91Ó°ÊÓ

The Impulse-Momentum Theorem is a fundamental concept in physics that helps us understand how forces affect the motion of objects. Whenever a force acts on an object for a certain period of time, it changes the object's momentum. This change in momentum is precisely what we call **impulse**. Impulse is calculated as the product of force and the time duration over which it acts. In equation form, this is represented as:\[F \cdot \Delta t = \Delta p\]where \(F\) is the force applied, \(\Delta t\) is the time duration, and \(\Delta p\) is the change in momentum. Momentum, in turn, is the product of the mass \(m\) and velocity \(v\) of an object, given by \(p = m \cdot v\). Thus, when you apply an impulse to an object, it results in a change in the object's velocity if its mass is constant.For our exercise, the pool stick exerted an impulse of \(1.50\, \mathrm{N} \cdot \mathrm{s}\) on the cue ball. By using the theorem, we relate the impulse to the change in velocity immediately following the impact, knowing that the mass is \(0.165 \, \mathrm{kg}\). The formula applied was:\[1.50 = 0.165 \cdot v_1\]Through calculation, the velocity \(v_1\) was found to be approximately \(9.09 \, \mathrm{m/s}\). This is how the once stationary cue ball gained momentum to move and eventually collide with the second ball.

Moments before and after a collision, the total momentum of the system remains constant if no external forces interfere. This principle is what we call the Conservation of Momentum. Especially in closed systems, the momentum before an event is equal to the momentum afterward.In our exercise, we observe this principle in action as the cue ball collides with the second ball, which is initially stationary. The situation simplifies to a classical problem where one moving body transfers momentum to another.First, calculate the total initial momentum. Before the collision, only the first ball moves:\[p_{\text{initial}} = m \cdot v_1 = 0.165 \cdot 9.09\]This must equal the total final momentum, which distributes between both balls:\[p_{\text{initial}} = 0.165 \cdot v_1' + 0.165 \cdot v_2'\]The exercise solved this by substituting known values to show that after the impact, the combined final velocities conform to the initial momentum. A solution using momentum conservation alone gave:\[9.09 = v_1' + v_2'\]Thus, the velocity of the second ball, \(v_2'\), ultimately equals \(9.09 \, \mathrm{m/s}\) once calculations are complete.

When a collision is elastic, another crucial principle applies: the Conservation of Kinetic Energy. In such collisions, not only is the momentum conserved, but the total kinetic energy before and after the collision remains constant as well.For a moving cue ball colliding head-on with a stationary second ball, the initial kinetic energy \(KE_{\text{initial}}\) is:\[KE_{\text{initial}} = \frac{1}{2} \cdot 0.165 \cdot 9.09^2\]After the collision, each ball acquires some of this energy as their velocities change. The final kinetic energies summed up must still equal the initial energy:\[\frac{1}{2} \cdot 0.165 \cdot (v_1')^2 + \frac{1}{2} \cdot 0.165 \cdot (v_2')^2\]The relationship given by conservation simplifies to:\[9.09^2 = (v_1')^2 + (v_2')^2\]Using this, alongside momentum conservation, helps us find the individual velocities post-collision. Thus, a unique solution arises where the velocity of the second ball equals the initial velocity of the first ball, keeping total kinetic energy consistent across the event.