In [1]:

```
#initiation of varible
v1=60.0; v2=40.0 #Velocities of cars wrt to observer in km/hr
#calculation
vr=v1-v2; #relative velocity
#result
print"The value of relative velocity in km/h. is",round(vr,3);
```

In [1]:

```
#initiation of variable
from math import atan, pi
import numpy as np
Va_w=[320.0,0.0]; Vw_g=[0.0, 65.0]; #Vp/q=[X Y]=>velocity of object p wrt q along X(east) and Y(north) directions.
#calculation
Va_g=Va_w + Vw_g; #net velocity
k=np.linalg.norm(Va_g) #magnitude
s=atan(Va_g[3]/Va_g[0])*180.0/pi; #angle in rad*180/pi for conversion to degrees
#result
print "the velocity in x direction in Km/h is", Va_w[0],"in y direction in km/h is",Vw_g[1]
print"The magnitude of velocity Va/g(airplane wrt ground) in Km/h is",round(k,3)," at ",round(s,3)," degrees north of east."
```

In [11]:

```
#initiation of variable
from math import sqrt
Lo=100.0*(10**3);c=3.0*(10**8); #Given values//all the quantities are converted to SI units
d=2.2*(10**-6); #time between its birth and decay
#calculation
t=Lo/c #where Lo is the distance from top of atmosphere to the Earth. c is the velocity of light. t is the time taken
u=sqrt(1-((d/t)**2)); # using time dilaion fromula for finding u where u is the minimum velocity in terms of c;
#result
print"Hence the minimum speed required in c is",round(u,6);
```

In [4]:

```
#intiation of variable
from math import sqrt
Lo=100.0*(10**3); #Lo is converted to Km
u=0.999978; #//u/c is taken as u since u is represented in terms of c.
#calculation
L=Lo*(sqrt(1-u**2)); # from the length contraction formula
#result
print"Hence the apparent thickness of the Earth's surface in metres. is",round(L,3)
print"answer is slightly different in the book"
```

In [9]:

```
#initiation of variable
from math import sqrt
L=65.0; c=3*10**8;u=0.8*c;
#calculation
t=L/u ; #The value of time taken as measured by the observer
#result
print"The time for rocket to pass a point as measured by O in musec is ",round(t*10**6,3); #The value of time taken as measured by the observer
#partb
Do=65.0; #given length
Lo= L/sqrt(1-(u/c)**2); #contracted length of rocket
#result
print"Actual length according to O is ",round(Lo,3);
#partc
D=Do*(sqrt(1-(u/c)**2)); #contracted length of platform.
#result
print"Contracted length according to O'' is",round(D,3);
#partd
t1=Lo/u; #time needed to pass according to O'.
print "Time taken according to O is ",t1
#part 3
t2=(Lo-D)/u; #time intervals between the two instancs
print"Time taken according to O'' is ",t2;
print'The value of t1 and t2 did not match';
```

In [23]:

```
#initiation of variable
v1=0.6; u=0.8; c=1.0; # all the values are measured in terms of c hence c=1
#calculation
v= (v1+u)/(1+(v1*u/c**2));
#result
print "The speed of missile as measured by an observer on earth in c is",round(v,3);
```

In [24]:

```
#initiation of variable
w1=600.0;w2=434.0; # w1=recorded wavelength;w2=actual wavelength
# c/w1 = c/w2 *(sqrt(1-u/c)/(1+u/c))
#calcualtion
k=w2/w1;
x=(1-k**2)/(1+k**2); #solving for u/c
#result
print"The speed of galaxy wrt earth in c is",round(x,3);
```

In [27]:

```
#initiation of variable
from math import sqrt
v1x=0.6;v1y=0.0;v2x=0.0;v2y=.8;c=1.0; # all the velocities are taken wrt c
v21x=(v2x-v1x)/(1-(v1x*v2x/c**2)); #using lorentz velocity transformation
v21y=(v2y*(sqrt(1-(v1x*c)**2)/c**2))/(1-v1y*v2y/c**2)
#result
print"The velocity of rocket 2 wrt rocket 1 along x and y directions is",round(v21x,3)," c &", round(v21y,3),"c respectively"
```

