Calculating Initial Velocity of an Electron with Constant Acceleration

In summary, an electron is accelerated by a positive charge and then travels 45mm in 35microseconds. The initial velocity is approximated to be 1290 ms-1.
  • #1
docbrown
2
0

Homework Statement



I am about to start AS Level Physics and have been given some questions/problems to solve as introductory homework. However, I am stuck on one and cannot seem to get the answer specified as the correct one.

An electron of mass 9.11x10-31kg is accelerated by a positive charge such that the electron experiences a constant force of 8.5x10-29N. The electron, which was already moving, then travels 45mm in 35microseconds. By first determining the acceleration of the electron, calculate its initial velocity.

Homework Equations

F = ma
s = ut + 1/2at2

F = Force (N)
m = mass (kg)
a = acceleration (ms-2)
s = distance traveled (m)
u = initial velocity (ms-1)
t = time (s)

The Attempt at a Solution



Converting Units

45mm = 0.045m
35microseconds = 0.000035s


Working Out Acceleration

F =ma
8.5x10-29 = 9.11x10-31 x a
8.5x10-29 / 9.11x10-31 = a
a = 93 + 277/911

Check: 9.11x10-31 x (93 + 277/911) = 8.5x10-29

Working Out Initial Velocity

Rearranging the Equation:

s = ut + 1/2at2
ut = s - 1/2at2
u = (s-1/2at2) / t

Putting in the Values

u = (0.045 - 1/2(93+277/911) x 0.0000352) / 0.000035
u = (0.045 - (1.142974755x10-7 x 1/2)) / 0.000035
u = (0.045 - 5.714873765x10-8) / 0.000035
u = 0.04499994285 / 0.000035
u = 1285.712653

According to the booklet, the answer should be 4290ms-1 and not the value I got.
I have been working at this for hours and cannot seem to find where I am going wrong. Help me out?
 
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  • #2
Your work looks fine. Could be the booklet has a misprint: 4290 → 1290.

With the given force the acceleration is on the order of 100m/s2. Over 35 microseconds that can only change the speed by a few thousandths of a meter per second (a*t). It won't significantly affect the initial velocity over that time period so a good approximation for the initial velocity is just Δs/Δt which is about 1290 m/s
 
  • #3
gneill said:
Your work looks fine. Could be the booklet has a misprint: 4290 → 1290.

That is definitely a possibility gneill. I never tried rounding the numbers so never spotted how similar they actually could be. Thanks!
 

Related to Calculating Initial Velocity of an Electron with Constant Acceleration

1. What is constant acceleration?

Constant acceleration refers to the situation in which an object's velocity changes at a constant rate over a period of time. This means that the object's acceleration remains the same throughout its motion.

2. How do you calculate constant acceleration?

Constant acceleration can be calculated by dividing the change in velocity by the change in time. This is represented by the equation a = (vf - vi) / t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval.

3. What is the difference between constant acceleration and uniform motion?

Constant acceleration refers to a situation in which an object's velocity changes at a constant rate, while uniform motion refers to a situation in which an object's velocity remains constant. In other words, constant acceleration involves a change in velocity, while uniform motion does not.

4. How does air resistance affect constant acceleration?

Air resistance can affect constant acceleration in several ways. In some cases, air resistance may cause an object to experience a change in acceleration, as the force of air resistance acts against the object's motion. In other cases, air resistance may have a negligible effect on an object's acceleration, depending on the object's mass and the strength of the force of air resistance.

5. What real-life examples demonstrate constant acceleration?

Some real-life examples of constant acceleration include a ball rolling down a ramp, a car accelerating from a stop sign, or a person jumping off a diving board. In all of these cases, the object's velocity changes at a constant rate, and therefore experiences constant acceleration.

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