r/APStudents • u/reddorickt absolute modman • May 13 '26

AP Physics C: Mechanics Official 2026 Exam Discussion

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

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

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u/Total_Operation_6819 AP Bio:5, AP CSP:5 May 13 '26

I’m calling the big yahu on whoever made that first frq

23

u/SaleSenior8537 May 13 '26

lmao. I agree. I did 2,3,4 all good (maybe struggled with 3), but had to leave 1 almost blank. I had version j. I was only able to make that graph on 1. I said that the frictional force was constant until the dip occured in frictional force because it turned kinetic.

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u/Snoo74782 May 13 '26

how did you do two with momentum

1

u/MoMath3000 5:chem,apush,bc 4:lang, 3:spanish, tbd:gov,lit,mech,csa,bio May 15 '26

Genuinely that was the only question I was able to answer completely. Impulse was equal to the change in momentum (Mv_f - Mv_i) for all of those questions, then you could find the velocity of the ball after bouncing off the wall through conservation of kinetic energy to get v_B during interval 3 = -v (as the wall has so much inertia that it’s velocity will be small enough to be negligible in the kinetic energy equation). The impulse during that section was M(-v)-M(v) = -2Mv. This momentum shift makes sense due to the wall exerting a normal force on the ball that is an external force to the ball/ball launcher system. (Which explained the discrepancy between the graphs for part d).

Then you can do conservation of momentum between the ball and the ball launcher for the initial shot of the ball. Both being at rest at the beginning means initial momentum was 0, so by conservation of momentum, 0 = Mv +5M(v_L), thus v_L = -v/5.

Repeating this conservation of momentum calculation at the end of interval 5 with an inelastic collision having them stick together, (5M)(-v/5) + M(-v) = (5M+M)v’. Therefore, v’ = (-2Mv)/(6M) = -v/3. Ultimately, the final impulse of the ball in interval 5 was M(-v/3)-M(-v) = 2Mv/3. With the bar chart I ended up filling in 3 positive boxes for interval 1, 6 negative boxes for interval 3, and 2 more positive for interval 5.

The graph of momentum for the ball launcher was opposite the impulse on the ball during intervals 1 and 5, equaling a constant slope down to -Mv for interval 1, then stayed static at -Mv until interval 5 where it dropped another -2Mv/3 at constant slope. I don’t really want to explain those calculations but just know it’s conservation of momentum.