To solve this problem, we can use the principles of conservation of momentum and the equations governing elastic collisions. In an elastic collision, both momentum and kinetic energy are conserved.
Let:
- \( m_1 = 48 \) kg (mass of the first train),
- \( m_2 = 52 \) kg (mass of the second train),
- \( u_1 = 0 \) m/s (initial velocity of the first train, at rest),
- \( u_2 = 24 \) m/s (initial velocity of the second train).
Let:
- \( v_1 \) be the final velocity of the first train after the collision,
- \( v_2 \) be the final velocity of the second train after the collision.
Step 1: Conservation of Momentum
The law of conservation of momentum states that the total momentum before the collision equals the total momentum after the collision:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Plugging in the known values:
\[ 48 \cdot 0 + 52 \cdot 24 = 48 v_1 + 52 v_2 \]
This simplifies to:
\[ 1248 = 48 v_1 + 52 v_2 \quad \text{(1)} \]
Step 2: Conservation of Kinetic Energy
In an elastic collision, kinetic energy is also conserved:
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Plugging in the values:
\[ \frac{1}{2} \cdot 48 \cdot 0^2 + \frac{1}{2} \cdot 52 \cdot 24^2 = \frac{1}{2} \cdot 48 v_1^2 + \frac{1}{2} \cdot 52 v_2^2 \]
This simplifies to:
\[ 0 + 1248 = 24 v_1^2 + 26 v_2^2 \quad \text{(2)} \]
Step 3: Solve the System of Equations
Now we have two equations to solve:
- \( 48 v_1 + 52 v_2 = 1248 \)
- \( 24 v_1^2 + 26 v_2^2 = 1248 \)
From equation (1), we can express \( v_1 \):
\[ v_1 = \frac{1248 - 52 v_2}{48} \]
We can substitute this expression for \( v_1 \) into equation (2):
\[ 24 \left(\frac{1248 - 52 v_2}{48}\right)^2 + 26 v_2^2 = 1248 \]
This substitution leads to a more complex equation, but it's easier to use the formulas for elastic collisions directly. The final velocities for an elastic collision can be directly calculated using:
\[ v_1 = \frac{(m_1 - m_2)u_1 + 2m_2u_2}{m_1 + m_2} \]
Substituting the values:
\[ v_1 = \frac{(48 - 52) \cdot 0 + 2 \cdot 52 \cdot 24}{48 + 52} = \frac{0 + 2496}{100} = 24.96 , \text{m/s} \]
And:
\[ v_2 = \frac{(m_2 - m_1)u_2 + 2m_1u_1}{m_1 + m_2} \]
Substituting again:
\[ v_2 = \frac{(52 - 48) \cdot 24 + 0}{100} = \frac{4 \cdot 24}{100} = \frac{96}{100} = 0.96 , \text{m/s} \]
Conclusion
Thus, the final velocity of the second train (52 kg) after the collision is approximately:
\[ \boxed{0.96 , \text{m/s}} \]