A metallic bob X of mass m is released from position A. It collides elastically with another identical bob Y placed at rest at position B on a horizontal frictionless table. The angle AOB is 30°. How high does the bob X rise immediately after the collision?
To the same height as that of position A on the other side in the same trajectory
To half the height as that of position A on the other side along the same trajectory
The same height at position A
It stops at position B
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC NDA-1 – 2024
– The colliding bobs are **identical**, meaning they have the same mass (m).
– Bob Y is **at rest** at position B before the collision. Bob X has a certain velocity (V) just before the collision at B due to its swing from A.
– For a head-on elastic collision between two objects of equal mass where one object is initially at rest, the moving object comes to rest, and the stationary object moves off with the initial velocity of the moving object.
Initial state: $v_{Xi} = V$, $v_{Yi} = 0$.
Conservation of Momentum: $m v_{Xi} + m v_{Yi} = m v_{Xf} + m v_{Yf} \implies m V + 0 = m v_{Xf} + m v_{Yf} \implies V = v_{Xf} + v_{Yf}$.
Conservation of Kinetic Energy: $\frac{1}{2} m v_{Xi}^2 + \frac{1}{2} m v_{Yi}^2 = \frac{1}{2} m v_{Xf}^2 + \frac{1}{2} m v_{Yf}^2 \implies \frac{1}{2} m V^2 + 0 = \frac{1}{2} m v_{Xf}^2 + \frac{1}{2} m v_{Yf}^2 \implies V^2 = v_{Xf}^2 + v_{Yf}^2$.
From the momentum equation, $v_{Yf} = V – v_{Xf}$. Substituting into the energy equation:
$V^2 = v_{Xf}^2 + (V – v_{Xf})^2 = v_{Xf}^2 + V^2 – 2Vv_{Xf} + v_{Xf}^2$
$0 = 2v_{Xf}^2 – 2Vv_{Xf} = 2v_{Xf}(v_{Xf} – V)$.
This gives two possible solutions for $v_{Xf}$: $v_{Xf} = 0$ or $v_{Xf} = V$.
If $v_{Xf} = V$, then $v_{Yf} = V – V = 0$, which means no collision occurred (they passed through each other, which is not the case).
Therefore, the physical solution is $v_{Xf} = 0$.
If $v_{Xf} = 0$, then $v_{Yf} = V – 0 = V$.
So, bob X stops ($v_{Xf}=0$) and bob Y moves off with velocity $V$ ($v_{Yf}=V$).
Since bob X stops at position B, it does not rise to any height immediately after the collision.