A ball of mass m moving with speed u undergoes a head-on elastic collision with a ball of mass nm initially at rest. The fraction of incident energy transferred to the second ball 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

$\frac{2 n}{{(1 + n)}^{2}}$

$\frac{4 n}{{(1 + n)}^{2}}$

$\frac{n}{1 + n}$

$\frac{n}{{(1 + n)}^{2}}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

As the collision is elastic, we can find

$v_{1} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} u_{1} v_{2} = \frac{2 m_{1}}{m_{1} + m_{2}} u_{1} s u b s t i t u t e g i v e n v a l l u e s i n a b o v e e q u a t i o n s, w e g e t v_{1} = \frac{1 - n}{1 - n} u v_{2} = \frac{2 u}{1 + n} - - - (1) H e n c e t h e r e q u i r e d f r a c t i o n o f K E t r a n s f e r e d b y m_{1} t o m_{2} i s \frac{K E_{2}}{K E_{I n i t i a l}} = \frac{\frac{1}{2} n m v_{2}^{2}}{\frac{1}{2} m u^{2}} \frac{K E_{2}}{K E_{I n i t i a l}} = n {(\frac{v_{2}}{u})}^{2} s u b s t i t u t e e q n (1) \frac{K E_{2}}{K E_{I n i t i a l}} = \frac{4 n}{{(1 + n)}^{2}}$