A) \[E/c\]
B) \[2E/c\]
C) \[Ec\]
D) \[E/{{c}^{2}}\]
Correct Answer: B
Solution :
Initial momentum of surface \[{{p}_{i}}=\frac{E}{c}\] where c = velocity of light (constant). Since, the surface is perfectly reflecting, so the same momentum will be reflected completely. Final momentum,\[{{p}_{f}}=\frac{E}{c}\] (with negative value) \[\therefore \]Change in momentum \[\Delta p={{p}_{f}}-{{p}_{i}}=-\frac{E}{c}-\frac{E}{c}=-\frac{2E}{c}\] Thus, momentum transferred to the surface is \[\Delta p'=|\Delta p|=\frac{2E}{c}\]
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