In [30]:

```
#initiation of variable
from math import sqrt
u=0.8*c;L=65.0;c=3.0*10**8; #all values are in terms of c
t=u*L/(c**2*(sqrt(1-((u/c)**2)))); #from the equation 2.31
#result
print"The time interval between the events is",t, "sec which equals",round(t*10**6,3),"musec."
```

In [34]:

```
#initiation of variable
from math import sqrt
m=1.67*10**-27;c= 3*10**8;v=0.86*c; #all the given values and constants
#calculation
p=m*v/(sqrt(1-((v/c)**2))); # in terms of Kgm/sec
#result
print"The value of momentum was found out to be in Kg-m/sec.\n",p;
#part 2
c=938.0;v=0.86*c;mc2=938.0 # all the energies in MeV where mc2= value of m*c^2
pc=(mc2*(v/c))/(sqrt(1-((v/c)**2))); #expressing in terms of Mev
#result
print"The value of momentum was found out to be in Mev.",round(pc,3);
```

In [1]:

```
#initiation of variable
from math import sqrt
pc=1580.0; mc2=938.0;E0=938.0; # all the energies in MeV mc2=m*c^2 and pc=p*c
#result
E=sqrt(pc**2+mc2**2);
K=E-E0; #value of possible kinetic energy
#result
print"The relativistic total energy in MeV. is",round(E,3); #value of Energy E
print"The kinetic energy of the proton in MeV.",round(K,3);
```

In [2]:

```
#initiation of variable
from math import sqrt
E=10.51; mc2=0.511; #all the values are in MeV
#calculation
p=sqrt(E**2-mc2**2); #momentum of the electron
v=sqrt(1-(mc2/E)**2); #velocity in terms of c
#result
print"The momentum of electron in MeV/c is",round(p,3);
print"The velocity of electron in c is",round(v,5);
```

In [3]:

```
#initiation of variable
from math import sqrt
k=50;mc2=0.511*10**-3;c=3.0*10**8; # all the values of energy are in GeV and c is in SI units
#calculation
v=sqrt(1-(1/(1+(k/mc2))**2)); #speed of the electron in terms of c
k=c-(v*c); #difference in velocities
#result
print"Speed of the electron as a fraction of c*10^-12 is.",round(v*10**12,3); # v=(v*10^12)*10^-12; so as to obtain desired accuracy in the result
print"The difference in velocities in cm/s.",round(k*10**2,3);
```

In [14]:

```
#initiation of variable
from math import sqrt, pi
r=1.5*10**11; I=1.4*10**3; #radius and intensity of sun
#calculation
s=4*pi*r**2 #surface area of the sun
Pr=s*I # Power radiated in J/sec
c=3.0*10**8; #velocity of light
m=Pr/c**2 #rate od decrease of mass
m=round(m,2)
#result
print"The rate of decrease in mass of the sun in kg/sec. is %.1e" %m;
```

In [49]:

```
#initiation of variable
from math import pi, sqrt
K=325; mkc2=498; #kinetic energy and rest mass energy of kaons
mpic=140.0; #given value
#calculation
Ek=K+mkc2;
pkc=sqrt(Ek**2-mkc2**2);
#consider the law of conservation of energy which yields Ek=sqrt(p1c^2+mpic^2)+sqrt(p2c^2+mpic^2)
#The above equations (4th degree,hence no direct methods)can be solved by assuming the value of p2c=0.
p1c=sqrt(Ek**2-(2*mpic*Ek));
#consider the law of conservation of momentum. which gives p1c+p2c=pkc implies
p2c=pkc-p1c;
k1=(sqrt(p1c**2+(mpic**2))-mpic); #corresponding kinetic energies
k2=(sqrt((p2c**2)+(mpic**2))-mpic);
#result
print"The corresponding kinetic energies of the pions are", k1," MeV and",round(k2,3)," MeV."
```

In [50]:

```
#initiation of variable
from math import sqrt
mpc2=938.0;c=3.0*10**8; #mpc2=mp*c^2,mp=mass of proton
#calculation
Et=4*mpc2; #final total energy
E1=Et/2;E2=E1; #applying conservation of momentum and energy
v2=c*sqrt(1-(mpc2/E1)**2); #lorentz transformation
u=v2;v=(v2+u)/(1+(u*v2/c**2));
E=mpc2/(sqrt(1-(v/c)**2));
K=E-mpc2;
#result
print"The threshold kinetic energy in Gev",round(K/10**3,3);
```