Solution to Question 6
Method of Unfolding:
The motion of a particle undergoing perfectly elastic collisions inside a cone can be analyzed by “unfolding” the cone’s surface into a 2D plane sector. In this unfolded frame, the reflection of the particle corresponds to the particle continuing in a straight line through the virtual image of the cone.
Let the apex of the cone $O$ be the origin. The particle starts at point $P$ at distance $x = 10 \text{ cm}$ from $O$ on the axis. The particle is projected at an angle $\alpha = 37^\circ$ with the axis. In the unfolded view, this path is a straight line.
Fig 1: Unfolded trajectory of the particle shown as a straight line in the first quadrant relative to P.
(a) Distance of Closest Approach
In the unfolded plane, the particle travels in a straight line starting from $P$. The distance of closest approach is the length of the perpendicular dropped from the apex $O$ to this line of motion.
From the geometry of the right-angled triangle formed by the origin $O$, the starting point $P$, and the point of closest approach: $$ d_{\text{min}} = OP \sin \alpha $$
Given: $$ OP = x = 10 \text{ cm} $$ $$ \alpha = 37^\circ $$
Substituting the values: $$ d_{\text{min}} = 10 \sin(37^\circ) $$ Using $\sin(37^\circ) \approx 0.6$: $$ d_{\text{min}} = 10 \times 0.6 = 6.0 \text{ cm} $$
(b) Number of Collisions
The number of collisions corresponds to how many “cone sectors” the particle traverses in the unfolded plane. This is given by the integer part of the total angular span available minus the initial offset, normalized by the sector angle: $$ n = \text{Integer Part of } \left[ \frac{360^\circ – 2\alpha + \beta}{2\beta} \right] $$
Substituting the values $\alpha = 37^\circ$ and $\beta = 20^\circ$: $$ n = \frac{360 – 2(37) + 20}{2(20)} $$ $$ n = \frac{360 – 74 + 20}{40} $$ $$ n = \frac{306}{40} $$ $$ n = 7.65 $$
Taking the integer part, the number of collisions is 7.