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Unformatted text preview: Chapter 8 209 Divide equation (2) by (1) tan 0 = = 0.633 2.43 From which
Then, either (I) or (2} gives V = (b) K, =%(9o.o kg)(5.oo m/s)2 +%(95.o kg)(3.oo m/s)2 = 1.55x 103 I
K, =%(185 kg)(2.88 111/5): = 7.67x 102 J
Thus, the kinetic energy lost is 783 Jinto internal energy.
p1]: = pxi: moo cos 19+ mvy cos(90.0°—0) = mo, 61
o ®
_ Before
00 cos 0+ vY sm9= v, (1) ©/io
pyf = pyi: moo sinB—mvy sin(90.0°—6’) =0 (/4;
_ . _  — _ q _ . _ _ _ _ ql...__—__..__.
v0 sin6=vycost9 (2) ‘m" _ \(_ 9
«From equation (2), u R
cost? I g;
= 3 Y
z’o “([3ng ( )
a FIG. P825
Substituting into equation (1), . 
v 0052 9 +0 sinﬂ :2
Y sinﬁ Y '
so oﬁcos2 9+ sin2 6)=v,sin6, and .
Then, from equation (3), 00 = v, cosﬂ .
We dld not need to wnte down an equation expressing conservation of mechanical energy. In the
problem situation, the requirement of perpendicular ﬁnal velocities is equivalent to the condition of
elasticity. ' 210
P826 P827 P828 Momentum and Coliisfons We use conservation of momentum for the system of two
vehicles for both northward and eastward components. For the eastward direction: M(13.0 m/s) = 21WJr cos 550° For the northward direction: 5.00 m/s + 0 = (4.33 III/S) cos 30.0°+
02!: = 1.25 m/s 0 =(4.33 m/s)sin30.0°+v2fy ‘0ny G 2.16 m/s 62f = 2.50 m/s at 6o.o° “Hzf: (a) §i=l3f 3° pn=paf
and 3% = PM‘
m7), = mv cos 9+ mvcos ¢ (1) O”
0=mvsin0+mosin¢ (2)
From (2.), sin0=~sin¢
so 6: ¢.
Furthermore, energy conservation for the syetem “6' P828
of two protons requires
émvi2 =§mv2 +—mv2 212 Momentum and Collisions (b) E=1 2 1 2
mlv1+
2 =21[(5.oox1027)(cs_00><10“)2 +( 13:439x10‘” 1 Section 8.5
P832 _ 2mg, _ (2.00
~37??— the middle square, and A3 .‘lCM 1ch = 2 The Center of Mass (250 kg)(0)+(4.oo ng—osoo In)
W %M(5.00 cm)+gM(15.D cm)+ @MXZSD cm)
W 1 2 may; 8.40x 10*”)(400 x105)2 +(3.60x 10”)(125 x10 : N kg)(3.00 m) +(3.00 kg)(2..50 m) + = 13.3 cm 220 Momentum and Collisions (d) person: my —me,. =60.0 kg(1.33—4.00)§ m/s=
cart: 120 kg(1.33 m/s)— 0 = 1 1
(e) x f —x,. = 3(vi + 3,): = E[(4,00+ 1.33) m/s]0.680 s = 1 1 '
(f) xf—xi =§(v,+vf)t=§(0+1.33 m/s}0.6805=m ‘PBAB Using conservation of momentum from just before to
just after the impact of the bullet with the block: 222 Momentum and Collisions
P851 (a) The initial momentum of the system is zero, which
remains constant throughout the motion. Therefore, when m1 leaves the wedge, we must
