Understanding Kinetic Energy Ratio on Inclined Plane

This problem involves understanding the concepts of potential energy, kinetic energy, and conservation of energy for objects moving on an inclined plane. We need to find the ratio of the kinetic energies of two blocks with different masses after they slide down a frictionless inclined plane starting from rest.

Physics Principles Applied

• Conservation of Energy: In a closed system where only conservative forces (like gravity) do work, the total mechanical energy (potential energy + kinetic energy) remains constant. Since the inclined plane is frictionless and we ignore air resistance, the potential energy lost by the blocks as they descend is converted into kinetic energy.
• Potential Energy (PE): The energy an object possesses due to its position relative to some zero point. For an object at height h, $PE = mgh$, where m is mass, g is the acceleration due to gravity.
• Kinetic Energy (KE): The energy an object possesses due to its motion. $KE = \frac{1}{2}mv^2$, where m is mass and v is velocity.

Step-by-Step Solution

1. Define Variables: Let the masses of the two blocks be m₁ and m₂. Let the height of the inclined plane be h. The blocks start from rest, so their initial velocity is 0. The plane is frictionless.
2. Mass Ratio: The problem states that the masses are in the ratio 2 : 3. We can write this as: $$ \frac{m_1}{m_2} = \frac{2}{3} $$ Let $m_1 = 2k$ and $m_2 = 3k$, where k is a constant.
3. Energy Calculation: When the blocks are at the top of the inclined plane (height h), their potential energy is maximum and kinetic energy is zero (since they start from rest). Potential Energy of block 1: $PE_1 = m_1gh = (2k)gh$ Potential Energy of block 2: $PE_2 = m_2gh = (3k)gh$ When the blocks reach the bottom (height 0), their potential energy is zero (relative to the bottom), and this potential energy is converted into kinetic energy. According to the conservation of energy: Kinetic Energy of block 1 at the bottom: $KE_1 = PE_1 = (2k)gh$ Kinetic Energy of block 2 at the bottom: $KE_2 = PE_2 = (3k)gh$
4. Ratio of Kinetic Energies: We need to find the ratio of their kinetic energies at the bottom: $$ \frac{KE_1}{KE_2} = \frac{(2k)gh}{(3k)gh} $$ The terms g, h, and k cancel out: $$ \frac{KE_1}{KE_2} = \frac{2}{3} $$

Conclusion

The kinetic energies of the two blocks at the bottom of the inclined plane are in the same ratio as their masses, which is 2 : 3. This happens because the potential energy difference (which determines the final kinetic energy) is directly proportional to the mass, given the same height change.