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Unformatted text preview: 188 P753 P154 _ P755 continued on next page ——————'* Potential Energy
(K+U),. =(K+U)f 0+(30.0 kg)(9.30 m/sz)(0.200 m)+;—(250 N/m)(0.200 m)2 _%(50_0 [(3)52 +(20.0 kg)(9.80 m/sz)(0.200 m)sin 400° 58.8 1+ 5.00 I = (25.0 kg)vz + 25.2 J (a) Between the second an/d the third picture, ABM = AK + AU —mgd = %mv,2 +§ch2 $50.0 N/m)d2 +0.250(1.00 kg)(9.80 111/52)
d = [—2.4512135] N 50.0 N/m =
(b) Between picture two and picture four, AE mech— =AK +AU , ,
—f(2d)=—mv2 __,m,2 j  {{3}}!
v— — =./_(3 00 m/s) — 500.45 N)(2)(O 373 m) l :{hﬁﬁ f m»: d [4— : .l/m i. :1 $0.00 kg)(3.00 m/s)z = 0 (c) For the motion from ' —f(D+ 2d) = —l(1.00 kg)(3.00 m/s)2 9.100
D: ~——————_______20.37s _ _03
2(0..2250)(100kg)(980m/s) ( m) m (a) Initial compression of spring: .1.ka = .1.va 2 2.
5(450 N/mXAxf =§(0soo nglzo m/sf =Ax continue 190 Potential! Energy f
K‘ +U, +AEmech =Kf +Uf: 0+(mlgyl +ngyz)‘. —Ifkdx=[;—mvz+mgy) f (sags + (3196.5 _ fag—0.01901: = as 1):): + (sign
0 5 5
40.0g+ 19.5g— 2.00gjdx+0.4003jxdx = 4.0002 +3203
0 0 5 2
27.5 g — 2.00311: + 0.40033‘?‘ = 4.00::2
0 27.53 — 2.00g(5.00) + 0.4003025) = 4.00:»2
215g = 4.0002 (22.5 9.80 s2
v=J “”100 m“ L— The geometry reveals D = L sin 9+ L sin 90, 50.0 m = 40.0 m(sin50°+ sin ¢), ¢= 289° P757 (a) From takeoff to alighting for the JaneEarth system K H W 2K H (+3).'+wmd (+3);
21mv,2 + mg(—L cos 6) + FD(—1)= 0 + mg(—L cos ¢)
1 2 50 kg viz +50 kg(9.8 n1/52)(40 mcosSO°)—110 N(50 m)=50 kg(9.8 111/5 %50 kg v?1.25x10‘ 1—5.5x103 I=1.72x10‘ I = ‘/ 294;” = (b) For the swing back émviz + mg(—L cos (0) + FD(+1) = 0 + mg(—L cos 0) 1 2 130 kg 0,? +130 kg(9.3 m/s2)(—40 mc0528.9°)+110 N(50 m)
= 130 kg(9.8 m/sz)(40 mcosSO") i130 kgv? —4.46x10‘ J+5500]=—3.23x10‘1 2}[40 mcos 28.“
I 2(6340D
via/ W = continue: ‘ I'7.61 (a) Energy is conserved in the swing of the pendulum,
and the stationary peg does no work. So the hall’s
speed does not change when the string hits or leaves
the peg, and the ball swings equally high on both
sides. (b) Relative to the point of suspension, U; = 0,
LI! = mg[d  (L — d)]. From this we ﬁnd that —mg(2d — L) +—;—mv2 = 0. Also for centripetal motion, my? ‘ F [6. P161
mg =—R— where R = L — d . Upon solvmg, we get I
5 I' 11.2 (a) At the top of the loop the car and riders are in
free fall: ZFy=may1 mgdown=£1;— down 0 = Rg forces 2*. Energy of the carridersEarth system is
conserved between release and top of loop: Ki+ugi=Kf+Ugfz G+mgh=§mv2+mg(2R) gh = gRg +g(2R) (1)) Let it now represent the height 2 2.5K of the
release point. At the bottom of the loop we have mgh=§mv§ or v? =2gh mv
ZFy=may: nb—mg=—Ri(up)
_ mg?!)
“I. — mg+ R At the top of the loop, mgh = i—mv? + mg(2R) of = Zgh‘4gR , {in nut! page 202 Momentum and Collisions P8.5 (a) For the system of two blocks Ap = 0, or P: =Pf' ﬁ‘d‘gﬁ Therefore, 0 = Mum + (3M)(2.00 m/s). Before .1? I)
Solving gives v," = (motion toward the left). Section 8.2 Impulse and Momentum l‘1’8.6 From the impulsemomentum theorem '1? (At)  Ap f=m(vxv:)=(12kgx0—6o mi/th Im/s J
(At) 0.0505—0 P8.8 0?) Work = K J, —K, =%(o.050 0)[(40.0)2 — (50.0)2] = ‘ continu Chapter 8 203 . [’83 A13 = FM
Apy = m(vfy  72%,) == m(v cos 600°)  mv cos 600°: 0
ApJr = m(v sin600°——v sin 600°) = —2mv sin 600° = —2(3.00 kg)(10.0 m/s)(0.866)
=—52.0 kgm/s
Fave = A5: = 4232:: 'Sm/s =W FIG. P83
Take xaids toward the pitcher
(a) pix + IJr = p fr: (0.200 kg)(150 m/s)(—cos 45.0°)+ IJr = (0.200 kg)(4ll.0 m/s)c0530.0°
1, = 9.05 N  5 pi}, +13, = pﬂ: (0.200 kg)(15.0 m/s)(—sin45.0°)+ 13, = (0.200 kg)(40.0 m/s)sin30.0° i: (9.053+6.12§)Ns (b) i=%(0+1'=,,, )(400 ms)+i,,, (20.0 ms)+;—im(400 ms) in, x 24.0 x 10‘3 s = (9.053 + 6.125) N  5
E = (377i + 2553) N
The force exerted on the water by the hose is _ Arm. _ mv: —mv. _ (0600 kgxzso m/s)—o _ .
