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Analytical Mech Homework Solutions 35
Course: PHYSICS 4360C, Fall 2011School: UNF
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Word Count: 123
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= v2 v2 2 ( + + ) m 1 2 2 v = v 2 ( + + ) m (b) v2 2 ( + + ) = 0 m 1 2 2 v = ( + + ) m (c) mx = Fx = V x mx = V my = = 2 y y V mz = = 3 z 2 z 4.5 (a) j F = ix + y dr = idx + dy j on the path x = y : (1,1) ( 0,0 ) 1 1 1 1 0 0 0 0 F dr = Fx dx + Fy dy = xdx + ydy = 1 dr = idx dr = dy j on the path along the x-axis: and on the line x = 1 : (1,1) ( 0,0 ) 1 1...
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= v2 v2
2
( + + )
m
1
2
2
v = v 2 ( + + )
m
(b)
v2
2
( + + ) = 0
m
1
2
2
v = ( + + )
m
(c)
mx = Fx =
V
x
mx =
V
my =
= 2 y
y
V
mz =
= 3 z 2
z
4.5
(a)
j
F = ix + y
dr = idx + dy
j
on the path x = y :
(1,1)
(
0,0 )
1
1
1
1
0
0
0
0
F dr = Fx dx + Fy dy = xdx + ydy = 1
dr = idx
dr = dy
j
on the path along the x-axis:
and on the line x = 1 :
(1,1)
(
0,0 )
1
1
0
0
F dr Fx = dx + Fy dy = 1
F is conservative.
(b)
j
F = iy x
on the path x = y :
(1,1)
(
0,0 )
1
1
0
1
0
1
0
0
F dr = Fx dx + Fy dy = ydx xdy
(1,1)
and, with x = y
(
on the x-axis:
(
and, with y = 0 on the x-axis
(
on the line x = 1 :
(
0,0 )
(1,0 )
0,0 )
(1,0 )
0,0 )
(1,1)
1, 0 )
3
1
1
0
0
F dr = xdx ydy = 0
1
1
0
0
F dr = Fx dx = ydx
F dr = 0
1
1
0
0
F dr = Fy dy = xdy
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UNF - PHYSICS - 4360C
(1,1)(and, with x = 1on this path1,0 )(1,1)(0,0 )1F dr = dy = 10F dr = 0 + 1 = 1F is not conservative.4.6From Example 2.3.2, V ( z ) = mgre2( re + z )zV ( z ) = mgre 1 + re From Appendix D,(1 + x )11= 1 x + x2 + z z2V ( z ) = mg
UNF - PHYSICS - 4360C
h=re 1 2 2re v 2zre g22h=re re2v 21 z22greFrom Appendix D,( h, z1(1 + x ) 2 = 1 +h=x x2 +28re re v 2zv4++ z +2 2 2 g 4 grehre )v 2z v 2z 1 +2 g 2 gre 1v2 v2 From Example 2.3.2, h =1 2 g 2 gre v2 v2 1And with (1 x ) 1
UNF - PHYSICS - 4360C
sin = gbv2cos 2 = 1 sin 2 =v 4 g 2b 2v4gb 2 v 4 g 2b 2 gb 2 v 2+= 2+v22 gv 22v2gMeasured from the ground,gb 2 v 2hmax = b + 2 +2v2ghmax = gb The mud leaves the wheel at = sin 1 2 v4.8x = R cos so t =andx = v x t = ( v cos ) tR
UNF - PHYSICS - 4360C
Now sin = cos = cos + 4 2 4 2 2 4 2 2v 2 Rmax =cos 2 + 2g cos 4 2Again using Appendix B, cos 2 = cos 2 sin 2 = 2 cos 2 1v2 2v 2 1 1Rmax =cos + + =cos 2 + + 1222g cos 2 2 g cos Using cos + = sin ,22vRmax =(1 sin )g (1
UNF - PHYSICS - 4360C
2sin 2 =2gh= 1 22csc v01gh 1 + 2 = v0 1 + ghv0 2gh Finally csc 2 = 2 1 + 2 v0 (b)Solving for Rmax Rmax =hhh==2sin 1 2 sin 1 2 csc2 Substituting for csc 2 and solving v2gh Rmax = 0 1 + 2 g v0 4.10 We can again use the results
UNF - PHYSICS - 4360C
