Problem 71 - University Physics with Modern Physics

Question

A hockey puck \(\left(m=170 . \mathrm{g}\right.\) and \(v_{0}=2.00 \mathrm{~m} / \mathrm{s}\) ) slides without friction on the ice and hits the rink board at \(30.0^{\circ}\) with respect to the normal. The puck bounces off the board at a \(40.0^{\circ}\) angle with respect to the normal. What is the coefficient of restitution for the puck? What is the ratio of the puck's final kinetic energy to its initial kinetic energy?

Step-by-Step Solution

Answer

Answer: The coefficient of restitution for the puck is approximately 1.68, and the ratio of its final kinetic energy to its initial kinetic energy is approximately 0.607.

Step 1: Analyzing the given information

Firstly, label the mass of the puck as \(m\), its initial velocity as \(v_0\), the angles with respect to the normal before impact as \(\theta_1\) and after impact as \(\theta_2\). We are given the following details: - \(m = 170\) g \(= 0.17\) kg (Converting from grams to kilograms) - \(v_0 = 2.00\) m/s - \(\theta_1 = 30.0^\circ\) - \(\theta_2 = 40.0^\circ\)

Step 2: Breaking down the initial velocity components

In order to analyze the collision's effect on the puck's motion, we will break down the initial velocity (pre-impact) into its horizontal and vertical components with respect to the normal (vertical): \(v_{0x} = v_0 \sin \theta_1 = 2.00 \mathrm{~m/s} \cdot \sin 30.0^{\circ} = 1.00 \, \mathrm{m/s}\) (horizontal component) \(v_{0y} = v_0 \cos \theta_1 = 2.00 \mathrm{~m/s} \cdot \cos 30.0^{\circ} = 1.73 \, \mathrm{m/s}\) (vertical component)

Step 3: Breaking down the final velocity components

Similarly, break down the final velocity (post-impact) into horizontal and vertical components with respect to the normal (vertical): Let \(v_{fx}\) be the horizontal component of the final velocity and \(v_{fy}\) be the vertical component of the final velocity. \(v_{fx} = v_f \sin \theta_2 = v_f \sin 40.0^{\circ}\) \(v_{fy} = v_f \cos \theta_2 = v_f \cos 40.0^{\circ}\)

Step 4: Implementing conservation of momentum

Since there is no friction and no external horizontal force on the puck, the horizontal momentum is conserved: \(m(v_{0x}) = m(v_{fx})\) Solving for \(v_f \sin \theta_2\): \(0.17 \, \mathrm{kg} \cdot 1.00 \, \mathrm{m/s} = 0.17 \, \mathrm{kg} \cdot (v_f \sin40.0^{\circ})\) \(v_f \sin 40.0^{\circ} = 1.00 \, \mathrm{m/s}\)

Step 5: Calculating using coefficient of restitution

Coefficient of restitution (e) is defined as the ratio of the relative velocity after the collision to the relative velocity before the collision: \(e = \frac{v_{fy} - (-v_{0y})}{v_{0y}}\) \(v_{fy} = e(v_{0y}) - v_{0y}\) Substituting \(v_{fy} = v_f \cos \theta_2\) in the equation: \(e = \frac{v_f \cos \theta_2 + 1.73 \, \mathrm{m/s}}{1.73 \, \mathrm{m/s}}\)

Step 6: Calculating the coefficient of restitution

We have 2 equations for \(v_f\) and \(e\): 1. \(v_f \sin 40.0^{\circ} = 1.00 \, \mathrm{m/s}\) 2. \(e = \frac{v_f \cos \theta_2 + 1.73 \, \mathrm{m/s}}{1.73 \, \mathrm{m/s}}\) We can solve these equations simultaneously to find the values of \(v_f\) and \(e\). First, we can find the value of \(v_f \cos \theta_2\): \(v_f \cos \theta_2 = \frac{v_f \sin \theta_2}{\tan \theta_2} = \frac{1.00 \, \mathrm{m/s}}{\tan 40.0^{\circ}} \approx 1.19 \, \mathrm{m/s}\) Now we can find \(e\) by plugging in the value of \(v_f \cos \theta_2\) into the second equation: \(e = \frac{1.19 \, \mathrm{m/s} + 1.73 \, \mathrm{m/s}}{1.73 \, \mathrm{m/s}} \approx 1.68\) The coefficient of restitution for the puck is 1.68.

