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 $\left(\frac{X}{W}\right)$ from equilibrium position, it hits another identical block and comes to rest momentarily, while the other block starts moving with a velocity of $3m{s}^{(-1)}$. The total initial energy of the spring is

• A. 0.5 J
• B. 0.6 J
• C. 0.3 J
• D. 0.8 J

Answer 1

The correct option is B 0.6 J

Applying principle of conservation of energy
$\frac{1}{2}k{x}^2=\frac{1}{2}k{\left(\frac{x}{2}\right)}^2+\frac{1}{2}m{u}^2$
Where u is the velocity before collision at $\frac{x}{2}$
$\Rightarrow u=\sqrt{\frac{3k{x}^2}{4m}}=\sqrt{\frac{3E}{2m}}$ where E is the initial energy of the spring
Now applying conservation of momentum
$mu=0+m\left(3\right)\Rightarrow \sqrt{\frac{3E}{2m}}=3\Rightarrow E=6m=0.6J$