A 139.6 MeV pion collides with and sticks to a stationary 938.3 MeV proton. The total energy of the composite particle is calculated to be 1270.68 MeV. The total momentum is found to be 301.14 MeV/c. Using the total energy and momentum, the rest energy of the composite particle is calculated to be 1234.48 MeV.

1. 1
A π meson of rest energy 139.6
MeV moving at a speed of 0.906c
collides with and sticks to a
proton of rest energy 938.3 MeV
that is at rest.
(a) Find the total relativistic
energy of the resulting
composite particle.
(b) Find the total linear
momentum of the composite
particle.
(c) Using the results of (a) and
(b), find the rest energy of the
composite particle.

2. 2
Helping Tools
No.1
No.2
No.3

3. 3
No.4
No.5
Solution
2
2
( )
( ) (1)
( ) (2)
E E (3)
p p p
p
a
Let
E K mc
E K mc
E
  

= + − − − −
= + − − − −
= + − − − − − − −

4. 4
2
2
2 2
2
2
2
2
,
( ) (1)
( )
( ) ( )
1
139.6 139.6
1 0.82
0.906 )
1
139.6
332.38
0.42
Now
E K mc
mc
mc mc
v
c
MeV MeV
c
c
MeV
MeV
  

 
= + − − − −
= − +
−
= =
−
−
= =
2
,
( ) (2)
0 938.3 938.3
&
E E (3)
E 332.38MeV 938.3MeV
1270.68
p p p
p
Now
E K mc
MeV MeV
E
MeV

= + − − − −
= + =
= + − − − − − − −
 = +
=
2
2
( )
Proton is at Rest
Total Momentum = p
1
b
mv
v
c

 =
−
2
2
2
1
( ) X
(0.906 )
1
1
139.6 0.906
0.42
301.14
v
mc
c c
c
c
MeVX X
c
MeV
c
=
−
=
=

5. 5
2 2 2 2
2 2 2 2
2 2 2
( )
( ) ( )
( ) ( )
( )
c
E pc mc
mc E pc
mc E pc
= +
 = −
 = −
2 2
2
2
(1270.68 ) (301.14 )
(1614627.66 90685.3)(MeV)
1523942.36( )
1234.48
MeV MeV
MeV
MeV
= −
= −
=
=