# Exam 1 - Problem 1(25 Points(a 5pts Which one of the...

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Unformatted text preview: Problem 1 (25 Points) (a 5pts) Which one of the following is not conclusive evidence of an interaction? Circle all that apply. 0 Change of velocity, either change of direction or change of speed. 0 Change of shape or conﬁguration without change of velocity. 0 Change of identity without change of velocity. 0 Change of temperature without change of velocity. (b 5pts) Write down the formula. for momentum valid at all speeds. Be sure to deﬁne any symbol or variable that you use. §=Km3 F5 is momerﬂ'um l X ‘ d 1:» (mt/sz I m \6 mass <7 \3 \Ie.\o eHﬁﬁ (c 5pts) Write down the momentum principle. ATS : Eel: A“: —k P? = JV Enet At t {5% : EL + Enet<t§-‘ti,3 The moon circles around the earth in a circular orbit at constant speed like the one shown in the figure at the right. Constant; Speedz \\QA\\:\\.\A’B\\: WA: \\-\jn\\: \I And \J<<C. (d 5pts) What is the direction of the change in momentum of the moon from joint B to point D? .i _\ 8 Va: \\\Ie\\/\\/B = <-\I)o ,o> <29 ﬁgs—F; <70 2 CID: <\/,0,0> A“; = a(m“\7)= m-AQ = m (VD—"QED = me <1,O.O> A z A 2 ______.______._____ : l P M?“ 1m, < .oio> (e 5pts) What is the direction of the change in momentum of the moon from point A to point D? \IA= <O)\1)o> or/ v ’, ._ * vbv Problem 2 (25 Points) An electron (me = 9 x 10‘31 kg) is observed to be moving in the +36 direction with a constant speed of 0.9960. Suddenly, at t : 1.5 x 10—9 s, the electron encounters a magnetic ﬁeld and experiences a constant force F = (0, -2 X10_12,0 > N. (a 9pts) What was the momentum of the electron before it encountered the magnetic ﬁeld? Since. \I N C; we need “to use. Tbm‘ét‘ﬁail. 1 l; 25: r“ : WW1. lire/ct '1” P = 11.19~ (qxxo‘a‘kgy {orsﬁwsmci O , Q>m/s 5 = mm? o , o> vim/s (b 4pts) At t = 3.0 X 10‘9 s, calculate the direction of the change in momentum. Be sure to show how you determined this. ._A Moman¥um PVXNQXEAQ. 1 Ala : Fﬁe—kA’t (c 4pts) At t = 3.0 x 10’9 s, calculate the magnitude of the change in momentum. Be sure to ShOW how you determined this. M3 = Em M; (d 8pts) Show how you can determine the direction and magnitude of the ﬁnal momentum graphically. \$2.: <5'lelézh Q ) o3 “gym/s Aw mm = <o,=s-o>n6“,o> Wis Problem 3 (25 Points) In the accompanying ﬁgure a 0.5 kg hockey puck sliding along the ice with velocity < 15,0,0 > m/s. As the puck slides past location < 1,0,3 > m on the rink, a player strikes the puck with a sudden constant force < 600,0,700 > N. (a)(10pts) In the space below, make a sketch of the path of the puck (as seen from above; from the y-azis) before and after it is hit. Note that the impact of the hockey stick is very sudden. When sketching the path of the puck you can treat this collision as if it occurs instantaneously. (b)(10pts) The hockey stick was in contact with the puck for only a brief moment Atl = 0.009 s. What is the position of the puck after this time has passed? __. -‘ _‘ ___‘ 9¥=§L+Fnﬁlxﬂ and \I<<C. so mev. mV‘F = mVL + Fne-l; Att "" i -‘ —“ _ L \ch -== \I-L + F; Fneilft‘ = <i5,c>,o‘>m;8 + 0.53% <(ooo,o [wows {0.00%} L V4: <‘5iO,O>‘“Is +<\O-8,O3 nigh/S = <25.