Problem Set 8 Solutions
Physics 11a
Fall 2010
Problem 1
1
a) The Lorentz factor is  
0.5c 
2
1

1
 1.155 .
1  0.25
0
2
0
c
The momentum is
p   mv  1.154 9.11 1031 kg 0.5 3.00  108 m/s  1.58  1022 kg m/s.







b) To double the momentum the product  v  1.154 0.5c0  0.577c0 must be doubled.


Therefore we want  v  2 0.577c0  1.154c0 . So
v
c0

v
c0
 v
1  
 c0 
2
 1.154 ,
v
1.1542

 0.756 , v = 0.756 c0 .
c0
1  1.1542
Problem 2
No. The relativistic momentum of the electron is given by p   mv 
mv
1  v 2 c2
. At low
speeds (compared to c) this reduces to the classical momentum, p = mv. As v approaches
c, γ approaches infinity so there is no upper limit to the electron’s momentum.
Problem 3
3.91 107 m/s (0.130c) toward the star
 1,i m1v1,i   2,i m2 v2,i   f (m1  m2 )v f



2
 1,i m1v1,i   2,i m2 v2,i / (m1  m2 ) (1  v 2f / c 2 )  v 2f

vf 
1,i
 



2
2


m1v1,i   2,i m2 v2,i / (m1  m2 ) /   1,i m1v1,i   2,i m2 v2,i / c(m1  m2 )  1


v f  0.130311c
Problem 4
Kinetic energy is given by Eq. (14.51), and the rest energy is mc02, so we
1
v2 1
 2 , 1 2  ,
want   1 mc02  mc02 ,   1  1 ,  
c0 4
v2
1 2
c0


v  c0 1 
1
 0.866 c0
4
Problem 5
For every Uranium nucleus we have a change in mass of 7.62  1030 kg which
corresponds to an energy of 6.858  10  13 J . In one kilogram of Uranium, one has
approximately 2.530 1024 nuclei . Hence, one gets a total energy released
of 6.858 1013 J  2.530 1024 nuclei  1.7349 1012 J
Problem 6
(a) For a particle of non-zero mass, we derive the following relationship between
kinetic energy
and momentum.
E  K  mc 2 ;
 pc 
2
 E 2   mc 2    K  mc 2    mc 2   K 2  2 K  mc 2 
2
2
2
2mc 2  4  mc 2   4  pc 
2
K  2 K  mc    pc   0  K 
2
2
2
2
For the kinetic energy to be positive, we take the positive root.
2mc 2  4  mc 2   4  pc 
2
K
2
2
 mc 2 
 mc    pc 
2 2
2
If the momentum is large, we have the following relationship.
2
K  mc2 
 mc    pc 
2 2
2
 pc  mc2
Thus there should be a linear relationship between kinetic energy and
momentum for large values of momentum.
If the momentum is small, we use the binomial expansion to derive the classical
relationship.
K  mc 2 
 mc    pc 
2 2
2
 pc 
  mc 2  mc 2 1   2 
 mc 
2
2
2

 pc   p
 mc2  mc 2 1  12  2   

 mc   2m

m0
m0
K
Thus we expect a quadratic relationship
for small values of momentum. The
adjacent graph verifies these
approximations.
(b)
For a particle of zero mass, the
relationship is
simply K  pc. See the graph.
p