A boy's catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and

Q: Why do we multiply by Volume in the potential energy formula? I thought the formula was just $\frac{1}{2} \times \text{Stress} \times \text{Strain}$.

The expression $\frac{1}{2} \times \text{Stress} \times \text{Strain}$ actually gives you the energy density, which is the energy stored per unit volume. To find the total energy stored in the entire cord, you must multiply this energy density by the total volume of the cord.

Q: A real catapult forms a V-shape when pulled back. Why didn't we use angles and trigonometry here?

That's a great observation! In a rigorous real-world analysis, we would indeed need to account for the angles and resolve the tension vectors. However, this problem implies a simplified 1D energy conversion model where we assume the cord acts like a single linear spring being stretched by $\Delta L$.

Q: Why did we neglect the mass of the rubber cord itself?

If the rubber cord had significant mass, some of the stored elastic potential energy would be used to give kinetic energy to the cord itself as it snaps back. By neglecting its mass, we assume 100% of the stored energy is transferred exclusively to the stone, simplifying our calculation.

Q: How did we get the area as $9\pi \times 10^{-6}$?

The diameter is given as $6\text{ mm}$, so the radius is half of that, which is $3\text{ mm}$. We must convert this to standard SI units (meters), so $r = 3 \times 10^{-3}\text{ m}$. The area of a circle is $\pi r^2$, which gives $\pi (3 \times 10^{-3})^2 = 9\pi \times 10^{-6}\text{ m}^2$.

Q: The exact answer is around $3 \times 10^6$, but the option is $10^6$. Is that normal?

Yes, in many competitive exams, questions ask for the value that is 'closest to' a set of options that differ by orders of magnitude (powers of 10). Since $3 \times 10^6$ is much closer to $10^6$ than it is to $10^8$ or $10^4$, $10^6$ is the correct choice.

Question

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

Options:

(A) $10^6	ext{ Nm}^{-2}$

(B) $10^4	ext{ Nm}^{-2}$

(C) $10^8	ext{ Nm}^{-2}$

(D) $10^3	ext{ Nm}^{-2}$

Sigma Percentile · Accepted Answer

Imagine you are standing in an open field, pulling back the pouch of a catapult. As you stretch the thick rubber cord, you can feel the tension building up. You are doing work against the elastic forces of the rubber, and all that effort is being stored inside the cord as **Elastic Potential Energy**.

When you finally let go, the cord snaps back to its original length, transferring all that pent-up energy into the stone. The stone shoots forward, now possessing **Kinetic Energy**. This beautiful conversion of energy is the core principle we will use to solve this problem.

### The Master Equation: Conservation of Energy

By the principle of conservation of energy, assuming no energy is lost to heat or sound, the elastic potential energy stored in the stretched cord must equal the kinetic energy gained by the stone.

$$U_{	ext{elastic}} = K_{	ext{kinetic}}$$

We know the formula for the kinetic energy of the stone is simply $\frac{1}{2}mv^2$. But what about the elastic potential energy of the cord? For a stretched wire or cord, the energy stored is given by:

$$U = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain} 	imes 	ext{Volume}$$

### Unpacking the Elasticity Terms

To find the Young's modulus ($Y$), we need to express Stress and Strain in terms of $Y$ and the physical dimensions of the cord. According to Hooke's Law:

$$	ext{Stress} = Y 	imes 	ext{Strain}$$

We also know that Strain is the fractional change in length, $\frac{\Delta L}{L}$. The volume of the cord is its cross-sectional area multiplied by its original length, $A \cdot L$. Substituting these into our potential energy formula gives:

$$U = \frac{1}{2} \left(Y \frac{\Delta L}{L}ight) \left(\frac{\Delta L}{L}ight) (A \cdot L) = \frac{Y A (\Delta L)^2}{2L}$$

### Equating and Solving for Young's Modulus

Now, let's bring our energy conservation equation back and equate the two energies:

$$\frac{Y A (\Delta L)^2}{2L} = \frac{1}{2} m v^2$$

We can elegantly cancel the $\frac{1}{2}$ from both sides. Rearranging the equation to isolate our target variable, Young's modulus ($Y$), we get:

$$Y = \frac{m v^2 L}{A (\Delta L)^2}$$

### The Final Calculation

Before we plug in the numbers, let's calculate the cross-sectional area $A$. The diameter is given as $6	ext{ mm}$, which means the radius $r = 3	ext{ mm} = 3 	imes 10^{-3}	ext{ m}$.

$$A = \pi r^2 = \pi (3 	imes 10^{-3})^2 = 9\pi 	imes 10^{-6}	ext{ m}^2$$

Now, we substitute all the given values into our master equation for $Y$:

$$Y = \frac{0.02 	imes (20)^2 	imes 0.42}{9\pi 	imes 10^{-6} 	imes (0.2)^2}$$

$$Y = \frac{0.02 	imes 400 	imes 0.42}{9\pi 	imes 10^{-6} 	imes 0.04}$$

$$Y = \frac{3.36}{1.13 	imes 10^{-6}} \approx 2.97 	imes 10^6	ext{ Nm}^{-2}$$

The calculated value is approximately $3 	imes 10^6	ext{ Nm}^{-2}$. Looking at the given options, the closest order of magnitude is **$10^6	ext{ Nm}^{-2}$**. Therefore, option (a) is the correct answer. The physics of a simple toy reveals the profound material properties hidden within it!

Select Answer:

Visualized Solution

$L \to L + Δ L$

$U_{elastic} = K_{kinetic}$

$U = \frac{1}{2} \times Stress \times Strain \times Volume$

$Stress = Y \cdot Strain$

$\frac{1}{2} (Y \frac{Δ L}{L}) (\frac{Δ L}{L}) (A \cdot L) = \frac{1}{2} m v^{2}$

$Y = \frac{m v ^{2} L}{A ( Δ L ) ^{2}}$

$A = π r^{2}$

$Y \approx 2.97 \times 1 0^{6} Nm^{- 2}$

$Closest Value$

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

The Master Equation

Unpacking the Elasticity Terms

Equating and Solving for Young's Modulus

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L→L+ΔL

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Uelastic​=Kkinetic​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U=21​×Stress×Strain×Volume

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Stress=Y⋅Strain

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 21​(YLΔL​)(LΔL​)(A⋅L)=21​mv2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y=A(ΔL)2mv2L​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=πr2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y≈2.97×106 Nm−2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Closest Value