Question

A striker is shot from a square carom board from a point A exactly at midpoint of one of the walls with a speed of $2m{s}^{-1}$ at an angle of ${45}^o$ with the x-axis as shown in the figure. The collisions of the striker with the walls of the fixed carom are perfectly elastic. The coefficient of kinetic friction between the striker and board is $0.2$. The coordinate of the striker when it stops (taking point $O$ to be origin) is (in SI units).

• A. $\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2}}$
• B. $0,\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{2\sqrt{2}},0$
• D. $\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}}$

Answer 1

Answer: (A)

$\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2}}$

The distance $x$ which the striker traverses during its motion=$\sqrt{(\frac{1}{2\sqrt{2}}{)}^2+(\frac{1}{2\sqrt{2}}{)}^2}=\frac{1}{2}=0.5m=x$

The initial kinetic energy of the striker goes into the heat emitted due to work done against friction.

$⟹\frac{1}{2}m{v}^2=\mu mgs$

$⟹s=$ Total distance traveled by the striker=$\frac{v^2}{2\mu g}$

$=\frac{4}{2\times 0.2\times 10}=1m=2x$

Thus the striker stops at the mid point of the top side.

The coordintates of that point with respect to the origin=$(\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2}})$.