Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 27 - Early Quantum Theory and Models of the Atom - General Problems - Page 801: 84

Answer

9.0 N.

Work Step by Step

Momentum is conserved. Dividing the photons’ change in momentum by the elapsed time tells us the force applied to the light. $$ \overline {F}=\frac{\Delta p}{\Delta t}$$ By Newton’s third law, this equals the force of the sunlight upon the sail. The photons bounce off the mirror, so the change in momentum (impulse) is twice their incident momentum. The reflected photons had momentum, given by equation 27-6. $$ \overline {F}=\frac{\Delta p}{\Delta t}=\frac{2(E/c)}{c\Delta t}$$ $$ =\frac{2(E/ \Delta t)}{c }$$ The photon power (energy divided by time) is the sunlight’s intensity multiplied by the sail’s area. Evaluate. $$ =\frac{2IA}{c }=\frac{2(1350W/m^2)(1.0\times10^3 m)^2}{3.00\times10^8 m/s}=9.0N $$
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