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

(D) Given: $m = 50\,g = 0.05\,kg$,$L = 1\,cm = 0.01\,m$,$v = 10\,cm/s = 0.1\,m/s$,$Y = 2 \times 10^{11}\,N/m^2$.
At maximum compression,the kinetic energy of the cubes is converted into elastic potential energy.
The effective spring constant $k$ for a cube is given by $F = k \Delta L = Y A \frac{\Delta L}{L}$.
Thus,$k = \frac{YA}{L} = \frac{Y L^2}{L} = YL$.
Total initial kinetic energy $KE = 2 \times (\frac{1}{2} m v^2) = m v^2 = 0.05 \times (0.1)^2 = 5 \times 10^{-4}\,J$.
Total elastic potential energy at maximum compression $\Delta L_{max}$ for two cubes is $PE = 2 \times (\frac{1}{2} k (\Delta L_{max})^2) = k (\Delta L_{max})^2$.
Equating $KE = PE$: $m v^2 = (YL) (\Delta L_{max})^2$.
$\Delta L_{max} = \sqrt{\frac{m v^2}{YL}} = \sqrt{\frac{5 \times 10^{-4}}{2 \times 10^{11} \times 0.01}} = \sqrt{2.5 \times 10^{-13}} \approx 5 \times 10^{-7}\,m$.