Question

A boy has a catapult made of rubber cord of length \[42\,{\text{cm}}\] and diameter \[6.0\,{\text{mm}}\]. The boy stretches the cord by \[20\,{\text{cm}}\] to catapult a stone of mass \[20\,{\text{g}}\]. The stone flies off with a speed of \[20\,{\text{m}}{{\text{s}}^{ - 1}}\]. Find young’s modulus for rubber. Ignore the change in the cross section of the cord in stretching.

Answer 1

Hint: First of all, find the cross-sectional area of the cord. Find the potential energy stored in the cord while stretching. This potential energy gets converted into kinetic energy for the stone, when it is released.

Complete step by step answer:
In the above problem, the given quantities are:
Diameter of the rubber cord is:
\[D = 6\,{\text{mm}}\]
Length of the cord is:
\[L = 42\,{\text{cm}}\]
Change in length is:
\[\Delta L = 20\,{\text{cm}}\]
Mass of the stone is:
\[m = 20\,{\text{g}}\]
Velocity of the stone is:
\[v = 20\,{\text{m}}{{\text{s}}^{ - 1}}\]
We know,
\[1\,{\text{mm}} = {10^{ - 3}}\,{\text{m}}\]
\[\Rightarrow 1\,{\text{cm}} = {10^{ - 2}}\,{\text{m}}\]
\[\Rightarrow 1\,{\text{g}} = {10^{ - 3}}\,{\text{kg}}\]
So, we convert the given units to S.I units as follows:
Diameter of the rubber cord is:
\[
D = 6\,{\text{mm}} \\
\Rightarrow D = 6 \times {10^{ - 3}}\,{\text{m}} \\
\Rightarrow D= 0.006\,{\text{m}} \\
\]
Radius of the rubber cord is:
\[
R = \dfrac{D}{2} \\
\Rightarrow R = \dfrac{{0.006}}{2}\,{\text{m}} \\
\Rightarrow R = 0.003\,{\text{m}} \\
\]
Length of the cord is:
\[
L = 42\,{\text{cm}} \\
\Rightarrow L= 42 \times {10^{ - 2}}\,{\text{m}} \\
\Rightarrow L= 0.42\,{\text{m}} \\
\]
Change in length is:
\[
\Delta L = 20\,{\text{cm}} \\
\Rightarrow\Delta L = 20 \times {10^{ - 2}}\,{\text{m}} \\
\Rightarrow\Delta L = 0.20\,{\text{m}} \\
\]
Mass of the stone is:
\[
  m = 20\,{\text{g}} \\
\Rightarrow m = 20 \times {10^{ - 3}}\,{\text{kg}} \\
\Rightarrow m = 0.02\,{\text{kg}} \\
\]
Now, we find the cross-sectional area \[\left( A \right)\] of the cord, which is given by the formula:
\[A = \pi {R^2}\] …… (1)
Substitute \[R = 0.003\,{\text{m}}\]
 in the equation (1), we get:
\[
  A = \pi {R^2} \\
\Rightarrow A = 3.14 \times {\left( {0.003} \right)^2}\,{{\text{m}}^2} \\
\Rightarrow A= 2.826 \times {10^{ - 5}}{\text{ }}{{\text{m}}^2} \\
 \]
We know,
Young’s modulus \[\left( Y \right)\] is given by the formula:
\[Y = \dfrac{{{\text{Stress}}}}{{{\text{Strain}}}}\]
\[{\text{Strain}} = \dfrac{{\Delta L}}{L}\]
Volume is the product of cross-sectional area and the length of the cord:
\[{\text{Volume}} = A \times L\]
Energy \[\left( E \right)\] stored in the catapult is a form of potential energy, which is given by the formula:
\[
\Rightarrow E = \dfrac{1}{2} \times {\text{Stress}} \times {\text{Strain}} \times {\text{volume}} \\
\Rightarrow E = \dfrac{1}{2} \times \dfrac{{{\text{Stress}}}}{{{\text{Strain}}}} \times {\left( {{\text{Strain}}} \right)^2} \times {\text{volume}} \\
\Rightarrow E = \dfrac{1}{2} \times Y \times {\left( {{\text{Strain}}} \right)^2} \times {\text{volume}} \\
\]
Now, substitute:
\[{\text{Strain}} = \dfrac{{\Delta L}}{L}\] and \[{\text{Volume}} = A \times L\] in the above expression.
\[
E = \dfrac{1}{2} \times Y \times {\left( {\dfrac{{\Delta L}}{L}} \right)^2} \times A \times L \\
\Rightarrow E= \dfrac{1}{2} \times Y \times \dfrac{{{{\left( {\Delta L} \right)}^2}}}{L} \times A \\
\]
The energy stored in the catapult is converted into kinetic energy.
Kinetic energy is given by:
\[K.E = \dfrac{1}{2}m{v^2}\]
Where,
\[m\] indicates the mass of the stone.
\[v\] indicates velocity at which the stone flies.
We can write:
\[
\Rightarrow \dfrac{1}{2} \times Y \times \dfrac{{{{\left( {\Delta L} \right)}^2}}}{L} \times A = \dfrac{1}{2}m{v^2} \\
\Rightarrow Y = m{v^2} \times \dfrac{L}{{A{{\left( {\Delta L} \right)}^2}}} \\
\]
Substituting the respective values in the above equation:
\[
Y = m{v^2} \times \dfrac{L}{{A{{\left( {\Delta L} \right)}^2}}} \\
\Rightarrow Y= 0.02 \times {\left( {20} \right)^2} \times \dfrac{{0.42\,}}{{2.826 \times {{10}^{ - 5}} \times {{\left( {0.2} \right)}^2}}} \\
\therefore Y= 2.97 \times {10^6}\,{\text{N}}{{\text{m}}^{ - 2}} \\
\]
Hence, the Young’s modulus is \[2.97 \times {10^6}\,{\text{N}}{{\text{m}}^{ - 2}}\].

Note: While solving this problem, many students tend to make mistakes while finding the area of the cross section. It should be remembered that the cross section is circular in shape. It is important to note that strain is not just change in length, rather it is the ratio of change in length and the original length.