zzmaxv0h0h1x0xx1RWe have reversed the trajectory so that h0 (= 9.8 ft), and x0 , the height and range withinwhich Mickey can catch the ball represent the starting point of the trajectory. h1 (=3.28ft) is the height of the ball when Mickey strik
UNF - PHYSICS - 4360C
u zmax = h1oru 2 zmax ( 2 ) = 2 zmax ( 2 .0475 ) = 3.9 zmax . This result is the correct one 3.9 zmax= 0.821 = 39.40x1Now solve for x0 using a relation identical to (4) Thus, tan =h0 = x0( x tan )tan 024 zmaxAgain we obtain a quadratic expres
UNF - PHYSICS - 4360C
(a) The maximum height occurs ator1v cos sin = gT2v21H=cos 2 sin 2 2g2or atdz=0dt1v cos sin 2T=gmaximum at fixed dH=0ddH v 2 11212= 2 cos sin cos cos sin sin = 0d 2 g 222Using the above trigonometric identities, we get1
UNF - PHYSICS - 4360C
projection on the xy-plane make an angle with the x-axis:x = s sin cos ,and Fx = Fr sin cos = mxy = s si n si n ,and Fy = Fr sin sin = myz = s cos ,and Fz = mg + Fr cos = mzSince Fr = c2 s = c2 ( x + y 2 + z 2 ) , the differential equations of moti
UNF - PHYSICS - 4360C
2 x z 8x z 2 +g3g 2z = v sin and 2 x z = v 2 sin 2 :xmax =Forxmax =v 2 sin 2 4v 3 sin 2 sin +g3g 24.16x = A cos ( t + ) ,xx = A sin ( t + )from x = 0 ,from x = A ,y = B cos ( t + ) ,y =0x = A cos ty = B sin ( t + )12121 2kB = ky +
UNF - PHYSICS - 4360C
V= 4 2 mxyy = B cos ( 2 t + )my = V= 9 2 mzzz = C cos ( 3 t + )mz = Since x = y = z = 0 at t = 0 , = = = 2x = A cos t = A sin t2x = A cos tSince v 2 = x 2 + y 2 + z 2 and x = y = z ,vx== A3vA=3vx=sin t3y = B sin 2 t ,y = 2 B c
UNF - PHYSICS - 4360C
tmin =4.182=2Equation 4.4.15 istan 2 =2 AB cos A2 B 2Transforming the coordinate axes xyz to the new axes xyz by a rotation aboutthe z-axis through an angle given, from Section 1.8:x = x cos + y sin ,y = x sin + y cosor,x = x cos y sin , and
UNF - PHYSICS - 4360C
1r1 = ( d x ) + y 2 + z 2 22 2x x2 + y2 + z 2 2+r = ( d 2 dx + x + y + z ) = d 1 dd21nFrom Appendix D, (1 + x ) = 1 + nx + n ( n 1) x 2 + 22 2x x + y2 + z2 1 r1 = d 1 + + 1d2 2 d 2 2 2 12222 22222222 2 x 2x3 2x x + y + z x
UNF - PHYSICS - 4360C
zkB(F = q E+vBEjiv0())jv B = ix + y + kz kB = iyB xBjF = iqyB + q ( E xB )jymx = Fx = qyBqBxx =ymxqBmy = Fy = qE qxB = qE qB x +my22qE qBx qB eE eBx eB y=+ y y=mm mmm meEeBy +2 y = +x ,=mm1 eEy = 2 + x + A co
UNF - PHYSICS - 4360C
hbh mv 2 mgmgR = mg = h ( b 2h ) = b ( 3h b )bbbthe particle leaves the side of the sphere when R = 0bbh = , i.e.,above the central plane33cos =4.2212mv + mgh = 02at the bottom of the loop, h = b12mv = mgb ,2v = 2 gbsohbmv 2
UNF - PHYSICS - 4360C
CHAPTER 5NONINERTIAL REFERENCE SYSTEMS5.1 (a) The non-inertial observer believes that he is in equilibrium and that the net forceacting on him is zero. The scale exerts an upward force, N , whose value is equal to thescale reading - the weight, W, of
UNF - PHYSICS - 4360C
