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

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

Answer 1

Answer: (A)

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