Question

Two equal spheres $A$ 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

• 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 $\frac{(2+e)}{e}$

e is the coefficient of restitution and let before collision the velocity of A is U and that of B is zero. After collision let the velocities becomes v₁ and v_{2 .}

A impinges on B for the first time i.e. $t=t_1$

$\therefore T_1=\frac{\pi R}{u}$ , R is the radius of groove.

Using the relation :

$v_1=\frac{(m_1-e{m}_2)u_1+(1+e)m_2{u}_2}{m_1+m_2}$

$=\frac{(m-em)u+(1+e)m\times 0}{2m}$

$=\frac{(1-e)u}{2}$

$v_2=\frac{(m_2-e{m}_1)u_2+(1+e)m_1{u}_1}{m_1+m_2}$

$v_2=\frac{(m-em)\times 0+(1+e)\times u}{2m}$

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

$T_2=t_1+T_1$

A covers a distance θ while B covers distance of $(\theta +\pi )$ to strike A.

$\therefore \theta =\frac{v_1}{R}{t}_{1}or\theta =\frac{(1-e)u}{2R}{t}_1........................\left(1\right)$

$and(\theta +\pi )=\frac{v_2}{R}{t}_{2}or(\theta +\pi )=\frac{(1+e)u}{2R}{t}_1..............\left(2\right)$

Solving equation (1) and (2)

$t_1=\frac{2\pi R}{ev}$

$\frac{t_1}{T_1}=\frac{2}{e}$

$\frac{T_2}{T_1}=\frac{T_1+t_1}{T_1}=1+\frac{t_1}{T_1}$

$\frac{T_2}{T1}=\frac{2+e}{e}$