GATE 2022 PH – Question 41
A particle of mass 1 kg is released from a height of 1 m above the ground. When it reaches the ground, what is the value of Hamilton’s action for this motion in J s? (g is the acceleration due to gravity; take gravitation potential to be zero on the ground)
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Correct answer: (D) $-\sqrt{2g}/3$
Explanation
**Hamilton's action** $S=\int L\,dt$ with $L=T-V$. Take $m=1$ kg, release from rest at height $h=1$ m, and the potential $V=mgy$ (zero on the ground).
**Motion:** $y=1-\tfrac12gt^2$ and $v=gt$. The particle reaches the ground when $y=0$:
$$t_f=\sqrt{\frac2g}.$$
**Lagrangian along the path:**
$$L=\tfrac12v^2-gy=\tfrac12g^2t^2-g\left(1-\tfrac12gt^2\right)=g^2t^2-g .$$
**Action:**
$$S=\int_0^{t_f}\left(g^2t^2-g\right)dt=\frac{g^2t_f^3}{3}-gt_f=\frac{g^2}3\left(\frac2g\right)^{3/2}-g\sqrt{\frac2g}.$$
$$=\frac{2\sqrt2}{3}\sqrt g-\sqrt2\sqrt g=\sqrt{2g}\left(\frac23-1\right)=-\frac{\sqrt{2g}}{3}\quad(\text{option D}).$$