Question

Two equal spheres $A$ and $B$ lie on a smooth horizontal circular groove at opposite ends of a diameter. At time $t=0$, $B$ is projected along the groove and it first impinges on $A$ at time $t=T_1$ and again at time $t=T_2$. If $e$ is the coefficient of restitution, the ratio $\frac{T_2}{T_1}$ is

• A. $\frac{2}{e}$
• B. $\frac{(2+e)}{2}$
• C. $\frac{2(e+1)}{e}$
• D. $\frac{(2+e)}{e}$

Answer 1

The correct option is D $\frac{(2+e)}{e}$


Let the ball B impinges ball A after time $T_1$ covering distance $\pi r$, where $r$ is the radius of the circular groove.
Then velocity of B w.r.t A will be,
$\begin{array}{}\overrightarrow{V_{B/A}}=\frac{\pi r}{T_1}& \text{(1)}\end{array}$
After the collision, ball B again impinges the ball A at time $T_2$ covering the whole circumference of the groove this time. Let $t^{\prime }$ be the time between 1st and 2nd collision.

velocity of B after 1st collision w.r.t A, $\begin{array}{}\overrightarrow{V_{B/A}^{\prime }}=\frac{2\pi r}{t^{\prime }}& \text{(2)}\end{array}$
Since $T_2=T_1+t^{\prime }$
$\frac{T_2}{T_1}+\frac{T_1+t^{\prime }}{T_1}=1+\frac{t^{\prime }}{T_1}$
Substituting the values of $T_1$ and $t^{\prime }$ from $\left(1\right)$ and $\left(2\right)$ we get,
$\begin{array}{}\frac{T_2}{T_1}=1+\frac{2V_{B/A}}{V_{B/A}^{\prime }}& \text{(3)}\end{array}$
Coefficient of restitution$\left(e\right)$ for the collision can be written as,
$e=\frac{\text{Velocity of separation}}{\text{Velocity of approach}}$
Let $v_A,v_B$ = Final velocity of A and B w.r.t ground after collision
$u_A,u_B$ = Initial velocity of A and B w.r.t ground before collision
$-e=\frac{v_B-v_A}{u_B-u_A}=\frac{-V_{B/A}^{\prime }}{V_{B/A}}$
$\Rightarrow e=\frac{V_{B/A}^{\prime }}{V_{B/A}}$
Substituting this value in $\left(3\right)$, we get
$\frac{T_2}{T_1}=\frac{2+e}{e}$

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