Answers to Hogg's problems:
1-1
(a) Relative speed is 120-100=20 km/s.
Time is 20/60=1/3 hour so your distance is d=rt=100/3=33.33 km  from pass point.
So is his, but it only took him t=d/r=33.33/120=0.2777 hour.
(b) He started 0.3333-0.2777=0.0555 hour (3.33 mins) after you passed.
(c) In 0.0555 hour you would travel 5.555 km.

1-3 
ball velocity is v wall velocity is -w
add w to get to wall frame:  ball velocity before impact is v+w (wall is 0)
ball velocity after  impact is -v-w
subtract w to get back to observer frame:
ball velocity is -v-2w  (wall is -w)
as w>>v ball velocity is just -2w after impact.

1-4
river vy=-0.5
swimr vx=+1.0 and resultant is vx=+1.0
tan theta = -0.5/1.0 so theta=26.57 degrees "south of east".
to go directly east you need to swim north to get a north component of 0.5:
sin theta=0.5/1 theta=30 degrees "north of east"
your penalty for doing this is an east component of 1*cos theta = 0.866

1-9
Light travels distance l in time t=l/c.
In time t objects fall a distance h=˝gt˛=˝g(l/c)˛
Angle is about 206265*h/l (206265 arcsec/radian) so for l=100m, g=9.8,
angle=103132gl/c˛=1.125E-09 arcsec (this is pretty small!)

2-1
Beta=1/2 so gamma=1/sqrt(1-1/4)=1.1547 

2-2
Gamma=10 so beta=sqrt(1-1/100)=0.995=1-0.005~1-0.5/gamma˛

2-3
Use binomial expansion like in equation (1.4) to get beta~1-0.5/gamma˛

2-4
Binomial again--(1+x)^n~1+nx--  gamma=sqrt(1-beta˛)~1-0.5beta˛.

2-6
(a) Beta=4/5 so gamma=1/sqrt(1-16/25)=1.66667 (=ticks of your clock)
(b) Distance=4c/5 
(c) Time to go 4c/5 is 4/5 of HIS tick
(d) Tick spacing is 1.66667(1+4/5)
In general it is gamma(1+beta) -- the law of redshifts!

2-7
Need gamma=3 so beta=0.942809

2-8
L=100m if traverse t=5E-06 then v=d/t=100/5E-06=2E+07 (beta=0.066667 gamma=1.00223)
Modify problem: let t=t*=3.849E-07 so beta=0.86667 and gamma=2.
The "moving nose clock" should tick off t*/2 but the moving nose guy saw what the
stationary guy saw -- didn't he?  what's wrong here?  It should take the same
amount of time, shouldn't it?  Yes, but you are asking two different things!
Case 1:  moving nose touches stationary tail then stationary nose.  t=t*.
Case 2:  the other ship seems to be going "backwards" of course...
moving ship tail touches stationary nose then the noses touch. t=t*/2.
But remember the moving ship is only half as long as the stationary ship so
another t*/2 elapses before moving ship tail touches stationary tail.
So the case 2 clock ticks over the same when the moving ship traverses the full length.

2-12
round trip distance to alpha Cen is 2*4.34=8.68ly.  We want the trip to take 20y.
8.68=beta*T=beta*20*gamma
gamma*beta=0.434
(gamma*beta)˛=beta˛/(1-beta˛)=0.434˛
beta˛=0.434˛/(1+0.434˛)=0.1585
beta=0.3981, gamma=1.090117, T=20*gamma=21.8 years and = 2*l/v too!

2-13
Gamma=18.8 shrinks 1624m to 1624/18.8=86.38m which takes 86.38/c=2.879E-07s
(OK, gamma=18.8, beta=0.998584 so t=2.875E-07s --just use c, close enough!)
---The muon lifetime is still 2.5E-06, but few decay in a tenth of a lifetime.
When gamma=6.3 it takes 3x longer -- still under half a lifetime????

2-14
2/3=2^(-t/2E-08)
ln(2/3)=-tln2/2E-08
t=0.40547*2E-08/0.693=1.17E-08s
To travel 30m in t/gamma=1.17E-08s takes (lets hope gamma is big so beta~1)
t=30/300000000=1.E-07=1.17E-08*gamma so gamma=8.547