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!
Step 1: Split the problem into two parts. First we find the speed at point B using energy conservation. Then we treat the motion after B as a flat horizontal launch, like a ball rolling off a table. Step 2: Find the speed at B by energy. The track is frictionless, so energy is conserved. The drop from start to B converts height into speed. With start height $h_1 = 1$ m and B height $h_2 = 0.5$ m: \[ g(h_1 - h_2) = \frac{1}{2}v_B^{2} \] Step 3: Solve for the launch speed. \[ v_B^{2} = 2g(1 - 0.5) = g \;\Rightarrow\; v_B = \sqrt{g} \] Step 4: Find the time to fall. After B the particle falls a height $h_2 = 0.5$ m straight down: \[ h_2 = \frac{1}{2}gt^{2} \;\Rightarrow\; 0.5 = \frac{1}{2}gt^{2} \;\Rightarrow\; t = \frac{1}{\sqrt{g}} \] Step 5: Find the horizontal distance. Horizontal speed stays the same while falling, so distance equals speed times time: \[ X = v_B \times t = \sqrt{g}\times\frac{1}{\sqrt{g}} \] Step 6: Simplify. The $\sqrt{g}$ terms cancel neatly: \[ \boxed{X = 1\ \text{m}} \]
