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 bar of length 6a& mass 8m lies on a smooth horizontal table. Two point masses 'm& 2m' moving in the same horizontal plane with speeds 2v & 'v' respectively strike the bar as shown & stick to the bar after collision then

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

Total kinetic energy just after collision is $\frac{3}{2} {mv}^{2}$

Velocity of the centre of mass is zero

Angular velocity about centre of mass $= \frac{v}{5 a}$

Total kinetic energy just after collision is $\frac{3}{5} {mv}^{2}$

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

From conservation of linear momentum

$2 mV - 2 mV = (11 m) V \Rightarrow V_{cm} = 0$
About C, from conservation of angular momentum

$(2 mV) a + m (2 V) 2 a + 0$ $= [\frac{8 m [6 a]^{2}}{12} + 2 {ma}^{2} + m (2 a)^{2}] ω \Rightarrow ω = \frac{v}{5 a}$

$K \cdot E = \frac{1}{2} [I_{system}] ω^{2} = \frac{1}{2} (30 {ma}^{2}) \frac{v^{2}}{25 a^{2}} = \frac{3}{5} {mv}^{2}$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!