gg = g A = g ij10For small oscillations of a simple pendulum:1T = 2g2gg = g + = 1.005 g 10 2T = 25.5 (a)11= 1.9951.005 ggf = mg is the frictional force acting on theA0box, sof mA0 = ma(b)(a)f( a is the acceleration of the box re
UNF - PHYSICS - 4360C
= i R sin t + R cos t x + i yjjx = y R sin ty = x + R cos there i = 1 !(c) Let u = x + iyu = x + iy = y R sin t i x + iR cos t= yi u =+i x u + i u = R sin t + iR cos t = i ReitTry a solution of the formu = Ae i t + Beitu = i Ae i t + iBeit
UNF - PHYSICS - 4360C
A, A are the accelerations of the asteroid and the Earth in the fixed, inertial frameof reference.1st : examine:A A r= R R ( R R )= ( ) R = ( 2 2 ) Rnote: = k , = kThus:a = ( 2 2 ) R 2 vTherefore: jyix + = ( 2 2 ) iR cos ( ) t R sin ( ) t 2 x +
UNF - PHYSICS - 4360C
= k , with = 0r = ber , so ( r ) = b 2 er v = ve , so v = ver From eqn 5.2.14,a = a + 2 v + ( r ) and putting in terms from abovev 2 2 v b 2ar = bFor no slipping F s mg , so a s gv 2+ 2 v + b 2 s gb2vm + 2 bvm + b 2 2 b s g = 0vm = b 2b 2
UNF - PHYSICS - 4360C
5.10(See Example 5.3.3)m 2 x = mxx ( t ) = Ae t + Be tx ( t ) = Ae t Be tBoundary Conditions:lx2 ( 0) = = A + B2x ( 0 ) = 0 = ( A B )A=l4B=lx ( t ) = cosh t2lllx (T ) = + = cosh T222(a)(b) cosh T = 2lx ( t ) = sinh t2when the bea
UNF - PHYSICS - 4360C
v = zv yi + zvx yvx k jj( v )horiz = zvyi + zvx ( v )horiz112222= ( z2v y + z2vx ) 2 = z ( vx + v y ) 2 = zvFcor = 2m v(F )cor5.13horiz= 2m ( v )horiz = 2m z v , independent of the direction of v .From Example 5.4.1 11 8h 3 2 = xh
UNF - PHYSICS - 4360C
da = r + r + r + 2 r + 2 r dt rot()() a = r + ( r ) + 2 ( r ) + ( r ) + ( r ) + r + r Now( r )is to and r . Let this define a direction n : r = r nSince n , ( r ) is in the plane defined by and r and ( r ) = n r = r .Since ( r ) ( r ) = 2
UNF - PHYSICS - 4360C
1z ( t ) = gt 2 + v t sin + v t 2 cos cos = 022v sin 2v sin t=org 2 v cos cos gWe have ignored the second term in the denominatorsince v would have to beimpossibly large for the value of that term to approach the magnitude gFor example, for = 4
UNF - PHYSICS - 4360C
(jHence: a = a 2 v = 3 2 x i 2 k ix + ySo5.19)ja = ix + y = 3 2 xi + 2 yi 2 xj2x 2 y 3 x = 0y + 2 x = 0(mr = qE + q v BEquation 5.2.14)r = r + r + 2 v + ( r )v = v + r Equation 5.2.13q =B so = 02mqmr q B v B ( r ) = qE + q ( v + r
UNF - PHYSICS - 4360C
Collecting terms and neglecting terms in 2 :gg x + x cos t + y + y sin t = 0llgg x + x sin t y + y cos t = 0ll24hourssin 24T== 73.7 hourssin 195.21T=5.22Choose a coordinate system with the origin at the center of the wheel, the x and
UNF - PHYSICS - 4360C
r = 0Vb VtVt VtVtVt Vt k sin cos + V sin i cos j k sin cosbbbbbb22V Vt Vt V Vt Vt+ k cos 2 j cos ( r ) = k sin 2+ i sinbb bbb22VV ( r ) = k nbSince the origin of the coordinate system is traveling in a circle of r
UNF - PHYSICS - 4360C
