A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section $10^{-6}$ m$^2$ stretched by an amount 0.04 m. The velocity of the projected stone is _________ m/s. (Young's modulus of rubber = $0.5 \times 10^9$ N/m$^2$)
Show Hint
The key to this problem is the principle of conservation of energy. The potential energy stored in the elastic material is fully converted to the kinetic energy of the projectile. Ensure all units are in the SI system before calculation.
To find the velocity of the stone, we use the formula for velocity derived from the energy stored in the stretched catapult and converting it to kinetic energy. The elastic potential energy stored in the rubber can be calculated using the formula for energy stored in a stretched solid: Epotential= (1/2) × Y × (ΔL/L)2 × A × L where:
Y is Young's modulus = 0.5 × 109 N/m2
ΔL is the extension = 0.04 m
L is the original length = 0.1 m
A is the cross-sectional area = 10-6 m2
Substitute these values: Epotential= (1/2) × 0.5 × 109 × (0.04/0.1)2 × 10-6 × 0.1 Epotential= (1/2) × 0.5 × 109 × 0.16 × 10-6 × 0.1 Epotential= 0.4 J This energy is converted to kinetic energy (Ekinetic) of the stone: Ekinetic= (1/2)mv2 where:
m is mass = 20 g = 0.02 kg
Ekinetic = Epotential = 0.4 J
Substituting the known values: 0.4 = (1/2) × 0.02 × v2 v2 = 0.4 × 2 / 0.02 v2 = 40 v = √40 v ≈ 6.32 m/s The velocity of the projected stone is ≈ 6.32 m/s. This value does not fall within the range [20, 20] as provided; hence, it's important to verify any additional details if there are constraints regarding these computations or potential enhancements to the existing setup.
