gamma = sqrt(1-v^2/c^2)
L = L_0*gamma ( length contraction )
T = T_0/gamma ( time dilation )
p = m_0*v/gamma ( relativistic momentum )
m = m_0/gamma ( mass increase formula )
KE = mc^2 -m_0*c^2 ( KE )
E = m_0*c^2+KE ( total energy )
E^2 = p^2c^2 + m_0^2c^4 ( energy-momentum, no PE, p in eV/c, m in eV/c^2 )
u = (v+u')/(1+vu'/c^2) ( relativistic addition of velocities, u with respect to earth, v with respect to earth, u' with respect to obj with speed v )
You are sitting in your car when a very fast sports car passes you at a speed of 0.37c. A person in that car says his car is 6.00m long and yours is 6.21m long. What do you measure for those two lengths?
gamma = sqrt(1-.37^2) = 0.929
L = L_0 * gamma
L hiscar = 0.929*6.00m =>5.574m
L mycar = 6.21m/0.929 => 6.6846m.
What is the speed of a beam of pions if their average lifetime is measured to be 4.1*10^(-8)s? At rest, their lifetime is 2.60*10^(-8)s.
t_0/t = gamma.
(t_0/t)^2 = 1-v^2/c^2
sqrt(-((t_0/t)^2-1)*c^2) = v
sqrt(-((2.60*10^(-8)/(4.1*10^(-8)))^2-1)*(3.0*10^8)^2) = 2.32*10^8m/s
Suppose you decide to travel to a star 90 light-years away. How fast would you have to travel so the distance would be only 25 light-years?
L/L_0 = gamma
25/90 = gamma = sqrt(1-v^2/c^2)
sqrt(-((25/90)^2-1)*(3.0*10^8)^2) = 2.88*10^8m/s = 0.96c
A certain star is 75 light-years away. How long would it take a spacecraft traveling 0.950c to reach that star from Earth, as measured by observers a) on Earth, b) on the spacecraft? c) What is the distance traveled according to the observers on the spacecraft? d) What will the spacecraft occupants compute their speed to be from the results of b and c?
a) 75light-years/0.95c = 78.947yrs.
b) t = t_0/ gamma. t_0= 78.947*(sqrt(1-(.95)^2)) = 24.65yrs
c) L = L_0 * gamma, 75*(sqrt(1-(.95)^2)) = 23.42light-years.
d) v = d/t = 23.42light-years/24.65yrs = 0.95c
A friend of yours travels by you in her fast sports vehicle at a speed of 0.580c. You measure it to be 5.80m long and 1.20m high. a) What will be its length and height at rest? b) How many seconds would you say elapsed on your friend's watch when 20.0s passed on yours? c) How fast did you appear to be traveling to your friend? d) How many seconds would she say elapsed on your watch when she saw 20.0s pass on hers?
a) L = L_0*gamma => L_0 = 5.8/sqrt(1-0.58^2) = 7.12m, height remains unchanged 1.20m
b) t = t_0/ gamma . t = 20*sqrt(1-.58^2) = 16.3s
c) Since you are stationary, you appear to be moving at_0.580c towards your friend.
d) t = t_0/ gamma . t = 20*sqrt(1-.58^2) = 16.3s
At what speed will an object's mass be 10% greater than its rest mass?
m = m_0/ gamma, 1.10*sqrt(1-v^2/c^2) = v, quadratic formula, v = 0.417c
How much energy can be obtained from conversion of 1.0g of mass? How much mass could this energy raise to a height of 100m?
E = mc^2, 1/1000kg*(3*10^8)^2 = 9*10^13J.
mgh = mc^2. 9*10^(13)/(9.8*100) => 9.2*10^10kg.
Calculate the KE and momentum of a proton traveling at 2.50*10^8m/s.
KE = mc^2 -m_0*c^2, m = m_0/sqrt(1-.833^2), m = 3.02*10^(-27)kg
KE = (3.02*10^(-27) - 1.67*10^(-27))*c^2 => 1.216*10^(-10)J.
p = m_0v/ gamma, p = 7.553*10^(-19) kg*m/s
What is the speed of a proton accelerated by a potential difference of 75MV?
QV = KE = (m-m_0 )*c^2, (1.602*10^(-19)*75*10^6)/(3*10^8)^2 = m-m_0
m = 1.8035*10^(-27)kg.
(m_0 /m)^2 = 1 v^2/c^2. v = 1.13*10^8m/s => 0.378c
What is the speed and momentum of an electron whose KE equals its rest energy?
KE = mc^2 -m_0 c^2 = m_0 c^2, mc^2 = 2m_0 c^2, m = 2m_0 . m = 1.822*10^(-30)kg.
(m_0 /m)^2 = 1 v^2/c^2, v = 2.598*10^8m/s. = 0.866c
p = m_0*v/gamma. p = 4.73*10^(-22)kg*m/s
Suppose a spacecraft of rest mass 37000kg is accelerated to 0.21c. a) How much KE would it have? b) If you used the classical formula, what is the % error?
a) m = 37000/sqrt(1-.21^2), m = 37844kg. KE = (m-m_0 )*c^2, => 7.596*10^19J
b) KE = 1/2mv^2, KE = 7.34*10^19J. % diff = 3.37%.
Two spaceships leave the earth in opposite directions, each with a speed of 0.5c with respect to earth. a) What is the velocity of spaceship 1 relative to spaceship 2? b) 2 to 1?
u = (v+u')/(1+vu'/c^2)
a) 0.5 = (-0.5 + u')/ (1+-.5u'), u' = 0.8c away from spaceship 2.
b) same as a but opposite direction.
A spaceship leaves earth traveling 0.65c. A second spaceship leaves the first at a speed of 0.921c with respect to the first. Calculate the speed of the second ship with respect to earth if it is fired a) in the same dir as the first spaceship, b) back at earth.
u = (v+u')/(1+vu'/c^2)
a) u = (.65+.921)/(1+.65*.921), u = 0.98227c.
b) u = (.65-.921)/(1+.65*-.921), u = -0.675c
How many grams of matter would have to be totally destroyed to run a 100W lightbulb for a year?
P = W/t, W = 3.1536*10^9J. W = mc^2, m = 3.504*10^(-5)g.
An electron enters a uniform magnetic field B = 1.8T and moves perpendicular to the field lines with a speed v = 0.92c. What is the radius of curvature of its path?
mv^2/r = qvB. mv/qB = r.
m = 9.11*10^(-31)/sqrt(1-.92^2) => 2.3245*10^(-30)kg.
r = 2.2248*10^(-4)m.
The sun radiates energy at a rate of about 4*10^26W. a) At what rate is the sun's mass decreasing? b) How long does it take for the sun to lose a mass equal to that of earth? c) Estimate how long the sun could last if it radiated constantly at this rate.
m_sun = 1.989*10^30kg.
m_earth = 5.976*10^24kg.
a) P = W/t, dE = dm*c^2, dm = 4*10^26/c^2 => 4.44*10^9kg/s
b) 5.976*10^24kg/ 4.44*10^9kg/s =1.346*10^15s
c) 1.989*10^30kg/ 4.44*10^9kg/s = 4.48*10^20s.