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. $\frac v2$
• C. $\frac v4$
• D. $\frac {v}{\sqrt 2}$

Answer 1

Answer: (B)

$\frac v2$

Answer 2

∵ Loss in Kinetic Energy = Gain in spring energy
$⇒\frac 12mv^2×2=\frac 12kx_m^2$
$⇒2×\frac 14×v^2=2×x_m^2$
$⇒x_m=\sqrt {\frac {v^2}{4}}$
$⇒x_m=\frac v2$

So, the correct option is (B): $\frac v2$