91Ó°ÊÓ

In the context of the exercise involving two ice skaters, kinetic friction plays a pivotal role in how the skaters eventually come to a stop. Initially, when the skaters push against each other, they move on the ice with minimal friction. Once they separate and begin to glide, kinetic friction starts acting on their skates, gradually slowing them down.

The kinetic frictional force is generally a resistive force that acts opposite to the direction of motion. It is calculated by the formula:

• Kinetic frictional force = coefficient of kinetic friction \( \times \) normal force.

Since the skaters are only affected by this force after they separate, the impact of kinetic friction is noticeable through their uniform deceleration. In this problem, it's crucial to know that the magnitudes of their accelerations from kinetic friction are equal, leading to one traveling twice the distance of the other before stopping.

Kinematic equations are used to describe the motion of objects, including those involved in constant acceleration, such as skaters coming to a stop due to friction. These equations allow us to calculate various parameters like velocity, acceleration, and displacement.

In our scenario, as both skaters slow down due to friction, the kinematic equations help relate the distance traveled to their initial velocities and accelerations. We use:

• \(v^2 = u^2 + 2as\), where \(v\) is the final velocity (zero, as they stop), \(u\) is the initial velocity, \(a\) is acceleration, and \(s\) is the distance traveled.

For Skater 1 and Skater 2 with distances \(d_1\) and \(d_2\) respectively, due to equal acceleration, we find the kinematic equation for each skater can be expressed as \(2ad_1 = v_1^2\) for Skater 1 and \(ad_2 = v_2^2\) for Skater 2. These equations help find the speed ratio and ultimately the mass ratio as one glides twice as far as the other.

The mass ratio in this problem provides insight into how different mass and speed relate due to the conservation of momentum. Given that Skater 1 travels twice the distance of Skater 2 and their accelerations remain the same due to friction, we can deduce their mass ratio from their velocities and distances.

• By using the conservation of momentum formula, \(m_1 v_1 = m_2 v_2\), the problem simplifies to finding \(\frac{m_1}{m_2}\).
• From the kinematic conclusions, we found \(\frac{v_1}{v_2} = 2\).

This speed ratio is then used in the momentum equation, showing that \(m_1 = \frac{m_2}{2}\), which clearly leads us to the conclusion that \(\frac{m_1}{m_2} = \frac{1}{2}\). Hence, Skater 1 is lighter compared to Skater 2, aligning with their respective velocities and distances traveled.