%,O ,\Z-Co>m/s Shag Fae}: "5 “Wei—031* in this “time. tniewcﬂ, VON “1 V1 "" \341 3 Z - “ms: mamas/s 9: J - ‘ ~ ~ M. 5: L‘s" ‘1 Z \ Vows At, <ZOH) O )Co.3>M/S (O'Coqs) J A A If 4\ 132,013.“? r “ , A =- t r +Ar‘ <1,0,3>m + <o.t83(o,o)o.osm>m (c)(5pts) After losing contact with the hockey stick, the puck travels across the ice for a time Atg = 4 s. What is the position of the puck after this time has passed? In this {\me 'm’texxmi Eu: 0‘ So 43\$ 2 Q Omct AK} :1 0' AG 1 it" At :4 m A A 0 \IL \ié * v; imam 90:): b) = <25.8)O)\'2..(o>mis: \1 v Q ‘35 F‘ = “ = m A L \iwgatz <25.s,o,iz.<o> /s (q s) "1‘ 2 ....\ .A _— I 4; r.L + m — (MW—MO ,3iogrg>m + <\OS.Z)O)SO.L\>M Cow—3%) o )55H57>m Problem 4 (25 Points) One day during lecture Dr. Greco gets overzealous with a spring during a demonstration and damages the windings. The effect of this damage is that the magnitude of the force supplied by the spring is now Iﬁl = M313 where k = 1113.6 N/m3 and [3| =[L — Lolis the stretch of the spring. The force of the spring still acts in a direction to restore the spring to its relaxed length L0 = 0.50 m. (a 15pts) You place a mass m = 0.5 kg on top of the vertical spring and compress the Spring so that its total length is L = 0.25 In. You make sure the block is at rest and then you quickly move your hand away. What is the position of the block 0.1 seconds later? m Foe-i = Fsphnﬁa + FaroN‘Gni X lem%‘ *kss L r— ’\\\3.(o<O-15’O-5\ «up» N (0,0,05m .._.._ <O)\‘7‘L\ )0» M A mex=nﬁ-org< Fspr‘xng 3 V ‘3 3 a ‘ 0"— q‘g )O> N .a :2 O _ Fﬁfcwl‘)“: < ) 4‘q 1 N ASSUME Rel: ‘5 QPPVOXimoA‘elu) CODSJVCUW‘SF in lin‘xs short“ time. \W‘z‘ervdt A“ =33 = p ﬁa at <O)EZ.S)O> {0.0 we, \l<<(; —~> “\$3ng a v-+\/ A ’ 9 V —-— t F . .- avg 2 emca EVA N coh5+an+. j qu v‘l‘ res‘i’ 3 "I _.i ‘— 4 - 0.5 d Vw3_ 1+ T =1 <Oil'1510>m/5 r‘l ‘- 40 17M A?:—QONSAJL 2 <Opi‘2510> (O'l>m 1 <O)O.‘25!O>m r =2 r +Ar =‘~<O,O.ZSEQ§m+A’-€ :' (b 10pts) Starting from your answer to part (a), calculate the ﬁnal velocity of the block after an additional 0.1 seconds have passed. a: it; {3er lbetore == <0,0.375: 01" r; =\4:)O'§,° 7 “4 _. 3 A V . FSPMS—s was L = _\\\3.(o (o.375~o.5)3<o,1,o> N or :: F " o <o)z.z‘s5io\>m ,P'Mal lolo7A/ Fﬂrauﬁ‘j 1: mg :1 <‘Q I 3Q? N 0" t um m «mi» M ) Fﬁé’u‘i w VsP‘mﬁ'k ﬁémwémigﬁ <0 )_Z~‘—)ZS) 0> N : 4ofﬁﬂlavlv M3 = Fm at = _<Q,»o.'z”225) o> ks-m/s ov “A J/ .A —-" ‘ \1; t: \l‘ 4» 7?? Fnatb‘t :1 <0) 2-5)0>m’s + <02"O'5L’510>m/5 STAG: <oi, M355, o>m/s {k a 40} \52’o7g ...
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Exam 1 - Problem 1(25 Points(a 5pts Which one of the...

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