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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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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.
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,
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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