TS EAMCET · Physics · Rotational Motion

A uniform rod of length \(8 a\) and mass \(6 m\) lies on a smooth horizontal surface. Two point masses \(m\) and \(2 m\) moving in the same plane with speed \(2 v\) and \(v\) respectively strike the rod perpendicularly at distances \(a\) and \(2 a\) from the mid point of the rod in the opposite directions and stick to the rod. The angular velocity of the system immediately after the collision is :

A \(\frac{6 v}{32 a}\)
B \(\frac{6 v}{33 a}\)
C \(\frac{6 v}{40 a}\)
D \(\frac{6 v}{41 a}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{6 v}{41 a}\)

Step-by-step Solution

Detailed explanation

Applying conservation of angular momentum about point \(O\), \(m(a)(2 v)+2 m(2 a)(v)=I \omega\) or \(\omega=\frac{6 m a v}{I}\) ...(i) Now, \(\quad I=\frac{6 m(8 a)^2}{12}+m\left(a^2\right)+2 m(2 a)^2\) \(=32 m a^2+m a^2+8 m a^2\) \(=41 \mathrm{ma}^2\) Hence, from Eq. (i)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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