JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in opposite directions with same speed of v and in the same plane as the bar. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency \( \omega \). The ratio of v and \( \omega \) is:
- A 33
- B \( 2\sqrt{88} \)
- C 66
- D 32
Answer & Solution
Correct Answer
(A) 33
Step-by-step Solution
Detailed explanation
Using angular momentum conservation about COM of rod: \( L_{i}=L_{f} \) \( m\times V\times4+2m\times V\times2=(\frac{20m(12)^{2}}{12}+m\times4^{2}+2m\times2^{2})\omega \) \( 8mV=(240m+24m)\omega \) \( 8V=264 \omega \) \( \frac{V}{\omega}=33 \)
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