have mszedge +mlvblock =0
or (3.00 1(3):;wedge +(0.500 kg)(+4.00 111/5) = 0 (b) Using conservation of energy for the block~wedge
Earth system as the block slides down the smooth (frictionless) wedge, we have FIG. P351
[Kbiock +usysteml. +[KwedgeL =[Kblock +usystem]f +[Kwedgelf I
or [0+ mlgh]+0 = [5415(400)2 + 0]+%m2 (—0.667)2 which gives . P852 T =d—m dt (Avmel)+%rt£(Avah)z T=(3.00 kg/s)(600 III/s)+(80.0 kg/s)(600 m/s—ZZB m/s) T=1800N+3.02x104 N= 3.20x104 N The delivered power is the force (or thrust) multiplied by the velocity, 3' 5P=Tv=(3.20x104 N)(2.23 m/s): *P8.53 It is interesting to compare this problem with Problem P650. The force exerted by the spting on ea
block is in magnitude IE4 = 1:2: = (3.85 N/m)(0.08 m) = 0.308 N. (a) With no friction, the elastic energy in the Spring becomes kinetic energy of the blocks, w ' have momenta of equal magnitude in opposite directions. The blocks move with constant
speed after they leave the spring. (K+LI)I.=(K+U)f 1' 2 1
~2—kx =Em1vff+§mzv§f l 1 ' 1
3(305 N/m)(0.08 m)z =E(0.25 1(3):]; +—2—(0.50 kg)v§,
"1161; + "1262' : m1?!” + "1262! 0 = (0.25 kg)vlf + (0.50 kg)vzfi
U” = 2102f 1 2 1 1
0.012 3 1:5(025 ngszf) +§(0.50 kg)v§, =§(1.5 1:90;, 1/2
0.123 I _. c
92! =[0I75 kg] =0.128 m/s v2, =0.128 m/s: v1, =2(0.128 m/s)=0.256 m/s 0” =0.256 m/s(—i) continued on next page Chapter 8 225
—1 "2y .1 1/50, 3
= 9: — =
(b) a tan (Uh) tan [ 3 Each object swings dOWn according to
ng =%mv12 MgR =§Mu12 01 = 1{2gR Swinging up: gun + "on; = (M + m)gR(1 — 005.35“) :12 = J 2gR(1 — cos 35°) I
1[233M] — cos 35°) (M + m). = (M — _m)1/2.gR 0.425M +0.425m = M— m
1.42511: = 0.575M P11“ = pn': mshell valtell cos 45‘0°+mca1mon Dread] = 0 (200K125) cos 45.0°+(5 000)!)recoil = 0 Use conservation of energy for the system of the cannon,
right after the cannon is ﬁred to the instant when the the carriage, and the spring from
cannon comes to rest. ‘ 1 1
1r<f+u‘ng +113). =Ki+um+um¢ owing” 2 m‘ kl: 2.00x10‘ m: Page 260 Rotational Motion SOLUTIONS TO PROBLEMS Section 10.1 Angular Position, Speed, and Acceleration P10.1 m 61H, = dB
011% =EL=O =10.0r4.00£1t=u = 10.0 rad/0
_ dd), _ 2
a,=0 — dt i=0 — 400 rad/s (b) 61Hum s = 5.00 + 30.0 + 18.0 =
d0
df=3£o s = dCL=3Im 5 = + 4.001‘Lz100 s = t
t d6?) 2
_ = ‘I = 4.00 re. s
“113.00 s d i=3”) 5 tics—The ngld Object Under Constant Angul
Acceleration P10.2 of = 251x104 rev/min = 2.63 x 103 rad/s I I
w a: .6 1 3 _ .
