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 the equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with velocity $3m{s}^{-1}$ . The total initial energy of the spring is:

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

Answer 1

Answer: (A)

By mechanical energy conservation between compression positions $x$ and $\frac{x}{2}$
$\frac{1}{2}k{x}^2=\frac{1}{2}k{\left(\frac{x}{2}\right)}^2+\frac{1}{2}m{v}^2$
$\frac{1}{2}k{x}^2-\frac{1}{2}k\frac{x^2}{4}=\frac{1}{2}m{v}^2$
$\frac{1}{2}k{x}^2\left(\frac{3}{4}\right)=\frac{1}{2}m{v}^2$
$v=\sqrt{\frac{3k{x}^2}{4m}}=\sqrt{\frac{3k}{m}}\frac{x}{2}$
On collision with a block at rest
$\because$ Velocities are exchanged $\Rightarrow$ elastic collision between identical masses.
$\therefore v=3=\sqrt{\frac{3k}{m}}\frac{x}{2}$
$\Rightarrow 6=\sqrt{\frac{3k}{m}}x$
$\Rightarrow x=6\sqrt{\frac{m}{3k}}$
$\therefore$ The initial energy of the spring is
$U=\frac{1}{2}k{x}^2=\frac{1}{2}k\times 36\frac{m}{3k}=6m$
$U=6\times 0.1=0.6J$