Question

Two masses are connected by a spring as shown in the figure. One of the masses was given velocity $v=2k$ as shown in figure where $k$ is the spring constant. Then maximum extension in the spring will be? (initially spring is in natural length)

• A. $2m$
• B. $m$
• C. $\sqrt{2mk}$
• D. $\sqrt{3mk}$

Answer 1

Answer: (C)

$\sqrt{2mk}$

The maximum elongation of the spring will be attained when the velocity of both blocks will attain the velocity of the center of mass.

Let, $x$ be the elongation of the spring.

Velocity of center of mass= $\frac{m_{2}v+m_{1}0}{m_1+m_2}=\frac{m_{2}v}{m_1+m_2}$

Change in kinetic energy= Potential energy stored in the spring.

$\frac{1}{2}{m}_2{v}^2-\frac{1}{2}(m_1+m_2){\left\{\frac{m_{2}v}{m_1+m_2}\right\}}^2=\frac{1}{2}k{x}^2$

$m_2{v}^2[1-\frac{m_2}{m_1+m_2}]=k{x}^2$

$x=v\sqrt{\frac{m_1{m}_2}{k(m_1+m_2)}}$---------- (1)

Here $m_1=m_2=m$

and v=2k

Putting the value of m and v in (1), x=$\sqrt{\frac{m}{2k}}.2k=\sqrt{2mk}$

So, the correct answer is 'C'.