CHAPTER 6GRAVITATIONAL AND CENTRAL FORCES6.14m = V = rs331 3m 3rs = 4 F=Gmm( 2rs )222Gm 2 4 3 G 4 3 43== m4 3m 4 3 2FFGm 2 4 3 3==mW mg4g 3 121F 6.672 1011 N m 2 kg 2 4 11.35 g cm 3 1 kg 106 cm3 3=3 (1 kg ) 3W4 9.8
UNF - PHYSICS - 4360C
The gravitational force on the particle is due only to the mass of the earth that isinside the particles instantaneous displacement from the center of the earth, r.The net effect of the mass of the earth outside r is zero (See Problem 6.2).4M = r334
UNF - PHYSICS - 4360C
6.56.6GMm mv 2GMv2 ==so2rrrfor a circular orbit r, v is constant.2 rT=v4 2 r 2 4 2 3T2 =r r3=2vGM2 rvFrom Example 6.5.3, the speed of a satellite in circular orbit is (a) T =1 gR 2 2v= e r3T=2 r 21g 2 Re1 T 2 gRe2 3r =
UNF - PHYSICS - 4360C
This is the same expression as derived in Problem 6.3 for a particle dropped into ahole drilled through the earth. T 1.4 hours.6.8The Earths orbit about the Sun is counter-clockwise as seen from, say, the northstar. Its coordinates on approach at the
UNF - PHYSICS - 4360C
4M d = r33GMm 4F ( r ) = 2 mGr3r6.101 1 k=errduk= e kdru=d 2u k 2 k= e = k 2ud 2 rFrom equation 6.5.10 d 2u1+ u = k 2u + u = 2 2 f ( u 1 )2duml u122f ( u ) = ml ( k + 1) u 3f (r ) = ml 2 ( k 2 + 1)r3The force varies as th
UNF - PHYSICS - 4360C
kIf 1 + 2 < 0 , ml kIf 1 + 2 = 0 , ml du= C1du = c1 + c21r=c1 + c2kIf 1 + 2 > 0 , ml u = A cos(c + d 2u cu = 0 ,d 2d 2u=0d 2)d 2u+ cu = 0 ,d 2kr = A cos 1 + 2 + ml6.12c > 0 , for which u = aeb is a solution.c>0111=r
UNF - PHYSICS - 4360C
For r = a , and the force to be central, try = bt nf ( ) = m 2ab 2 n 2t 2 n 2 + ab 2 n ( n 1) t 2 n 2 For a central field f ( ) = 02n + ( n 1) = 0n=131 = bt 3(a)6.14Calculating the potential energy 4 a2 dv = f ( r ) = k 3 + 5 drrr 2 a2
UNF - PHYSICS - 4360C
1r = a cos ( r = a @ = 0 )33(b) at =r 0 the origin of the force. To find how long it takes 2avvl= 2==11ra 2 cos 2 a cos 2 33a1dt = cos 2 dv3ThusT=320a3a 23 a212cos d = cos d = 4vv3v01 9k 2Since v = 2 2a 113 2
UNF - PHYSICS - 4360C
dV V 1 r (b) dr1 = 2r1 dV = 2 60 1% = 120% !(a)() dr1 2 r1 r1 r V r1 The approximation of a differential has broken down a correct result can be obtained bycalculating finite differences, but the implication is clear a 1% error in boost causes
UNF - PHYSICS - 4360C
6.17From Example 6.10.1 12vr2 2 = 1 + q 2 ( qd sin ) where q =and d =dveaeare dimensionless ratios of the comets speed and distance from the Sun in terms of theEarths orbital speed and radius, respectively (q and d are the same as the factor
UNF - PHYSICS - 4360C
But the total orbital energy iskkE=SoE = 2 a2a2aSince planetary orbits are nearly circularkkV ~andT~a2aaE aand=Thus, E TaTaav=2We obtainav6.20 (a) V =1Vdt0krFrom equation 6.5.4, l = r 2dlr 2 d= 2 or dt =dt rl2 kr
UNF - PHYSICS - 4360C
Integrate LHS by parts111 mr r mr 2 dt = F rdt0 00The first term is zero since the quantity has the same value at 0 and .denote time average of the quantity within brackets.Thus 2 T = F r wherebut r F = r V = rdVk== Vdrrhence 2 T = Vbu
