Question

A block of mass $m=0.1kg$ is connected to a spring of unknown spring constant $k$. It is compressed to a distance $x$ from its equilibrium position and released from rest. After approaching half the distance $\left(\frac{x}{2}\right)$ from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity $3m{s}^{-1}$. The total initial energy of the spring is :

• A. 1.5 J
• B. 0.6 J
• C. 0.3 J
• D. 0.8 J

Answer 1

Answer: (B)

At $\frac{x}{2}$, it hits another identical block, applying conservation of energy, where $u$ is the speed
$\frac{1}{2}K{x}^2=\frac{1}{2}K{\left(\frac{x}{2}\right)}^2+\frac{1}{2}m{u}^2;$ just before collision
$\frac{3}{8}K{x}^2=\frac{1}{2}m{u}^2$
$\Rightarrow u=\frac{x}{2}\sqrt{\frac{3k}{m}}$
Conservation of momentum gives, $mu=mv$
where $v$ is the speed of the second block just after collision, therefore, $v=u=3m/s$
Hence, $3=\sqrt{\frac{3K}{m}}\frac{x}{2}$
$\Rightarrow x=6\sqrt{\frac{m}{3K}}$
Initial energy of the spring
$=\frac{1}{2}K{x}^2=\frac{1}{2}\cdot \frac{36m}{3K}=6m,U_i=0.6J$