91Ó°ÊÓ

Impulse is a crucial concept in physics that describes the effect of a force applied over a time interval. The impulse-momentum theorem provides a solid bridge between forces experienced by an object and its motion. This principle states that the impulse applied to an object equals its change in momentum. Mathematically, this is expressed as: \[ J = \Delta p = m\Delta v \] where \( J \) is the impulse, \( \,m\) is the mass of the object, and \( \Delta v \) is the change in its velocity.
### Example in ActionLet's consider the cue ball in our problem: The pool stick gives an impulse of \(+1.50 \, \text{N} \cdot \text{s}\) to the stationary ball. Given that the ball's mass is \(0.165 \, \text{kg}\), we can use this relationship to determine how its velocity changes.
By rearranging the formula to solve for \( \Delta v \), we get:\[ \Delta v = \frac{J}{m} = \frac{1.50}{0.165} \approx 9.09 \, \text{m/s} \] So, the cue ball's velocity increases to \(9.09 \, \text{m/s}\) due to the applied impulse.

Momentum is often called "quantity of motion". Conservation of momentum is a fundamental principle used especially in analyzing collisions. It states that in the absence of external forces, the total momentum of a system remains constant before and after a collision. ### Momentum in Elastic CollisionsIn the exercise, our cue ball and the second ball, both having the same mass (\(0.165 \, \text{kg}\)), collide elastically. This means both momentum and kinetic energy are conserved. Before the collision, only the cue ball has momentum:\[ p_{\text{initial}} = m_1v_1 + m_2v_2 \] where \( v_2 = 0 \). After the collision, both balls have momentum, and the equation becomes:\[ m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' \] Given that the initial velocity of the second ball, \( v_2 \), is zero, and the masses are equal, this simplifies to \( v_1'+v_2' = v_1 \). Thus, if the cue ball stops (\( v_1' = 0 \)), then\[ v_2' = v_1 = 9.09 \, \text{m/s} \] Post-collision, the second ball heads off with a velocity of \(9.09 \, \text{m/s}\) as momentum shifts efficiently.

When two objects undergo an elastic collision, another important conservation law applies: the conservation of kinetic energy. Elastic collisions are special because they result in no net loss of the total kinetic energy of the system.
### Understanding Kinetic Energy in CollisionsFor the collision concerning the cue ball problem, kinetic energy initially is:\[ KE_{\text{initial}} = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \] Since \( v_2 = 0 \), the expression simplifies for the cue ball alone:\[ KE_{\text{initial}} = \frac{1}{2} \times 0.165 \times (9.09)^2 \] After the collision, both balls move, and the formula becomes:\[ KE_{\text{final}} = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2 \]To satisfy the condition for an elastic collision, \( KE_{\text{initial}} = KE_{\text{final}} \).
Due to the symmetry and equal masses, if the cue ball comes to rest (\( v_1' = 0\)), the second ball absorbs its kinetic energy, moving precisely at the speed needed to balance both equations:\[ v_2' = 9.09 \, \text{m/s} \] This showcases the perfect transfer of kinetic energy seen in an ideal elastic collision scenario.