Mass of cube, m = 50g = 5.0 × 10-2 kg
Speed of cube, v = 10 cm/s = 1.0 × 10-1 m/s
Young’s modulus Y = 2.0 × 1011 N/m2
Side of cube(L) = 1 cm = 1.0 × 10-2 m
Apply Hooke’s Law,
Young modulus Y = \(\frac{\frac{F}{A}}{\frac{ΔL}{L}}\)
So, \(\frac{F}{ΔL} = \frac{YA}{L} ........(i)\)
Or, \(\frac{F}{ΔL} = k ........(ii)\)
From eqn (i) & eqn (ii)
\(k=\frac{YA}{L} =\frac{YL^2}{L},\)
(Here A = L2)
K = YL
Initial KE = \(2\times \frac{1}{2}mv^2=5.0\times 10^{-4}J\)
Final PE = \(2\times \frac{1}{2}kΔI^2\)
= kΔI2 = YLΔI2
Apply Law of conservation of energy
YLΔI2 = 5.0 × 10-4
or, ΔL = \(\sqrt \frac{5.0\times 10^{-4}}{YL}\)
= \(\sqrt \frac{5.0\times 10^{-4}}{(2.0\times 10^{11}\times 10^{-2})}m\)
Δl = 5.0 × 10-7 m