Question

A hockey puck with mass \(0.170 \mathrm{~kg}\) traveling along the blue line (a blue-colored straight line on the ice in a hockey rink) at \(1.50 \mathrm{~m} / \mathrm{s}\) strikes a stationary puck with the same mass. The first puck exits the collision in a direction that is \(30.0^{\circ}\) away from the blue line at a speed of \(0.750 \mathrm{~m} / \mathrm{s}\) (see the figure). What is the direction and magnitude of the velocity of the second puck after the collision? Is this an elastic collision?

Short answer

Answer: To find the magnitude and direction of the final velocity of the second hockey puck, use the x and y-components of the final velocity found in Step 4: Magnitude: v2_final = sqrt(v2_final_x^2 + v2_final_y^2) Direction: theta = atan(v2_final_y / v2_final_x), converted to degrees if needed. Determine the nature of the collision by comparing the initial and final kinetic energies (Step 5). If KE_initial = KE_final, the collision is elastic. If not, the collision is inelastic.

Step-by-step solution

Analyze the Problem and Identify Known and Unknown Variables

We have the following known variables: - Mass of the first puck (m1): 0.170 kg - Velocity of the first puck before collision (v1_initial): 1.50 m/s - Mass of the second (stationary) puck (m2): 0.170 kg - Velocity of the first puck after collision (v1_final): 0.750 m/s - Angle between the initial direction of the first puck and its final direction after collision: 30.0 degrees - Direction and magnitude of the velocity of the second puck after the collision: Unknown We can use the conservation of momentum to find the unknowns.

Determine the initial and final momenta of each puck

Let's first determine the total initial and final momenta of each puck. Initial momentum of first puck (p1_initial) = m1 * v1_initial Initial momentum of second puck (p2_initial) = m2 * 0 (since the second puck is stationary) Final momentum of first puck (p1_final) = m1 * v1_final Final momentum of second puck (p2_final) = m2 * v2_final (unknown) Total initial momentum (p_initial) = p1_initial + p2_initial Total final momentum (p_final) = p1_final + p2_final

Apply conservation of momentum to find the velocity of the second puck

According to the conservation of momentum: p_initial = p_final p1_initial + p2_initial = p1_final + p2_final Since the second puck was initially stationary: p1_initial = p1_final + p2_final Now let's break the final momenta into x and y-components: p1_initial_x = p1_final_x + p2_final_x m1 * v1_initial = m1 * v1_final * cos(30) + m2 * v2_final_x p1_initial_y = p1_final_y + p2_final_y 0 = m1 * v1_final * sin(30) + m2 * v2_final_y Now we can solve each equation for v2_final_x and v2_final_y: v2_final_x = (m1 * v1_initial - m1 * v1_final * cos(30)) / m2 v2_final_y = (-m1 * v1_final * sin(30)) / m2

Calculate the magnitude and direction of the final velocity of the second puck

Now that we have the x and y-components of the final velocity of the second puck, we can find its magnitude and direction. Magnitude: v2_final = sqrt(v2_final_x^2 + v2_final_y^2) Direction: theta = atan(v2_final_y / v2_final_x) Remember to convert theta to degrees if needed.

Check if the collision is elastic

An elastic collision conserves both momentum and kinetic energy. We have already shown that momentum is conserved in this collision, so let's now check if the kinetic energy is conserved as well. The initial and final kinetic energies of the pucks are as follows: Initial kinetic energy (KE_initial) = 0.5 * m1 * v1_initial^2 Final kinetic energy (KE_final) = 0.5 * m1 * v1_final^2 + 0.5 * m2 * v2_final^2 If KE_initial = KE_final, then the collision is elastic. Otherwise, it is inelastic. Compare the two kinetic energy values to determine the nature of the collision.

Key concepts

Elastic Collision

When two objects collide and bounce off each other without any loss in total kinetic energy, we describe the event as an elastic collision. It is a type of collision where both momentum and kinetic energy are conserved. In the context of the hockey puck problem, our aim is to determine if the collision between the two pucks is elastic. After establishing that momentum is conserved in the collision, as seen in the step-by-step solution, we must also check for the conservation of kinetic energy. To do this, we calculate the kinetic energy before and after the collision and compare the values.

In an ideal elastic collision, no energy is dissipated as heat, sound, or deformation. Objects involved in such collisions typically have very little interaction time and rebound quickly, making them approximately elastic. Common examples include collisions between billiard balls or steel bearings. Understanding elastic collisions is crucial for problems involving particle physics, engineering, and materials science, where the preservation of energy is essential.

Kinetic Energy

Kinetic energy (KE) represents the energy that an object possesses due to its motion. It is directly proportional to the mass of the object and the square of its velocity. Given by the formula,
KE = \(\frac{1}{2} m v^2\),
kinetic energy quantifies the work required to accelerate an object from rest to its current velocity. In a closed system, the total kinetic energy can change form through interactions but, if the system is isolative, its total energy remains constant. This is a core concept behind momentum and energy conservation laws applied in collision problems, such as the hockey puck exercise. When assessing whether a collision is elastic or inelastic, comparing the total kinetic energies before and after the event reveals whether kinetic energy is conserved. For the collision to be considered elastic, the total kinetic energy before collision (initial kinetic energy) should equal the total kinetic energy after collision (final kinetic energy).

Vector Components

In physics, vectors represent quantities that have both a magnitude (size) and a direction, such as velocity, acceleration, or force. Vectors can be broken down into vector components, which represent the effect of the vector in specific, usually perpendicular, directions—traditionally labeled as the x-axis (horizontal) and y-axis (vertical) components. These components allow us to analyze multidimensional motion on a one-dimensional basis, making calculations more manageable.

For example, in the hockey puck problem, we decompose the final momentum of the first puck into x and y components to solve for the velocity of the second puck after the collision. This decomposition uses trigonometric functions, such as sine and cosine, to find the components parallel (x-axis) and perpendicular (y-axis) to the blue line.

Application in Elastic Collisions

Decomposing vectors into their components is crucial when solving for velocities in elastic collisions, as it allows us to apply the conservation laws in each directional component separately. Once the components are found, we then use Pythagorean theorem and trigonometry to determine the overall magnitude and direction of the resultant vector.