Ncert Exercises & Solutions

Q.47 Two identical steel cubes (masses 50g, 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 $= Y = 2 \times 10^{11} N / m^{2}$ .

Hint: All KE will be converted to elastic potential energy.

Step 1: Find the Initial kinetic energy of the system of two cubes.

Let

$\begin{array}{l} m = 50 g = 50 \times 10^{- 3} kg \\ Side = L = 1 cm = 0 .01 m \\ Speed = v = 10 cm / s = 0 .1 m / s \\ Young's modulus = Y = 2 \times 10^{11} N / m^{2} \\ Maximum compression ΔL = ? \end{array}$

In this case, all KE will be converted to PE
By Hooke's law,
where A is the surface area and L is the length of the side of the cube If k is spring or compression constant. then

$Initial KE = 2 \times \frac{1}{2} {mv}^{2} = 5 \times 10^{- 4} J$
$Step 2 : Find the expression of elastic energy stored in the system of two cube .$
$By Hook' s law$
$\frac{F}{A} = Y \frac{ΔL}{L}$
$where A is the surface area and L is the length of the side of the cube$
$If k is spring constant, then$
$force F = kΔL$
$∴ k = Y \frac{A}{L} = YL$
$Final PE = 2 \times \frac{1}{2} k (ΔL)^{2}$
$Step 3 : Find maximum compression of each of the cubes by equating total kinetic energy with total PE .$
$∴ ΔL = \sqrt{\frac{KE}{k}} = \sqrt{\frac{KE}{γL}} = \sqrt{\frac{5 \times 10^{- 4}}{2 \times 10^{11} \times 0 .1}} = 1 .58 \times 10^{- 7} m [∵ PE = KE]$

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