A hockey puck sliding along frictionless ice with speed v to the right collides with a horizontal spring and compresses it by 1.0 cm before coming to a momentary stop. What will be the spring's maximum compression if the same puck hits it at a speed of 2v?

There are two energies in this problem, kinetic energy [tex] K= \frac{mv^{2}}{2} [/tex] and elastic potential energy [tex] U= \frac{kx^{2}}{2} [/tex] (with m the mass, v the velocity, x the compression and k the spring constant. ) so the total mechanical energy at every moment is the sum of the two energies:

[tex] E=K+U [/tex]

Here we have a situation where the total mechanical energy of the system is conserved because there are no dissipative forces (there's no friction), so:

[tex] E_{i}=E_{f} [/tex]

[tex]K_{i}+U_{i}=K_{f}+U_{f} [/tex]

Note that at the initial moment where the hockey puck has not compressed the spring all the energy of the system is kinetic energy, but for a momentary stop all the energy of the system is potential elastic energy, so we have:

[tex]K_{i} = U_{f} [/tex]

[tex]\frac{mv^{2}}{2}=\frac{kx^{2}}{2} [/tex] (1)

Due conservation of energy the equality (1) has to be maintained, so if we let k and m constant x has to increase the same as v to maintain the equality. Therefore, if we increase velocity to 2v we have to increase compression to 2x to conserve the equality. This is 2(1.0) = 2.0 cm