91Ó°ÊÓ

In physics, the law of conservation of momentum states that the total momentum of a closed system remains constant if it is not influenced by external forces.

In our collision problem, we're dealing with two particles that stick together after colliding. This type of collision is known as a perfectly inelastic collision. The initial total momentum of the system is the vector sum of the momenta of both particles.

Before the collision:

• Particle A's momentum: \( \vec{p}_{A1} = m_A \times \vec{v}_{A1} \)
• Particle B's momentum: \( \vec{p}_{B1} = m_B \times \vec{v}_{B1} \)

After the collision, the combined momentum:
\[ \vec{p}_{total} = \vec{p}_{A1} + \vec{p}_{B1} \]
The momentum remains the same before and after the collision because no external forces are acting on the system. This forms the basis for calculating the final velocity of the combined mass.

Vector notation helps in representing physical quantities that have both magnitude and direction.

In our context, we use unit vectors \( \hat{\text{i}} \) and \( \hat{\text{j}} \) to represent the directions along the x-axis and y-axis respectively.

For example:

• Velocity of Particle A: \( \vec{v}_{A1} = 4.0 \mathrm{~m/s} \hat{\text{i}} \)
• Velocity of Particle B: \( \vec{v}_{B1} = 2.0 \mathrm{~m/s} \hat{\text{j}} \)

After the collision, the total momentum in vector notation becomes a crucial part of finding the final velocity:

\[ \vec{p}_{total} = 8 \mathrm{~kg \cdot m/s} \hat{\text{i}} + 8 \mathrm{~kg \cdot m/s} \hat{\text{j}} \]

Collisions can be classified into various types based on how the colliding entities interact and the resulting effects. In this case, we have a perfectly inelastic collision where the particles stick together.

Key points to note:

• The kinetic energy is not conserved in inelastic collisions, but the total momentum always is.
• During a perfectly inelastic collision, the two objects move together with a common velocity after the impact.

The combined mass after the collision is the sum of the individual masses: \( m_{total} = m_A + m_B = 6.0 \mathrm{~kg} \)

To find the final velocity of the combined mass, use the formula for momentum:

\[ \vec{v}_{final} = \frac{\vec{p}_{total}}{m_{total}} \]
By substituting the respective values:

\[ \vec{v}_{final} = \frac{8 \hat{\text{i}} + 8 \hat{\text{j}}}{6.0} = 1.33 \mathrm{~m/s} \hat{\text{i}} + 1.33 \mathrm{~m/s} \hat{\text{j}} \]
This velocity vector indicates that the combined mass post-collision will move with equal components in both the x and y directions.

To determine the direction of the final velocity, you can calculate the angle it makes with a reference direction. Typically, the x-axis is chosen as the reference.

The angle \( \theta \) can be calculated using the arctangent function:

\[ \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) \]
For our problem, the final velocities in the x and y directions are equal:

\[ \theta = \tan^{-1}\left(\frac{1.33}{1.33}\right) = 45^\circ \]
Thus, the final velocity vector forms a 45-degree angle with the x-axis, indicating that it moves diagonally in the combined direction of the initial velocities.