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momentum A proton interacts electrically with a neutral HCl molecule located at the origin. At a certain time t, the proton’s position is h1.6 × 10−9 , 0, 0i m and the proton’s velocity is h3200, 800, 0i m/s. The force exerted on the proton by the HCl molecule is h−1.12 × 10−11 , 0, 0i N. At a time t + (2 × 10−14 s), what is the approximate velocity of the proton? answer

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Answer:

[tex]<3068.2352, 800, 0>\ m/s[/tex]

Explanation:

F = Force = [tex]<-1.12\times 10^{-11}, 0, 0>[/tex]

m = Mass of proton = [tex]1.7\times 10^{-27\ kg[/tex]

t = Time taken = [tex]2\times 10^{-14}\ s[/tex]

Acceleration is given by

[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{<-1.12\times 10^{-11}, 0, 0>}{1.7\times 10^{-27}}\\\Rightarrow a=<-6.58824\times 10^{15}, 0, 0>\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow v=<3200, 800, 0>+<-6.58824\times 10^{15}, 0, 0>\times 2\times 10^{-14}\\\Rightarrow v=<3200, 800, 0>+<-6.58824\times 10^{15}, 0, 0>\times 2\times 10^{-14}\\\Rightarrow v=<3200, 800, 0>+<-131.7648, 0, 0>\\\Rightarrow v=<3068.2352, 800, 0>\ m/s[/tex]

The velocity of the proton is [tex]<3068.2352, 800, 0>\ m/s[/tex]

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