Q 11-05-043NEETJEE MainMedium
A $2$ kg ball moving at $6$ m/s collides head-on elastically with a $4$ kg ball at rest. The velocities of the two balls after the collision are
Answer: (B) $-2$ m/s and $4$ m/s
$$v_1 = \frac{m_1 - m_2}{m_1 + m_2}u = \frac{-2}{6}(6) = -2\ \text{m/s}, \qquad v_2 = \frac{2m_1}{m_1 + m_2}u = \frac{4}{6}(6) = 4\ \text{m/s}$$
The lighter ball bounces back. Check: momentum $12 = -4 + 16$ ✓; KE $36 = 4 + 32$ ✓.
Solution by Sreeraj P, M.Sc Physics