1.

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 × 1011 N/m2

Answer»

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 = \(2\times \frac{1}{2}k​​ΔI^2\)

= 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


Discussion

No Comment Found

Related InterviewSolutions