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kg.m.s-2
The speed of light does not accelerate. How can you compare any acceleration to the speed of light?
Ratch
I know it's hard, but please try to use the same notation that chopnhack is required to use.
Do you mean use the milli-second abbreviation ms for meter-sec? No way, I am sticking to standard notation.
Ratch
LoL - you can use what you like. I can convert (most of the times). I believe the reason he taught us this way has to do with algebraic handling of the units. Perhaps its how they were taught. In any event, no worries.
Correct, which is what I had stated in the qualitative description in post 1.
It's only been done that way for at least the last 40 years, and all you ever see are units so m is always metres not milli.
If you want to confuse things I am happy to remove your posts.
So what must the final velocity be?
I'm talking about the velocity when the particle reaches the starting point again.
In my mind there are two final velocities - the first is when the particle reaches zero velocity, when the field arrests its initial velocity.
The second vf would be the field's acceleration on the proton times time. To get time we would need distance. To get distance traveled we would need to know how long ti takes to stop the particle after it enters the field.
To use Vf=Vi+at to solve for t I would need to know acceleration. To get acceleration I thought I could use F=ma, but I got that large acceleration that everyone is hinting at being wrong...
True, but it shows a lack of understanding.
We'll take the longer route so you can see that the final answer is really intuitive and trivially easy.
Do you have a formula which involves 2 velocities (typically u and v - for initial and final velocity), acceleration, and time?
Two semesters in less than 12 weeks will do that....
The only one that I know of is Vf = Vi + a*t (that reads final velocity equal to initial velocity plus acceleration multiplied by time).
Yeah, that's commonly also written as
v = u + at
But I'll stick with your formula.
Can you tell me what Vi, Vf, and a are (the sign of a is important)?
And then can you calculate t?
At this point you are one simple step from your answer, but unless you can spot it, we'll take a detour through some more equations.
v = u + at
But I'll stick with your formula.
Can you tell me what Vi, Vf, and a are (the sign of a is important)?
And then can you calculate t?
At this point you are one simple step from your answer, but unless you can spot it, we'll take a detour through some more equations.
For the first leg of the trip:
Vi = the given 6.2x10^5 ms-1
Vf = 0
a = -5.95x10^12 ms-2
-6.2x10^5 ms-1 = -5.95x10^12 ms-2 * t
divide by -5.95x10^12 ms-2
t = 1.04x10-7 sec to stop particle in the field
Now the particle is repelled in the opposite direction.
Since we have time we should calculate distance traveled.
d = Vi*t + 1/2a t^2
d = 0.06448m -0.03217
d = 0.032m
We can now calculate the second Vf:
Vf^2 = Vi^2 + 2ax
Vf^2 = 0 - 2*5.95x10^12 ms-2*0.032m
Vf = -6.17x10^5
Time to travel the 0.032m = 5.1x10^-8 seconds
Total time = 1.55x10^-7
Vi = the given 6.2x10^5 ms-1
Vf = 0
a = -5.95x10^12 ms-2
-6.2x10^5 ms-1 = -5.95x10^12 ms-2 * t
divide by -5.95x10^12 ms-2
t = 1.04x10-7 sec to stop particle in the field
Now the particle is repelled in the opposite direction.
Since we have time we should calculate distance traveled.
d = Vi*t + 1/2a t^2
d = 0.06448m -0.03217
d = 0.032m
We can now calculate the second Vf:
Vf^2 = Vi^2 + 2ax
Vf^2 = 0 - 2*5.95x10^12 ms-2*0.032m
Vf = -6.17x10^5
Time to travel the 0.032m = 5.1x10^-8 seconds
Total time = 1.55x10^-7
OK, now remember this value
You have lost some precision in your calculations. The result should be -6.2*105ms-1.
Note this velocity is precisely the same magnitude as the initial velocity, but in the opposite direction.
This should make sense. The particle has been subject to a constant force, resulting in a constant acceleration. So the trip back to the starting point is *exactly* the mirror image of the trip to the point at which its velocity was zero.
To do that, the velocity would have to be 6.27x105ms-1 but that is the FINAL velocity, the proton did not travel at that velocity the whole time (we know that it started from rest) .
There are many formulae that you could use, but using one you've used before:
d = Vi.t + ½at²
(I would call that s = ut + ½at² -- s is displacement, it appears as d in your formulae)
-0.032 = 0.t + ½(-5.95x1012)t²
t = ?
Compare that to the value I suggested you remember above.
This should lead to a very simple step after the first calculation to yield the answer.
It would also probably be best to retain a few more significant digits in your intermediate values.
It makes sense after the fact, but its not intuitive. From the outset of the problem I would not have guessed that the time back would have been equal to the time to travel the first leg of the trip. The second leg's time would depend completely on the acceleration. If the field was stronger, the time back would have been shorter and vice versa. So, no, I don't see from the outset how I would have changed the direction I took.
For the first leg of the trip:
Vi = the given 6.2x10^5 ms-1
Vf = 0
a = -5.95x10^12 ms-2
-6.2x10^5 ms-1 = -5.95x10^12 ms-2 * t
divide by -5.95x10^12 ms-2
t = 1.04x10-7 sec to stop particle in the field
Now the particle is repelled in the opposite direction.
Since we have time we should calculate distance traveled.
d = Vi*t + 1/2a t^2
d = 0.06448m -0.03217
d = 0.032m
We can now calculate the second Vf:
Vf^2 = Vi^2 + 2ax
Vf^2 = 0 - 2*5.95x10^12 ms-2*0.032m
Vf = -6.2x10^5
At this point I can use a different equation - d = [(Vi+Vf)/2]*t to solve for t.
0.03231m = [(0-6.2x10^5)/2] * t
t = 1.04x10-7 sec
Add the two values of t - both legs of the trip = total t = 2.08x10-7 sec
It's not intuitive until you see it happen.
However the next time you see a problem like this, you should be able to take the short cut.
It's really useful to be doing calculus at the same time as learning this stuff because many of these equations are simple derivatives or integrals with (respect to time) of each other.
There first derivative of displacement is velocity and the second is acceleration.
However the next time you see a problem like this, you should be able to take the short cut.
It's really useful to be doing calculus at the same time as learning this stuff because many of these equations are simple derivatives or integrals with (respect to time) of each other.
There first derivative of displacement is velocity and the second is acceleration.
Yes, but being that I took calculus so long ago, and did well in both sections they wouldnt let me take it again. I have to audit the courses at my own expense which I can not do until this fall. Luckily I was able to work through some derivations and integrals, simple stuff, but when I see a problem I usually work through it using kinematics and algebra. I occasionally recognize some uses of calculus and got an A in physics 1, but this go around I am at a B and sinking...
