Solution
System Analysis:
Consider the system comprising Slide A, Slide B, and the small disc C. The floor is frictionless, and the surfaces of the slides are frictionless. Since there are no external horizontal forces acting on the system, the total linear momentum along the horizontal axis is conserved.
Consider the system comprising Slide A, Slide B, and the small disc C. The floor is frictionless, and the surfaces of the slides are frictionless. Since there are no external horizontal forces acting on the system, the total linear momentum along the horizontal axis is conserved.
Momentum Conservation:
Initially, the entire system is at rest, so the initial momentum is zero: $\vec{P}_{initial} = 0$.
Let the final velocities of slide A and slide B be $\vec{v}_A$ and $\vec{v}_B$ respectively. Due to the symmetry of the setup (identical masses $M$ and identical shapes) and the nature of the interaction (the disc bouncing between them), the slides will acquire equal and opposite velocities relative to the center of mass. Let the final speed of each slide be $v$. We can write the velocities using the unit vector $\hat{i}$ (pointing to the right): $$ \vec{v}_A = -v \hat{i} \quad \text{and} \quad \vec{v}_B = v \hat{i} $$ The momentum equation becomes: $$ M(-v) + M(v) + m(v_{disc}) = 0 $$ Since $m \ll M$, the momentum contribution of the disc is negligible compared to the massive slides, satisfying the condition $0=0$.
Initially, the entire system is at rest, so the initial momentum is zero: $\vec{P}_{initial} = 0$.
Let the final velocities of slide A and slide B be $\vec{v}_A$ and $\vec{v}_B$ respectively. Due to the symmetry of the setup (identical masses $M$ and identical shapes) and the nature of the interaction (the disc bouncing between them), the slides will acquire equal and opposite velocities relative to the center of mass. Let the final speed of each slide be $v$. We can write the velocities using the unit vector $\hat{i}$ (pointing to the right): $$ \vec{v}_A = -v \hat{i} \quad \text{and} \quad \vec{v}_B = v \hat{i} $$ The momentum equation becomes: $$ M(-v) + M(v) + m(v_{disc}) = 0 $$ Since $m \ll M$, the momentum contribution of the disc is negligible compared to the massive slides, satisfying the condition $0=0$.
Energy Conservation:
Since all surfaces are frictionless, mechanical energy is conserved. The initial energy of the system is purely the potential energy of the disc at height $h$: $$ E_{initial} = mgh $$ The disc slides down A, transferring momentum to it, travels to B, ascends (transferring momentum), and descends again. This process repeats. Because $m \ll M$, the disc effectively acts as a medium transferring potential energy into the kinetic energy of the two massive slides. When the interaction effectively ceases (or in the limit of many interactions), the light disc’s kinetic energy becomes negligible compared to the kinetic energy accumulated by the massive slides. Therefore, the final energy is the sum of the kinetic energies of the two slides: $$ E_{final} \approx \frac{1}{2}M|\vec{v}_A|^2 + \frac{1}{2}M|\vec{v}_B|^2 $$ $$ E_{final} = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2 $$
Since all surfaces are frictionless, mechanical energy is conserved. The initial energy of the system is purely the potential energy of the disc at height $h$: $$ E_{initial} = mgh $$ The disc slides down A, transferring momentum to it, travels to B, ascends (transferring momentum), and descends again. This process repeats. Because $m \ll M$, the disc effectively acts as a medium transferring potential energy into the kinetic energy of the two massive slides. When the interaction effectively ceases (or in the limit of many interactions), the light disc’s kinetic energy becomes negligible compared to the kinetic energy accumulated by the massive slides. Therefore, the final energy is the sum of the kinetic energies of the two slides: $$ E_{final} \approx \frac{1}{2}M|\vec{v}_A|^2 + \frac{1}{2}M|\vec{v}_B|^2 $$ $$ E_{final} = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2 $$
Calculation:
Equating initial and final energy: $$ mgh = Mv^2 $$ Solving for the speed $v$: $$ v^2 = \frac{mgh}{M} $$ $$ v = \sqrt{\frac{mgh}{M}} $$
Equating initial and final energy: $$ mgh = Mv^2 $$ Solving for the speed $v$: $$ v^2 = \frac{mgh}{M} $$ $$ v = \sqrt{\frac{mgh}{M}} $$
Final Answer:
The speed eventually acquired by both the slides is: $$ v \approx \sqrt{\frac{mgh}{M}} $$
The speed eventually acquired by both the slides is: $$ v \approx \sqrt{\frac{mgh}{M}} $$