Solution to Question 13

Physics Solution: Projectile with Multiple Collisions

1. Analyzing the Path

The particle starts at the center, undergoes 3 collisions (Wall → Ceiling → Wall), and returns to the center. Since collisions are elastic, the magnitude of horizontal and vertical velocity components remains unchanged (directions reverse).

Horizontal Motion: We can “unfold” the path into a straight line.

Center to First Wall: distance $ r $
Wall to Second Wall (across diameter): distance $ 2r $
Second Wall to Center: distance $ r $

Total horizontal distance covered $ D = r + 2r + r = 4r $.

The horizontal speed is $ u_x = u \cos \theta $. The total time is $ T $.

$$ u \cos \theta \times T = 4r \implies u = \frac{4r}{T \cos \theta} $$

This matches option (b).

2. Analyzing the Ceiling Collision Condition

The problem states the particle hits the ceiling. This implies the trajectory is interrupted. If the ceiling were not there, a projectile with launch speed $ u $ and angle $ \theta $ would stay in the air for time $ T_0 = \frac{2u \sin \theta}{g} $.

Because the particle hits the ceiling and bounces down, its vertical path is effectively shortened (the top of the parabola is “chopped off”). Therefore, the actual time of flight $ T $ is less than the theoretical time of flight $ T_0 $.

$$ T < \frac{2u \sin \theta}{g} $$

Substitute $ T = \frac{4r}{u \cos \theta} $ into this inequality:

$$ \frac{4r}{u \cos \theta} < \frac{2u \sin \theta}{g} $$ $$ 4rg < 2 u^2 \sin \theta \cos \theta $$ $$ 4rg < u^2 \sin 2\theta $$ $$ u^2 > \frac{4rg}{\sin 2\theta} \implies u > 2 \sqrt{\frac{gr}{\sin 2\theta}} $$

This matches option (d).

Final Answer

The correct options are (b) and (d).