Question

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

Answer 1

Given,
Height of the starting point of the track, H = 1 m
Height of the ending point of the track, h = 0.5 m
Let v be the velocity of the particle at the end point on the track.
Applying the law of conservation of energy at the starting and ending point of the track,we get

$$\text{ mgH }=\frac{1}{2}\text{m}{\nu }^2+\text{mgh }$$

$$\Rightarrow g-\left(\frac{1}{2}\right){\nu }^2=0.5\text{g}$$

$$\Rightarrow {\nu }^2=2\left(\text{ g - 0 . 5 g }\right)=\text{g}$$

$$\Rightarrow \nu =\sqrt{\text{g}}=3.1\text{ m/s}$$

After leaving the track, the body exhibits projectile motion for which,

$$\theta =0$$

$$y=-0.5$$

$$\text{ Using equation of motion along the horozontal direction, }$$

$$-0.5=(\text{ u }\sin \theta )t-\left(\frac{1}{2}\right){\text{ gt}}^2$$

$$\Rightarrow 0.5=4.9\times {\text{t}}^2$$

$$\Rightarrow \text{t}=0.31\text{ sec }$$

$$\text{ So, x }=(\nu \cos \theta )\text{ t }$$

$$=3.1\times 0.31=1\text{ m }$$

So, the particle will hit the ground at a horizontal distance of 1 m from the other end of the track.