**Solution:**

To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.

**Conservation of Momentum:**
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

The momentum of the first particle before the collision is given by:
\(p_1 = m_1 \cdot v_1\)

The momentum of the second particle before the collision is zero since it is at rest:
\(p_2 = 0\)

After the collision, the momentum of the first particle is given by:
\(p_1' = m_1 \cdot v_1'\)

The momentum of the second particle after the collision is given by:
\(p_2' = m_2 \cdot v_2'\)

Since the two particles have equal masses, \(m_1 = m_2\), we can simplify the equation to:
\(p_1' = p_2'\)

Using the conservation of momentum, we can write the equation as:
\(m_1 \cdot v_1 = m_1 \cdot v_1' + m_1 \cdot v_2'\)

**Conservation of Kinetic Energy:**
According to the principle of conservation of kinetic energy, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

The kinetic energy of the first particle before the collision is given by:
\(KE_1 = \frac{1}{2} \cdot m_1 \cdot v_1^2\)

The kinetic energy of the second particle before the collision is zero since it is at rest:
\(KE_2 = 0\)

After the collision, the kinetic energy of the first particle is given by:
\(KE_1' = \frac{1}{2} \cdot m_1 \cdot v_1'^2\)

The kinetic energy of the second particle after the collision is given by:
\(KE_2' = \frac{1}{2} \cdot m_2 \cdot v_2'^2\)

Since the two particles have equal masses, \(m_1 = m_2\), we can simplify the equation to:
\(KE_1 = KE_1' + KE_2'\)

Using the conservation of kinetic energy, we can write the equation as:
\(\frac{1}{2} \cdot m_1 \cdot v_1^2 = \frac{1}{2} \cdot m_1 \cdot v_1'^2 + \frac{1}{2} \cdot m_1 \cdot v_2'^2\)

**Solving the Equations:**
We now have two equations:
\(m_1 \cdot v_1 = m_1 \cdot v_1' + m_1 \cdot v_2'\)
\(\frac{1}{2} \cdot m_1 \cdot v_1^2 = \frac{1}{2} \cdot m_1 \cdot v_1'^2 + \frac{1}{2} \cdot m_1 \cdot v_2'^2\)

Dividing both sides of the first equation by \(m_1\), we get:
\(v_1 = v_1' + v_2'\)

Simplifying the second equation,