Question

A boy's catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and of negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms^-1. Neglect the change in area of cross-section of the cord while stretched. The Young's modulus of rubber is closest to:
1. 10⁶ Nm^-2
2. 10⁴ Nm^-2
3. 10⁸ Nm^-2
4. 10³ Nm^-2

Answer 1

Correct Answer - Option 1 : 10⁶ Nm^-2

Concept:

Elastic potential energy: When a force is applied to deform an elastic object, the energy is stored in the object until the force is removed and it gets back to its original shape. Such energy is called as elastic potential energy. Example: catapult.

Kinetic energy: Kinetic energy is the energy that possessed by an object due to its motion.

Young’s modulus: Young’s modulus is the ratio of longitudinal stress to the longitudinal strain.

Calculation:

Given, Length of the cord, l = 42 cm

Diameter of cross-section, d = 6 mm

Mass of the stone, m = 0.02 kg

Change in length of the cord, Δl = 20 cm

Velocity of the stone, v = 20 ms^-1

When the kinetic energy acting on the stone includes energy excreted by the catapult on the stone.

The energy of the catapult (elastic potential energy) is:

\({\rm{E}} = \frac{1}{2} \times {\rm{stress}} \times {\rm{strain}} \times {\rm{volume}}\)

\({\rm{E}} = \frac{1}{2} \times \left( {{\rm{Y}}\left( {\frac{{{\rm{\Delta }}l}}{l}} \right){\rm{A}}l} \right) \times \left( {\frac{{{\rm{\Delta }}l}}{l}} \right)\)

The kinetic energy acting on the stone is:

\({\rm{K}}.{\rm{E}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2}\)

Now, the kinetic energy of the stone is equal to the energy of the catapult.

\(\frac{1}{2} \times \left( {{\rm{Y}}\left( {\frac{{{\rm{\Delta }}l}}{l}} \right){\rm{A}}l} \right) \times \left( {\frac{{{\rm{\Delta }}l}}{l}} \right) = \frac{1}{2}{\rm{m}}{{\rm{v}}^2}\)

Where, Y = Young’s modulus

A = Area of the stone

i.e. A = πr² = π × (3 × 10^-3)² = π × 9 × 10^-6 m² \(\left[ {r = \frac{d}{2}} \right]\)

\(\Rightarrow \frac{1}{2}\times \left( \text{YA}l \right)\times {{\left( \frac{\Delta l}{l} \right)}^{2}}=\frac{1}{2}\text{m}{{\text{v}}^{2}}\)

\(\Rightarrow \frac{1}{2}\times \text{YA}\times \frac{{{\left( \text{ }\!\!\Delta\!\!\text{ }l \right)}^{2}}}{l}=\frac{1}{2}\text{m}{{\text{v}}^{2}}\)

Keeping the unknown value on one side and bringing rest of the variables on another side,

\(\Rightarrow {\rm{Y}} = \frac{{{\rm{m}}{{\rm{v}}^2}l}}{{{{\left( {{\rm{\Delta }}l} \right)}^2} \times {\rm{A}}}}\)

On substituting the values, we get

\({\rm{Y}} = \frac{{0.02 \times {{20}^2} \times 42 \times {{10}^{ - 2}}}}{{{{\left( {20 \times {{10}^{ - 2}}} \right)}^2} \times {\rm{\pi }} \times 9 \times {{10}^{ - 6}}}}\)

\(\Rightarrow {\rm{Y}} = \frac{{3.36}}{{1.13 \times {{10}^{ - 6}}}} = 2.97 \times {10^6} \approx 3 \times {10^6}{\rm{\;N}}/{{\rm{m}}^2}\)

Thus, the order of young’s modulus is 10⁶ Nm^-2