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 VI- v2/2 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
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 VI- v2/2 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
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Transcribed Image Text:We have f "(x) =
%3D
and we are given that | v | < 60 m/s, so
3moc2
3moc2
| f "(x) | =
(= M)
4(1 - v²/c2)5/2 4(1 - 60²/c²,5/2
Thus, with c = 3.00 × 10° m/s,
3moc?
604
| R1(x) | <
4(1 - 60²/²,5/2 A
mo
So when | v| < 60 m/s, the magnitude of the error in using the Newtonian
expression for kinetic energy is at most
mo.
![In Einstein's theory of special relativity the mass of an object
EXAMPLE 3
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
m =
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 = ½mov².
(b) Use Taylor's Inequality to estimate the difference in these expressions for K
when | v|s 60 m/s.
SOLUTION (a) Using the expression given for K and m, we get
moc?
K = mc² - moc² =
moc
V1 - v/2
%3D
(글),
moc2t||1-
1
With x = -v/2, the Maclaurin series for (1 + x) 12 is most easily computed as
a binomial series with k =
. (Notice that | x | < 1 because v < c.) Therefore we have
(1 + x)-1/2 = 1 -
(-1/2)(-3/2)
x²+
x³ + .
2!
3!
1
5
= 1 --x +
x3
16
12
K = moc² [(1 +.
5 v6
--+ ...) - 1]
16 6
and
+
5 v6
+ ...)
16 6
mọc²(-
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 = moc²(--) =
22
(b) If x = -v²/², f(x) = moc²[(1 + x)*1/2 - 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!](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb13ad392-c0f2-4a1f-8c28-c5aad8f5d77d%2Fb076122d-fdd9-4539-9dfa-a796c14b30f9%2Fbzt1gx_processed.png&w=3840&q=75)
Transcribed Image Text:In Einstein's theory of special relativity the mass of an object
EXAMPLE 3
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
m =
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 = ½mov².
(b) Use Taylor's Inequality to estimate the difference in these expressions for K
when | v|s 60 m/s.
SOLUTION (a) Using the expression given for K and m, we get
moc?
K = mc² - moc² =
moc
V1 - v/2
%3D
(글),
moc2t||1-
1
With x = -v/2, the Maclaurin series for (1 + x) 12 is most easily computed as
a binomial series with k =
. (Notice that | x | < 1 because v < c.) Therefore we have
(1 + x)-1/2 = 1 -
(-1/2)(-3/2)
x²+
x³ + .
2!
3!
1
5
= 1 --x +
x3
16
12
K = moc² [(1 +.
5 v6
--+ ...) - 1]
16 6
and
+
5 v6
+ ...)
16 6
mọc²(-
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 = moc²(--) =
22
(b) If x = -v²/², f(x) = moc²[(1 + x)*1/2 - 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!
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