Is the AAPT 2023 Physics Bowl Answer Key Incorrect?

[kuruman]

I was looking at the AAPT 2023 Physics Bowl questions (and answers) found here
https://aapt.org/programs/physicsbowl/currentexam.cfm
and multiple choice question 6 caught my eye:

Consider one horizontal circular path that a ball makes on the inside surface of a cone. In this situation, the normal force on the ball
a. is mg.
b. is always greater than mg.
c. may be greater or less than mg.
d. is always less than mg.
e. depends upon the speed of the ball.

The posted correct answer is B
The normal vector has a greater magnitude than its vertical component (magnitude of mg). The magnitude of the normal vector is also greater than the horizontal component, the net force directed to the center of circular motion.

Only one answer is allowed which implies that the rest are incorrect. Not so. Since the ball moves in a horizontal circle, the vertical component of the normal force must be equal to the weight and the horizontal component provides the centripetal acceleration.
Let = the half-angle of the cone. Then $$\begin{align} & N\sin\!\alpha=mg \nonumber \\ & N\cos\!\alpha =\frac{mv^2}{r} \nonumber \\ & \sqrt{N^2\sin^2\!\alpha+N^2\cos^2\!\alpha}=N=\sqrt{(mg)^2+\left(\frac{mv^2}{r}\right)^2}.
\nonumber \end{align}$$It looks like both B and E are correct answers. I expected better from the AAPT.

(Edited to fix the equations' format.)

 

[Vanadium 50]

Only one answer is allowed which implies that the rest are incorrect.

I believe the instructions say "Choose the best answer."

I also don't think the distractors are good. If you just gave me the answers and not the question, I could exclude C and E as likely possibilities.

 

[kuruman]

I believe the instructions say "Choose the best answer."

I didn't see this instruction anywhere here. Maybe I have the wrong document. In any case, I really don't see why the normal force "depends upon the speed of the ball" is not as good as "is always greater than the weight". The mathematical expression for the normal force says it all.

I agree that the distractors could be better. For example, replace E with

"The normal force is equal to $mg\cos\!\alpha$ where $\alpha$ is the cone half-angle."

Then the centripetal is component is cut out and the question is entirely focused on the comparison between the normal force and the weight.