Question

A small disc of mass $ m $ slides down with initial velocity zero from the top (A) of a smooth hill of height $ H $, having a horizontal portion (BC) as shown in the figure. If the height of the horizontal portion of the hill is $ h $, then the maximum horizontal distance covered by the disc from the point D is

• A. \(\dfrac{H}{2}\)
• B. \(2H\)
• C. \(H\)
• D. \(3H\)

Hint:

Use energy conservation to find horizontal speed and projectile motion equations to determine range when objects fall off elevated surfaces.

Answer 1

The Correct Option is C

When the disc slides down from the height \( H \), all of its potential energy converts into kinetic energy at the horizontal section.
Using conservation of energy: \[ \text{Initial potential energy at A} = m g H \\ \text{Kinetic energy at point B} = \frac{1}{2}mv^2 \Rightarrow m g H = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gH} \] When the disc moves off the edge at point C (height \( h \) above ground), it performs projectile motion. Time of fall from height \( h \): \[ t = \sqrt{\frac{2h}{g}} \] Horizontal distance covered (Range) from point D: \[ R = v \cdot t = \sqrt{2gH} \cdot \sqrt{\frac{2h}{g}} = \sqrt{4Hh} = 2\sqrt{Hh} \] Now, from the figure, for the disc to reach a maximum horizontal distance of \( H \), we must have: \[ 2\sqrt{Hh} = H \Rightarrow \sqrt{Hh} = \frac{H}{2} \Rightarrow Hh = \frac{H^2}{4} \Rightarrow h = \frac{H}{4} \] Therefore, the correct answer (for this specific case) when height \( h = \frac{H}{4} \), the range \( R = H \). Thus, the maximum horizontal distance is: \[ \boxed{H} \]