M & 2M are connected to an elastic spring (k) and placed on ground vertical in equilibrium. By how much M should be compressed so that 2M leaves contact after release?

Steps to Solve

1. Identify the Forces Acting on Mass 2M

When mass $2M$ is released from rest after being compressed against mass $M$, the forces acting on $2M$ are gravity ($2Mg$ acting downwards) and the spring force ($F_s$ acting upwards).

2. Applying Hooke's Law

The spring force can be described by Hooke's Law, which states that $F_s = kx$, where $k$ is the spring constant and $x$ is the compression. We need to express the condition for mass $2M$ to just lose contact with mass $M$.

3. Setting up the Equation for Forces

To determine when mass $2M$ loses contact, we need to set up the inequality that describes the balance of forces:

At the point of losing contact, $$ F_s = 2Mg $$

Substituting Hooke's law into this gives: $$ kx = 2Mg $$

4. Rearranging the Equation for Compression

Now we can solve for the compression $x$: $$ x = \frac{2Mg}{k} $$

5. Final Expression for Compression

This means the required compression of the spring to ensure mass $2M$ just loses contact with mass $M$ when released is given by: $$ x = \frac{2Mg}{k} $$