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,

\[\theta = 0\]

\[y = - 0 . 5\]

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

\[ - 0 . 5 = \left( \text{ u } \sin \theta \right) t - \left( \frac{1}{2} \right) \text{ gt}^2 \]

\[ \Rightarrow 0 . 5 = 4 . 9 \times \text{t}^2 \]

\[ \Rightarrow \text{t} = 0 . 31 \text{ sec }\]

\[\text{ So, x } = \left( \nu \cos \theta \right) \text{ t }\]

\[ = 3 . 1 \times 0 . 31 = 1 \text{ m }\]

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