UNF - PHYSICS - 4360C
k 2 1m RE a and solving for aREa=noting that 2 RE2 mvkkGM E== gRE(3)mRERERERa== E = 4.49 103 km2v 1.4262gRE v2 =The eccentricity can be found from the angular momentum per unit mass, l, equation6.5.19, and the data on ellipses de
UNF - PHYSICS - 4360C
6.23From Problem 6.9, f ( r ) = GMm 4 mGr3r22GMm 4 mG3r344 a 32GMma 3 mG2 +f (a)33M==4f ( a ) GMma 2 mGa 4 a 3 a 1 +33M f (r ) =f (a) From equation 6.14.3, = 3 + af (a) 4 a 2 + 3M = 3 +4 a 31+3M3121 1+ c 2 = , 1 +
UNF - PHYSICS - 4360C
6.25f (r ) =2k 4+r3 r5From equation 6.13.7, the condition for stability is f ( a ) +k a 2k 4 4 + 3 + 5 <02aa3 aak 2 + 4 <03a 3aaf (a) < 03a2<k1 2a> k6.26 (a)e brr2e br b 2f ( r ) = ke br 2 3 = k 2rrrf ( r ) = k2b + r
UNF - PHYSICS - 4360C
a (1 2 )1 + cos and the data on ellipses in Figure 6.5.1 p = a (1 ) so(1 + )r=p1 + cos For a parabolic orbit, = 1The comet intersects earths orbit at r = a .2pa=1 + cos 2pcos = 1 +a6.27(See Figure 6.10.1) From equation 6.5.18a r =6.28(See
UNF - PHYSICS - 4360C
3T=2 p2a32d+ (1 + cos )From a table of integrals,2yrdx (1 + cos x )2=1x1xtan + tan 3226232 p 2 1 3 T=tan + tan yr232 a 1x 1 cos x 2tan = 2 1 + cos x 12 p 21 2 a a p 2tan = =2 2p p a1332 p 2 a p 2 1 a p 2 T
UNF - PHYSICS - 4360C
dV k 3k k 2=+= 4 ( r + 3 )dr r 2 r 4r2k 12k2k 2f ( r ) = 3 5 = 5 ( r + 6 )rrr2f (a)2 r + 6 = 2f (a)a r + 3 f (r ) = f (a) From equation 6.14.3, = 3 + af (a) ForFor6.30121 r 2 + 6 2r 2 + 3 = 3 2 2 = 2r 3 r + 3 2 = RR , R =
UNF - PHYSICS - 4360C
E1 + m c 2 =1k 1 + m c 2 E + a m c2 + E =3+ af ( a ) m c2 + E + k af (a)f (a) = 3 + af (a) 121k2a = 1 +2 mc +Ea (1 2 )(Here is the polar angle of conic6.31 From equation 6.5.18a r =1 + cos section trajectories as illustrated
UNF - PHYSICS - 4360C
1212sin = 2 ( RV 2 sin 2 ) + 2 ( RV 2 sin 2 ) 1Again from Example 6.5.3 122 2 = 1 + V 2 ( RV sin ) R 2 = 1 + ( RV 2 sin ) 2 RV 2 sin 2 212221sin = ( RV 2 sin ) ( RV 2 sin 2 ) 1sin = RV 2 sin cos cos RV 2 sin 2 1=sin RV 2 sin cos 2
UNF - PHYSICS - 4360C
Letting r1 = RE + h1 and r2 = RE + h 2 where h1 and h2 are the heights of the 2 satellitesabove the ground. Inserting these into the above gives 33h1 + h 2 2h1 + h 2 22+Tt =RE 2 +=RE RE RE gRE g From Example 6.6.2, RE = 6371 km, h1 = 200 mi
UNF - PHYSICS - 4360C
Before evaluating this integral, we need to find umax ( = rmin 1 ) , in other words, thedistance of closest approach to the scattering center.11k1E = T ( rmin ) + V ( rmin ) = mv 2 += mv0 2222 rmin2But, the angular momentum per unit mass l is
UNF - PHYSICS - 4360C
Chapter 7Dynamics of Systems of Particles7.1z1 mi rimiFrom eqn. 7.1.1, rcm =(3)11jj( r1 + r2 + r3 ) = i + + + k + k331rcm = i + 2 + 2kj3d11vcm = rcm = ( v1 + v2 + v3 ) = 2i + + i + + kjjdt331vcm = 3i + 2 + kj3From eqn 7.1.
