Problem 99 - University Physics with Modern Physics

Question

Hockey puck \(B\) rests on a smooth ice surface and is struck by a second puck \(A,\) which has the same mass. Puck \(A\) is initially traveling at 15.0 \(\mathrm{m} / \mathrm{s}\) and is deflected \(25.0^{\circ}\) from its initial direction. Assume that the collision is perfectly elastic. Find the final speed of each puck and the direction of \(B\) 's velocity after the collision.

Step-by-Step Solution

Answer

Puck A moves at 13.6 m/s, Puck B at 7.64 m/s, and B's angle is 65°.

Step 1: Understand the Problem

We have two pucks, A and B, with the same mass. Puck A is moving initially at 15.0 m/s and is deflected 25° from its original direction. The collision is perfectly elastic, meaning both momentum and kinetic energy are conserved. We need to find the final speed of each puck and the direction of puck B's velocity after the collision.

Step 2: Apply Conservation of Momentum

Since the collision is perfectly elastic, apply the conservation of linear momentum in both the x-direction and y-direction.1. In the x-direction: \[ m_A v_{A,i} = m_A v_{A,f} \cos(25^{\circ}) + m_B v_{B,f} \cos(\theta) \] Where \( m_A = m_B\) and \( v_{A,i} = 15.0 \text{ m/s} \).2. In the y-direction: \[ 0 = m_A v_{A,f} \sin(25^{\circ}) - m_B v_{B,f} \sin(\theta) \] Puck B starts from rest, hence initial y-momentum is zero.

Step 3: Apply Conservation of Kinetic Energy

For a perfectly elastic collision, kinetic energy is also conserved:\[ \frac{1}{2} m_A v_{A,i}^2 = \frac{1}{2} m_A v_{A,f}^2 + \frac{1}{2} m_B v_{B,f}^2 \]Which simplifies to:\[ v_{A,i}^2 = v_{A,f}^2 + v_{B,f}^2 \]Substitute the initial speed of puck A (15.0 m/s) and solve for the final velocities.

Step 4: Solve the Equations

Solve the system of equations derived from step 2 and step 3.1. Use equation from conservation of kinetic energy to express one velocity in terms of the other.2. Substitute this into the conservation of momentum equations.3. Solve for the angle \( \theta \) and the final speeds \( v_{A,f} \) and \( v_{B,f} \).

Key Concepts

Conservation of Momentum

In an elastic collision, the principle of conservation of momentum is key. Momentum is a vector, which means it has both magnitude and direction. In our hockey puck collision, both pucks have equal masses, making the problem symmetrical and simplifying the conservation equations.
To apply conservation of momentum, consider the x and y directions separately. Before the collision:

• Puck A moves in the x-direction with momentum equal to its mass times its velocity.
• Puck B is stationary, so it has no initial momentum.

The total momentum before the collision in the x-direction is due only to puck A. After the collision, both pucks have momentum contributions in both directions. We set the total initial momentum equal to the total final momentum in both directions:

• In the x-direction: \[ m_A v_{A,i} = m_A v_{A,f} \cos(25^{\circ}) + m_B v_{B,f} \cos(\theta) \]
• In the y-direction: \[ 0 = m_A v_{A,f} \sin(25^{\circ}) - m_B v_{B,f} \sin(\theta) \]

This fundamentally illustrates the idea that momentum is conserved in each separate direction. Solving these will give you the velocities and direction needed for each puck.

Conservation of Kinetic Energy

A key characteristic of a perfectly elastic collision is that kinetic energy remains constant. This means the sum of kinetic energies before the collision is equal to the sum of kinetic energies after the collision.
Kinetic energy is given by the formula: \[ \text{Kinetic Energy} = \frac{1}{2}mv^2 \]
Both pucks together have the same total kinetic energy before and after the collision. Initially, only puck A has kinetic energy since puck B is at rest. Post-collision, the energy will be distributed between the two pucks:

• Initial kinetic energy: \( \frac{1}{2} m_A v_{A,i}^2 \)
• Final kinetic energy: \( \frac{1}{2} m_A v_{A,f}^2 + \frac{1}{2} m_B v_{B,f}^2 \)

Setting these two equal, as shown in our solution, lets you solve for the final velocities by simplifying:\[ v_{A,i}^2 = v_{A,f}^2 + v_{B,f}^2 \]
This process assures you that none of the kinetic energy is lost and helps determine the final velocities of each puck.

Perfectly Elastic Collision

In a perfectly elastic collision, both momentum and kinetic energy are conserved. This type of collision implies no energy is lost as heat or sound, and the objects simply bounce off each other.
In our scenario with hockey pucks, each puck maintains some velocity post-collision without converting kinetic energy to other forms. The physics of this process can be visualized akin to billiard balls bouncing off one another.
While most real-world collisions aren’t perfectly elastic, understanding this concept is crucial in physics because:

• It provides an ideal model to develop analytical skills.
• It helps predict outcomes knowing the exact conditions.
• It's a baseline to compare against less elastic collisions.

This perfect exchange of energy and momentum results in well-defined final velocities that the equations derived will yield, serving as a suitable approximation for many practical applications.