L N _ At _ 1.005 ’EzE' According to Newton’s third law, the water exerts a force of equal magnitud. 2 back on the hose.
Thus, the gardener must apply a 15.0 N force (in the direction of the velocity of the exiting water stream) to hold the hose stationary.
(a) Energy is conserved for the springmass system:
1 1
K, +Us, =Kf +1.15}: 0+Ekxz 23min2 +0 k
v=x —
m k
(b From the e nation, a smalle value of m makes = — lar er.
> e: v a}. g .7+ '7 cm mar! page 16 (a), (b) Let Us and trP be the velocity of the girl and the plank
relative to the' ICE surface. Then we may say that vg — up is
the velocity of the girl relative to the plank, so that v —u,, =1.50 (1) 6i 8 But also we must have m go 8, + m ”U = 0 since total momentum of the girlplank system is zero relative to the
ice surface. Therefore 45.0193 +1500? =0, or Us =.':l.?.’o3vP Putting this into the equation (1) above gives —3.33vP  Up = 1.50 or up = v 3 = ——3.33(—0.346) = Chapter 8 205 01, speed of m1 at B before collision. A]. m]
lmﬂ’f = "‘13" 5 m
2 L m:
_____ ‘ —_l._——_—— = " 2(9.80)(5.00) '—" 9.90 I'll/S  B C
v, , s eed of m1 at B 'ust after collision. 1’ P 1 FIG P8 16 m] — m2 1
= = —— 9.90 s = —3.30 s v}, m1+m201 3( )m/ 111/ At the highest point (after collision)
—3.30 s
m.gh....= EMA330) hm =(2(—9'——80 ng For the cartruck—driveﬂdriver system, momentum is conserved:
131.62.: 13:16:21: (4000 ngs m/s)i + (800 kgxs m/s)(—i) = (4 800 kg)vfi 25600 kg m/s=
”I: 4300kg 533111/5 For the driver of the truck, the impulse—momentum theorem is m = 131. 43,. : Fame 5) = (30 kg)(5.33 m/s)i — (so kg)(8 m/s)i  = 1.78 x 103 N(—i) on the truck driver “ For the driver of the car, 13(0120 s) = (80 kg)(5.33 m/s)i — (80 kg)(8 m/sN—i) F = 8.89 x 103 Ni on the car driver , 5 times la rger. 206 Momentum and Collisions P8.18 Energy is conserved for the bobEarth s ystem between bottom and top of .  ~.. 
SWing. At the top the stiff rod is in com pression and the bob nearly at rest. I Kj+ui=Kf+uﬁ ng§+O=O+MgZE 7;: =34! so 0,, =2J§ # '
Momentum of the bobbullet system is conserved in the collision: v 4M FIG. P8.
News H P8.19 (a) According to the Example in the chapter text, the fraction of total kinetic energy  —
to the moderator is where m2 is the moderator nucleus and in this case, 7112 =12ml f2=W=fﬁg=— of the neutron energy is transferred to the carbon nucleus. (b) Kc = (0.2s4)(1.e xm‘” I)=
Kn =(0.716)(1.6x 10"3 I)= P8.” We assume equal ﬁring speeds v and equal forces F required for the two bullets to push wood ﬁ
apart These equal forces act backward on the two bullets.
For the first, Ems,” =Kf E(7'00x10 3 kg)v2—F(800x10'2 m)=[
For the second, p =pf (700x10 3 kg)v=(1014 kg)vf
(7 00 x 10'3 )2:
I)! =
1.014
. ‘ _ _ 1 _3 2 __ 1 2
Again, Ki + new, _ K ,. E(7.00 x 10 kg)v — Pd _ 50.014 1(3):: f
2
. . 1  1 7.00 x 10—32;
Subshtutm fo 0, — 7.00x10 3 k 2—Fd=— 1.014k
g r 1’ zl 3)” 2l 3 1.014
‘ 2
7.00 x 10*3
Pd =l(7.oo x 10312;2 —1(————_)_v2
2 2 1.014
.3
Substituting for v, Pd = F(8.00x 10"2 m)[1— 70:}ng J d = 7.94 cm continue ...
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 Fall '10
 Koskeshian
 Physics

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