As shown in figure, a particle slides on a frictionless track which terminates in a straight line horizontal section B. If the particle starts slipping from A, then the horizontal distance to be covered by the particle before it hits the ground after crossing B is:
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Notice how the gravitational acceleration constant \( g \) cancels out completely during the substitution step. This elegant cancellation means the final horizontal range is completely independent of which planet the experiment takes place on!
Concept:
We break this problem into two stages: using conservation of mechanical energy to find the horizontal launch speed at point \( B \), and then applying horizontal projectile formulas to determine the landing distance.
Step 1: Finding horizontal velocity at point B using energy conservation.
The track is frictionless, so mechanical energy is conserved:
\[
mgh_1 = mgh_2 + \frac{1}{2}m v_B^2 \implies g(h_1 - h_2) = \frac{1}{2}v_B^2
\]
Given heights are \( h_1 = 1 \, \text{m} \) and \( h_2 = 0.5 \, \text{m} \):
\[
v_B^2 = 2g(1 - 0.5) = 2g(0.5) = g \implies v_B = \sqrt{g}
\]
Step 2: Calculating time of flight for the drop from height \( h_2 \).
The vertical drop motion from point B is governed by:
\[
h_2 = \frac{1}{2}gt^2 \implies 0.5 = \frac{1}{2}gt^2 \implies 1 = gt^2 \implies t = \frac{1}{\sqrt{g}}
\]
Step 3: Determining the total horizontal range distance.
Since the horizontal velocity component remains constant during free fall:
\[
X = v_B \cdot t = \sqrt{g} \cdot \frac{1}{\sqrt{g}} = 1 \, \text{m}
\]
Thus, the horizontal distance covered is exactly \( 1 \, \text{m} \).
