Question

Figure (8-E4) 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?
Figure

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

$$\begin{gathered} mgH=\frac{1}{2}m{\nu }^2+mgh \\ \Rightarrow g-\left(\frac{1}{2}\right){\nu }^2=0.5g \\ \Rightarrow {\nu }^2=2\left(g-0.5g\right)=g \\ \Rightarrow \nu =\sqrt{g}=3.1\mathrm{m}/\mathrm{s} \end{gathered}$$

After leaving the track, the body exhibits projectile motion for which,
$$\begin{gathered} \mathrm{\theta }=0 \\ y=-0.5 \\ \mathrm{Using}\mathrm{equation}\mathrm{of}\mathrm{motion}\mathrm{along}\mathrm{the}\mathrm{horozontal}\mathrm{direction}, \\ -0.5=\left(u\sin \theta \right)t-\left(\frac{1}{2}\right)g{t}^2 \\ \Rightarrow 0.5=4.9\times t^2 \\ \Rightarrow t=0.31\sec \\ \mathrm{So},\mathit{}x=\left(\mathrm{\nu }\cos \mathrm{\theta }\right)t \\ =3.1\times 0.31=1\mathrm{m} \end{gathered}$$

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