(a) a=—f—t—_'=z—§ig—2—r§_/i_9= 8.21x102 ta 52
. S . 1 (b) 0f=oJ,.r+§.spr"=0+—;_}(8.21x102 rad/52)(3.2.s)2= 421x103 rad 100 rev 1min erad 10::
P103 w=———__ ____ ._.__ I =0
‘ 1.00min(60.0sII.00rev) 3 rad/s mi w 0103
(a) f: f I __ az+w 10 1
0» 9r=ar=[—’z——']t=(z’5mws )= a), = 3 600 rev/min =3.77>< 102 rad/s "PIOA 49:50.0 rev=3.14><102 rad and a1, =0 50% = a),2 +2a9 Chapter 10 261 a; = 5.00 rev/s = 10.01: rad/s. We wiil break the motion into two stages: (1) a period during which the
tub speeds up and (2) a period during which it slows down. 0 + 10.07: rad/s . P105 While speeding up, 91 = a = 2 (8.00 s) = 40.0:rrad
While slowing down, 92 = ‘ait = WOZD s) = 60.0lrrad
so, em = 01 + 62 = 1001'rrad = 9r —9, =a),t+%arz and to}: =a1,+at aretwo equationsintwo unknowns mi and a:
— ~m a —9 (m ~at)t+latz~wt lmz
wi'wf  f i" f 2  f ‘2 2.1: rad
1 rev 61.5rad '
232 d=294 d 4.50 2a: a: =
ra ra ( s) 45,052 m
At? lrev anad _5
== = = 7.27x10
(3) w M May 864005
A0 107° anad 4 _
At=——=——— = 2.57x10 3 or428mm
(b) . a, 7.27x10'5 rad/Si 350°] M 37.0 rev( J=9s.o rad/s (3.00 s)—%a(3.00 s)2 $0.3 Relations Between Rotational and Translational Quantities 2 Estimate the tire's radius at 0.250 m and miles driven as 10 000 per year. 9_§_1.00x104 mi[1609m =6.44 107
r 0.250111 J x rad/yr lmi lrev
644x rad/yr[2”rad] 10 x 0 rev/yr or 0 rev/yr _Zﬂrad[1200rev v a at = (126 rad/s)(3.00 x 10'2 m) = =w2r=<126>2(800x10'2)=1260 m/sz soar =
s 2 r6: m = (126 rad/s)(8.00 x10‘2 m)(2.oo s) = (d) (6) Chapter 10 263 _ v_1.30 In/s_
.—;————0_023m —
1.30 m/s 1
“i=m=
‘ 2.2.4 rad/s—56.5 rad/s —34J rad/s _3 2
=w+at:a= =————= —7.63x10 5
“'1' ' (74(60)+33)s 44735 rad/
—1 —1 565 224 4473 —1 105 d
Bf—Bi—E(a)f+a)i)t—E(( . + .)rad/s)( s)— x =0: =(1.30 m/s)(4 473 s): 5.81 x103 m = 400 kg, r1 =y1'= 3.00 m;
= 2.00 kg, 72 =ly2] = 2.00 m; =300 kg; r3 =Iy3 = 400 m;
_;_:2.00 rad/s about the xaxis Ix=m1 I, = 40003.00)2 + 200(200)2 + 3.000400)2 = page 2_1 2_
2wa §(9z.0)(2.00) _ The force of static friction must act forward and then Inore and more inward on the tires, to produce
both tangential and centripetal acceleration. Its tangential component is m(1.70 III/32). Its radially 2 . . mv . . ‘
mward component 15 —. 'I'l'us takes the maxunum value 1‘ mwﬁr = mr(aJ,2 +2m56) = mr(0 + 20%] = mind: mmt = mrr(1.70 m/sz). With skidding impending we have 2F}, = may, +n — mg = 0, n = mg f5 zﬂsn =ﬂsmg=1/m2(1.70 m/sz)2 +m27rz(1.70 m/sz)2 1.70 2
ﬂs=———;‘/s  Rotational Kinetic Energy 1'12 4' marzz + M3732 264 Rotational Motion 1 1
(b) 121 = r150: 3.00(2.00) = KI = Em! of = Emmxaoof = 720 1
U2 =T2w= K2 25171123: =
03 =r3w=4.00(2.00)= K3 =21—m31)32 1<=I<I +1<2 +K3 =72.0+16.0+96.0==211wa P10.15 From conservation of energy for the objectmmtablecylinder
Earth system,
1 v 2 1 2 +3"!!! —mgh
U2 2
I;— = 2mgh — my
P10.16 The moment of merha of a ﬂun rod aboutan 3x15 through one end IS I §ML2.
kmehc energy is 3mm as K R =zllhw§ +2154»:
2 2
“nth bl = :11th = 60.0 kg(32.70 n1) =146 kg m2
2 2
and 1,. ELIng = 100 kg: 50m) =675 kg'm2
erad 1 h
In ddm ————. =1 1 4
a on 0,, 12h [36003] 45x 0 rad/s
anad 1 h _3
= ——— = 1 0
whﬂe 0),” 1h {36005) 75x1 rad/s
Therefore, K R = 5(146)(1.45 x 10“)2 +— N
'3‘
{31‘
1—:
E‘n
X
H
Q
d;
M
II ...
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This note was uploaded on 01/24/2011 for the course PHYSICS 6A taught by Professor Koskeshian during the Fall '10 term at UCLA.
 Fall '10
 Koskeshian
 Physics

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