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Step-by-Step Solution
Step 1: Identify the Physical Principles Involved
The question involves the concept of Young’s modulus and the energy stored in an elastic cord (rubber) when stretched. This stored potential energy is converted into kinetic energy of the stone when released.
Step 2: Write Down the Energy Conservation Relationship
According to the problem, the potential energy stored by stretching the rubber cord is equal to the kinetic energy gained by the stone:
$ \frac{1}{2} Y \left(\frac{\Delta L}{L}\right)^2 A \, L \;=\; \frac{1}{2} m v^2 $
Here:
$Y$ is the Young’s modulus of the rubber.
$L$ is the original length of the cord.
$\Delta L$ is the extension of the cord.
$A$ is the cross-sectional area of the cord.
$m$ is the mass of the stone.
$v$ is the velocity of the stone after release.
Step 3: List All Known Values
Original length of cord, $L = 42\,\text{cm} = 0.42\,\text{m}$
Extension of cord, $\Delta L = 20\,\text{cm} = 0.2\,\text{m}$
Mass of stone, $m = 0.02\,\text{kg}$
Velocity of stone, $v = 20\,\text{m/s}$
Diameter of cord, $d = 6\,\text{mm} = 6 \times 10^{-3}\,\text{m}$
Step 4: Calculate the Cross-sectional Area of the Cord
The cross-sectional area $A$ of the (cylindrical) cord is:
$ A \;=\; \pi \left(\frac{d}{2}\right)^2
\;=\; \pi \left(\frac{6 \times 10^{-3}}{2}\right)^2
\;=\; \pi \times 3^2 \times 10^{-6}
\;=\; 9\pi \times 10^{-6} \,\text{m}^2.$
Step 5: Substitute Known Values into the Energy Equation
From the relation
$ \frac{1}{2} Y \left(\frac{\Delta L}{L}\right)^2 A \, L
\;=\; \frac{1}{2} m v^2, $
rearrange to solve for $Y$:
$ Y \;=\; \frac{m v^2 \, L}{A \, (\Delta L)^2 }. $
Substitute the values:
$
Y \;=\;
\frac{\,0.02 \times (20)^2 \times 0.42\,}
{\,\bigl(9\pi \times 10^{-6}\bigr) \times (0.2)^2 \,}.
$
Step 6: Numerical Evaluation
Calculate step by step (approximate):
Numerator: $ 0.02 \times 400 \times 0.42 = 0.02 \times 168 = 3.36 $
Denominator: $ 9\pi \times 10^{-6} \times 0.04
= 0.36\pi \times 10^{-6}
= 3.6\pi \times 10^{-7} $
Hence,
$
Y \approx \frac{3.36}{3.6\pi \times 10^{-7}}
= \frac{3.36}{3.6\pi} \times 10^{7}.
$
Numerically, this is on the order of $10^6 \,\text{N/m}^2$.
Step 7: Conclusion
The nearest order of magnitude for the Young’s modulus of the rubber cord is
$10^6 \,\text{N/m}^2.$
