Question

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?

Answer 1

Answer:

The spring's maximum compression will be 2.0 cm

Explanation:

There are two energies in this problem, kinetic energy $K= \frac{mv^{2}}{2}$ and elastic potential energy $U= \frac{kx^{2}}{2}$ (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:

$E=K+U$

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:

$E_{i}=E_{f}$

$K_{i}+U_{i}=K_{f}+U_{f}$

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:

$K_{i} = U_{f}$

$\frac{mv^{2}}{2}=\frac{kx^{2}}{2}$ (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