Question

A $2$ Kg block moving with $10$m/s strikes a spring of constant ${\pi }^{2}N/m$ attached to 2 Kg block of rest kept on a smooth floor. The time for which rear moving block remain in contact with spring will be

• A. $\sqrt{2}sec$
• B. $\frac{1}{\sqrt{2}}sec$
• C. $1sec$
• D. $\frac{1}{2}$ sec

Answer 1

The correct option is A $\sqrt{2}sec$

$\text{Given:-}$ Mass of first block = 2 Kg

Velocity of first block = 10 m/s

Mass of second block attached to spring = 2 Kg

Spring constant (k) = ${\pi }^2$

$\text{Solution:-}$

Here, $m_1$ stays in contact with the spring for $\frac{T}{2}$

Now we have taken the center of mass $\left(COM\right)$ frame,

Reduced mass, $\mu =\frac{m_1{m}_2}{m_1+m_2}$

$\Rightarrow \mu =\frac{2\times 2}{2+2}$

$\mu =1Kg$

Now,

$T=2\pi \sqrt{\frac{\mu }{k}}$

$T=2\pi \sqrt{\frac{1}{{\pi }^2}}$

$\Rightarrow T=2sec.$

$\therefore$ $m_1$ stays in contact with the spring for $\frac{T}{2}$

$\Rightarrow \frac{2}{2}=1sec$

$\text{Hence the correct option is C}$