Answer:
The velocity with which the stone leaves the catapult is 4 m/s
Explanation:
The mass of the stone released, m = 50 g = 0.05 kg
The length by which the rubber catapult is stretched, Δx = 4 cm = 0.04 m
The force constant of the rubber, k = 500 N/m
By the conservation of energy formula, we have;
K₁ + U₁ = K₂ + U₂
Where;
K₁ = The initial kinetic energy of the stone and the catapult system = 0 J (The stone and the rubber of the catapult are held back stationary)
U₁ = The initial potential energy of the stone and the catapult system = 1/2·k·(Δx)²
K₂ = The final kinetic energy of the stone and the catapult system = 1/2·m·v²
Where;
v = The velocity with which the stone leaves the catapult
U₂ = The final potential energy of the stone and the catapult system = 0 J (The rubber of the catapult returns to the relaxed state)
Therefore, by substitution of the above values and equivalent expressions, we have;
0 J + 1/2·k·(Δx)² = 1/2·m·v² + 0 J
1/2·k·(Δx)² = 1/2·m·v²
We substitute the given known values as follows;
1/2 × 500 N/m × (0.04 m)² = 1/2 × 0.05 kg × v²
∴ v² = (1/2 × 500 N/m × (0.04 m)²)/(1/2 × 0.05 kg) = 16 m²/s²
v = √(16 m²/s²) = 4 m/s
The velocity with which the stone leaves the catapult = v = 4 m/s.
