An electron is projected horizontally at a speed of 1.2x10^7 m/s into an electric field that exerts a constant vertical force of 4.5x10^-16 N on it. The mass of the electron is 9.11x10^-31 kg. Determine the vertical distance the electron is deflected during the time it has moved forward 33 mm horizontally.
If the answer is 1.9 mm, how would I get there?
~thank you.
If the answer is 1.9 mm, how would I get there?
~thank you.
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Let initial velocity = v, force = f, mass = m, distance = d (note that distance is 0.033m, dont use 33)
Equations of motion are:
y'' = f/m (since vertical acceleration is given by newton's law f=ma)
y' = (f/m)t (this is vertical velocity)
y = ((f/m)t^2)/2 (vertical displacement)
x''= 0 (since no horizontal accel.)
x' = v (horizontal velocity)
x = vt (horizontal displacement)
Now the question asks us to find y when x = d. Simply sub x = d into the horizontal displacement equation, solve for t, and sub t into our vertical displacement equation.
d = vt
t = d/v
sub
y = ((f/m)*(d/v)^2)/2 = 1.87 * 10^(-3) meters or 1.87 mm. (answers to 2 decimal places)
Don't forget to choose this as best answer! Hope it helped!
Equations of motion are:
y'' = f/m (since vertical acceleration is given by newton's law f=ma)
y' = (f/m)t (this is vertical velocity)
y = ((f/m)t^2)/2 (vertical displacement)
x''= 0 (since no horizontal accel.)
x' = v (horizontal velocity)
x = vt (horizontal displacement)
Now the question asks us to find y when x = d. Simply sub x = d into the horizontal displacement equation, solve for t, and sub t into our vertical displacement equation.
d = vt
t = d/v
sub
y = ((f/m)*(d/v)^2)/2 = 1.87 * 10^(-3) meters or 1.87 mm. (answers to 2 decimal places)
Don't forget to choose this as best answer! Hope it helped!