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

`<3068.2352, 800, 0>\ m/s`

Step-by-step explanation:

F = Force =
`<-1.12* 10^(-11), 0, 0>`

m = Mass of proton =
`1.7* 10^{-27\ kg`

t = Time taken =
`2* 10^(-14)\ s`

Acceleration is given by

`a=(F)/(m)\\\Rightarrow a=(<-1.12* 10^(-11), 0, 0>)/(1.7* 10^(-27))\\\Rightarrow a=<-6.58824* 10^(15), 0, 0>\ m/s^2`

`v=u+at\\\Rightarrow v=<3200, 800, 0>+<-6.58824* 10^(15), 0, 0>* 2* 10^(-14)\\\Rightarrow v=<3200, 800, 0>+<-6.58824* 10^(15), 0, 0>* 2* 10^(-14)\\\Rightarrow v=<3200, 800, 0>+<-131.7648, 0, 0>\\\Rightarrow v=<3068.2352, 800, 0>\ m/s`

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