Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 39, Problem 39.49P

A proton in a high-energy accelerator moves with a speed of c/2. Use the work-kinetic energy theorem to find the work required to increase its speed to (a) 0.750c and (b) 0.995c

(a)

Expert Solution
Check Mark
To determine

The work required to increase the speed of the particle to 0.750c .

Answer to Problem 39.49P

The work required to increase the sped of the particle is 335MeV .

Explanation of Solution

Given Info: The initial speed of the particle is c2 .

The rest mass of the proton is 1.67×1027kg .

The speed of the light is 3×108m/s .

The final kinetic energy of the proton is,

Kf=(11(ufc)21)m(c)2

Here,

uf is the final velocity

m is the mass of the proton.

c is the speed of the light.

The initial kinetic energy of the proton is,

Ki=(11(uic)21)m(c)2

Here,

ui is the initial velocity.

Form the work energy theorem.

W=KfKi

Substitute (11(ufc)21)m(c)2 for Kf and (11(uic)21)m(c)2 for Ki in the above equation.

W=(11(ufc)21)m(c)2(11(uic)21)m(c)2=(11(ufc)2)m(c)2mc2(11(uic)2)m(c)2+mc2=(11(ufc)2)m(c)2(11(uic)2)m(c)2

Substitute c2 for ui and 0.750c for uf in the above equation.

W=(11(0.75cc)2)m(c)2(11(c2c)2)m(c)2=(10.661)mc2(10.75)mc2=0.357(mc2)

Substitute 1.67×1027kg for m and 3×108m/s for c in the above equation.

W=0.357(1.67×1027kg(3×108m/s)2)=5.37×1010J×(1eV1.602×1019J)=335.20×106eV335MeV

Conclusion:

Therefore, the work required to increase the sped of the particle is 335MeV .

(b)

Expert Solution
Check Mark
To determine

The work required to increase the speed of the particle to 0.995c .

Answer to Problem 39.49P

The work required to increase the sped of the particle is 8.31GeV .

Explanation of Solution

Given Info: The initial speed of the particle is c2 .

The rest mass of the proton is 1.67×1027kg .

The speed of the light is 3×108m/s .

The work done is,

W=(11(ufc)2)m(c)2(11(uic)2)m(c)2

Substitute c2 for ui and 0.995c for uf in the above equation.

W=(11(0.995cc)2)m(c)2(11(c2c)2)m(c)2=(10.099)mc2(10.75)mc2=8.85(mc2)

Substitute 1.67×1027kg for m and 3×108m/s for c in the above equation.

W=8.85(1.67×1027kg(3×108m/s)2)=1.33×109J×(1eV1.602×1019J)=8.303×109eV8.31GeV

Conclusion:

Therefore, the work required to increase the sped of the particle is 8.31GeV .

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Chapter 39 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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Length contraction: the real explanation; Author: Fermilab;https://www.youtube.com/watch?v=-Poz_95_0RA;License: Standard YouTube License, CC-BY