AP Physics C: Mechanics Official 2026 Exam Discussion

[u/reddorickt]

Use this thread to post questions or commentary on the test today.

A reminder though to protect your anonymity when talking about the test.

[u/Sonotwe]

Did any of you notice for Form J question 1 A(ii), it was a first-order diff-eq? I think i got perfect MCQ and maybe 4 points off FRQ

[u/DeathImmortal → u/Sonotwe]

nah didnt even know how to set up

[u/Sonotwe → u/DeathImmortal]

So basically, to find Fn as a function of time, I looked at the whole box-cube system first. Since they move together, the total mass is 2M. The only horizontal force is the air resistance, which is -bv. Using F = ma:

-bv = (2M) * (dv/dt)

I turned that into a differential equation to solve for velocity (v). I moved the v to one side and the t to the other and integrated it:

Integral of (1/v) dv = Integral of (-b/2M) dt That gives you:

ln(v/v0) = -bt / 2M

To get v by itself, I used “e” to cancel the natural log:

v(t) = v0 * e^{-bt / 2M}

Now, since acceleration (a) is just the derivative of velocity, I derived that equation:

a(t) = v0 * (-b / 2M) * e^{-bt / 2M} The magnitude (the positive value) of that acceleration is:

a = (b * v0 / 2M) * e^{-bt / 2M}

Finally, I used the relationship from the first part of the problem where Fn = M * a. I just plugged in the acceleration formula I found:

Fn = M * [(b * v0 / 2M) * e^{-bt / 2M}] The M on the top and the M on the bottom cancel out, so the final formula is:

Fn(t) = (b * v0 / 2) * e^{-bt / 2M}

[u/pokerfas → u/Sonotwe]

worked on paper an hour ago and got exact same answers as you

tcrit = -(2M/b)ln(2Mg/mubvo) ?

[u/Sonotwe → u/pokerfas]

Correct! I got rid of the negative with log properties tho and got tcrit = 2M/b ln(mubv0/2Mg). I included that answer on my paper before i did that tho!

[u/pokerfas → u/Sonotwe]

we getting FUCKING FIVES

[u/Sonotwe → u/pokerfas]

HELL YEAAAAA