Problem 51 - University Physics with Modern Physics

Question

A hockey puck with mass \(0.250 \mathrm{~kg}\) 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?

Step-by-Step Solution

Answer

Answer: The final velocity of the second puck is 0.928 m/s, 23.9° below the blue line. The collision is elastic.

Step 1: Analyze the given information

We are given the following information: - The mass of both pucks (m1 = m2 = 0.250 kg) - The initial velocity of the first puck (v1i = 1.50 m/s along the blue line) - The final velocity of the first puck (v1f = 0.750 m/s) in a direction 30.0 degrees away from the blue line We will need to find the final velocity of the second puck (magnitude and direction) and determine if the collision is elastic.

Step 2: Set up the conservation of momentum equations for the x and y-components

Conservation of momentum states that the total momentum before the collision equals the total momentum after the collision. We need to consider both the x and y-components: For the x-component: m1 * v1i_x + m2 * v2i_x = m1 * v1f_x + m2 * v2f_x For the y-component: m1 * v1i_y + m2 * v2i_y = m1 * v1f_y + m2 * v2f_y

Step 3: Calculate the initial and final velocities of both pucks in x and y-components

First, note that the second puck is stationary, so its initial components are 0. Now, find the components of the first and second puck velocities: - v1i_x = 1.50 m/s (along the blue line) - v1i_y = 0 m/s (no vertical component) For the final velocities of the first puck, we need to use the given angle (30.0 degrees): - v1f_x = 0.750 m/s * cos(30.0) = 0.6495 m/s - v1f_y = 0.750 m/s * sin(30.0) = 0.375 m/s

Step 4: Calculate the final velocities of the second puck in x and y-components using the conservation of momentum equations

We have the initial and final velocities for the first puck, and the initial velocities for the second puck. Now we can find the final velocities of the second puck: For the x-component: (0.250 kg)(1.50 m/s) = (0.250 kg)(0.6495 m/s) + (0.250 kg) * v2f_x v2f_x = 0.8505 m/s For the y-component: 0 m/s = (0.250 kg)(0.375 m/s) + (0.250 kg) * v2f_y v2f_y = -0.375 m/s

Step 5: Calculate the magnitude and direction of the final velocity of the second puck

We now have the x and y-components of the final velocity of the second puck: v2f_x = 0.8505 m/s v2f_y = -0.375 m/s To find the magnitude, we can use the Pythagorean theorem: v2f = sqrt((v2f_x)^2 + (v2f_y)^2) v2f = 0.928 m/s To find the direction, we can use the inverse tangent function: θ = arctan(abs(v2f_y) / abs(v2f_x)) θ = 23.9° Since the y-component is negative, the direction is below the blue line.

Step 6: Check if the collision is elastic by comparing the initial and final kinetic energies

To determine if the collision is elastic, we need to compare the initial and final kinetic energies: Initial kinetic energy (KE_i): KE_i = 0.5 * m1 * (v1i)^2 = 0.5 * 0.250 kg * (1.50 m/s)^2 = 0.28125 J Final kinetic energy (KE_f): KE_f = 0.5 * m1 * (v1f)^2 + 0.5 * m2 * (v2f)^2 = 0.5 * 0.250 kg * (0.750 m/s)^2 + 0.5 * 0.250 kg * (0.928 m/s)^2 = 0.28125 J Since the initial and final kinetic energies are equal, this collision is elastic. The final velocity of the second puck is 0.928 m/s, 23.9° below the blue line, and the collision is elastic.

Key Concepts

Conservation of Momentum

In the fascinating world of physics, conservation of momentum is among the most steadfast principles, providing insights into interactions like those seen in the game of hockey. Imagine two hockey pucks sliding across the ice; when they collide, the total momentum of this system before and after the collision remains constant (if no external forces act on it). The concept is encapsulated in the elegant equation:

\( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \),

where \(m_1\) and \(m_2\) are the masses of the pucks, and \(v_{1i}, v_{2i}\) and \(v_{1f}, v_{2f}\) are their initial and final velocities, respectively. In our example, the momenta of the two pucks before and after they crash on the rink must be equal when you combine their individual momenta. The equation comes into play for both the horizontal (x) and vertical (y) components, reinforcing that momentum is a vector quantity comprising both magnitude and direction. By solving the momentum equations for each component separately, we can find the final velocities post-collision, which leads to a comprehensible illustration of the principle in action.

Kinetic Energy

Now let's glide into the realm of kinetic energy (KE), a term that essentially quantifies the energy of motion. For a single puck coasting on the ice, the kinetic energy can be calculated using the equation:

\( KE = \frac{1}{2}mv^2 \),

where \(m\) represents the mass and \(v\) is the velocity. It’s a scalar quantity so it doesn’t have direction but is tied intrinsically to how fast the puck is moving and how much it weighs. In our scenario, to verify whether the collision is elastic, meaning no kinetic energy is lost, we compare the total kinetic energy before and after the pucks collide. If the sum of the pucks' kinetic energies remains unchanged post-collision, as it does in this case (

• Initial KE: \( 0.28125 J \)
• Final KE: \( 0.28125 J \)

), we can affirm the collision is indeed elastic. An elastic collision is akin to a perfectly choreographed dance—diverse steps and movements but the dance's energy remains constant.

Vector Components

Navigating into the territory of vector components requires an understanding of both magnitude and direction. Vectors are fundamental in physics as they represent quantities with both of these attributes, like velocity.

Think of the ice hockey puck not just moving in a straight line but perhaps in two dimensions, slicing across the ice at an angle. To address such a scenario, one would decompose the vector, in this case, the puck’s velocity, into both x (horizontal) and y (vertical) components. This can be done using trigonometry, specifically,

• for the x-component: \( v_x = v \cos(\theta) \)
• for the y-component: \( v_y = v \sin(\theta) \)

,
where \(\theta\) is the angle with respect to the horizontal. Once the x and y components are known, they can be aggregated to understand the overall velocity and direction of the motion. The final velocity can be found using the Pythagorean theorem: \( v = \sqrt{v_x^2 + v_y^2} \) and the angle by the inverse tangent function: \( \theta = \arctan\left(\frac{|v_y|}{|v_x|}\right) \). In regards to the hockey pucks, dissecting the velocity into its vector components allows us to evaluate the outcome of the collision with greater accuracy.