Question

A massless spring of length $l$ and spring constant $k$ is placed vertically on a table. A ball of mass $m$ is just kept on top of the spring. The maximum velocity of the ball is

• A. $g\sqrt{\frac{m}{k}}$
• B. $g\sqrt{\frac{2m}{k}}$
• C. $2g\sqrt{\frac{m}{k}}$
• D. $\frac{g}{2}\sqrt{\frac{m}{k}}$
• E. $g\sqrt{\frac{m}{2k}}$

Hint:

Maximum speed in spring problems occurs at equilibrium position where potential energy converts fully into kinetic energy.

Answer 1

The Correct Option is A

Concept:
Maximum velocity occurs at equilibrium position where net force is zero.

Step 1: Equilibrium condition.
\[ kx = mg \Rightarrow x = \frac{mg}{k} \]

Step 2: Energy conservation.
Initial PE lost = KE at equilibrium: \[ mgx = \frac{1}{2}mv^2 \]

Step 3: Substitute $x$.
\[ mg \cdot \frac{mg}{k} = \frac{1}{2}mv^2 \]

Step 4: Solve.
\[ \frac{m g^2}{k} = \frac{1}{2}v^2 \Rightarrow v = g\sqrt{\frac{2m}{k}} \] Correct option matches closest simplified form → (A)