Question

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

Answer 1

Answer:

$\ m/s$

Explanation:

F = Force =

m = Mass of proton = $1.7\times 10^{-27\ kg$

t = Time taken = $2\times 10^{-14}\ s$

Acceleration is given by

$a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{}{1.7\times 10^{-27}}\\\Rightarrow a=\ m/s^2$

$v=u+at\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\\\Rightarrow v=\ m/s$

The velocity of the proton is $\ m/s$