91Ó°ÊÓ

In physics, momentum refers to the quantity of motion an object has. Momentum is the product of an object's mass and its velocity and is conserved in a closed system without external forces. This means that the total momentum before and after an event, such as particles being released from a spring, remains constant.
For particle A and particle B, since there are no external forces, the momentum of the system is conserved. Since they start from rest, their initial momentum is zero. Therefore, their final momentum must also sum to zero. Let the mass of particle B be \(m\) and the mass of particle A be \(2m\). If the velocity of particle A is \(v_A\) and the velocity of particle B is \(v_B\), the conservation of momentum gives us:
\[2m v_A = m v_B\]
Rearranging this equation, we find that: \[v_B = 2 v_A\]
This relationship implies that particle B moves twice as fast as particle A in the opposite direction.

Kinetic energy is the energy that an object possesses due to its motion. The formula for kinetic energy (\(K\)) is given by:
\[K = \frac{1}{2} mv^2 \]
When the spring releases its stored energy, this energy is transferred to particles A and B as kinetic energy. With particle A having mass \(2m\) and particle B having mass \(m\), we can calculate their respective kinetic energies using their velocities:

• For particle A: \[K_A = \frac{1}{2} \times 2m \times v_A^2 = m v_A^2\]
• For particle B: \[K_B = \frac{1}{2} \times m \times (2v_A)^2 = 2 m v_A^2\]

Using the provided information that the total kinetic energy is 60 J, we sum the kinetic energies of both particles:
\[K_A + K_B = 60 J\]
Substituting the expressions from above, we get:
\[m v_A^2 + 2m v_A^2 = 60 J\]
Simplifying, we find:
\[3m v_A^2 = 60 J\]
Hence:
\[m v_A^2 = 20 J\]
Therefore, the kinetic energy of particle A (\(K_A\)) is 20 J, and the kinetic energy of particle B (\(K_B\)) is 40 J.

The relationship between mass and velocity is fundamental in understanding momentum and kinetic energy. In our problem, particle A has a mass twice that of particle B (\(m_A = 2m, m_B = m\)). Because momentum is conserved, the velocities of these particles are related inversely to their masses. We found that:
\[v_B = 2 v_A\]
This relationship shows that particle B, which has a smaller mass, will move faster than particle A. This inverse relationship is crucial when solving problems involving conservation of momentum. When dealing with kinetic energy, this mass-to-velocity relationship helps us determine how energy is distributed between the particles.
Since kinetic energy depends on the square of the velocity, even a small change in velocity leads to a significant change in kinetic energy. For particles in this system:

• The kinetic energy of particle A (\(K_A = 20 J\)) comes from its mass and its lower velocity.
• The kinetic energy of particle B (\(K_B = 40 J\)) is higher because, despite having half the mass of particle A, its velocity is doubled, making the velocity term in the kinetic energy equation larger.

Thus, the combined principles of mass, velocity, and energy conservation allow us to determine the behaviors and properties of moving particles.