Midterm exam

Physics 111, University of British Columbia, Okanagan, Name:

/40

Student number:

1 _152 _153. lS4 _155 _ls6. _ln

-AJq- Ko b ow- tL'' So I r n g

T,-Lt- Wb^rski gi,n(2

For multipie choice questions, please show all work for part marks. Correctanswers will get full marks, incorrect choices will get part marks for any correctwork, irrcluding crossing off incollect answersi provided tire correct answer is notcrossed off. Please choose only one correct answer per question.

FIG. 1: t-component of position versus time graph, g-component of velocity versus time graph

graphs is consistent with projectile motion? l51. Which of the above sets of( (.i}raphs (ii) and (v).

$craphs (ii) and (vi).(c) Graphs (i) and (iv).(d) Graphs (iii) and (vi).(e) Graphs (i) and (v).

iav pto itt,h[r- r*tlo^ i

x= + V,t d Vx'ce* St&

Vg=V

sivat/nl [,* a/ pds cr *J fhf* Q'ey'^'lf 'f Vx rf fui'or *1)C i,')

- ft-(

(iii)

tlra,XW t'w lol nofi, g{y , (v)

J",ke- &o*\sLr fil'nr3

2. A charged particle moving in a uniform magnetic field experiences a constantmagnitude acceleration perpendicular to its velocity vector at ali times. This means

l5(a) The particle will move in a straight line.(b) The particle will travel at constant velocity.

-lg) The particle will spiral inwards.(1d}te particle will travel at constant speed.

Y.i The particle will spirai outward"s.

\"lJt---- vc hav7.e

g1n&dfV,tf'/at chnryrt cluyni( cha'rtVt,

D{"i '-t ctrcl,Jar l,uolgn .

FIG. 2:

3. A particle's path follows the trajectory shown in Fig. 2. Which of the followingstatements is correct? 15

(a) The particle's instantaneous velocity at r : 0 is zero.

(b) lt{o information about the particle's velocity or acceleration can be found fromits tra.jector.y.

a"@>he particle's instantaneous velocity at r -- 0 points in either the e or the -?\-dir6tion.(d) The particle's instantaneous acceleration at tr :0 is zero.

(e) The particle's instantaneous velocity at r : 0 is positive.

S.si W$uu X=o 4irtr{ af*ris d ttc,t! iA /*dir,ttt . cu'*'( +ctl'

iftcwt s,^j, dterreor,^6, sY c6 n6+*l

x<D / ?c,/{t{{l n^n{tuq

tf pu,r{/Ll('t sF€d r !

c 6-ns+^"+ b*t tl n.ArLft h",',L

4- tMt c*3 w c,x,rng tl* K-qxt wlu,r. Y =o,

{ala hLawsb, Qal'n s

4

--+ -+ --+4. Given A :b3_l6Jt B *i_gj, C : _2i+3j and d _ 2, find thc magnirudc

and direction of A - B + dC. Expressing this vector in the form (r,0) where d ismeasured relative to the.pos$iveE-axis,rthe closest to your result is:

-s rs

H, o^:_ru;lll il =,q - V * ad : (S f* 6il _ gf $i)(\/450, o : _82)qile,(zf

)i ,3F(") (Ju,9 : 31')(d) (\^ss, 9 :80.5')(") (Ju, o:211")

0=10*F=19,1'

: (s-+-q)i t; -3?+l_t f

( (+1*(

2/rd\= -- -:

---* \ tts> #*F(-sJ'*

.dvst

lr =cowfi

6

y4(m)

a_L

a-.L

-4

-4

54(m/s)'2

Srrl*- Brbw(k,' -Sal'n f

5.4

t (s)

(b)

t6

5=L0 t)lw

5t/':--

(s)

slopt = g

('v'1

?^udv dq ,

&= sl"pu d V vs t

awcn o7 V vs {,

_t

V r'5 S,(o( e l

:" rl-^ != lwr

V vs t ru,rrt h^ea

SloV< l,w I f

3

i

i-i

Wl,,m !s - 3 wt I 5 " Cu-^,tt' 3y 4

y rls + Yrnns:l V"N

skw -jv(r

FIG. 3:

6. In Fig. 3 (b), you are given a graph of the g-component of the velocitrr of aparticle as a function of time in seconds. Find the gr-component of the position vectorstarting at A - 0 when t : 0, and the y-component of the acceleration as a functionof time, given the y component of the velocity. The g-component of the position

AJ frn,^ @ h& I S tS ew6of v,J vs twh''c h t(

"r:

from t : 3 - 4s is also provided on the graph. 111

FORMULA SHEET

1D kinematics (constant acceleration):As : s(t/) - s(ti) : ui\t + |a(Lt)2- A,uu: Lt

'?:ul+zaLsGeneral 1D kinematics equations:u(t) : #, a(t) : #, As : s(t/) - s(ti) : ft u1t1at,

Au: jj' o1t1at

,(t) : #, ,(t) : #, Lo : [], ,1t1at, Lu :[," *1t]dt.

Circular motion:a": *S:fQ U1 :T1J A1 :T(tConstant aL0:wiL,t+ja(Lt)2Q.f:wi*atC aliiean transformations :

___-! r J ___-!?:R+r' ?:ut+VNewton's eouations:----\ r ----\ ----\ ----\

Frrl:mi Frrr :Fr+Fz+...f ,: Frfr fn: Fnn f, : FrfrD x jAu2Fc : mg Fc.nrz : Ati-Calculus rules for functiorls u1(t) : ct",u2(t): d'ebt, and z3(t) : A(t)+ B(t).Derivatives:

du1(t) , tn_1: (:'ll,l,dt

duzft);' ' : d,bebt exponential law (2)d,t

49 : o7l! + dB(t) distributive law (3)dt-dt'dt

power law (1)

A(t)dt * 1,"

1,,"

t,

1,,"

-+nlIu1(t)d,t-=l',nl I

t s d"ebt .

u2(t)ctt : Ti'

u3(t)dt: 1,,"

: ,(tf)"*' _ r(tu)"*'n*I n*I

d,ebtt i,ebtn

power larv

exponential law (5)

(4)

B(t)dt distributive law (6)