1. Kinetic Energy: K = ( - 1)mc2
Rest Energy (includes internal kinetic and potential energy): ER  mc2
For an object moving in an inertial frame),
Total energy : E = K + ER = mc2
Problem 1: A mosquito has a mass of 2.5 mg.
a) What is its rest energy?
b) By what fraction does it change its energy if it is flying at 20 m/s ( 40 miles/hour)
m = 2.5 x 10-6 kg
ER = (2.5 x 10-6 kg) (3 x 108 m/s)2 = 2.25 x 1011 J
[This is the energy output of Big Sandy power plant in 3.5 minutes  days.
Therefore, if you could completely convert the mass of  400 mosquitoes/day to energy
(we can’t !!), you could match the energy output of Big Sandy!]
b) Since v <<c, so can use K  ½ mv2 = 0.5 (2.5 x 10-6 kg) (20 m/s)2 = 5 x 10-4 J
So K/ER = 5 x 10-4 / 2.25 x 1011 = 2.2 x 10-15
For everyday objects at ordinary speeds, relativistic corrections to energy are negligible – classical mechanics works fine.

2. E = mc2 In any process, the total energy is conserved!
Kinetic Energy: K = ( - 1)mc2
Rest Energy (includes internal kinetic and potential energy): ER  mc2
For an object moving in an inertial frame,
Total energy : E = K + ER
E = mc2
In any process, the total energy is conserved!
And: if no external forces, momentum is also conserved!

3. Consider a general reaction in which some species (Aj) react to form new species (Bj):
A1 + A2 + A3 + ….  B1 + B2 + B3 + …
Conservation of Energy: c2 A’s (m) = c2 B’s (m)
Conservation of Momentum: A’s (mv) = B’s (mv)
[Today: Because masses will be given in atomic mass units (u), we’ll use v’s, not u’s, for particle velocities.]
(1u) c2 = 931 MeV

4. Consider the fusion reaction:
2D1 +3T1  4He2 + n,
where 2D1 is a deuterium atom, with m(D) = u,
3T1 is a tritium atom, with m(T) = u,
4He2 is a helium atom, with m(He) = u
and n is a neutron, with m(n) = u.
The incoming mass = m(D) + m(T) = u.
The outgoing mass = m(He) + m(n) = u.
Therefore, exothermic with m c2= ( u) c2
Use (1 u)c2 = 931 MeV
 m c2 = 17.6 MeV
Problem 2: Of this energy, K(He) = 3.5 MeV and K(n) = 14.1 MeV.
Show that these values are consistent with conservation of momentum when the incoming deuterium and tritium have v << c (i.e. p(D)  p(T)  0).

5. m(He) = 4.0026 u = 3726 MeV/c2 , m(n) = 1.0087 u = 939 MeV/c2,
m c2= 17.6 MeV
Of this energy, K(He) = 3.5 MeV and K(n) = 14.1 MeV.
Show that these values are consistent with conservation of momentum:
Because the incoming particles are  at rest, the total momentum  0, so the neutron and helium atom should have equal and opposite momenta.
K(n) = 14.1 MeV  ((n) – 1) m(n)c2 = 14.1 MeV
(n) -1 = 14.1 MeV / 939 MeV =
(n) = 1.015
 v(n) = 0.17c
 p(n) =  m v = (1.015) (939 MeV/c2) (0.017c) =162 MeV/c
K(He) = 3.5 MeV  ((He) -1) mHec2 = 3.5 MeV
(He) -1 = 3.5 MeV / 3726 MeV = 9.4 x 10-4
(He) =
 v(He) = 0.043c
 p(He) = mv = ( ) (3726 MeV/c2) (.043c) = 162 MeV/c 
= [1-(v/c)2]-1/2

6. Problem 3: A particle of mass M0 at rest divides into two particles, with masses m1 = 1u and m2 = 4u. The 1u particle has speed v1 = 0.8c.
What is the speed of the 4u particle?
What is the value of M0?
What is the total kinetic energy of the outgoing particles?
pinitial = 0 so p2 = -p1
2m2v2 = 1m1v1
4v2/[1-(v2/c)2]1/2 = v1/[1-(v1/c)2]1/2 = 0.8c / [1-0.64]1/2 = 0.8c/0.6
v2 /[1-(v2/c)2]1/2 = c/3
(3v2/c)2 = 1-(v2/c)2
(v2/c)2 = 1/10
v2 = c

7. Problem 3: A particle of mass M0 at rest divides into two particles, with masses m1 = 1u and m2 = 4u. The 1u particle has speed v1 = 0.8c.
What is the speed of the 4u particle?
What is the value of M0?
What is the total kinetic energy of the outgoing particles?
v2 = c
Since M0 is at rest, its energy = M0c2
M0c2 = 1m1c2 + 2m2c2
M0 = 1u [1/(1-0.82)1/2 + 4/(1 – )1/2
M0 = 1u [1/0.361/2 + 4/0.91/2] = ( )u = 5.89 u
c) Method 1: Ktotal = K1 + K2 = (1-1) m1c2 + (2-1)m2c2 = ( (1-1/0.91/2) uc2
= ( ) x 931 MeV = 829 MeV
Method 2: Since M0 is at rest,
Ktotal = (M0-m1-m2) c2 = ( )uc2 = 0.89 x 931 MeV = 829 MeV

