A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one and and is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed V0 strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one-sixth of the mass of the rod. The final angular velocity of the rod is (ω)and the ratio of the kinetic energy of the system after collision to the kinetic energy of the bullet before collision is 1x where x is _________

About A, angular momentum of the system (rod + bullet) is conservedLi=Lf⇒m6v0L2=mL23+m6L22ω⇒ω=2v09L Ki=12m6v02=mv0212Kf=12IAω2=12mL23+mL224ω2=mv02108∴KfKi=mv02/108mv02/12=19

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one and and is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed V₀ strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one-sixth of the mass of the rod. The final angular velocity of the rod is ( $ω$ )and the ratio of the kinetic energy of the system after collision to the kinetic energy of the bullet before collision is $\frac{1}{x}$ where $x$ is _________

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 9.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

About A, angular momentum of the system (rod + bullet) is conserved
$\begin{array}{l} Li = Lf \Rightarrow \frac{m}{6} (v_{0}) (\frac{L}{2}) = [\frac{{mL}^{2}}{3} + \frac{m}{6} {(\frac{L}{2})}^{2}] ω \\ \Rightarrow ω = \frac{2 v_{0}}{9 L} \end{array}$

$\begin{array}{l} Ki = \frac{1}{2} (\frac{m}{6}) v_{0}^{2} = \frac{{mv}_{0}^{2}}{12} \\ K_{f} = \frac{1}{2} I_{A} ω^{2} = \frac{1}{2} [\frac{{mL}^{2}}{3} + \frac{{mL}^{2}}{24}] ω^{2} = \frac{{mv}_{0}^{2}}{108} \\ ∴ \frac{K_{f}}{K_{i}} = \frac{{mv}_{0}^{2} / 108}{{mv}_{0}^{2} / 12} = \frac{1}{9} \end{array}$