Question

A striker is shot from a square carrom board from a point exactly at the mid point of one of the walls with a speed $2\text{m/s}$ at an angle of ${45}^{\circ }$ with the $x-$axis as shown in the figure. The collisions of the striker with the walls of the carrom are perfectly elastic. The coefficient of kinetic friction between the stricker and board is $0.2$. Then,
[Take $g=10{\text{m/s}}^2$]

• A. $x-$coordinate of the striker when it stops (taking point $\text{O}$ to be the origin and neglect the friction between wall and striker) is $\frac{1}{2\sqrt{2}}.$
• B. $y-$coordinate of the stricker when it stops (taking point $\text{O}$ to be the origin and neglect the friction between wall and striker) is $\frac{1}{2}.$
• C. $x-$coordinate of the striker when it stops (taking point $\text{O}$ to be the origin and neglect the friction between wall and striker) is $\frac{1}{\sqrt{2}}.$
• D. $y-$coordinate of the stricker when it stops (taking point $\text{O}$ to be the origin and neglect the friction between wall and striker) is $\frac{1}{2\sqrt{2}}.$

Answer 1

The correct option is A $x-$coordinate of the striker when it stops (taking point $\text{O}$ to be the origin and neglect the friction between wall and striker) is $\frac{1}{2\sqrt{2}}.$

First of all we will find the total distance travelled by the striker.
Here, $u=2\text{m/s}$
and, retardation due to friction is $a=\frac{\mu mg}{m}=\mu g$
$\Rightarrow a=0.2\times 10=2{\text{m/s}}^2$
Also, $v=0$
Thus, from equation of motion we have
$v^2=u^2+2as$
$0=(2{)}^2-2\times 2\times s$ (As it is retarding.)
$\Rightarrow S=1\text{m}$


Since, collision is perfectly elastic,
$\Rightarrow \text{velocity of seperation}=\text{velocity of approach}$

Now, at point $B$



It is clear from the diagram that speed of the striker will be same after the collision and angle of reflection will be ${45}^{\circ }$

Thus, distance travelled by stricker to reach from point $A$ to $B$ is
$AB=\sqrt{{\left(\frac{1}{2\sqrt{2}}\right)}^2+{\left(\frac{1}{2\sqrt{2}}\right)}^2}=\sqrt{\frac{1}{8}+\frac{1}{8}}=\frac{1}{2}$
Similarly, $BC=\sqrt{{\left(\frac{1}{2\sqrt{2}}\right)}^2+{\left(\frac{1}{2\sqrt{2}}\right)}^2}=\sqrt{\frac{1}{8}+\frac{1}{8}}=\frac{1}{2}$

Hence, from $A$ to $B$ and $B$ to $C$ striker will cover $1\text{m}$ distance and stop at $C$.
So, the co-ordinates of point $C$ is $(\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2}})=(x,y)$

Also, at point B:



Along the y-axis, component of velocity will not change because momentum perpendicular to the $\text{LOI}$ will be conserved.
and along the line of impact, collision is elastic hence velocity of seperation is equal to velocity of approach, only direction will reverse

Hence, after the collision velocity will be $u$ making an angle of reflection as ${45}^{\circ }$ .