Question 1: Define elastic potential energy and explain how it is stored in an elastic object.

Answer: Elastic potential energy is the energy stored in an object when it is deformed elastically, such as stretched or compressed. It is stored due to the work done against the restoring force of the material during deformation.

Question 2: Describe the mechanism by which a stretched spring stores elastic potential energy.

Answer: When a spring is stretched, work is done to extend it, causing the molecules within to be displaced from their equilibrium positions. The spring's restoring force stores energy as elastic potential energy proportional to the amount of extension.

Question 1: A spring with a spring constant of 200 N/m is stretched by 0.05 m. Calculate the elastic potential energy stored in the spring.

Answer: E = ½ × 200 N/m × (0.05 m)² = ½ × 200 × 0.0025 = 0.25 Joules

Question 2: A rubber band is stretched to a length of 0.10 m beyond its original length. If the elastic potential energy stored is 0.15 Joules, and the elastic constant is 30 N/m, find the extension x of the rubber band.

Answer: Rearranged formula: x = √(2E / k) = √(2 × 0.15 / 30) = √(0.01) ≈ 0.10 m

Question 1: Describe a simple experiment to measure the elastic constant (k) of a spring. Include the variables involved and safety precautions.

Answer: Attach a known mass to the spring and measure its extension with different masses. Plot extension against force to find the slope, which equals k. Variables: mass, extension. Safety precautions include ensuring the spring is securely fixed and handling weights carefully to prevent injury.

Question 2: Identify the independent and dependent variables in an experiment where a spring's extension is measured at different applied forces.

Answer: Independent variable: applied force (or mass). Dependent variable: extension of the spring.

Question 1: A spring is stretched with forces of 2 N, 4 N, and 6 N, causing extensions of 0.01 m, 0.02 m, and 0.03 m respectively. What does this data suggest about the spring's elastic behaviour? Explain your reasoning.

Answer: The data shows a proportional relationship between force and extension, indicating the spring obeys Hooke's Law within this range. The elastic constant can be determined from the slope of force versus extension.

Question 1: Explain why elastic potential energy increases when a spring is stretched further. Include the relevant formula in your explanation. (6 marks)

Answer: Elastic potential energy increases with the square of the extension (x), as given by E = ½ k x². When a spring is stretched further, more work is done against the restoring force, increasing stored energy quadratically. Thus, doubling the extension quadruples the energy stored, explaining why energy increases rapidly with greater stretch.

Question 1: Describe how elastic potential energy is utilised in the design of a trampoline. What safety considerations are involved?

Answer: A trampoline stores elastic potential energy when a person jumps onto it, allowing a rebound. Safety considerations include ensuring the frame and mat are strong enough to withstand repeated stretching, and that the elastic components are secure to prevent breakage.

Question 2: Discuss how understanding elastic potential energy can improve the design of energy-absorbing materials in sports equipment such as helmets or pads.

Answer: Knowledge of elastic potential energy helps in designing materials that deform elastically to absorb impact energy, reducing injury. Materials with high elastic energy storage capacity can dissipate energy efficiently during collisions.