UNF - PHYSICS - 4360C
vg = 7.4v1+ v mv = m + Mvblk21mvblk = v=M2221 2 1 v 1 v Ti T f = mv m + M 22 2 2 2 Momentum is conserved:1 2 1 1 2 = mv 1 M24 4mTi T f 3 =44Ti7.5v0 / 2At the top of the trajectory:vv = iv cos 60 = i2Momentum is conserved
UNF - PHYSICS - 4360C
Total distance = h + 2 2 h + 2 4 h + = h 1 + 2 2 n n=0a, r <1.Now ar n =1 rn=021+ 21 + 2 2 n = 1 +=1 2 1 2n=0 1+ 2 total distance = h 2 1 For the first fall,122hgt = h , so t =2g2h2h=ggAccounting for equal rise and fall times:
UNF - PHYSICS - 4360C
7.81112From eqn. 7.2.15, T = mi vi2 = m1v12 + m2 v222i2Meanwhile:21 2 1 2 1 m1v1 + m2 v2 1 m1m22mvcm + v = m ( v1 v2 )+m222 2 m1 + m21222 m12 v12 + m2 v2 + 2m1m2 v1 v2 + m1m2 ( v12 + v2 2v1 v2 ) =2m112= m1v12 + m2 v2221212
UNF - PHYSICS - 4360C
2m1 2v14T1 T1m==1T1mm1v1227.11From eqn. 7.2.14, L = rcm mvcm + ri mi vir m viiii= r1 m1v1 + r2 m2 v2i m m + m2 m1From eqn. 7.3.2, R = r1 1 + 1 = r1 1r1= m2 m2 mSince from eqn. 7.3.1, r1 = 2 r2m1mR = 2 r2r m viiii R m1
UNF - PHYSICS - 4360C
(1 )b) V ( x , y ) = 12( x ) + y ( x + 1 ) + y V (1 ) ( x ) ( x + 1 )=+x33x2222[]2212x2 + y 22(7.4.13)[]Now x = 0.5 at L4 and L5also, each bracket term in the denominator equals 1 at L4 , L5V= (1 ) ( 0.5 ) + ( 0.5 + 1 ) ( 0.5
UNF - PHYSICS - 4360C
2v 2v 2 + 60v 2 v2 62=1010vp > 0 , so the positive square root is used.(vp =)vp = 0.9288 vvpx = vpy =vp2= 0.657 v1112( v vp2 ) 2 = v2 (1 .9288 2 ) 22v = 0.1853vvpvp.92882=tan ==vp2v vp2 .9288v21 = tan 1.9134 = 62.41v x = v
UNF - PHYSICS - 4360C
= tan 1 1.2638 = 51.65v x = v cos = 0.110 vv y = v sin = 0.140 v7.16sin 1 and are the scattering angles in + costhe Lab and C.M. frames respectively.mFrom eqn. 7.6.16, for Q = 0, = 1m2sin tan 45 =1 =1+ cos 41+ cos = sin and squaring 4
UNF - PHYSICS - 4360C
P17.18Conservation of momentum:P = P cos + P2 cos ( 11P1)0 = P sin P2 sin ( )1From Appendix B for sin ( + ) and cos ( + ) :P2P = P cos + P2 ( cos cos + sin sin )110 = P sin P2 ( sin cos cos sin )1P 2 = P2 cos 2 + P22 ( cos 2 cos 2 + 2 cos
UNF - PHYSICS - 4360C
vcm1m1==Equation 7.6.11v1 m2 + m1 1 + Thus1212v1212r2r=+=+22v1 (1 + )2 (1 + )2 (1 + )1 + ) (1 + ) (1 + )(Simplifying2 1 1 rr2 +=01+ 1+ Let x 2 = r and solving the resulting quadratic for x11 2 (1 2 ) 2+x=1+ 1+ Squar
UNF - PHYSICS - 4360C
1=m1m2( 2T m1 ) 212( 2 m1 ) (1 + m1Tm2 ) Q (1 + m1 m2 )1212Finally=7.21m1m211 Q (1 + m1 m2 ) 21 TThe time of flight, = constantso =rbut from problem 7.19 abovev1v1 + 2 + 2 11+As an example, let v1 = 1 and we haver1 =
UNF - PHYSICS - 4360C
7.234m = r33m = 4 r 2 r r 2 zwhere v = zr = kzk a constant of proportionalityr = r + kzFrom eqn. 7.7.6, mg = mv + vm434 r g = r 3 z + 4 r 2 ( kz ) z3323kzg = z+r3z 2z=grz+k3z 2For r = 0 , z = g zA series solution is used for thi