Question

Block $A$ & $B$ of mass $m$ each are connected with spring of constant $k$, both blocks lie on frictionless ground and are imparted horizontal velocity $v$ as shown. Initially spring is unstretched, find the maximum extension of the spring.

• A. $v\sqrt{\frac{m}{k}}$
• B. $2v\sqrt{\frac{m}{k}}$
• C. $3v\sqrt{\frac{m}{k}}$
• D. $0.5v\sqrt{\frac{m}{k}}$

Answer 1

The correct option is A $v\sqrt{\frac{m}{k}}$

Reduced mass of the system $\mu =\frac{m_1{m}_2}{m_1+m_2}$
$\Rightarrow \mu =\frac{m\times m}{m+m}$
$\Rightarrow \mu =\frac{m}{2}$

velocity can be resolved as,


Using Work Energy theorem, in COM reference frame
$K_i+U_i=K_f+U_f$
Initial kinetic energy
$K_i=\frac{1}{2}\mu V_{rel}^2=\frac{1}{2}\times \mu {(\frac{v}{\sqrt{2}}+\frac{v}{\sqrt{2}})}^2=\frac{1}{2}\times \frac{m}{2}\times 2v^2=\frac{1}{2}m{v}^2$
Initially spring is unstretched, so
$U_i=0$
Finally extension is maximum, so there should be no relative velocity between the blocks.
Hence, final kinetic energy
$K_f=0$
Final potential energy, let maximum extension be $x$
$U_f=\frac{1}{2}k{x}^2$

$K_i+U_i=K_f+U_t$
$\Rightarrow \frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2$
$\Rightarrow x=v\sqrt{\frac{m}{k}}$