Solution 1: Elastic Cord Projection
Dynamics of constrained particles
1. Analysis of the Mechanism
The system consists of a disc $A$ connected to a pivot $B$ by an inextensible elastic cord of length $l$. Initially, $A$ is at distance $l$ from $B$. When $A$ is projected, it travels in a straight line until the cord becomes taut.
The constraint effectively acts as a circular wall of radius $l$ centered at $B$. The disc undergoes an elastic reflection at point $P$ on this circle and must subsequently hit the trap $C$.
2. Geometric Construction
Let the projection angle be $\alpha$. Based on the geometry of the isosceles triangle $\Delta ABP$ (since $AB = BP = l$):
- Incident Path: $\angle PAB = \alpha$. Consequently, $\angle BPA = \alpha$.
- Pivot Angle: The exterior angle at the center $B$ is the sum of interior opposites: $\angle PBC = 2\alpha$.
- Reflection: By the law of reflection relative to the radius $BP$, the angle of reflection $\angle BPC = \alpha$.
3. Trigonometric Solution
Applying the Sine Rule to $\Delta PBC$:
The angle at $C$ is given by the sum of angles in the triangle:
$$ \angle PCB = 180^\circ – (2\alpha + \alpha) = 180^\circ – 3\alpha $$Using the Sine Rule with lengths $PB=l$ and $BC=\eta l$:
$$ \frac{PB}{\sin(180^\circ – 3\alpha)} = \frac{BC}{\sin(\alpha)} $$ $$ \frac{l}{\sin(3\alpha)} = \frac{\eta l}{\sin(\alpha)} $$4. Solving for $\alpha$
We expand $\sin(3\alpha)$ using the identity $\sin(3\alpha) = 3\sin\alpha – 4\sin^3\alpha$:
$$ \frac{1}{3\sin\alpha – 4\sin^3\alpha} = \frac{\eta}{\sin\alpha} $$Assuming $\alpha \neq 0$, we cancel $\sin\alpha$:
$$ \frac{1}{3 – 4\sin^2\alpha} = \eta \implies 1 = 3\eta – 4\eta\sin^2\alpha $$ $$ 4\eta\sin^2\alpha = 3\eta – 1 $$ $$ \sin\alpha = \sqrt{\frac{3\eta – 1}{4\eta}} $$5. Numerical Result
Given the parameter $\eta = 0.5$:
$$ \sin\alpha = \sqrt{\frac{3(0.5) – 1}{4(0.5)}} = \sqrt{\frac{0.5}{2.0}} = \sqrt{0.25} = 0.5 $$