Final answer:
The question is about energy conservation during a collision between a hockey puck and a spring on frictionless ice. When the speed of the puck is doubled before hitting the spring, the energy quadruples, resulting in twice the initial compression distance, making the correct answer 2.8 cm.
Step-by-step explanation:
The question relates to the concept of conservation of energy and spring compression from classical mechanics in physics, where a hockey puck collides with a spring on frictionless ice. When the puck with velocity v compresses the spring by 2.0 cm (0.02 m), we can assume conservation of energy implies kinetic energy (KE) is converted into potential energy stored in the spring (elastic potential energy).
KE can be expressed as (1/2)mv2 and spring potential energy as (1/2)kx2, where m is mass, v is velocity, k is the spring constant, and x is the compression distance. Since the spring constant and the mass of the puck remain the same, when the hockey puck hits the spring at speed 2v, the kinetic energy will be (1/2)m(2v)2, which is four times the initial kinetic energy because (2v)2 = 4v2.
To conserve energy, the potential energy in the spring must be equal to this new kinetic energy, or (1/2)kx2 = 4*(1/2)mv2. If the initial compression x is known (2.0 cm), then the final compression can be solved by taking the square root of 4 times the initial compression, giving a final compression of x = 2*sqrt(2) cm, which equals approximately 2.8 cm, so the correct answer is B. 2.8 cm.