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Constant Acceleration Problem

  • Thread starter docbrown
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  • #1
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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 travelled (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?
 

Answers and Replies

  • #2
gneill
Mentor
20,793
2,773
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
2
0
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!
 

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