1. A train travels in the +x direction with a speed of β=0.80 with respect to the ground. At a certain time, two balls
are ejected, one traveling in the +x direction with x-velocity of +0.60 with respect to the train and the other traveling
in the -x direction with x-velocity of -0.40 with respect to the train.
(a) What are the x-velocities of the balls with respect to the ground?
VBall 1 WRT Ground = VBall 1 WRT Train + VTrain WRT Ground = 0.6+0.8=1.4 x VTrain
VBall 2 WRT Ground = VBall 2 WRT Train + VTrain WRT Ground = -0.4+0.8=0.4 x VTrain
(b) what is the x-velocity of the first ball with respect to the second? [hint: the frames you will choose to be the
Home Frame and the Other Frame for part (a) will not be the same as your choices for part (b).]
VBall 1WRT Ball 2 = VBall 1 WRT Train - VBall 2 WRT Train = 0.6-(-0.4)=1.0 x VTrain or
VBall 1 WRT Ground - VBall 2 WRT Ground =1.4-0.4=1.0 x VTrain
2. You are the captain of a spaceship that is moving through an asteroid belt on impulse power at a speed of (4/5)
relative to the asteroids. Suddenly you see an asteroid dead ahead a distance of only 24s away, according sensor
measurements in your ship's reference frame. You immediately shoot off a missile, which travels forward at a speed
of (4/5) relative to your ship. The missile hits the asteroid and detonates, pulverizing the asteroid into gravel.
However, you learned in Starfleet Academy that it is not safe to pass through such a debris field (even with shields
on full) sooner than 8s (measured in the asteroid frame) after the detonation. Are you safe?
DistanceShip to Asteroid= VShip WRT Asteroid x TimeShip to hit Asteroid = (0.8 lengths/s)(24s)=19.2 lengths
[Length is an arbitrary measure of distance since no actual speed was given. You will see below that this term will
cancel out anyway when solving the problem]
VMissile WRT Asteroid = VMissile WRT Ship + VShip WRT Asteroid = 0.8+0.8=1.6 lengths/s
TimeMissile to hit Asteroid = DistanceShip to Asteroid / VMissile WRT Asteroid = 19.2 lengths / 1.6 lengths/s = 12s
Asteroid is destroyed 12 seconds after you detect it, leaving 12 more seconds before hitting the debris field.
Bottom Line—you are safe with 4 seconds to spare
3. Imagine that an object of mass 5.0 kg is observed in a certain inertial frame to have velocity components of Vx= 0.866 and Vy=Vz=0. Evaluate the following in that reference frame:
(a) the object's total energy E
Kinetic Energy = ½mv2 = (0.5)(5.0 kg)[(-0.866 m/s)2+(0)2+(0)2] = 1.875 Joules
(since the next question refers to relativistic KE, perhaps Vx may have been-0.866c? If so, then (a) becomes:
(0.5)(5.0 kg)[[(-0.866 x 3x108 m/s)2+(0)2+(0)2]1/2]2 = 1.6875x1017 Joules)
(d) its relativistic kinetic energy K. (assuming Vx=-0.866c) then
Ek=mc2[(1-(v/c)2)-1/2-1] =(5.0 kg)(3x108m/s)2[(1-(-0.866c/c)2)-1/2-1] = 4.5x1017 Joules
Question 4 is on the next page
4. If electrical energy costs about $0.04 per 10 6 J (the approximate current price in southern California) and you
have $1.5 million at your disposal to spend on energy to convert to kinetic energy, about how fast can you make a
1.0-g object travel?
Total Energy = ($1.5x106)(106 J/$0.04) = 3.75x1013 Joules
v=(2E/m)1/2 = (2 x 3.75x1013 Joules/0.001 kg)1/2 = 2.74x108 m/s
(note that at these speeds, you will get relativistic effects, taking this into account, the equation for v becomes:
v=c(1-(mc2/(mc2+E))2) where mc2 = (0.001 kg)(3x108 m/s)2 = 9x1013 Joules
v=c(1-(9x1013/(9x1013+3.75x1013))2)=c(1-0.5)=1.5x108 m/s
Using the binomial approximation, show that the relativistic expression for kinetic energy reduces to the classical
one at low speeds (v << c). (Don't set v = 0!)
The binomial expansion means (1+x)n can be written as:
(1+x)n = 1+[n]x+[n(n-1)/2!]x2+[n(n-1)(n-2)/3!]x3+…and so on
It applies to Kinetic energy in the following way: Ek=mc2[(1-(v/c)2)-1/2-1].
If we expand the (1-(v/c)2)-1/2 term using the binomial expansion and substitute x=-(v/c)2, it can be written as:
(1-(v/c)2)-1/2 = (1+x)-1/2 = 1+[-1/2]x+[(-1/2)(-1/2-1)/(2*1)]x2+[(-1/2)(-1/2-1)(-1/2-2)/(3*2*1)]x3+… and so on.
But at low speeds where v<<c, the x=-(v/c)2 term is a very small number. The x2, x3 and all other xn terms can be
approximated to equal zero so they can be dropped. The binomial expansion becomes the binomial approximation
and (1+x)-1/2 ≈ 1-(1/2)x. Substituting x=-(v/c)2 back into the original equation for Ek we get:
Ek=mc2[(1-(v/c)2)-1/2-1] ≈ mc2[(1-(1/2)[-(v/c)2]-1] ≈ mc2[(1/2)v2/c2] ≈ ½mv2 which is the classical definition of
kinetic Energy.