Two identical steel cubes (masses $50\,g$, side $1\,cm$) collide head-on face to face with a space of $10\,cm/s$ each. Find the maximum compression of each. Young’s modulus for steel $Y = 2 \times 10^{11}\,Nm^{-2}$.

Let $m=50 \mathrm{~g}=50 \times 10^{-3} \mathrm{~kg}$ $\mathrm{~L}=1 \mathrm{~cm}=0.01 \mathrm{~m}$

$v=10 \mathrm{~cm} / \mathrm{s}=0.1 \mathrm{~m} / \mathrm{s}$

$\mathrm{Y}=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$

Here, KE will be converted to $PE$

$\mathrm{F}=\frac{\mathrm{YAL}}{\mathrm{L}}$ (Hooke's Law)

$\therefore$ Also, $\mathrm{F}=k \Delta \mathrm{L} \quad(\mathrm{K}=$ spring constant $)$

$\therefore k=\mathrm{Y} \frac{\mathrm{A}}{\mathrm{L}}=\mathrm{YL} \quad\left[\because \mathrm{A}=\mathrm{L}^{2}\right]$

Initial $\mathrm{KE}=2 \times \frac{1}{2} m v^{2}=5 \times 10^{-4} \mathrm{~J}$

Final $\mathrm{PE}=2 \times \frac{1}{2} k(\Delta \mathrm{L})^{2}=k(\Delta \mathrm{L})^{2}$

$\therefore k(\Delta \mathrm{L})^{2}=5 \times 10^{-4}=\sqrt{\frac{5 \times 10^{-4}}{2 \times 10^{11} \times 0.1}}=1.58 \times 10^{-7} \mathrm{~m}$