Question

A block of mass $m \,=\, 0.1\, kg$ 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 $3 \,ms^{-1}$. 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
$\frac{3}{8} K x^{2}=\frac{1}{2} m u^{2}$
$\Rightarrow u=\frac{x}{2} \sqrt{\frac{3 k}{m}}$
Conservation of momentum gives, $m u=m v$
where $v$ is the speed of the second block just after collision, therefore, $v=u=3\, m / s$
Hence, $3=\sqrt{\frac{3 K}{m}} \frac{x}{2}$
$\Rightarrow x=6 \sqrt{\frac{m}{3 K}}$
Initial energy of the spring
$=\frac{1}{2} K x^{2}=\frac{1}{2} \cdot \frac{36 m }{3 K}=6\, m , U_{i}=0.6\, J$