UPSC NDA I 2024 General Ability Paper

Concept:In an elastic collision between two identical masses, the moving mass comes to rest and the stationary mass moves with the original speed.Explanation:Bob X is released from position A and gains speed $u_1$ at position B. Its speed just before collision is $u_1 = \sqrt{2gh}$, where $h = L(1 - \cos 30^{\circ})$. Bob Y is at rest, so $u_2 = 0$. The collision is perfectly elastic and both bobs have the same mass $m$. Applying conservation of momentum: $m u_1 + 0 = m v_1 + m v_2$ gives $u_1 = v_1 + v_2$. Applying conservation of kinetic energy: $\frac{1}{2} m u_1^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2$ gives $u_1^2 = v_1^2 + v_2^2$. Solving these equations yields $v_1 = 0$ and $v_2 = u_1$. Thus bob X stops at B, while bob Y moves away with speed $u_1$. Therefore bob X does not rise at all after the collision.Answer:Option D – It stops at position B.