A particle of mass 20 g is released with an initial velocity 5 m/s
By Problem: A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h = 10 m from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about the point O will be (g = 10 m/s2) (JEE Mains 2019)
- 2 kg-m2/s
- 8 kg-m2/s
- 6 kg-m2/s
- 3 kg-m2/s
Solution: The particle of mass m = 0.02 kg is released with a velocity u = 5 m/s from the point A. The particle travels a vertical height h = 10 m as it moves from the point A to the point B.
Let reference level for the gravitational potential energy be at the height of point B. The potential and kinetic energy of the particle at the point A are given by \begin{align} U_A&=mgh,\\ K_A&=\frac{1}{2}mu^2.\nonumber \end{align} Let $v$ be speed of the particle at the point B. The potential and kinetic energy of the particle at this point are \begin{align} U_B&=0, \\ K_B&=\frac{1}{2}mv^2.\nonumber \end{align} The mechanical energy of the particle is conserved because the surface is frictionless. By conservation of mechanical energy, $U_A+K_A=U_B+K_B$, we get \begin{align} v&=\sqrt{u^2+2gh}\\ &=\sqrt{5^2+2(10)(10)}\\ &=15\; \mathrm{m/s}.\nonumber \end{align} The velocity of the particle at the point B is in the horizontal direction because it is tangential to the surface at this point. The magnitude of angular momentum of the particle about the point O is \begin{align} L&=m r_\perp v\\ &=0.02(10+10)15\\ &=6\;\mathrm{kg\,m^2/s}.\nonumber \end{align}
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