Solution to Question 29
1. Initial State (Before Impact)
The system (Container $M$ + Gas $m$) falls from height $h_0 = 12.1 \, \text{m}$. Speed just before impact: $$ v_0 = \sqrt{2gh_0} $$ Both $M$ and $m$ are moving downwards with $v_0$.
2. The Impact (Key Logic)
The problem states the collision of the container with the ground is perfectly elastic. The gas is suspended inside.
- Container ($M$): Hits ground and rebounds elastically. Velocity changes from $v_0$ (down) to $v_0$ (up).
- Gas ($m$): Does not hit the ground. Immediately after the container bounces, the gas is still moving $v_0$ (down).
3. Center of Mass Velocity
Just after impact, we have a system with two parts moving in opposite directions.
- $v_M = +v_0$ (up)
- $v_m = -v_0$ (down)
4. Energy Dissipation & Final Height
The problem states “internal fluctuations… damp out”. This means the relative motion between gas and container dissipates into heat. The only energy remaining to lift the system against gravity is the Kinetic Energy of the Center of Mass. $$ K_{CM} = \frac{1}{2} (M+m) v_{CM}^2 $$ This energy converts to Potential Energy $(M+m)gh$ at the peak height $h$. $$ (M+m)gh = \frac{1}{2} (M+m) v_{CM}^2 $$ $$ h = \frac{v_{CM}^2}{2g} $$
5. Final Calculation
Substitute $v_{CM}$: $$ h = \frac{1}{2g} \left[ v_0 \left( \frac{M-m}{M+m} \right) \right]^2 $$ Since $v_0^2 = 2gh_0$: $$ h = \frac{2gh_0}{2g} \left( \frac{M-m}{M+m} \right)^2 = h_0 \left( \frac{M-m}{M+m} \right)^2 $$
Plugging in the numbers: $M = 10$, $m = 1$, $h_0 = 12.1$. $$ h = 12.1 \times \left( \frac{10-1}{10+1} \right)^2 $$ $$ h = 12.1 \times \left( \frac{9}{11} \right)^2 $$ $$ h = 12.1 \times \frac{81}{121} $$ Note that $12.1 = \frac{121}{10}$. $$ h = \frac{121}{10} \times \frac{81}{121} = \frac{81}{10} = 8.1 \, \text{m} $$