8. Problem 4: A particle of mass M0 = 15 u at rest decays into three particles. The first particle has mass m1 = 1 u and velocity v1 = 0.8 c i. The second particle has mass m2 = 3 u and velocity v2 = - v1. What are the mass and velocity of the third particle?
1 = 2 = 1/[1 –v12/c2]1/2 = 1/[1 – 0.82]1/2 = 1/ [1 – 0.64]1/2 = 1/0.361/2 = 1/0.6 = 5/3
Consvtn. of Energy: M0c2 = 1m1c2 + 2m2c2 + 3m3c2
3m3 = [15 u – 5/3 (1 u + 3 u] = (25/3) u
Consvtn. of momentum: initial momentum = final momentum
0 = 1m1 v1 + 2m2 v2 + 3m3 v3
0 = 5/3 (1 u) (0.8 c)i – 5/3 (3 u) (0.8c) i + (25/3 u) v3
v3 = (3/25) (5/3) (1-3) (0.8) i = c i (= 9.6 x 107 m/s in +x direction)
Therefore 3 = 1/[1-(v3/c)2]1/2 = 1/[ ]1/2 = 1.056
Therefore m3 = 25/3 u / = 7.90 u
(This doesn’t correspond to a real particle.)

9. Problem 5: An object with mass m1 = 900 kg and traveling at speed v1 = 0.85c collides with a stationary object with mass m2 = 1400 kg and the two objects stick together. What are the speed and mass of the composite object (#3)?
Energy conservation: 1m1c2 + m2c2 = 3m3c2
Momentum conservation: 1m1v1 = 3m3v3
Energy: 1 = 1/( )1/2 =  3m3 = (1.90 x ) kg = 3110 kg
Momentum: 3m3v3 = x 900 kg * 0.85c = 1454 kgc
 v3 = (1454/3110) c = c
 3 = 1/( )1/2 = 1.13
m3 = 3110 kg/1.13 = 2750 kg

10. Conservation of Energy: (1 + P)MPc2 = XMXc2
Problem 6: A 6.6 GeV (= 6600 MeV) proton (MP = u) collides with a proton at rest to create a new particle, X. What are the mass and energy of the new particle?
Conservation of Energy: (1 + P)MPc2 = XMXc2
Conservation of Momentum: PMP vP = XMXvX
The rest energy of the proton ERP = x 931 MeV = 938 MeV.
Therefore the moving proton has P = 6600/938 =
Since 7.04 = P = 1/(1-vP2/c2)1/2, the moving proton has speed vP/c =
Conservation of Energy: XMX = 8.04MP = 8.10 u
 EX = 8.10 x 931 MeV = 7540 MeV = 7.54 GeV
Conservation of Momentum: vX = 7.04 MP (0.990 c) / 8.04 MP = c
Then X = 1/( )1/2 = 2.01
MX = 8.10 u / 2.01 = 4.0 u
stationary proton
moving proton

11. For a particle with velocity v, p = mv and E = mc2
E2 – p2c2 = (mc)2 (c2-v2)/ [1 – v2/c2] = (mc2)2
E2 = p2c2 + (mc2)2
This equation is true for all particles, even those with mass = 0
(e.g. photons, gravitons, gluons).
For a particle with m=0:
ER = mc2 = 0, so all the energy is kinetic.
K = mc2  0   =   v = c (Massless particles must travel at speed c.)
Then p = (mv) = E/c.
For example, a photon of frequency f and wavelength  = c/f has energy E = hf. Therefore it has momentum p = hf/c = h/.

12. Problem 7: A pion (mc2 = 140 MeV) at rest decays to a muon (mc2 = 106 MeV) and antineutrino (m  0). Find the energy of the antinuetrino.
Since the antineutrino’s mass  0, essentially all of its energy is kinetic: E = pc.
Conservation of energy: mc2 = mc2 + pc
Conservation of momentum: mv = p
mc2 = mc2 (1 + v/c)
m/ m = (1 + v/c)/ [1 – (v/c)2]1/2
But [1 – (v/c)2] = (1 – vu/c) (1 + vu/c)
 m/ m = [(1 + v/c)/ (1 – v/c)]1/2
140/106 = 1.32 = [(1+vu/c)/ (1 – vu/c) ]1/2
1.74 = [(1+vu/c)/ (1 – vu/c) ]
1.74 – 1.74 vu/c = 1 + vu/c
2.74 vu/c = 0.74
vu/c = 0.27
 = 1.039
E = pc = mc2 - mc2 = [140 – (1.039)106] MeV
E = 29.9 MeV

13. Recommended reading: Feynman’s Lectures on Physics, Volume I, Chapters 15-17.
Available (for free !) online at
(or hardcopy for ~ $40).
Caution: Many older texts (e.g. Feynman) use different names and notation:
What we call mass, m, is often called the “rest mass” and denoted m0.
What we denote as m0 is often called the relativistic mass, m.
Quantity
Us (Serway/Jewett)
Feynman
m = mass
m0 = rest mass m
m = (relativistic) mass
Momentum (p)
mu
m0u = mu
kinetic energy (K)
(-1) mc2
(-1) m0c2 = (m-m0)c2
rest energy (ER)
mc2
m0c2
total energy (E)
mc2
m0c2 = mc2