Question

an electron with a speed of 1.20×10^7 m/s moves horizontally into a region where a constant vertical force of 4.00×10^(-16) n acts on it. the mass of the electron is 9.11×10^(-31) kg. determine the vertical distance the electron is deflected during the time it has moved 49.0 mm horizontally.

Explanation:

Step1: Calculate the time taken to move horizontally

The horizontal motion is a uniform - motion. The formula for time in uniform motion is $t=\frac{d}{v}$, where $d = 49.0\ m$ and $v=1.20\times 10^{7}\ m/s$.
$t=\frac{49.0}{1.20\times 10^{7}}\ s$

Step2: Calculate the vertical acceleration

According to Newton's second law $F = ma$, where $F = 4.00\times 10^{-16}\ N$ and $m = 9.11\times 10^{-31}\ kg$. The acceleration $a=\frac{F}{m}$.
$a=\frac{4.00\times 10^{-16}}{9.11\times 10^{-31}}\ m/s^{2}$

Step3: Calculate the vertical displacement

The vertical motion is a uniformly - accelerated motion with an initial vertical velocity $u_y = 0\ m/s$. The formula for displacement in uniformly - accelerated motion is $y=u_y t+\frac{1}{2}at^{2}$. Since $u_y = 0\ m/s$, $y=\frac{1}{2}at^{2}$.
First, $t=\frac{49.0}{1.20\times 10^{7}}\ s\approx4.083\times 10^{-6}\ s$, $a=\frac{4.00\times 10^{-16}}{9.11\times 10^{-31}}\ m/s^{2}\approx4.39\times 10^{14}\ m/s^{2}$.
$y=\frac{1}{2}\times(4.39\times 10^{14})\times(4.083\times 10^{-6})^{2}$
$y=\frac{1}{2}\times4.39\times 10^{14}\times1.667\times 10^{-11}$
$y=\frac{4.39\times1.667\times 10^{14 - 11}}{2}$
$y=\frac{7.32\times 10^{3}}{2}=3660\ m$

Answer:

$3660\ m$