Topic: Dynamics
Concepts: Equivalence principle, Projectile motion

Solution:

In the accelerating frame of the box, we introduce the (fictitious) inertial force $-m\vec{a}$ on the particle. This has the same form as gravity so combining them results in an effective gravity $g_{\text{eff}}$ directed in the lower left direction.

The particle now behaves effectively as a projectile with gravity pointing in a new direction. Thus, the path is a rotated parabola so the answer is C.

Topic: Dynamics
Concepts: Equivalence principle, Projectile motion

Solution:

From the previous problem, we found that (in the box frame) the particle behaves as a projectile with gravity $g_{\text{eff}}$ pointing in the lower left direction.

For a projectile to return to its starting position, it must be launched in the opposite direction to gravity. The particle travels in a straight line to the peak “height” and then comes back down. Thus, the answer is A.