A) \[6.4\times {{10}^{-7}}kg-m/{{s}^{2}}\]
B) \[4.8\times {{10}^{-8}}kg-m/{{s}^{2}}\]
C) \[3.2\times {{10}^{-9}}kg-m/{{s}^{2}}\]
D) \[1.6\times {{10}^{-10}}kg-m/{{s}^{2}}\]
Correct Answer: D
Solution :
Momentum transferred in one second \[p=\frac{2U}{c}=\frac{2{{S}_{av}}A}{c}=\frac{2\times 6\times 40\times {{10}^{-4}}}{3\times {{10}^{8}}}\] \[=1.6\times {{10}^{-10}}\] kg-m/s2.
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