Two identical balls A and B having velocities of 0.5 m s-¹ and -0.3 m s-¹ respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be

Two identical balls A and B having velocities of 0.5 m/s and -0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be:

(a) -0.5 m/s and 0.3 m/s

(b) 0.5 m/s and -0.3 m/s

(c) -0.3 m/s and 0.5 m/s

(d) 0.3 m/s and 0.5 m/s

Detailed Answer:

For an elastic collision in one dimension between two identical balls, the velocities after the collision can be found using the following formulas: \[ v_A' = \frac{(m_A - m_B) v_A + 2 m_B v_B}{m_A + m_B} \] \[ v_B' = \frac{(m_B - m_A) v_B + 2 m_A v_A}{m_A + m_B} \] Where: - \( v_A \) and \( v_B \) are the initial velocities of balls A and B respectively, - \( v_A' \) and \( v_B' \) are the final velocities of balls A and B respectively, - \( m_A \) and \( m_B \) are the masses of the two balls (since the balls are identical, \( m_A = m_B \)). In this case, the balls are identical, so the equations simplify to: \[ v_A' = v_B \] \[ v_B' = v_A \] Given: - \( v_A = 0.5 \, \text{m/s} \), - \( v_B = -0.3 \, \text{m/s} \). After the collision: - \( v_A' = v_B = -0.3 \, \text{m/s} \), - \( v_B' = v_A = 0.5 \, \text{m/s} \). Therefore, the velocities of B and A after the collision are: - Velocity of B: \( -0.3 \, \text{m/s} \), - Velocity of A: \( 0.5 \, \text{m/s} \). The correct answer is: \[ \boxed{(c) \, -0.3 \, \text{m/s} \, \text{and} \, 0.5 \, \text{m/s}}. \]