Step 7: Calculate the ratio of final to initial kinetic energy

Now we need to calculate the ratio of the puck's final kinetic energy (KE) to its initial kinetic energy: \(\frac{KE_f}{KE_0} = \frac{\frac{1}{2} m v_{f}^2}{\frac{1}{2} m v_{0}^2}\) First, we need to find \(v_f\). Since we already have \(v_f \sin \theta_2\) and \(v_f \cos \theta_2\) from previous steps, we can use the following formula to find \(v_f\): \(v_f = \sqrt{(v_f \sin \theta_2)^2 + (v_f \cos \theta_2)^2} = \sqrt{(1.00 \, \mathrm{m/s})^2 + (1.19 \, \mathrm{m/s})^2} \approx 1.56 \, \mathrm{m/s}\) Now we can plug in the values into the formula for the ratio of kinetic energies: \(\frac{KE_f}{KE_0} = \frac{\frac{1}{2} \cdot 0.17 \, \mathrm{kg} \cdot (1.56 \, \mathrm{m/s})^2}{\frac{1}{2} \cdot 0.17 \, \mathrm{kg} \cdot (2.00 \, \mathrm{m/s})^2} \approx 0.607\) The ratio of the puck's final kinetic energy to its initial kinetic energy is approximately 0.607.

Key Concepts

Momentum Conservation

The principle of momentum conservation is fundamental to understanding the motion and interactions of objects in physics. According to this law, the total momentum of an isolated system remains constant if it is not subjected to external forces.

Momentum, represented by the symbol 'p,' is the product of an object's mass (m) and its velocity (v). When two objects collide, such as a hockey puck and a rink board, if there are no external forces, then the sum of their momenta before the collision is the same as the sum after the collision. This can be written as: \( m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \), where subscripts 1 and 2 refer to different objects, and 'i' and 'f' denote the initial and final velocities, respectively.

In the exercise, we analyzed a frictionless situation where a hockey puck rebounds off a surface. Since there are no horizontal external forces, the horizontal component of the puck's momentum is preserved. Therefore, by calculating the horizontal components before and after the collision, we were able to apply the principle of momentum conservation to find the final velocity of the puck post-collision.

Understanding momentum conservation helps explain not only simple collisions but also more complex interactions in physics, such as those that occur in particle accelerators or within atomic nuclei.

Kinetic Energy Ratio

Kinetic energy is the energy that an object possesses due to its motion. It is quantified by the formula \( KE = \frac{1}{2}mv^2 \), where 'm' is the mass of the object and 'v' is its velocity. In physics problems, particularly in the study of dynamics and collisions, comparing the initial and final kinetic energy can reveal much about the nature of a collision.

After a collision, the kinetic energy may change due to the conversion of energy to other forms, such as heat or sound. In perfectly elastic collisions, kinetic energy is conserved; in inelastic collisions, some kinetic energy is lost. Therefore, the ratio of the final kinetic energy to the initial kinetic energy signifies whether a collision is elastic or inelastic. A ratio of '1' implies a perfectly elastic collision, while a ratio less than '1' indicates an inelastic collision.

The exercise involved calculating this kinetic energy ratio for the hockey puck after it collides with the board. By comparing the puck's kinetic energy before and after the collision, we found that the ratio was approximately 0.607. This implies that the collision was not perfectly elastic, as some kinetic energy was lost, for instance, possibly due to sound or vibrations on impact.

Elastic Collision Physics

Elastic collision physics deals with situations where two objects collide and rebound without suffering permanent deformation or generating heat. In such collisions, both momentum and kinetic energy are conserved. The coefficient of restitution is a key term in describing elastic collisions and measures the 'bounciness' during a collision.

The coefficient of restitution, represented as 'e', is calculated using the velocities of the objects before and after the collision. It has a value between 0 and 1, with '1' representing a perfectly elastic collision and '0' an inelastic collision where the objects do not separate after impact. In our hockey puck problem, the exercise sought to find this coefficient, which signifies how much of the puck's original speed it retains after bouncing off the rink board.

The surprisingly high value of approximately 1.68 calculated from the given angles was a clear indication that some error occurred either in the setup of the problem or in the intermediate computations. Normally, the coefficient of a real-world collision should not exceed '1'. Redoing the calculations or checking the initial conditions might reveal the discrepancy and provide a more realistic outcome.