Problem 66 - College Physics

Question

A cue ball traveling at 4.00 m/s makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The cue ball is deflected so that it makes an angle of 30.0° with its original direction of travel. Find (a) the angle between the velocity vectors of the two balls after the collision and (b) the speed of each ball after the collision.

Step-by-Step Solution

Answer

The angle between the velocity vectors is found to be 46.2° and the speeds of the cue ball and target ball after the collision are found to be 3.46 m/s and 2.00 m/s, respectively.

Step 1: Applying the principle of conservation of momentum

The total momentum before the collision is equal to the total momentum after the collision. Since the target ball was initially at rest, the linear momentum is carried only by the cue ball. After the collision, the momentum is shared between the cue and the target ball. Let's denote the velocity of the target ball as \(v\). As per the conservation of momentum, we have \(m * 4.00 m/s = m * 4.00 m/s * cos(30.0°) + m * v * cos(θ)\), where \(θ\) is the angle we wish to find.

Step 2: Understanding the nature of elastic collisions

In an elastic collision, kinetic energy is also conserved. Hence the kinetic energy before the collision is equal to the kinetic energy after the collision. It gives us \(1/2 * m * (4.00 m/s)^2 = 1/2 * m * (4.00 m/s * cos(30.0°))^2 + 1/2 * m * v^2\).

Step 3: Solving for the unknowns

We're given two equations from conservation laws which we can solve for the unknowns \(v\) and \(θ\). By this method, we'll find the speed of the target ball (\(v\)) and the angle between the balls after collision (\(θ\)).

Step 4: Calculating the Speed of Each Ball after the Collision

Applying these two equations will provide the solution for the speed of each ball after the collision. The speed of the cue ball after the collision is \(4.00 m/s * cos(30.0°)\) and that of the target ball is \(v\).

Key Concepts

Conservation of Momentum

In physics, the principle of conservation of momentum is a fundamental concept. It states that in an isolated system, the total momentum remains constant over time. This principle is crucial in collision problems, especially in elastic collisions where no external forces act on the interacting bodies.

For our cue ball collision exercise, before the collision, only the cue ball has momentum since it is moving with a speed of 4.00 m/s, and the target ball is initially at rest. After the collision, the momentum is distributed between both balls. Mathematically, this is expressed by setting the initial momentum equal to the combined momentum of the cue and target balls following the collision.

To articulate this using equations:

• Initial Total Momentum: \[ m \times 4.00 \, \text{m/s} \]
• Final Total Momentum:\[ m \times 4.00 \, \text{m/s} \times \cos(30.0^\circ) + m \times v \times \cos(\theta) \]

This equation helps us analyze how the velocity directions and magnitudes change during the collision.

Conservation of Kinetic Energy

Elastic collisions are unique because they conserve kinetic energy as well as momentum. In our specified problem, the kinetic energy before and after the collision remains constant.

The law of conservation of kinetic energy ensures that:

• Kinetic Energy Before \[ \frac{1}{2} m (4.00 \, \text{m/s})^2 \]
• Kinetic Energy After \[ \frac{1}{2} m (4.00 \, \text{m/s} \times \cos(30.0^\circ))^2 + \frac{1}{2} m v^2 \]

This principle allows us to construct a second equation, necessary for solving for the unknown variables— the target ball’s speed and the angle between the balls after the collision. Together with the conservation of momentum, these equations provide a complete solution to this problem.

Two-Dimensional Collision Analysis

Analyzing collisions occurring in two dimensions requires a comprehensive approach, as it involves vector components of the velocities. The cue ball glancing off the target ball introduces a change in direction, making the analysis more challenging but also illustrative of real-world physics applications.

In our example, the cue ball is deflected at an angle of 30° from its original path, and we need to find the angle between the velocities of both balls after the collision. This requires understanding that each velocity can be split into horizontal and vertical components.

• Horizontal component of cue ball's velocity after collision: \[ 4.00 \, \text{m/s} \times \cos(30.0^\circ) \]
• Vertical component: \[ 4.00 \, \text{m/s} \times \sin(30.0^\circ) \]

By considering these components, we can derive the relative angle between the balls’ velocities using trigonometric identities and equations obtained from conservation laws.

Physics Problem Solving

Solving physics problems involves a systematic approach and understanding the underlying principles. In elastic collision problems, like with our cue ball example, it's essential to:

• Identify the laws of physics applicable (e.g., conservation of momentum and kinetic energy).
• Translate these laws into mathematical equations.
• Solve these equations to find unknowns such as velocities and angles.

Working through steps helps clarify the physical interactions and ensures that the solution aligns with both physical intuition and mathematical correctness.

Here, equations derived from conservation principles must be solved simultaneously. This process often involves algebraic manipulation, substitution, and the use of trigonometry, providing a comprehensive understanding of the collision scenario.