1. Impact Analysis

When the front cylinder collides with the slope, the collision is “perfectly inelastic” with the slope surface. This implies the velocity component perpendicular to the slope becomes zero (or the particle is constrained to move along the slope). Crucially, the problem states that the total momentum component along the inclined plane is conserved during the impact.

Let $u$ be the initial velocity. Just after the front cylinder hits the slope:

• The front cylinder (mass $m$) moves along the incline with velocity $v$.
• The rear cylinder (mass $m$) is still on the horizontal floor. Due to the rigid rod, its velocity component along the rod must equal the front cylinder’s velocity component along the rod.

Geometric Constraint:
Velocity of front cylinder: $v$ (at angle $\theta$). Component along rod (horizontal): $v \cos \theta$.
Velocity of rear cylinder: $v’$ (horizontal). Component along rod: $v’$.
Therefore, $v’ = v \cos \theta$.

2. Conservation of Momentum

We apply conservation of momentum along the direction of the incline ($\hat{e}_{\parallel}$).

Initial Momentum ($P_i$):
Both cylinders move horizontally with $u$. $$ P_i = m(u \cos \theta) + m(u \cos \theta) = 2mu \cos \theta $$

Final Momentum ($P_f$):
Front cylinder moves along incline: $p_1 = mv$.
Rear cylinder moves horizontally: $p_2 = m v’ \cos \theta$ (projection of horizontal velocity onto incline). $$ P_f = mv + m(v \cos \theta) \cos \theta = mv (1 + \cos^2 \theta) $$

Equating $P_i = P_f$: $$ 2mu \cos \theta = mv (1 + \cos^2 \theta) $$ $$ v = \frac{2u \cos \theta}{1 + \cos^2 \theta} $$

3. Energy Condition for Rising

For the toy to “rise completely,” it must possess enough kinetic energy immediately after the first impact to lift its center of mass until the rear cylinder also reaches the slope.

Kinetic Energy after impact ($K_f$): $$ K_f = \frac{1}{2} m v^2 + \frac{1}{2} m (v’)^2 $$ $$ K_f = \frac{1}{2} m v^2 + \frac{1}{2} m (v \cos \theta)^2 = \frac{1}{2} m v^2 (1 + \cos^2 \theta) $$ Substituting $v$: $$ K_f = \frac{1}{2} m \left( \frac{2u \cos \theta}{1 + \cos^2 \theta} \right)^2 (1 + \cos^2 \theta) = \frac{2m u^2 \cos^2 \theta}{1 + \cos^2 \theta} $$

Potential Energy Increase ($\Delta U$):
The critical position is when the toy is fully on the slope. The center of mass rises from $h=0$ to $h_{cm} = \frac{l \sin \theta}{2}$. $$ \Delta U = (2m) g \left( \frac{l \sin \theta}{2} \right) = mgl \sin \theta $$

Condition: $K_f \ge \Delta U$ $$ \frac{2m u^2 \cos^2 \theta}{1 + \cos^2 \theta} \ge mgl \sin \theta $$ $$ 2u^2 \cos^2 \theta \ge gl \sin \theta (1 + \cos^2 \theta) $$ $$ u \ge \sqrt{\frac{gl \sin \theta (1 + \cos^2 \theta)}{2 \cos^2 \theta}} $$