Question

Two identical bodies of mass 1 kg each are moving towards each other with velocities of 5 m/s and 3 m/s, respectively. They collide elastically. What will be the velocity of the body initially moving at 5 m/s after the collision?

• A. -3 m/s
• B. 3 m/s
• C. -5 m/s
• D. 5 m/s

Hint:

For elastic collisions of identical masses, the bodies exchange velocities. Use conservation of momentum and energy to solve.

Answer 1

The Correct Option is A

To find the velocity of the body initially moving at 5 m/s after an elastic collision with an identical body moving at 3 m/s, we use the principles of conservation of momentum and kinetic energy, since the collision is elastic.
Step 1: Define variables 
- Mass of both bodies \( m_1 = m_2 = 1 \, \text{kg} \)
- Initial velocity of the first body \( u_1 = 5 \, \text{m/s} \) (moving right)
- Initial velocity of the second body \( u_2 = -3 \, \text{m/s} \) (moving left, opposite direction)
- Final velocities are \( v_1 \) (for the first body) and \( v_2 \) (for the second body)
Step 2: Conservation of momentum 
For an elastic collision, momentum is conserved: \[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \] Since \( m_1 = m_2 = 1 \, \text{kg} \): \[ u_1 + u_2 = v_1 + v_2 \] Substitute the initial velocities: \[ 5 + (-3) = v_1 + v_2 \] \[ 2 = v_1 + v_2 \quad \cdots (1) \] 
Step 3: Conservation of kinetic energy 
For an elastic collision, kinetic energy is also conserved: \[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \] Since \( m_1 = m_2 = 1 \, \text{kg} \), cancel \( \frac{1}{2} \) and \( m \): \[ u_1^2 + u_2^2 = v_1^2 + v_2^2 \] Substitute the initial velocities: \[ 5^2 + (-3)^2 = v_1^2 + v_2^2 \] \[ 25 + 9 = v_1^2 + v_2^2 \] \[ 34 = v_1^2 + v_2^2 \quad \cdots (2) \] 
Step 4: Solve the equations 
From equation (1): \[ v_2 = 2 - v_1 \] Substitute into equation (2): \[ v_1^2 + (2 - v_1)^2 = 34 \] Expand: \[ v_1^2 + (4 - 4v_1 + v_1^2) = 34 \] \[ 2v_1^2 - 4v_1 + 4 = 34 \] \[ 2v_1^2 - 4v_1 - 30 = 0 \] Divide by 2: \[ v_1^2 - 2v_1 - 15 = 0 \] Solve this quadratic equation using the quadratic formula \( v_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), \( c = -15 \): \[ \text{Discriminant} = (-2)^2 - 4 \cdot 1 \cdot (-15) = 4 + 60 = 64 \] \[ v_1 = \frac{2 \pm \sqrt{64}}{2} = \frac{2 \pm 8}{2} \] \[ v_1 = \frac{2 + 8}{2} = 5 \quad \text{or} \quad v_1 = \frac{2 - 8}{2} = -3 \] 
Step 5: Determine corresponding \( v_2 \) 
- If \( v_1 = 5 \): \[ v_2 = 2 - 5 = -3 \] - If \( v_1 = -3 \): \[ v_2 = 2 - (-3) = 5 \] 
Step 6: Interpret the results 
The two solutions represent the possible exchanges of velocities due to the elastic collision of identical masses. Since the bodies are identical and moving toward each other, they exchange velocities:
- The body initially at 5 m/s will move at -3 m/s (reversing direction to the left).
- The body initially at -3 m/s will move at 5 m/s (reversing direction to the right).
Step 7: Answer the question 
The velocity of the body initially moving at 5 m/s after the collision is \(-3 \, \text{m/s}\). 
Step 8: Verify 
- Momentum: \( 5 \cdot 1 + (-3) \cdot 1 = 2 \), \( (-3) \cdot 1 + 5 \cdot 1 = 2 \) (conserved).
- Kinetic energy: \( \frac{1}{2} \cdot 1 \cdot 5^2 + \frac{1}{2} \cdot 1 \cdot (-3)^2 = 12.5 + 4.5 = 17 \), \( \frac{1}{2} \cdot 1 \cdot (-3)^2 + \frac{1}{2} \cdot 1 \cdot 5^2 = 4.5 + 12.5 = 17 \) (conserved). 
Step 9: Conclusion 
The velocity is \(-3 \, \text{m/s}\), matching option (a).