Solution to Question 17
System Analysis:
The system consists of a light bar (almost inertia-less) suspended by springs and counterweights. A small disc ($0.01m$) is on the bar, and a heavy block ($1.99m$) hangs from it.
Step 1: Initial State (Equilibrium)
Before the cord is cut, the system is in equilibrium. The tension in the cords connected to the counterweights ($m$) is $T = mg$. This tension acts on the springs.
The total upward force provided by the two springs balances the total downward weight of the central assembly (Bar + Disc + Block).
$$ 2 \times F_{spring} = W_{total} $$
$$ 2(mg) = (0.01m + 1.99m)g $$
$$ 2mg = 2mg $$
This confirms the system is balanced. The extension $x_i$ in each spring (stiffness $k$) is determined by the tension $mg$:
$$ k x_i = mg \implies x_i = \frac{mg}{k} $$
Total elastic potential energy stored in the two springs is:
$$ U_{initial} = 2 \times \frac{1}{2} k x_i^2 = k \left( \frac{mg}{k} \right)^2 = \frac{m^2 g^2}{k} $$
Step 2: Dynamics After Cutting the Cord
When the cord holding the $1.99m$ block is cut, the load on the bar drops instantly from $2.00m$ to $0.01m$.
- Inertial approximation: The counterweights $m$ are 100 times heavier than the disc ($0.01m$). We assume they remain effectively stationary during the initial rapid acceleration of the disc.
- Energy Transfer: The potential energy stored in the springs is released and converted into the kinetic energy of the light disc. Since the bar is inertia-less and the disc is very light, this energy transfer happens almost instantaneously as the springs relax to their natural length.
Let $v$ be the velocity of the disc immediately after the springs relax.
$$ \text{Initial Elastic PE} = \text{Kinetic Energy of Disc} $$
$$ \frac{m^2 g^2}{k} = \frac{1}{2} (0.01m) v^2 $$
$$ v^2 = \frac{2 m^2 g^2}{k (0.01m)} = \frac{200 m g^2}{k} $$
Step 3: Maximum Height
The disc now acts as a projectile moving upward against gravity with initial velocity $v$. The maximum height $h$ attained is:
$$ h = \frac{v^2}{2g} $$
Substituting $v^2$:
$$ h = \frac{1}{2g} \left( \frac{200 m g^2}{k} \right) $$
$$ h = \frac{100 m g}{k} $$