A hockey puck sliding along frictionless ice with speed v to the right collides with a horizontal spring and compresses it by 2.0 cm before coming to a momentary stop. What will be... A hockey puck sliding along frictionless ice with speed v to the right collides with a horizontal spring and compresses it by 2.0 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?
Understand the Problem
The question involves a hockey puck compressing a spring. It initially compresses the spring by 2.0 cm at speed v. The problem then asks for the maximum compression of the spring when the puck hits it at a velocity of 2v. This requires an understanding of the physics behind kinetic energy and potential energy in springs.
Answer
The maximum compression is approximately $2.83 \, \text{cm}$.
Answer for screen readers
The maximum compression of the spring when the puck hits it at a speed of $2v$ is approximately $2.83 , \text{cm}$.
Steps to Solve
- Understand Energy Conservation
The initial kinetic energy of the puck is converted into potential energy in the spring at maximum compression.
- Write the Initial and Final Energy Equations
The initial kinetic energy when the puck is moving with speed $v$ is given by: $$ KE_i = \frac{1}{2} mv^2 $$
At maximum compression ($x_{\text{max}} = 2.0 , \text{cm} = 0.02 , \text{m}$), the potential energy stored in the spring is: $$ PE_f = \frac{1}{2} k x_{\text{max}}^2 $$
- Set Up Energy Equations for Initial Compression
Setting initial kinetic energy equal to potential energy gives us: $$ \frac{1}{2} mv^2 = \frac{1}{2} k (0.02)^2 $$
- Express Potential Energy for $2v$
When the puck hits the spring at speed $2v$, the initial kinetic energy is: $$ KE_i' = \frac{1}{2} m (2v)^2 = 2mv^2 $$
At maximum compression $x'{\text{max}}$, the potential energy is: $$ PE_f' = \frac{1}{2} k x'{\text{max}}^2 $$
- Set Up New Energy Equation
Now, set the kinetic energy at $2v$ equal to the potential energy at maximum compression: $$ 2mv^2 = \frac{1}{2} k x'_{\text{max}}^2 $$
- Relate the Two Situations
Using the relationship from the first situation, where: $$ mv^2 = \frac{1}{2} k (0.02)^2 $$ we find that: $$ k = \frac{mv^2}{0.02^2} $$
Substituting this back into the equation for the maximum compression: $$ 2mv^2 = \frac{1}{2} \left(\frac{mv^2}{0.02^2}\right) x'_{\text{max}}^2 $$
- Solve for (x'_{\text{max}})
Simplifying gives us: $$ 2 = \frac{x'_{\text{max}}^2}{0.02^2} $$
Thus, $$ x'{\text{max}}^2 = 2 \times (0.02)^2 $$ $$ x'{\text{max}} = \sqrt{2} \times 0.02 \approx 0.0283 , \text{m} $$
Converting to centimeters: $$ x'_{\text{max}} \approx 2.83 \text{ cm} $$
The maximum compression of the spring when the puck hits it at a speed of $2v$ is approximately $2.83 , \text{cm}$.
More Information
In this scenario, the energy conservation principle highlights how kinetic energy converts to potential energy in a compressed spring. The maximum spring compression is influenced by the initial speed of the puck.
Tips
- Not converting units correctly, especially between meters and centimeters.
- Forgetting to square the compression distance when calculating potential energy.
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