Solution: Ball Colliding with Spring-Platform
The ball of mass \( m \) falls from height \( h \) and hits the platform. The collision is elastic, and since the platform is light (massless), the ball effectively attaches to the spring and compresses it until it stops, then bounces back. The entire motion while in contact is Simple Harmonic Motion.
Equilibrium Position: Where net force is zero: \( mg = k x_{eq} \implies x_{eq} = mg/k \). Impact Position: The spring is relaxed (x=0). This is at distance \( d = mg/k \) above the equilibrium position. Velocity at Impact: \( v_0 = \sqrt{2gh} \).
Using the velocity-position relation for SHM \( v^2 = \omega^2 (A^2 – y^2) \): At impact, displacement from equilibrium is \( y = mg/k \). $$ 2gh = \frac{k}{m} \left( A^2 – \left(\frac{mg}{k}\right)^2 \right) $$ $$ A^2 = \frac{2mgh}{k} + \frac{m^2 g^2}{k^2} $$
Max Velocity (\( v_{max} = \omega A \)): $$ v_{max}^2 = \frac{k}{m} A^2 = \frac{k}{m} \left( \frac{2mgh}{k} + \frac{m^2 g^2}{k^2} \right) = 2gh + \frac{mg^2}{k} $$ $$ v_{max} = \sqrt{g \left( 2h + \frac{mg}{k} \right)} $$
Max Acceleration (\( a_{max} = \omega^2 A \)): $$ a_{max} = \frac{k}{m} A = \frac{k}{m} \sqrt{\frac{m^2 g^2}{k^2} + \frac{2mgh}{k}} = g \sqrt{1 + \frac{2kh}{mg}} $$