A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section 10⁻⁶ m² stretched by an amount 0.04 m. The velocity of the projected stone is ______________ m/s.
(Young's modulus of rubber = 0.5 × 10⁹ N/m²)

Solution

Physics sir · 4 steps

All the energy stored in the stretched rubber turns into the stone's kinetic energy. Equate the two.

1. Energy stored. A wire or cord stretched by Δ l stores elastic potential energy U = 1/2 × stress × strain × volume. With stress = Y Δ l / L and volume = AL, this becomes the formula below.

   U = (Y A (Δ l)²)/2L = 1/2mv²

2. Put in numbers. U = (0.5 × 10⁹ × 10⁻⁶ × (0.04)²)/(2 × 0.1) = (500 × 0.0016)/0.2 = 4 J.
3. Solve for v. With m = 20 g = 0.02 kg: 1/2 × 0.02 × v² = 4, so v² = 400.
4. So. The stone leaves the catapult at v = 20 m s⁻¹.

The common slip: 28.3. Dropping the 1/2 in the stored energy gives U = 8 J, so v² = 800 and v ≈ 28.3, not 20.

The idea to keep. For any stretched elastic body, stored energy is 1/2 × stress × strain × volume, or YA(Δ l)²/2L. In a launch, set it equal to 1/2mv².

Answer from the NTA final answer key. Solution by PrepWiser.