We spent an entire semester talking about Newtonian mechanics ( F=ma, KE=1/2 mv
ID: 1470073 • Letter: W
Question
We spent an entire semester talking about Newtonian mechanics (F=ma, KE=1/2 mv2) so that knowledge must be important. The truth is that those equations are almost precisely correct except for the fastest particles. Let's see if we can find out how fast something must be going for those approximations to be off by a signficant amount. Assume we have a object with a mass of mo = 1gram.
1)
What if the object is going 0.05 c. What is the object's velocity?
v =
2) That's pretty fast, right? What is the object's relativistic kinetic energy at this velocity?
K=
3)
What is the object's Newtonian kinetic energy at this velocity?
K=
4) Look to see how much the two values of kinetic energy differ. Now the object is going 0.14 c. What is the object's velocity?
v =
5) What is the object's relativistic kinetic energy at this velocity?
K=
6) What is the object's Newtonian kinetic energy at this velocity?
K=
7) Now we're going really fast. Look to see how much the two values differ now. Next, the object is going 0.58 c. What is the object's velocity?
v =
8) What is the object's relativistic kinetic energy at this velocity?
K=
9) What is the object's Newtonian kinetic energy at this velocity?
K=
10) Next, the object is going 0.93 c. What is the object's velocity?
v =
11) What is the object's relativistic kinetic energy at this velocity?
K=
12) What is the object's Newtonian kinetic energy at this velocity?
K=
Explanation / Answer
mo = 1g = 1*10^-3 kg
1.)
If the object is going at 0.05 c
Velcoity = 0.05 * 3.0 * 10^8 m/s
Velocity , v = 1.5 * 10^7 m/s
2.)
Relativistic K.E = m0*c^2 *[1/sqrt(1 - v^2/c^2) - 1]
Relativistic K.E = 1*10^-3 * (3.0 * 10^8)^2 * [1/sqrt(1 - 0.05^2 - 1)]
Relativistic K.E = 1.127 * 10^11 J
3.)
Newtonian K.E = 1/2 * m*v^2
Newtonian K.E = 1/2 * 1*10^-3 *(1.5 * 10^7)^2
Newtonian K.E = 1.125 * 10^11 J
4.)
If the object is going at 0.14 c
Velcoity = 0.14 * 3.0 * 10^8 m/s
Velcoity = 4.2 * 10^7 m/s
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