 1012 N/kg. B) (10 pts) What is the gravitational force, 13. due to Ml acting on ME?
F = §M2 = (—1.8% + 2.443" + 7.321;) x 1011 N. C) (10 pts) How much work would YOU have to. do to move M“; from its position
to a point inﬁnitely far from Ml? Assume that M1 does not move during this operation.
The initial potential energy is RE. 2 —G;1‘“:qﬂ = —1.03 X 10'9 Joules. YOu must do that amount of ” positive” work to get the particle free
—> W = —P.E. = +1.03 x 10—9 Joules. . kg
3. A 70 kg man falls from a platform onto a trampoline 10 meters '3' m below. Consider the trampoline to be a spring with a spring “Lister: A” Lion _‘\“.«'m constant Of araznpoime E, A) (10 pts) What is the man’s speed when he ﬁrst touches the trampoline? _. l I
Use conservation of energy and put the y = 0 origin at the top of the trampoline.
The initial position of the person is ygniﬁag : +10 meters. ﬁrm)? = mgymmg —> v = 14.14 ni/scc. B) (10 pts) How far does the trampoline stretch before the man stops his downward
motion?
Call the position of the trampoline when it is fully stretched “ystop”. It should be
a negative position, is. below the trampoline’s original position which is y = 0.
Again use conservation of energy, with no kinetic energy when the man is at yaw.
mgy'initial z mgystop + _) ystop = _3‘7 meters
The trampoline stretches by 3.7 meters. C) (10 pts) What is the man’s location when he has his maximum speed during his
journey from the platform to being stopped by the trampoline? Draw a picture to make your answer clear.
The man continues to accelerate after he hits the trampoline until the upward force of the trampoline equals the downward force of gravity. That is the equilibrium
position of the trampoline with the man on it. So his speed is when he
is at that equilibrium position. Kyeqmnmum + m9 = 0 —* yequélébréum = —05 111 u i : : 0.5 m
The man’s highest speed occurres when he is 0.5 m '
below the original position of the trampoline. D) (10 pts) What is his maximum speed during his journey from the platform to being
stopped by the trampoline?
Again use conservation of energy at tho yeqmubrmm point:
mgyim'tial : mgyequﬂi'briam ‘l‘ %K(yequﬂibrium)2 ‘l— Zlmﬂg "—> ’L' = 14.318 III/SEC. !//////f ...
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This test prep was uploaded on 04/15/2008 for the course PHYSICS 171.101 taught by Professor Professorhenry during the Fall '08 term at Johns Hopkins.
 Fall '08
 ProfessorHenry
 Physics

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