Question

In a game of Carrom board, the Queen ( a wooden disc of radius $2cm$ and mass $50gm$) is placed at the exact center of the horizontal board. The striker is a smooth plastic disc of radius $3cm$ and mass $100gm$. The board is frictionless. The striker is given an initial velocity '$u$' parallel to the sides $BC$ and $AD$ so that it hits the Queen inelastically with coefficient of restitution$=2/3$. The impact parameter for the collision is '$d$' (as shown in the figure). The Queen rebounds from the edge $AB$ of the board inelastically with the same coefficient of restitution$=2/3$ and enters the hole $D$ following the dotted path as shown. The side of the board is $L$. Find the value of impact parameter '$d$' and the time which the Queen takes to enter hole $D$ after collision with the striker :

• A. $\frac{8}{\sqrt{17}}cm,53L/30u$
• B. $\frac{8}{\sqrt{17}}cm,153L/80u$
• C. $\frac{5}{\sqrt{17}}cm,153L/80u$
• D. $\frac{5}{\sqrt{17}}cm,153L/30u$

Answer 1

The correct option is D $\frac{5}{\sqrt{17}}cm,153L/30u$

Let O be the point where the queen stribes on AD after collision.

Velocity of queen after collision is $V_y+V_x$ with $V_x$ along the x-axis and $V_y$ along the y-axis.

After collision from edge AB, $V_x$ remains same, velocity in y direction becomes $\frac{2V_y}{3}$

$\frac{tan\alpha }{tan\theta }=\frac{\frac{V_x}{V_y}}{\frac{3V_x}{2V_y}}=\frac{2}{3}.............\left(i\right)$

Using trigonometry,

$\frac{L}{2}=\frac{L}{2}tan\alpha +Ltan\theta$

$\frac{1}{2}=\frac{1}{2}tan\alpha +\frac{3}{2}tan\alpha$

$\tan \alpha =\frac{1}{4}$

$\therefore \cos \alpha =\frac{4}{\sqrt{17}}$

Impact parameter is given by:

$d=5sin\alpha =\frac{5}{\sqrt{17}}$

Now conservation of momentum along the line of collision

$2mucos\alpha =m{u}_2+2m{u}_1$ ---- (i)

where $2m=100g$ is the mass of the striker and $m=50g$ is the mass of the queen, $u$ is the velocity of the striker before collision, $u_2$ is the velocity of the queen along the collision line and $u_1$ is the velocity of striker after collision along the line of collision.

and from co-efficient of restitution equation we get

$\frac{2}{3}=\frac{u_2-u_1}{u}$ ----- (ii)

from (i) and (ii) we get $u_2=\frac{}{}$

Time $=\frac{\frac{L}{2}}{U_{2}sin\alpha }=\left(\frac{153L}{30u}\right)$