To determine the final velocity of a body that initially moves at \( 5 \, \text{m/s} \) after an elastic collision with an identical body moving at \( 3 \, \text{m/s} \), we apply the principles of conservation of momentum and kinetic energy.
Step 1: Variable Definition
- Mass: \( m_1 = m_2 = 1 \, \text{kg} \)
- Initial velocity of body 1: \( u_1 = 5 \, \text{m/s} \) (positive direction)
- Initial velocity of body 2: \( u_2 = -3 \, \text{m/s} \) (negative direction)
- Final velocities: \( v_1 \) and \( v_2 \)
Step 2: Momentum Conservation
\( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \). Given \( m_1 = m_2 \), this simplifies to \( u_1 + u_2 = v_1 + v_2 \).
Substituting initial velocities: \( 5 + (-3) = v_1 + v_2 \), resulting in \( 2 = v_1 + v_2 \quad \cdots (1) \).
Step 3: Kinetic Energy Conservation
For an elastic collision with identical masses, \( u_1^2 + u_2^2 = v_1^2 + v_2^2 \).
Substituting initial velocities: \( 5^2 + (-3)^2 = v_1^2 + v_2^2 \), yielding \( 34 = v_1^2 + v_2^2 \quad \cdots (2) \).
Step 4: Equation Solving
From equation (1), \( v_2 = 2 - v_1 \). Substituting into equation (2):
\( v_1^2 + (2 - v_1)^2 = 34 \)
\( v_1^2 + 4 - 4v_1 + v_1^2 = 34 \)
\( 2v_1^2 - 4v_1 - 30 = 0 \)
Dividing by 2: \( v_1^2 - 2v_1 - 15 = 0 \).
Using the quadratic formula, \( v_1 = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)} = \frac{2 \pm \sqrt{64}}{2} = \frac{2 \pm 8}{2} \).
The solutions for \( v_1 \) are \( v_1 = 5 \) or \( v_1 = -3 \).
Step 5: Determine \( v_2 \)
- If \( v_1 = 5 \), then \( v_2 = 2 - 5 = -3 \).
- If \( v_1 = -3 \), then \( v_2 = 2 - (-3) = 5 \).
Step 6: Result Interpretation
The two sets of solutions represent the exchange of velocities between the identical bodies during the elastic collision.
- The body initially at \( 5 \, \text{m/s} \) moves at \( -3 \, \text{m/s} \) (reversing direction).
- The body initially at \( -3 \, \text{m/s} \) moves at \( 5 \, \text{m/s} \) (reversing direction).
Step 7: Final Answer
The velocity of the body initially moving at \( 5 \, \text{m/s} \) after the collision is \( -3 \, \text{m/s} \).
Step 8: Verification
- Momentum: \( 5 + (-3) = 2 \) and \( (-3) + 5 = 2 \) (conserved).
- Kinetic energy: \( \frac{1}{2}(5^2) + \frac{1}{2}((-3)^2) = 17 \) and \( \frac{1}{2}((-3)^2) + \frac{1}{2}(5^2) = 17 \) (conserved).
Step 9: Conclusion
The final velocity of the body initially moving at \( 5 \, \text{m/s} \) is \( -3 \, \text{m/s} \).