Question

A body sliding on a frictionless track which terminates in a straight horizontal section CD. The body starts slipping from A and reaches point D. If h1 = 1 m, then how far from point D, will the body hit the ground?Take, g = 10 ms^-2.

• (A) 1 m
• (B) 2 m
• (C) 3 m
• (D) 4 m

Answer 1

The Correct Option is A

Explanation:
Given:Height of point A, h1 = 1 mLet u be the speed of the body at point D.Total energy of the body at point A = Potential energy of the body at point A.Total energy of the body at point D = [ kinetic energy + potential energy] at point D.Applying the law of conservation of energy at point A and D, we haveTotal energy of the body at point A = Total energy of the body at point D$$\begin{array}{l}\Rightarrow mg{h}_1=\frac{1}{2}m{u}^2+mg\frac{h_1}{2}\\ \Rightarrow h_1=\frac{1}{2}\frac{u_2}{g}+\frac{h_1}{2}\\ \Rightarrow 1=\frac{1}{2}\frac{u_2}{10}+\frac{1}{2}\\ \Rightarrow u^2=\frac{1}{2}\times 20\\ \Rightarrow u=\sqrt{10}m/s\end{array}$$The body falls from point D at a height of 0.5 m, towards the ground with velocity u, following a parabolic trajectory.

Since the body is projected horizontally at point D, its initial velocity component along the downward direction, i.e. $u_y$ is zero.$u=\sqrt{10}\mathrm{m}/s$at point $D$  horizantal componant of $u=0$$v^2=u^2+2gs$So$V^2={\left(0\right)}^2+2\times 10\times s$$(\sqrt{10}{)}^2=2\times 10\times s$$s=\frac{10}{2\times 10}=\frac{1}{2}$$s=\frac{1}{2}$Hence, the correct option is (A).