EXAMPLE 3 In Einstein's theory of special relativity the mass of an object moving with velocity v is given below, where mo is the mass of the object when at rest and c is the speed of light. The kinetic energy of the object is the difference between its total energy and its energy at rest: K = mc2 - moc²
EXAMPLE 3 In Einstein's theory of special relativity the mass of an object moving with velocity v is given below, where mo is the mass of the object when at rest and c is the speed of light. The kinetic energy of the object is the difference between its total energy and its energy at rest: K = mc2 - moc²
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Transcribed Image Text:3moc?
604
| R1(x) | <
4(1 - 60/2,5/2 4
<
mo
So when | v| < 60 m/s, the magnitude of the error in using the Newtonian expression for kinetic energy is at most
mo.
![EXAMPLE 3
In Einstein's theory of special relativity the mass of an object moving with velocity v is given below, where mo is the mass of the object when at rest and c is
the speed of light.
mo
1 – v2/c2
The kinetic energy of the object is the difference between its total energy and its energy at rest:
K = mc2 - moc2
(a) Show that when v is very small compared with c, this expression for K agrees with classical Newtonian physics: K =
Vamov?.
(b) Use Taylor's Inequality to estimate the difference in these expressions for K when | v | < 60 m/s.
SOLUTION
(a) Using the expression given for K and m, we get
moc?
K = mc2 - moc2 :
moc?
V1 - v/c
(금),
moc²t1
C
With x = -v/c2, the Maclaurin series for (1 + x)-1/2 is most easily computed as a binomial series with k =
. (Notice that | x| < 1 because v < c.) Therefore we have
(-1/2)(-3/2)
(1 + x)-
-1/2
= 1 -
+
+
+ ...
2!
3!
= 1 --x +
-x³ +
2
16
1v2
and K = moc2 [(1 +--+
22
5 v6
-+ ...) - 1]
16 6
1v2
mọc? (–
22
5 v6
+ ...)
16 6
If v is much smaller than c, then all terms after the first are very small when compared with the first term. If we omit them, we get
1v2
K =
moc?(-
-) =
2 2
(b) If x = -v/c, f(x) = moc²[(1 + x)-2 - 1], and M is a number such that | f "(x) | < M, then we can use Taylor's Inequality to write
M
| R1(x) | <–x?
2!
We have f "(x) =
and we are given that | v | < 60 m/s, so
Зтос
Зтос?
| f "(x) | =
-(= M)
4(1 - v/c2)5/2 4(1 - 602/c2,5/2
Thus, with c = 3.00 × 108 m/s,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F25befd20-0d58-44de-b948-d9335e17825d%2F2389c2ad-4472-405d-80e5-a6eb01ba0c74%2Fua24vwb_processed.png&w=3840&q=75)
Transcribed Image Text:EXAMPLE 3
In Einstein's theory of special relativity the mass of an object moving with velocity v is given below, where mo is the mass of the object when at rest and c is
the speed of light.
mo
1 – v2/c2
The kinetic energy of the object is the difference between its total energy and its energy at rest:
K = mc2 - moc2
(a) Show that when v is very small compared with c, this expression for K agrees with classical Newtonian physics: K =
Vamov?.
(b) Use Taylor's Inequality to estimate the difference in these expressions for K when | v | < 60 m/s.
SOLUTION
(a) Using the expression given for K and m, we get
moc?
K = mc2 - moc2 :
moc?
V1 - v/c
(금),
moc²t1
C
With x = -v/c2, the Maclaurin series for (1 + x)-1/2 is most easily computed as a binomial series with k =
. (Notice that | x| < 1 because v < c.) Therefore we have
(-1/2)(-3/2)
(1 + x)-
-1/2
= 1 -
+
+
+ ...
2!
3!
= 1 --x +
-x³ +
2
16
1v2
and K = moc2 [(1 +--+
22
5 v6
-+ ...) - 1]
16 6
1v2
mọc? (–
22
5 v6
+ ...)
16 6
If v is much smaller than c, then all terms after the first are very small when compared with the first term. If we omit them, we get
1v2
K =
moc?(-
-) =
2 2
(b) If x = -v/c, f(x) = moc²[(1 + x)-2 - 1], and M is a number such that | f "(x) | < M, then we can use Taylor's Inequality to write
M
| R1(x) | <–x?
2!
We have f "(x) =
and we are given that | v | < 60 m/s, so
Зтос
Зтос?
| f "(x) | =
-(= M)
4(1 - v/c2)5/2 4(1 - 602/c2,5/2
Thus, with c = 3.00 × 108 m/s,
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