PHYSICS FOR SCI.AND ENGR W/WEBASSIGN
PHYSICS FOR SCI.AND ENGR W/WEBASSIGN
10th Edition
ISBN: 9781337888462
Author: SERWAY
Publisher: CENGAGE L
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Question
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Chapter 38, Problem 26P

(a)

To determine

The fine for travel at the speed of 1090km/h .

(a)

Expert Solution
Check Mark

Answer to Problem 26P

The fine for travel at the speed of 1090km/h is $800 .

Explanation of Solution

Given info: The speed limit is 90.0km/h and the fine for driving at 190E is $80.0 .

Write the expression for the total fine in the excess momentum.

F=pup90km/h (1)

Here,

F is the total fine.

pu is the momentum of the vehicle.

p90km/h is the speed limit momentum.

Write the expression for fine to travel.

$80.0=p190km/hp90km/h (2)

Divide equation (1) and equation (2).

F$80.0=pup90km/hp190km/hp90km/h (3)

Write the expression for the relativistic momentum of the vehicle.

pu=mu1(u/c)2

Here,

m is the mass of vehicle.

u is the speed of vehicle.

c is the speed of the light.

Write the expression for the relativistic momentum of the vehicle at speed of 90.0km/h .

p90km/h=m(u90km/h)1(u90km/h/c)2

Here,

u is the speed of vehicle travel at speed of 90.0km/h .

Write the expression for the relativistic momentum of the vehicle at speed of 190.0km/h .

p190km/h=m(u190km/h)1(u190km/h/c)2

Here,

u is the speed of vehicle travel at speed of 190.0km/h .

Substitute mu1(u/c)2 for pu , m(u90km/h)1(u90km/h/c)2 for p90km/h and m(u190km/h)1(u190km/h/c)2 for p190km/h in equation (3).

F$80.0=mu1(u/c)2m(u90km/h)1(u90km/h/c)2m(u190km/h)1(u190km/h/c)2m(u90km/h)1(u90km/h/c)2 (4)

Substitute 1090.0km/h for u , 90.0km/h for u90.0km/h and 190.0km/h for u190.0km/h in equation (4).

F$80.0=m(1090km/h)1(1090km/h/c)2m(90km/h)1(u/c)2m(190km/h)1(190km/h3×108m/s×3.6km/h1m/s)2m(90km/h)1(u/c)2F$80.0=1090km/h1(0.999)290km/h100km/hF$80.01090km/h90km/h100km/hF$800

Conclusion:

Therefore, the fine for travel at the speed of 1090km/h is $800 .

(b)

To determine

The fine for travel at the speed of 1000000090km/h .

(b)

Expert Solution
Check Mark

Answer to Problem 26P

The fine for travel at the speed of 1000000090km/h is $2.12×109 .

Explanation of Solution

Given info: The speed limit is 90.0km/h and the fine for driving at 190E is $80.0 .

From equation (4), the equation is given as,

F$80.0=mu1(u/c)2m(u90km/h)1(u90km/h/c)2m(u190km/h)1(u190km/h/c)2m(u90km/h)1(u90km/h/c)2

Substitute 1000000090.0km/h for u , 90.0km/h for u90.0km/h and 190.0km/h for u190.0km/h in equation (4).

F$80.0=m(1000000090.0km/h)1(1000000090.0km/h/c)2m(90km/h)1(u/c)2m(190km/h)1(190km/h3×108m/s×3.6km/h1m/s)2m(90km/h)1(u/c)2F$80.0=1000000090.0km/h1(0.999)290km/h100km/hF$80.01000000090km/h90km/h100km/hF$2.12×109

Conclusion:

Therefore, the fine for travel at the speed of 1000000090km/h is $2.12×109 .

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Chapter 38 Solutions

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN

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