Two particles of equal mass m are moving with velocities $v \hat{i}$ and $v (\frac{\hat{i} + \hat{j}}{2})$ respectively. If the particles collide completely inelastically, then the energy lost in the process is $\frac{{mv}^{2}}{n}$ ,then the value of ‘ n ’ is _____

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 8.

(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

Conserving momentum $mv \hat{i} + m (\frac{v}{2} \hat{i} + \frac{v}{2} \hat{j}) = 2 m (v_{1} \hat{i} + v_{2} \hat{j})$

On solving $V_{1} = \frac{3 v}{4}$ and $V_{2} = \frac{v}{4}$ Change in K.E

$[\frac{1}{2} {mv}^{2} + \frac{1}{2} m {(\frac{v}{2} \sqrt{2})}^{2}] - [\frac{1}{2} (2 m) (\frac{9 v^{2}}{16} + \frac{v^{2}}{16})]$

$= \frac{3 {mv}^{2}}{4} - \frac{5 {mv}^{2}}{8} = \frac{{mv}^{2}}{8}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE