Two equal sphere A and B lie on a smooth horizontal circular groove at opposite ends of a diameter. At time t=0, A is projected along the groove and it first impinges on B at time $t = T_{1}$ and again at time $t = T_{2}$ If e is the coefficient of restitution, the ratio $T_{2} / T_{1}$ 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

$\frac{2}{e}$

$\frac{(2 + e)}{2}$

$\frac{2 (e + 1)}{e}$

$\frac{(2 + e)}{e}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} time = \frac{distance of opposite end}{velocity} \\ \Rightarrow T_{1} = \frac{πR}{u_{1}} - - - (1); here u_{2} = 0 \\ coefficient of restitution = \frac{v_{2} - v_{1}}{u_{1} - 0} = e \\ \Rightarrow v_{2} - v_{1} = {eu}_{1} - - - (2) \end{array}$

Time taken to collide A and B again is
$T_{2} - T_{1} = \frac{2 π R}{v_{2} - v_{1}}$

substitute (2)
$\Rightarrow T_{2} - T_{1} = \frac{2 π R}{e u_{1}}$ -------(3)
Dividing (3) by (1) , we get $\frac{T_{2}}{T_{1}}$ = $\frac{(2 + e)}{e}$