Its a joined question We have f "(x) =%3Dand we are given that | v | < 60 m/s, so3moc23moc2| f "(x) | =(= M)4(1 - v²/c2)5/2 4(1 - 60²/c²,5/2Thus, with c = 3.00 × 10° m/s,3moc?604| R1(x) | <4(1 - 60²/²,5/2 AmoSo when | v| < 60 m/s, the magnitude of the error in using the Newtonianexpression for kinetic energy is at mostmo. In Einstein's theory of special relativity the mass of an objectEXAMPLE 3moving with velocity v is given below, where mo is the mass of the object whenat rest and c is the speed of light.mom =The kinetic energy of the object is the difference between its total energy andits energy at rest:K = mc2 - moc²(a) Show that when v is very small compared with c, this expression for Kagrees with classical Newtonian physics: K = ½mov².(b) Use Taylor's Inequality to estimate the difference in these expressions for Kwhen | v|s 60 m/s.SOLUTION (a) Using the expression given for K and m, we getmoc?K = mc² - moc² =mocV1 - v/2%3D(글),moc2t||1-1With x = -v/2, the Maclaurin series for (1 + x) 12 is most easily computed asa 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!15= 1 --x +x31612K = moc² [(1 +.5 v6--+ ...) - 1]16 6and+5 v6+ ...)16 6mọc²(-If v is much smaller than c, then all terms after the first are very small whencompared with the first term. If we omit them, we get1v2K = 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 writeM| R1(x) | <_x2!

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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. 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

Modern Physics

3rd Edition

ISBN:9781111794378

Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer

Publisher:Raymond A. Serway, Clement J. Moses, Curt A. Moyer

Chapter2: Relativity Ii

Section: Chapter Questions

Problem 16P

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