Question

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 $0.5\, m / s$

Answer 1

The Correct Option is B

This problem involves an elastic collision between two identical balls, A and B, moving in a straight line. Given their initial velocities, we need to determine their velocities after the collision.

In an elastic collision, both momentum and kinetic energy are conserved. Let's use these conservation laws to solve the problem:

1. Conservation of Momentum:
   The initial momentum of the system is equal to the final momentum of the system. Mathematically, this can be expressed as:
   m \cdot v_{A1} + m \cdot v_{B1} = m \cdot v_{A2} + m \cdot v_{B2}
   Given that the masses (m) are identical and can be canceled from the equation, we get:
   v_{A1} + v_{B1} = v_{A2} + v_{B2}
   Substituting the known initial velocities:
   0.5 \, m/s + (-0.3 \, m/s) = v_{A2} + v_{B2}
   0.2 = v_{A2} + v_{B2}
2. Conservation of Kinetic Energy:
   The initial kinetic energy of the system is equal to the final kinetic energy:
   0.5 \cdot m \cdot v_{A1}^2 + 0.5 \cdot m \cdot v_{B1}^2 = 0.5 \cdot m \cdot v_{A2}^2 + 0.5 \cdot m \cdot v_{B2}^2
   Cancelling out (0.5 \cdot m) from each term, we get:
   v_{A1}^2 + v_{B1}^2 = v_{A2}^2 + v_{B2}^2
   Substituting the initial velocities:
   (0.5)^2 + (-0.3)^2 = v_{A2}^2 + v_{B2}^2
   0.25 + 0.09 = v_{A2}^2 + v_{B2}^2
   0.34 = v_{A2}^2 + v_{B2}^2
3. Solving the Equations:
   We have two equations:
   0.2 = v_{A2} + v_{B2}
   0.34 = v_{A2}^2 + v_{B2}^2
   From the first equation, express v_{A2} in terms of v_{B2}:
   v_{A2} = 0.2 - v_{B2}
   Substitute into the second equation:
   0.34 = (0.2 - v_{B2})^2 + v_{B2}^2
   Expand and simplify:
   (0.2)^2 - 2 \cdot 0.2 \cdot v_{B2} + v_{B2}^2 + v_{B2}^2 = 0.34
   0.04 - 0.4v_{B2} + 2v_{B2}^2 = 0.34
   2v_{B2}^2 - 0.4v_{B2} + 0.04 = 0.34
   2v_{B2}^2 - 0.4v_{B2} = 0.30
   v_{B2}^2 - 0.2v_{B2} = 0.15
   Solve the quadratic equation:
   
   Two Root Possible
   v_{B2} = 0.5\, m/s or v_{B2} = -0.3\, m/s
4. Conclusion:
   With v_{B2} = 0.5\, m/s, substitute back to find v_{A2}:
   v_{A2} = 0.2 - 0.5 = -0.3\, m/s
   Thus, the velocities of B and A after the collision are 0.5\, m/s and -0.3\, m/s respectively.

Therefore, the correct answer is:

0.5\, m/s and -0.3\, m/s