1. Objective

To create a sound bang of maximum loudness, Block B must collide with Block C with the maximum possible velocity.

2. Dynamics of A and B

When A (velocity \(u\)) hits the spring attached to B (at rest), the system of A and B starts moving. Center of Mass Velocity (\(V_{cm}\)): \[ V_{cm} = \frac{mu + m(0)}{2m} = \frac{u}{2} \] Block B oscillates relative to the Center of Mass. Its velocity in the lab frame is: \[ v_B = V_{cm} + v_{rel} \] \(v_B\) is maximum when \(v_{rel}\) is maximum and in the direction of motion. This happens when the spring returns to its natural length (fully relaxed) after the first compression. At this point, \(v_{B, max} = u\).

3. Timing

This maximum velocity occurs after exactly half a time period of the oscillation (\(T/2\)). Reduced mass \(\mu = m/2\). Period \(T = 2\pi \sqrt{\frac{\mu}{k}} = 2\pi \sqrt{\frac{m}{2k}}\). Time to collision: \(t = T/2 = \pi \sqrt{\frac{m}{2k}}\).

4. Length of Cord

During this time \(t\), the Center of Mass has traveled a distance \(d_{cm} = V_{cm} \times t\). Since B starts at the COM and ends at the COM relative position (in the relaxed state), the distance B travels is determined by the COM shift + relative shift? Actually simpler: B moves from velocity 0 to \(u\). The distance covered is \(d = \frac{u}{2} \times t\) (average velocity of A+B system). \[ L = \frac{u}{2} \times \pi \sqrt{\frac{m}{2k}} = \frac{\pi u}{2} \sqrt{\frac{m}{2k}} \]

Answer: \(\frac{\pi u}{2} \sqrt{\frac{m}{2k}}\)