A particle of mass 2kg moving horizontally with a speed of 2m/s strikes perpendicularly to the uniform rod of mass 10 kg & length 4m lying on a smooth horizontal surface. Particle strikes the rod at a distance of 1m from its centre and comes to the rest after the collision. The coefficient of restitution between the particle & the rod is

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

3/7

7/20

1/2

7/15

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

Apply linear momentum conservation & angular momentum conservation.
$\begin{array}{l} {mv}_{0} = Mv \\ {mv}_{0} \times 1 = \frac{{ML}^{2}}{12} \times ω \end{array}$
Solving above equations we get, v=0.4 m/s and $ω$ =0.3 rad/s
$e = \frac{[v + (ω \times 1)] - 0}{v_{0}} = \frac{7}{20}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET