91Ó°ÊÓ

In the realm of physics, force and motion are two fundamental concepts that are intertwined. Force is essentially any interaction that, when unopposed, will change the motion of an object. It can be a push or a pull on an object, resulting in changes to the object’s motion. In the context of our problem, the cue stick applies a force to the pool ball, initiating its motion. This force is quantified as the product of mass and acceleration, as described by Newton's Second Law: \[ F = ma \] Where:

• \( F \) is the force applied,
• \( m \) is the mass of the object,
• \( a \) is its acceleration.

This relationship implies that force directly affects how an object accelerates in response to a given interaction, dictating the resultant motion of the object. Thus, understanding the role of force in changing an object's state of motion is essential to predict and analyze movements in various scenarios.

Momentum can be thought of as the quantity of motion an object possesses. It is directly influenced by the object's mass and velocity, described mathematically as: \[ p = mv \] Here, \( p \) represents momentum, \( m \) stands for mass, and \( v \) is velocity. When an external force is applied over a period, it results in a change in momentum, known as impulse. This relationship is encapsulated by the impulse-momentum theorem: \[ F\Delta t = \Delta p \] With \( \Delta p \) representing the change in momentum, \( F \) the applied force, and \( \Delta t \) the time duration of the force. In our example, this theorem helps us calculate how the force applied by the cue stick over a specific time interval changes the pool ball's momentum. The impulse effectively dictates the new momentum state of the ball, allowing us to infer changes in its speed.

Velocity quantifies how fast an object is moving in a specific direction. In our exercise, calculating the pool ball’s speed post-impact hinges on understanding its change in velocity. Once the change in momentum is known, we can relate it back to the change in velocity using: \[ \Delta p = m \Delta v \] Rewriting this gives us the formula to find the change in velocity:\[ \Delta v = \frac{\Delta p}{m} \] By substituting the values we've already determined (\( \Delta p = 0.448\, \mathrm{kg\cdot m/s} \) and \( m = 0.20\, \mathrm{kg} \)), we calculate: \[ \Delta v = \frac{0.448}{0.20} = 2.24 \, \mathrm{m/s} \] Thus, because the ball was initially at rest, its final velocity is essentially this change. Velocity calculation is crucial in deducing the resultant speed and direction of objects affected by forces.