Figure shows a particle sliding on a frictionless track which terminates in a straight horizontal section. If the particle starts slipping from the point A, how far away from the track will the particle hit the ground?

The correct option is C. 1 m.

Particle slips from point A and will reach the ground after doing projectile motion.

If we know the velocity of particle at B then it's easy to find out the distance where it will fall.

As the block is sliding down from A to B forces acting on it gravity & Normal. Normal won't do any work as at every small displacement Normal is perpendicular so $\theta ={90}^{\circ }$& cos ${90}^{\circ }$ = 0.

Block at A has gravitational potential energy of $U_i=mgh=mg$......(1) ($\because$ h = 1m)

At A kinetic energy of block will be 0 as it has no velocity.

$K{E}_i$ = 0

When block reaches B its potential energy= $U_f=mgh=\frac{mg}{2}$

At B its velocity will be $v$

So its $K{E}_f=\frac{1}{2}m{v}^2$

From conservation of mechanical energy we have

$U_i+K{E}_i=U_f+K{E}_f$

$mg+0=\frac{1}{2}mg+\frac{1}{2}m{v}^2$

$v^2=g$

$v=\sqrt{g}$ ----------------------(ii)

Now, we have

$u_y=0$

$s_y=-\frac{1}{2}$

$a_y=-g$

From second equation of motion we get,

$-\frac{1}{2}=0\times t+\frac{1}{2}(-g)t^2$

$t^2=\frac{1}{g}$

$t=\frac{1}{\sqrt{g}}$

$u_x=v$

$t=\frac{1}{\sqrt{g}}=1m$

$a_y=0$

$\therefore s_x=v\times t=\sqrt{g}\times \frac{1}{\sqrt{g}}=1m$.