Question

The diagram shown represents two objects at rest on a frictionless horizontal surface with a spring compressed between them. When the compressed spring is released, the two objects are pushed apart. What is the total momentum of the two-object system that is shown after the expansion of the spring? Mass of object 1 is 2kg and Mass of object 2 is 1kg

Answer 1

To determine the total momentum of the two-object system after the spring is released, we can use the principle of conservation of momentum. According to this principle, the total momentum of a closed system remains constant if no external forces act on it.

Initially, both objects are at rest, so their initial velocities are zero. This means the initial momentum of the system is:

\[
\text{Initial momentum} = m_1 \cdot v_1 + m_2 \cdot v_2 = (2\, \text{kg} \cdot 0\, \text{m/s}) + (1\, \text{kg} \cdot 0\, \text{m/s}) = 0\,
\text{kg m/s}
\]

After the spring is released and the objects are pushed apart, the total momentum of the system must still equal the initial momentum, which is zero. Therefore:

\[
\text{Final momentum} = m_1 \cdot v_1' + m_2 \cdot v_2' = 0
\]

Where:
- \( m_1 = 2\, \text{kg} \) (mass of object 1)
- \( m_2 = 1\, \text{kg} \) (mass of object 2)
- \( v_1' \) is the final velocity of object 1
- \( v_2' \) is the final velocity of object 2

Since the total momentum is conserved, we can express this as:

\[
2\, \text{kg} \cdot v_1' + 1\, \text{kg} \cdot v_2' = 0
\]

This means that \( 2 v_1' + v_2' = 0 \). From this equation, we can conclude that the momentum in one direction will be equal in magnitude and opposite in direction to the momentum in the other direction, thereby ensuring that the total momentum remains zero.

Thus, the total momentum of the two-object system after the expansion of the spring is:

\[
\text{Total momentum} = 0\, \text{kg m/s}
\]