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

The key idea to solve this problem is the Conservation of Mechanical Energy.

1. The initial total kinetic energy of the two cubes is $KE_{initial} = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.

   Given $m = 50 \text{ g} = 0.05 \text{ kg}$ and $v = 10 \text{ cm/s} = 0.1 \text{ m/s}$.

   $KE_{initial} = (0.05 \text{ kg})(0.1 \text{ m/s})^2 = 0.05 \times 0.01 = 0.0005 \text{ J}$.

2. At the instant of maximum compression, the cubes momentarily come to rest relative to each other, and all the initial kinetic energy is converted into elastic potential energy stored in the deformed cubes. The elastic potential energy stored in one cube is $U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}$.

   Using Stress $= Y \times \text{Strain}$ and Strain $= \frac{\Delta L}{L}$, where $L$ is the original length ($1 \text{ cm} = 0.01 \text{ m}$) and $A$ is the cross-sectional area ($L^2 = (0.01 \text{ m})^2 = 10^{-4} \text{ m}^2$), the potential energy in one cube is $U = \frac{1}{2} Y \left(\frac{\Delta L}{L}\right)^2 (A L) = \frac{1}{2} Y A \frac{(\Delta L)^2}{L}$.

3. Since there are two identical cubes, the total elastic potential energy stored is $U_{total} = 2U = Y A \frac{(\Delta L)^2}{L}$. …

When two objects collide and deform, their kinetic energy is converted into elastic potential energy. The maximum compression occurs at the instant when all the kinetic energy associated with the deformation has been converted into elastic potential energy. For a head-on collision of identical objects with equal and opposite speeds, the system's center of mass remains stationary. At the point of maximum compression, the relative speed between the cubes is momentarily zero, and since the center of mass is stationary, the cubes themselves are momentarily at rest. Therefore, all the initial kinetic energy of the system is converted into elastic potential energy.

Here's how we can determine the maximum compression:

1. Convert all given values to SI units:

   • Mass of each cube, $m = 50 \text{ g} = 0.050 \text{ kg}$.
   • Side length of each cube, $L = 1 \text{ cm} = 0.01 \text{ m}$.
   • Speed of each cube, $v = 10 \text{ cm/s} = 0.1 \text{ m/s}$.
   • Young's modulus for steel, $Y = 2 \times 10^{11} \text{ N/m}^2$.
   • The cross-sectional area of the face of each cube, $A = L^2 = (0.01 \text{ m})^2 = 1 \times 10^{-4} \text{ m}^2$.

2. Calculate the total initial kinetic energy of the system:

   Since there are two identical cubes, each moving with speed $v$, the total initial kinetic energy ($KE_{initial}$) is the sum of their individual kinetic energies:

Substituting the values:

3. Determine the elastic potential energy stored at maximum compression:

   When the cubes collide, both are compressed. Let $\Delta L$ be the maximum compression of each cube. The elastic potential energy ($U$) stored in a material is given by:

   The elastic potential energy $U$ stored in a material of Young's modulus $Y$, cross-sectional area $A$, and original length $L$, when compressed or stretched by $\Delta L$, is given by:

   $U = \frac{1}{2} \frac{YA}{L}(\Delta L)^2$

   Since both cubes are identical and experience the same compression $\Delta L$, the total elastic potential energy stored in the system ($U_{total}$) at maximum compression is twice the energy stored in one cube:

   $U_{total} = 2 \times \left( \frac{1}{2} \frac{YA}{L}(\Delta L)^2 \right) = \frac{YA}{L}(\Delta L)^2$ …

Concept: Elastic Energy Storage (Young's Modulus) + Energy Conservation

Step 1: Find the total initial kinetic energy

Both cubes move toward each other at $v = 10$ cm/s $= 0.1$ m/s, mass $m = 50$ g $= 0.05$ kg each:

Step 2: Argue that both cubes are momentarily at rest at maximum compression

By symmetry (equal masses, equal and opposite speeds) the centre of mass is stationary; at maximum compression the relative speed between the cubes is zero, so both cubes are momentarily at rest — all the KE has converted to elastic PE.

Step 3: Write the elastic PE stored in one cube

Treating the cube as a bar of length $L$, cross-section $A=L^2$, compressed by $\Delta L$, using $U=\tfrac12\,\text{stress}\times\text{strain}\times\text{volume}=\tfrac12\frac{YA}{L}(\Delta L)^2$:

Step 4: Total elastic PE for both (identical) cubes

$U_{total} = 2U_{one} = \frac{YA}{L}(\Delta L)^2$ …