Question

As per the given figure, two blocks each of mass $250 \,g$ are connected to a spring of spring constant $2 \,Nm ^{-1}$ If both are given velocity $v$ in opposite directions, then maximum elongation of the spring is :

• A. $\frac{ V }{2 \sqrt{2}}$
• B. $\frac{v}{2}$
• C. $\frac{v}{4}$
• D. $\frac{ v }{\sqrt{2}}$

Answer 1

The Correct Option is B

To determine the maximum elongation of the spring, we use the principle of conservation of energy. Initially, the blocks have only kinetic energy, and at maximum elongation, all this kinetic energy is converted into potential energy stored in the spring.

Given:

• Mass of each block, \(m = 250 \, \text{g} = 0.25 \, \text{kg}\)
• Spring constant, \(k = 2 \, \text{Nm}^{-1}\)
• Velocity of each block, \(v\) (in opposite directions) 

Initial Kinetic Energy:

The total initial kinetic energy of the system is the sum of the kinetic energies of both blocks:

\(K_{initial} = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2\)

Potential Energy in the Spring at Maximum Elongation:

The potential energy stored in the spring when it is elongated by a length \( x \) is:

\(U = \frac{1}{2}kx^2\)

Using Conservation of Energy:

At maximum elongation, all kinetic energy is converted to potential energy:

\(mv^2 = \frac{1}{2}kx^2\)

Solve for \(x\):

\(x^2 = \frac{2mv^2}{k}\)

\(x = \sqrt{\frac{2mv^2}{k}}\)

Substitute the values:

\(x = \sqrt{\frac{2 \times 0.25 \times v^2}{2}}\)

\(= \sqrt{0.25 \times v^2}\)

\(= \frac{v}{2}\)

Conclusion: The maximum elongation of the spring is \(\frac{v}{2}\).

The correct answer is:

\(\frac{v}{2}\)