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. 0.8 J
• C. 0.3 J
• D. 0.6 J
• E. 0.7 J

Answer 1

The correct option is D 0.6 J

Apply principle of conservation of momentum and energy

Momentum before collision = Momentum after collision

$0.1u+m\times 0=0.1\times 0+m\times 3$

$\frac{1}{2}\times 0.1\times u^2=\frac{1}{2}\times m\times 3^2$

Solving these two equations, $u=3$

$\frac{1}{2}k{x}^2=\frac{1}{2}k(\frac{x}{2}{)}^2+\frac{1}{2}\times 0.1\times 3^2$

$\frac{3}{4}k{x}^2=0.9$

$\frac{1}{2}k{x}^2=0.6J$