Question

An electron with speed of 104 m/s enters a ""forbidden"" region where an electric force tries to push it back along its path with a constant acceleration of 107 m/s2 . How far will the electron go into the ""forbidden"" region? How long will it be in that region?

Answer 1

Answer:

The distance travelled is 151.22m and it took 0.97s

Explanation:

Well, this is an ARM problem, so we will need the following formulas

$x(t)=x_{0} +v_{0} *(t-t_{0} )+0.5*a*(t-t_{0} )^{2}$

$v(t)=v_{0} +a*(t-t_{0} )$

where $x_{0}$ is the initial position (we can assume is zero), $v_{0}$ is the initial speed of 104 m/s, $t_{0}$ is the initial time (we also assume is zero), a is the acceleration of 107 m/s2, v is speed, x is position and t is time.

Now that we have the formulas, we know that when the electron stops it has no speed. Then we calculate how much time it takes to stop.

$0=104m/s-107m/s^{2} *t\\t=0.97s$

Finally, we calculate the distance travelled in this time

$x(0.97s)=104m/s*0.97s+0.5*107m/s^{2}*(0.97s)^{2}=151.22m$

Answer 2

Answer:

Part a)

$d = 5 m$

Part b)

$T = 2\times 10^{-3} s$

Explanation:

Part a)

The electron will move in this forbidden region till its speed will become zero

So here we will have

$v_f = 0$

$v_i = 10^4 m/s$

also its deceleration is given as

$a = - 10^7 m/s^2$

so we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - (10^4)^2 = 2(-10^7) d$

$d = 5 m$

Part b)

Now the time till its speed is zero

$v_f - v_i = at$

$0 - 10^4 = -10^7 t$

$t = 10^{-3} s$

so total time that it will be in the region is given as

$T = 2 t$

$T = 2\times 10^{-3} s$