A hockey puck sliding along frictionless ice with speed v collides with a horizontal spring and compresses it by 2.0 cm before stopping. What will be the spring's maximum compressi... A hockey puck sliding along frictionless ice with speed v collides with a horizontal spring and compresses it by 2.0 cm before stopping. What will be the spring's maximum compression if the same puck hits at speed of 2v? Read more

Steps to Solve

1. Initial Energy Conservation Principle

When the puck compresses the spring by $2.0 \text{ cm}$ at speed $v$, the kinetic energy of the puck is transformed into elastic potential energy stored in the spring. The relationship is given by:

$$ \frac{1}{2}mv^2 = \frac{1}{2}kx^2 $$

Where $m$ is the mass of the puck, $k$ is the spring constant, and $x$ is the compression of the spring.

2. Using New Speed

Now, we consider the case where the puck hits the spring at double the speed $2v$. The kinetic energy at this speed is:

$$ KE = \frac{1}{2}m(2v)^2 = 2mv^2 $$

3. Equating Energies

At the new maximum compression $x'$, the equation for energy conservation can be established as:

$$ 2mv^2 = \frac{1}{2}kx'^2 $$

Since we know from the earlier case:

$$ \frac{1}{2}mv^2 = \frac{1}{2}kx^2 \implies kx^2 = mv^2 $$

Substituting back gives:

$$ 2mv^2 = 2kx^2 \implies kx'^2 = 4kx^2 $$

4. Solving for New Compression

From this, we have:

$$ x'^2 = 4x^2 $$

Taking the square root of both sides results in:

$$ x' = 2x $$

Substituting the initial compression:

$$ x' = 2(2.0 \text{ cm}) = 4.0 \text{ cm}$$